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122
|
A
|
Lucky Division
|
PROGRAMMING
| 1,000
|
[
"brute force",
"number theory"
] | null | null |
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya calls a number almost lucky if it could be evenly divided by some lucky number. Help him find out if the given number *n* is almost lucky.
|
The single line contains an integer *n* (1<=≤<=*n*<=≤<=1000) — the number that needs to be checked.
|
In the only line print "YES" (without the quotes), if number *n* is almost lucky. Otherwise, print "NO" (without the quotes).
|
[
"47\n",
"16\n",
"78\n"
] |
[
"YES\n",
"YES\n",
"NO\n"
] |
Note that all lucky numbers are almost lucky as any number is evenly divisible by itself.
In the first sample 47 is a lucky number. In the second sample 16 is divisible by 4.
| 500
|
[
{
"input": "47",
"output": "YES"
},
{
"input": "16",
"output": "YES"
},
{
"input": "78",
"output": "NO"
},
{
"input": "48",
"output": "YES"
},
{
"input": "100",
"output": "YES"
},
{
"input": "107",
"output": "NO"
},
{
"input": "77",
"output": "YES"
},
{
"input": "477",
"output": "YES"
},
{
"input": "480",
"output": "YES"
},
{
"input": "1",
"output": "NO"
},
{
"input": "3",
"output": "NO"
},
{
"input": "4",
"output": "YES"
},
{
"input": "49",
"output": "YES"
},
{
"input": "56",
"output": "YES"
},
{
"input": "124",
"output": "YES"
},
{
"input": "1000",
"output": "YES"
},
{
"input": "999",
"output": "NO"
},
{
"input": "298",
"output": "NO"
},
{
"input": "274",
"output": "NO"
},
{
"input": "998",
"output": "NO"
},
{
"input": "42",
"output": "YES"
},
{
"input": "788",
"output": "YES"
},
{
"input": "70",
"output": "YES"
},
{
"input": "444",
"output": "YES"
},
{
"input": "777",
"output": "YES"
},
{
"input": "799",
"output": "YES"
},
{
"input": "25",
"output": "NO"
},
{
"input": "882",
"output": "YES"
},
{
"input": "88",
"output": "YES"
},
{
"input": "11",
"output": "NO"
},
{
"input": "2",
"output": "NO"
},
{
"input": "7",
"output": "YES"
},
{
"input": "8",
"output": "YES"
},
{
"input": "94",
"output": "YES"
},
{
"input": "477",
"output": "YES"
},
{
"input": "141",
"output": "YES"
}
] | 1,692,732,086
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 36
| 92
| 0
|
lucky = [4, 7, 44, 47, 74, 77, 444, 447, 474, 744, 477, 747, 774, 777]
n = int(input())
for luck in lucky:
if n % luck == 0:
print('YES')
exit()
print('NO')
|
Title: Lucky Division
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya calls a number almost lucky if it could be evenly divided by some lucky number. Help him find out if the given number *n* is almost lucky.
Input Specification:
The single line contains an integer *n* (1<=≤<=*n*<=≤<=1000) — the number that needs to be checked.
Output Specification:
In the only line print "YES" (without the quotes), if number *n* is almost lucky. Otherwise, print "NO" (without the quotes).
Demo Input:
['47\n', '16\n', '78\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
Note that all lucky numbers are almost lucky as any number is evenly divisible by itself.
In the first sample 47 is a lucky number. In the second sample 16 is divisible by 4.
|
```python
lucky = [4, 7, 44, 47, 74, 77, 444, 447, 474, 744, 477, 747, 774, 777]
n = int(input())
for luck in lucky:
if n % luck == 0:
print('YES')
exit()
print('NO')
```
| 3
|
|
312
|
A
|
Whose sentence is it?
|
PROGRAMMING
| 1,100
|
[
"implementation",
"strings"
] | null | null |
One day, liouzhou_101 got a chat record of Freda and Rainbow. Out of curiosity, he wanted to know which sentences were said by Freda, and which were said by Rainbow. According to his experience, he thought that Freda always said "lala." at the end of her sentences, while Rainbow always said "miao." at the beginning of his sentences. For each sentence in the chat record, help liouzhou_101 find whose sentence it is.
|
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=10), number of sentences in the chat record. Each of the next *n* lines contains a sentence. A sentence is a string that contains only Latin letters (A-Z, a-z), underline (_), comma (,), point (.) and space ( ). Its length doesn’t exceed 100.
|
For each sentence, output "Freda's" if the sentence was said by Freda, "Rainbow's" if the sentence was said by Rainbow, or "OMG>.< I don't know!" if liouzhou_101 can’t recognize whose sentence it is. He can’t recognize a sentence if it begins with "miao." and ends with "lala.", or satisfies neither of the conditions.
|
[
"5\nI will go to play with you lala.\nwow, welcome.\nmiao.lala.\nmiao.\nmiao .\n"
] |
[
"Freda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nOMG>.< I don't know!\n"
] |
none
| 500
|
[
{
"input": "5\nI will go to play with you lala.\nwow, welcome.\nmiao.lala.\nmiao.\nmiao .",
"output": "Freda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nOMG>.< I don't know!"
},
{
"input": "10\nLpAEKiHVJrzSZqBVSSyY\nYECGBlala.\nUZeGpeM.UCwiHmmA\nqt_,.b_.LSwJtJ.\nFAnXZtHlala.\nmiao.iapelala.\nCFPlbUgObrXLejPNu.F\nZSUfvisiHyrIMjMlala.\nmiao. lala.\nd,IWSeumytrVlala.",
"output": "OMG>.< I don't know!\nFreda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nFreda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nFreda's\nOMG>.< I don't know!\nFreda's"
},
{
"input": "10\nmiao.,taUvXPVlala.\nmiao.txEeId.X_lala.\nLZIeAEd JaeBVlala.\ncKPIsWpwIlala.\nfYp.eSvn,g\nKMx,nFEslala.\nmiao.QtMyxYqiajjuM\nDutxNkCqywgcnCYskcd\ngFLKACjeqfD\n,Ss UmY.wJvcX",
"output": "OMG>.< I don't know!\nOMG>.< I don't know!\nFreda's\nFreda's\nOMG>.< I don't know!\nFreda's\nRainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!"
},
{
"input": "10\nmiao.Plala.\nDVm,VYslala.\nmiao.rlala.\nmiao.,KQNL.fO_.QRc\nUBLCKEUePlala.\nIouS.Alala.\nmiao.lala.\nmiao.rlala.\nEJZwRJeKlala.\nmiao.Olala.",
"output": "OMG>.< I don't know!\nFreda's\nOMG>.< I don't know!\nRainbow's\nFreda's\nFreda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nFreda's\nOMG>.< I don't know!"
},
{
"input": "10\nmiao.grFTpju.jCLRnZ\ng.pVHYA_Usnm\nlloWONolcMFElala.\nAW,n.JJkOTe.Nd\n.bP.HvKlala.\nGziqPGQa,lala.\nmiao.,QkOCH.vFlala.\n.PUtOwImvUsoeh \nmiao.Z,KIds.R\nmiao.,_MDzoaAiJlala.",
"output": "Rainbow's\nOMG>.< I don't know!\nFreda's\nOMG>.< I don't know!\nFreda's\nFreda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nOMG>.< I don't know!"
},
{
"input": "10\nmiao.xWfjV\nHFVrGCDQXyZ,Sbm\nLMDS.xVkTCAY.vm\nmiao.lLBglala.\nnl,jRPyClala.\nFYnHoXlala.\nmiao. oxaHE\n.WTrw_mNpOQCa\nHOk..wHYoyMhl\nQX,XpMuPIROM",
"output": "Rainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nFreda's\nFreda's\nRainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!"
},
{
"input": "10\nJBQqiXlala.\npUNUWQRiMPCXv\nAiLnfNHWznwkC.lala.\nmiao.Dl_Oy\nxJJJkVkdfOzQBH_SmKh\nfgD_IHvdHiorE,W\nmiao.usBKixglala.\nwCpqPUzEtD\nmiao.rlala.\nmiao.JylcGvWlala.",
"output": "Freda's\nOMG>.< I don't know!\nFreda's\nRainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!"
},
{
"input": "10\nmiao..FLhPl_Wjslala.\nmiao. tdEGtfdJlala.\nGAzEUlala.\nKCcmOa .aKBlZyYsdu.V\nmiao.lala.\njKylnM,FXK\nmiao.GBWqjGH.v\nmiao.RefxS Cni.\nOxaaEihuHQR_s,\nmiao.a,Axtlala.",
"output": "OMG>.< I don't know!\nOMG>.< I don't know!\nFreda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nRainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!"
},
{
"input": "10\nNo.I_aTXlala.\nmiao.JKSCoRZS\nnOBMIlala.\nmiao.nlala.\nmiao._xqxoHIIlala.\nmiao.NJPy SWyiUDWc\nmiao.cCnahFaqqj.Xqp\nnreSMDeXPPYAQxI,W\nAktPajWimdd_qRn\nmiao.QHwKCYlala.",
"output": "Freda's\nRainbow's\nFreda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nRainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!"
},
{
"input": "10\n \n,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ \n \nmiao.miao.miao.\nlala.lala.lala.\nlala.miao.\nmiaolala. \nmiao.lala\nmiaolala_\n,.._ abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ",
"output": "OMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nFreda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!"
},
{
"input": "10\nduClyjMIPsEuWmx_Ce.byVoizYlTM,sF\nuZHsNip_,Mwtg,FZjM_LzPC,_pSvEOyTHfAOvoZXvxCZdgYDTCDdCAoSVZWyxXGcLgWlala.\nEGtJFPAvTEcqjkhaGxdduaQ_rmUzF.WaU, EIuX B,aVzFFpFrxpwADXuayRD azDfj \n_tJqYzXyqc.,u.F,mUYukveBPWnPq,f,dJnPHuBazdnbRHfzwNUdRbheAIjcoaPcnLvocrzcioxCapb R\n.YUBeb_zmwUt.QQuUdQIiOXtqshcsycEe,HLytHlala.\ndJndLqGBHt.GfpN.BgvsbXoLh_DIzAJOtFDmLSCYEztvPcS_GHPxivzV,NPMmSAtfk.Mg.w,A UcCt_lCD.csEzyJJBYtSMkzqiA\nmiao.qlala.\nmiao.FmDlY\nmiao.UQI.aJmnanNvRLskuVaMybDMsOlala.\nmiao.lala.",
"output": "OMG>.< I don't know!\nFreda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nFreda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!"
},
{
"input": "10\nmiao.vyscfysAtWcPkpFHdwZqAQ,UPPcjhKQTlala.\nmiao.KESqus DybUuYFoWVpo..LWZh.UqEdUsTHFlKfzqkThAUPklala.\nUNoE vfZIAdxkiWKhsHPfsqRPTNQoHgAxooVLYxRzugHjo jaEHWQFF\nCCmdIwr.UkoiYWK.Z,,ZesMpISTXNgnpYnJaWquCyL,gO\n.JvOayhXK_bgoYbfAtnXg\nbvdSzRrXoGxVgWvdXnsjEnEfxDzIQo_aZVGDGrzwuAMtzVAHioMBx_DHuTxyieGbGuSRNUojOREqxBBxvCgqAOMzwIWT\nMBuaWduZmRaOGyIPzWOsBVeqtDrblAbXxmM_uRfqMvnVlLEuhVKlhidN_aigiXyq,ZEDqQAx\nmiao.wCHVCuVKNePKmIUFLL_lala.\nmiao.iAqstXHUv\n pMO yvPkNtnNwmUCao W,wW.OvIMVaEeVYHmqaniWq.ivlala.",
"output": "OMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nFreda's"
},
{
"input": "10\nmiao.\nmiao.jrwLBCpNaDCjyoK.PFzbwWU.h.. wfQquG_P..lala.\nmiao.LGlYdKjw__.Chlala.\nW.wtr qG KDOHj.xWxPbXIXjD_,GJZDaAZ,JBHphsjWJwSKcZAIAi\nmiao.pHsGAZQDWPJQwKC.zHjJituLgp.eUrzObTI.wrpect.FMUJqu,Zuslala.\nmiao.YVlOpXccUA_YU igbsbZbhOVwyYTyOjnWqgiTmxwAuFa.flCHn.,MtVbqxZQl_BGHXWkwijGjuL, ,ezyNlala.\nmiao.xCrVSz.aMv UOSOroDlQxWeBmlWe.FA.ZfUmviMlala.\nxebAlala.\nmiao.qVSxqf vOTlala.\nD.oBUwsLQRgXAoNkQJhQN.w.oMhuvtujnmiwgQYMfjlNTSHh .lSKgI.OEp",
"output": "Rainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nFreda's\nOMG>.< I don't know!\nOMG>.< I don't know!"
},
{
"input": "10\nZXXzYlTiQU\nkXE.DdcbOojSaSgjMcFBPubKHefEVAzbi,PDFgSZIz,lala.\nxEfrTCjKhhwBC.UNmJXgTGUdkQeVDlala.\nLfaEw.jvMmuOBWtfoiJNtDIlQAVWNU,xWK_efBBtfkM\nqtBpqKZMWZMX_NKrUAEKYyQcLZWQlqbM\nmiao.PrJEbUtInremuaKRItqXOrfQEjQcAak VQ\nMpGCq awvQaHRvDr uvtVMKsvZI\nmiao.A.RVGu.szCEp.pXQJwL EuTltlN.WradoTvWHJyhcNSoulala.\nmiao.rzlUHzUdxtDRpWRuc,QZwEBfsKKGHMLGtFymPPQdptLFlzZ_ORWqrlfOrlntuDkpXEvz.CxwAsFYUvpnOnFWG\nmiao.VXUoNBwlgBwcna_n.CgAAcKKUuiVA.doOJKHpMdwNwlHAcLpdfN.Awa SthrlEWpUcuOonUTxIQNszYcHDXxnhArrM..A",
"output": "OMG>.< I don't know!\nFreda's\nFreda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nRainbow's"
},
{
"input": "10\nmiao.qbxBFzrjtWv.yOk\nDBgi,loApO AACrGnwssCHN\nmiao.LV.wbQEE_V.BSAtdTIHTQOJVJ_nGOthbL,nJvQ.UeWFpsa.GGsK_Uv,HQxHS,AN_bkrolala.\nmiao.tBEqk rIQuByGKhfq_iP.BW,nySZEfrfySEcqnnIzxC,lrjIiivbxlkoVXJFiegGFRn NO,txGPhVBcv.CVhMmNO zlala.\nmiao.aBZWDWxk.wkR ,NyCzGxJnJDqBZpetdUPAmmBZDXl_Tbflala.\nmiao. XN,uMwWm. VqloYr..jTLszlala.\n.rshcgfZ.eZOdMu_RMh\nmiao.ahiwpECEe.lala.\nLeoUSroTekQAMSO__M L_ZEeRD_tUihYvQETFB,RzJmFtFiKrU\nBtygQG_OoFEFBL.KsVWTYbtqtalXoStFCZ RINHda.NuLmlkRB.vAQJFvelbsfoJ.T,M sJn",
"output": "Rainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!"
},
{
"input": "10\nYoYBCcaqhXLfvKKf.UYMODTHyPZlala.\ncxgWn J.Q\nmiao.nwH.IHntgKYDhdsjU DMTHXEVRyeJP ZaAecCIBJXuv.YjhEmtbjvjKnK.U,oc,x\nmiao.EcQ.FDtRJgmpAzxhq.RwXBLxjyC,IeMqaFoheMPFCGWBcwUAFnbiwlbz_fcsEGPfJaeryCtFocBNEWTlala.\nmiao.W\nmiao. ZQpIeyCXJSnFgAIzu.THfrmyoogYWQzFqblala.\nmiao.ifzdCwnTDcxpvdr OTC.YqPv.MKDp..utICtAsbfYyGlala.\nmiao.\nmiao.tS.U.wH.s,CxORZJsBAHLi,fXeoDJWVBH\nrcUMpeupOVRKrcIRAvU.rP kgUEfoeXcrFPQOBYG.BNvAQPg.XHMWizhLpZNljXc .LQmVXCi",
"output": "Freda's\nOMG>.< I don't know!\nRainbow's\nOMG>.< I don't know!\nRainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nRainbow's\nOMG>.< I don't know!"
},
{
"input": "10\nlala.\nmiao.milalala.lmmialamiao.la.o.iao.a.ao.\nmialala.o.\nmiao.millala.allala.amiao..miao.miao.lala.ao.miammiao.iao.o.\nmiao.miaomiao..\nlalmiao.amiao..\nmiao.lala.lamiamiaolala..o.lalala.miao..\nmlala.iao.lalamiao..\nlmlala.iao.alalamiao.lmialala.lala.miao.o.alala..lala..lalmiaomiao..lalmiao.a.lalamiao..miao.alala..\nlalllamiao.la.lala.alamiao.lalalala.lala..miao.lamiao.la.lallalamiao..a..a.",
"output": "Freda's\nRainbow's\nOMG>.< I don't know!\nRainbow's\nRainbow's\nOMG>.< I don't know!\nRainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!"
},
{
"input": "10\nlalllala.ala.lala.a.mmimiao.aomiao.lllala.ala.amiao.la.mialalala.la.o..imiao.miao.amlala.iao.o.\nmilala.alllala.ala.amiao.lamiao..o.\nlala.lalalala..lalalala..\nlala.miao.\nmimiao.ao.lala.\nlalmiao.amlala.iamialala.o.o..\nlalammlala.iaolammiao.imiao.ao.la..iao..\nmiao.mialala.omiao..mlala.iaolala..\nmiamiao.o.llallala.ala.la.miao.ala.miao.mimialmiao.ala.o.alala.miaomiao..olala..\nmialala.lamiao.la.lala.miao.ollala.allala.ala.lmiaommiao.imiao.ao.lallallala.a.miao.a..a..",
"output": "OMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nFreda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!"
},
{
"input": "10\nlamiao.lamiao.mimiao.ao..\nllala.almiaomiao..lala.miao.a.\nlalala.lala.lalala.lala.lalala..la.\nlalala.la.miao.\nmiao.llalallala.miao.a.la.lala.almimiao.ao.a.\nmiao.mlala.ilala.aomilala.ao..\nmilala.ao.\nmmiao.iao.\nmlala.lala.ialalmiao.alala..olala.lala..\nlala.lmlalalalalala...iao.milala.aommiao.iao..alamiaolallala.milala.ao.a..lalalmiao.a..llala.amiao..",
"output": "OMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nRainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!"
},
{
"input": "4\n \n miao. \n lala. \n ",
"output": "OMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!"
},
{
"input": "1\nMiao.",
"output": "OMG>.< I don't know!"
},
{
"input": "1\nm",
"output": "OMG>.< I don't know!"
},
{
"input": "1\nMiAo.sdsdlala.",
"output": "Freda's"
},
{
"input": "1\nLalA.",
"output": "OMG>.< I don't know!"
},
{
"input": "3\n.\nm\nl",
"output": "OMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!"
}
] | 1,555,267,012
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 36
| 140
| 0
|
n = int(input())
for i in range(n):
s = input()
st = s.startswith('miao.')
end = s.endswith('lala.')
if (st == end):
print('OMG>.< I don\'t know!')
elif (st):
print('Rainbow\'s')
else:
print('Freda\'s')
|
Title: Whose sentence is it?
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day, liouzhou_101 got a chat record of Freda and Rainbow. Out of curiosity, he wanted to know which sentences were said by Freda, and which were said by Rainbow. According to his experience, he thought that Freda always said "lala." at the end of her sentences, while Rainbow always said "miao." at the beginning of his sentences. For each sentence in the chat record, help liouzhou_101 find whose sentence it is.
Input Specification:
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=10), number of sentences in the chat record. Each of the next *n* lines contains a sentence. A sentence is a string that contains only Latin letters (A-Z, a-z), underline (_), comma (,), point (.) and space ( ). Its length doesn’t exceed 100.
Output Specification:
For each sentence, output "Freda's" if the sentence was said by Freda, "Rainbow's" if the sentence was said by Rainbow, or "OMG>.< I don't know!" if liouzhou_101 can’t recognize whose sentence it is. He can’t recognize a sentence if it begins with "miao." and ends with "lala.", or satisfies neither of the conditions.
Demo Input:
['5\nI will go to play with you lala.\nwow, welcome.\nmiao.lala.\nmiao.\nmiao .\n']
Demo Output:
["Freda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nOMG>.< I don't know!\n"]
Note:
none
|
```python
n = int(input())
for i in range(n):
s = input()
st = s.startswith('miao.')
end = s.endswith('lala.')
if (st == end):
print('OMG>.< I don\'t know!')
elif (st):
print('Rainbow\'s')
else:
print('Freda\'s')
```
| 3
|
|
421
|
A
|
Pasha and Hamsters
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"implementation"
] | null | null |
Pasha has two hamsters: Arthur and Alexander. Pasha put *n* apples in front of them. Pasha knows which apples Arthur likes. Similarly, Pasha knows which apples Alexander likes. Pasha doesn't want any conflict between the hamsters (as they may like the same apple), so he decided to distribute the apples between the hamsters on his own. He is going to give some apples to Arthur and some apples to Alexander. It doesn't matter how many apples each hamster gets but it is important that each hamster gets only the apples he likes. It is possible that somebody doesn't get any apples.
Help Pasha distribute all the apples between the hamsters. Note that Pasha wants to distribute all the apples, not just some of them.
|
The first line contains integers *n*, *a*, *b* (1<=≤<=*n*<=≤<=100; 1<=≤<=*a*,<=*b*<=≤<=*n*) — the number of apples Pasha has, the number of apples Arthur likes and the number of apples Alexander likes, correspondingly.
The next line contains *a* distinct integers — the numbers of the apples Arthur likes. The next line contains *b* distinct integers — the numbers of the apples Alexander likes.
Assume that the apples are numbered from 1 to *n*. The input is such that the answer exists.
|
Print *n* characters, each of them equals either 1 or 2. If the *i*-h character equals 1, then the *i*-th apple should be given to Arthur, otherwise it should be given to Alexander. If there are multiple correct answers, you are allowed to print any of them.
|
[
"4 2 3\n1 2\n2 3 4\n",
"5 5 2\n3 4 1 2 5\n2 3\n"
] |
[
"1 1 2 2\n",
"1 1 1 1 1\n"
] |
none
| 500
|
[
{
"input": "4 2 3\n1 2\n2 3 4",
"output": "1 1 2 2"
},
{
"input": "5 5 2\n3 4 1 2 5\n2 3",
"output": "1 1 1 1 1"
},
{
"input": "100 69 31\n1 3 4 5 6 7 8 9 10 11 12 14 15 16 17 18 19 20 21 24 26 27 29 31 37 38 39 40 44 46 48 49 50 51 53 55 56 57 58 59 60 61 63 64 65 66 67 68 69 70 71 72 74 76 77 78 79 80 81 82 83 89 92 94 95 97 98 99 100\n2 13 22 23 25 28 30 32 33 34 35 36 41 42 43 45 47 52 54 62 73 75 84 85 86 87 88 90 91 93 96",
"output": "1 2 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 2 2 1 2 1 1 2 1 2 1 2 2 2 2 2 1 1 1 1 2 2 2 1 2 1 2 1 1 1 1 2 1 2 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 2 1 2 1 1 1 1 1 1 1 1 2 2 2 2 2 1 2 2 1 2 1 1 2 1 1 1 1"
},
{
"input": "100 56 44\n1 2 5 8 14 15 17 18 20 21 23 24 25 27 30 33 34 35 36 38 41 42 44 45 46 47 48 49 50 53 56 58 59 60 62 63 64 65 68 69 71 75 76 80 81 84 87 88 90 91 92 94 95 96 98 100\n3 4 6 7 9 10 11 12 13 16 19 22 26 28 29 31 32 37 39 40 43 51 52 54 55 57 61 66 67 70 72 73 74 77 78 79 82 83 85 86 89 93 97 99",
"output": "1 1 2 2 1 2 2 1 2 2 2 2 2 1 1 2 1 1 2 1 1 2 1 1 1 2 1 2 2 1 2 2 1 1 1 1 2 1 2 2 1 1 2 1 1 1 1 1 1 1 2 2 1 2 2 1 2 1 1 1 2 1 1 1 1 2 2 1 1 2 1 2 2 2 1 1 2 2 2 1 1 2 2 1 2 2 1 1 2 1 1 1 2 1 1 1 2 1 2 1"
},
{
"input": "100 82 18\n1 2 3 4 5 6 7 8 9 10 11 13 14 15 16 17 18 19 20 22 23 25 27 29 30 31 32 33 34 35 36 37 38 42 43 44 45 46 47 48 49 50 51 53 54 55 57 58 59 60 61 62 63 64 65 66 67 68 69 71 72 73 74 75 77 78 79 80 82 83 86 88 90 91 92 93 94 96 97 98 99 100\n12 21 24 26 28 39 40 41 52 56 70 76 81 84 85 87 89 95",
"output": "1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 2 1 1 2 1 2 1 2 1 1 1 1 1 1 1 1 1 1 2 2 2 1 1 1 1 1 1 1 1 1 1 2 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 2 1 1 1 1 2 1 1 2 2 1 2 1 2 1 1 1 1 1 2 1 1 1 1 1"
},
{
"input": "99 72 27\n1 2 3 4 5 6 7 8 10 11 12 13 14 15 16 17 20 23 25 26 28 29 30 32 33 34 35 36 39 41 42 43 44 45 46 47 50 51 52 54 55 56 58 59 60 61 62 67 70 71 72 74 75 76 77 80 81 82 84 85 86 88 90 91 92 93 94 95 96 97 98 99\n9 18 19 21 22 24 27 31 37 38 40 48 49 53 57 63 64 65 66 68 69 73 78 79 83 87 89",
"output": "1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 2 2 1 2 2 1 2 1 1 2 1 1 1 2 1 1 1 1 1 2 2 1 2 1 1 1 1 1 1 1 2 2 1 1 1 2 1 1 1 2 1 1 1 1 1 2 2 2 2 1 2 2 1 1 1 2 1 1 1 1 2 2 1 1 1 2 1 1 1 2 1 2 1 1 1 1 1 1 1 1 1 1"
},
{
"input": "99 38 61\n1 3 10 15 16 22 23 28 31 34 35 36 37 38 39 43 44 49 50 53 56 60 63 68 69 70 72 74 75 77 80 81 83 85 96 97 98 99\n2 4 5 6 7 8 9 11 12 13 14 17 18 19 20 21 24 25 26 27 29 30 32 33 40 41 42 45 46 47 48 51 52 54 55 57 58 59 61 62 64 65 66 67 71 73 76 78 79 82 84 86 87 88 89 90 91 92 93 94 95",
"output": "1 2 1 2 2 2 2 2 2 1 2 2 2 2 1 1 2 2 2 2 2 1 1 2 2 2 2 1 2 2 1 2 2 1 1 1 1 1 1 2 2 2 1 1 2 2 2 2 1 1 2 2 1 2 2 1 2 2 2 1 2 2 1 2 2 2 2 1 1 1 2 1 2 1 1 2 1 2 2 1 1 2 1 2 1 2 2 2 2 2 2 2 2 2 2 1 1 1 1"
},
{
"input": "99 84 15\n1 2 3 5 6 7 8 9 10 11 12 13 14 15 16 17 19 20 21 22 23 24 25 26 27 28 29 30 31 32 34 35 36 37 38 39 40 41 42 43 44 47 48 50 51 52 53 55 56 58 59 60 61 62 63 64 65 68 69 70 71 72 73 74 75 77 79 80 81 82 83 84 85 86 87 89 90 91 92 93 94 97 98 99\n4 18 33 45 46 49 54 57 66 67 76 78 88 95 96",
"output": "1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 2 2 1 1 2 1 1 1 1 2 1 1 2 1 1 1 1 1 1 1 1 2 2 1 1 1 1 1 1 1 1 2 1 2 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 2 2 1 1 1"
},
{
"input": "4 3 1\n1 3 4\n2",
"output": "1 2 1 1"
},
{
"input": "4 3 1\n1 2 4\n3",
"output": "1 1 2 1"
},
{
"input": "4 2 2\n2 3\n1 4",
"output": "2 1 1 2"
},
{
"input": "4 3 1\n2 3 4\n1",
"output": "2 1 1 1"
},
{
"input": "1 1 1\n1\n1",
"output": "1"
},
{
"input": "2 1 1\n2\n1",
"output": "2 1"
},
{
"input": "2 1 1\n1\n2",
"output": "1 2"
},
{
"input": "3 3 1\n1 2 3\n1",
"output": "1 1 1"
},
{
"input": "3 3 1\n1 2 3\n3",
"output": "1 1 1"
},
{
"input": "3 2 1\n1 3\n2",
"output": "1 2 1"
},
{
"input": "100 1 100\n84\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100",
"output": "2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2"
},
{
"input": "100 100 1\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100\n17",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1"
},
{
"input": "98 51 47\n1 2 3 4 6 7 8 10 13 15 16 18 19 21 22 23 25 26 27 29 31 32 36 37 39 40 41 43 44 48 49 50 51 52 54 56 58 59 65 66 68 79 80 84 86 88 89 90 94 95 97\n5 9 11 12 14 17 20 24 28 30 33 34 35 38 42 45 46 47 53 55 57 60 61 62 63 64 67 69 70 71 72 73 74 75 76 77 78 81 82 83 85 87 91 92 93 96 98",
"output": "1 1 1 1 2 1 1 1 2 1 2 2 1 2 1 1 2 1 1 2 1 1 1 2 1 1 1 2 1 2 1 1 2 2 2 1 1 2 1 1 1 2 1 1 2 2 2 1 1 1 1 1 2 1 2 1 2 1 1 2 2 2 2 2 1 1 2 1 2 2 2 2 2 2 2 2 2 2 1 1 2 2 2 1 2 1 2 1 1 1 2 2 2 1 1 2 1 2"
},
{
"input": "98 28 70\n1 13 15 16 19 27 28 40 42 43 46 53 54 57 61 63 67 68 69 71 75 76 78 80 88 93 97 98\n2 3 4 5 6 7 8 9 10 11 12 14 17 18 20 21 22 23 24 25 26 29 30 31 32 33 34 35 36 37 38 39 41 44 45 47 48 49 50 51 52 55 56 58 59 60 62 64 65 66 70 72 73 74 77 79 81 82 83 84 85 86 87 89 90 91 92 94 95 96",
"output": "1 2 2 2 2 2 2 2 2 2 2 2 1 2 1 1 2 2 1 2 2 2 2 2 2 2 1 1 2 2 2 2 2 2 2 2 2 2 2 1 2 1 1 2 2 1 2 2 2 2 2 2 1 1 2 2 1 2 2 2 1 2 1 2 2 2 1 1 1 2 1 2 2 2 1 1 2 1 2 1 2 2 2 2 2 2 2 1 2 2 2 2 1 2 2 2 1 1"
},
{
"input": "97 21 76\n7 10 16 17 26 30 34 39 40 42 44 46 53 54 56 64 67 72 78 79 94\n1 2 3 4 5 6 8 9 11 12 13 14 15 18 19 20 21 22 23 24 25 27 28 29 31 32 33 35 36 37 38 41 43 45 47 48 49 50 51 52 55 57 58 59 60 61 62 63 65 66 68 69 70 71 73 74 75 76 77 80 81 82 83 84 85 86 87 88 89 90 91 92 93 95 96 97",
"output": "2 2 2 2 2 2 1 2 2 1 2 2 2 2 2 1 1 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 1 2 2 2 2 1 1 2 1 2 1 2 1 2 2 2 2 2 2 1 1 2 1 2 2 2 2 2 2 2 1 2 2 1 2 2 2 2 1 2 2 2 2 2 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2"
},
{
"input": "97 21 76\n1 10 12 13 17 18 22 25 31 48 50 54 61 64 67 74 78 81 86 88 94\n2 3 4 5 6 7 8 9 11 14 15 16 19 20 21 23 24 26 27 28 29 30 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 49 51 52 53 55 56 57 58 59 60 62 63 65 66 68 69 70 71 72 73 75 76 77 79 80 82 83 84 85 87 89 90 91 92 93 95 96 97",
"output": "1 2 2 2 2 2 2 2 2 1 2 1 1 2 2 2 1 1 2 2 2 1 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 1 2 2 1 2 2 1 2 2 2 2 2 2 1 2 2 2 1 2 2 1 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2"
},
{
"input": "96 10 86\n2 5 31 37 68 69 80 82 90 91\n1 3 4 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 32 33 34 35 36 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 70 71 72 73 74 75 76 77 78 79 81 83 84 85 86 87 88 89 92 93 94 95 96",
"output": "2 1 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 2 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 1 1 2 2 2 2 2"
},
{
"input": "95 4 91\n58 65 70 93\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 59 60 61 62 63 64 66 67 68 69 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 94 95",
"output": "2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 1 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2"
},
{
"input": "98 88 10\n1 2 4 5 6 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 33 34 35 36 38 39 40 41 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 62 63 64 65 66 67 68 69 71 72 73 74 75 76 77 79 80 81 83 84 85 86 87 88 89 90 92 93 94 95 96 97 98\n3 7 32 37 42 61 70 78 82 91",
"output": "1 1 2 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 2 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 2 1 1 1 2 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1"
},
{
"input": "98 96 2\n1 2 3 4 5 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98\n6 7",
"output": "1 1 1 1 1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1"
},
{
"input": "97 97 1\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97\n94",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1"
},
{
"input": "97 97 1\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97\n20",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1"
},
{
"input": "96 96 1\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96\n48",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1"
},
{
"input": "95 95 1\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95\n55",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1"
}
] | 1,626,588,069
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 2
| 108
| 20,172,800
|
n,a,b=map(int,input().split())
ar=list(map(int,input().split()))
br=list(map(int,input().split()))
for i in range (1,n+1):
if(i in ar):
print(1,end=' ')
for i in range (1,n+1):
if(i not in ar):
print(2,end=' ')
|
Title: Pasha and Hamsters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pasha has two hamsters: Arthur and Alexander. Pasha put *n* apples in front of them. Pasha knows which apples Arthur likes. Similarly, Pasha knows which apples Alexander likes. Pasha doesn't want any conflict between the hamsters (as they may like the same apple), so he decided to distribute the apples between the hamsters on his own. He is going to give some apples to Arthur and some apples to Alexander. It doesn't matter how many apples each hamster gets but it is important that each hamster gets only the apples he likes. It is possible that somebody doesn't get any apples.
Help Pasha distribute all the apples between the hamsters. Note that Pasha wants to distribute all the apples, not just some of them.
Input Specification:
The first line contains integers *n*, *a*, *b* (1<=≤<=*n*<=≤<=100; 1<=≤<=*a*,<=*b*<=≤<=*n*) — the number of apples Pasha has, the number of apples Arthur likes and the number of apples Alexander likes, correspondingly.
The next line contains *a* distinct integers — the numbers of the apples Arthur likes. The next line contains *b* distinct integers — the numbers of the apples Alexander likes.
Assume that the apples are numbered from 1 to *n*. The input is such that the answer exists.
Output Specification:
Print *n* characters, each of them equals either 1 or 2. If the *i*-h character equals 1, then the *i*-th apple should be given to Arthur, otherwise it should be given to Alexander. If there are multiple correct answers, you are allowed to print any of them.
Demo Input:
['4 2 3\n1 2\n2 3 4\n', '5 5 2\n3 4 1 2 5\n2 3\n']
Demo Output:
['1 1 2 2\n', '1 1 1 1 1\n']
Note:
none
|
```python
n,a,b=map(int,input().split())
ar=list(map(int,input().split()))
br=list(map(int,input().split()))
for i in range (1,n+1):
if(i in ar):
print(1,end=' ')
for i in range (1,n+1):
if(i not in ar):
print(2,end=' ')
```
| 0
|
|
729
|
A
|
Interview with Oleg
|
PROGRAMMING
| 900
|
[
"implementation",
"strings"
] | null | null |
Polycarp has interviewed Oleg and has written the interview down without punctuation marks and spaces to save time. Thus, the interview is now a string *s* consisting of *n* lowercase English letters.
There is a filler word ogo in Oleg's speech. All words that can be obtained from ogo by adding go several times to the end of it are also considered to be fillers. For example, the words ogo, ogogo, ogogogo are fillers, but the words go, og, ogog, ogogog and oggo are not fillers.
The fillers have maximal size, for example, for ogogoo speech we can't consider ogo a filler and goo as a normal phrase. We should consider ogogo as a filler here.
To print the interview, Polycarp has to replace each of the fillers with three asterisks. Note that a filler word is replaced with exactly three asterisks regardless of its length.
Polycarp has dealt with this problem in no time. Can you do the same? The clock is ticking!
|
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the length of the interview.
The second line contains the string *s* of length *n*, consisting of lowercase English letters.
|
Print the interview text after the replacement of each of the fillers with "***". It is allowed for the substring "***" to have several consecutive occurences.
|
[
"7\naogogob\n",
"13\nogogmgogogogo\n",
"9\nogoogoogo\n"
] |
[
"a***b\n",
"***gmg***\n",
"*********\n"
] |
The first sample contains one filler word ogogo, so the interview for printing is "a***b".
The second sample contains two fillers ogo and ogogogo. Thus, the interview is transformed to "***gmg***".
| 500
|
[
{
"input": "7\naogogob",
"output": "a***b"
},
{
"input": "13\nogogmgogogogo",
"output": "***gmg***"
},
{
"input": "9\nogoogoogo",
"output": "*********"
},
{
"input": "32\nabcdefogoghijklmnogoopqrstuvwxyz",
"output": "abcdef***ghijklmn***opqrstuvwxyz"
},
{
"input": "100\nggogogoooggogooggoggogggggogoogoggooooggooggoooggogoooggoggoogggoogoggogggoooggoggoggogggogoogggoooo",
"output": "gg***oogg***oggoggoggggg******ggooooggooggooogg***ooggoggoogggo***ggogggoooggoggoggoggg***ogggoooo"
},
{
"input": "10\nogooggoggo",
"output": "***oggoggo"
},
{
"input": "20\nooggooogooogooogooog",
"output": "ooggoo***o***o***oog"
},
{
"input": "30\ngoggogoooggooggggoggoggoogoggo",
"output": "gogg***ooggooggggoggoggo***ggo"
},
{
"input": "40\nogggogooggoogoogggogooogogggoogggooggooo",
"output": "oggg***oggo***oggg***o***gggoogggooggooo"
},
{
"input": "50\noggggogoogggggggoogogggoooggooogoggogooogogggogooo",
"output": "ogggg***ogggggggo***gggoooggoo***gg***o***ggg***oo"
},
{
"input": "60\nggoooogoggogooogogooggoogggggogogogggggogggogooogogogggogooo",
"output": "ggooo***gg***o***oggooggggg***gggggoggg***o***ggg***oo"
},
{
"input": "70\ngogoooggggoggoggggggoggggoogooogogggggooogggogoogoogoggogggoggogoooooo",
"output": "g***ooggggoggoggggggoggggo***o***gggggoooggg*********ggogggogg***ooooo"
},
{
"input": "80\nooogoggoooggogogoggooooogoogogooogoggggogggggogoogggooogooooooggoggoggoggogoooog",
"output": "oo***ggooogg***ggoooo******o***ggggoggggg***ogggoo***oooooggoggoggogg***ooog"
},
{
"input": "90\nooogoggggooogoggggoooogggggooggoggoggooooooogggoggogggooggggoooooogoooogooggoooogggggooooo",
"output": "oo***ggggoo***ggggoooogggggooggoggoggooooooogggoggogggooggggooooo***oo***oggoooogggggooooo"
},
{
"input": "100\ngooogoggooggggoggoggooooggogoogggoogogggoogogoggogogogoggogggggogggggoogggooogogoggoooggogoooooogogg",
"output": "goo***ggooggggoggoggoooogg***ogggo***gggo***gg***ggogggggogggggoogggoo***ggooogg***oooo***gg"
},
{
"input": "100\ngoogoogggogoooooggoogooogoogoogogoooooogooogooggggoogoggogooogogogoogogooooggoggogoooogooooooggogogo",
"output": "go***oggg***ooooggo***o*********oooo***o***oggggo***gg***o******oooggogg***oo***ooooogg***"
},
{
"input": "100\ngoogoggggogggoooggoogoogogooggoggooggggggogogggogogggoogogggoogoggoggogooogogoooogooggggogggogggoooo",
"output": "go***ggggogggoooggo******oggoggoogggggg***ggg***gggo***gggo***ggogg***o***oo***oggggogggogggoooo"
},
{
"input": "100\nogogogogogoggogogogogogogoggogogogoogoggoggooggoggogoogoooogogoogggogogogogogoggogogogogogogogogogoe",
"output": "***gg***gg******ggoggooggogg******oo***oggg***gg***e"
},
{
"input": "5\nogoga",
"output": "***ga"
},
{
"input": "1\no",
"output": "o"
},
{
"input": "100\nogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogog",
"output": "***g"
},
{
"input": "99\nogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo",
"output": "***"
},
{
"input": "5\nggggg",
"output": "ggggg"
},
{
"input": "6\ngoogoo",
"output": "go***o"
},
{
"input": "7\nooogooo",
"output": "oo***oo"
},
{
"input": "8\ngggggggg",
"output": "gggggggg"
},
{
"input": "9\nogggogggg",
"output": "ogggogggg"
},
{
"input": "10\nogogoggogo",
"output": "***gg***"
},
{
"input": "11\noooggooggog",
"output": "oooggooggog"
},
{
"input": "12\nogggooooggog",
"output": "ogggooooggog"
},
{
"input": "13\nogoggogogooog",
"output": "***gg***oog"
},
{
"input": "15\nggooogoooggoggg",
"output": "ggoo***ooggoggg"
},
{
"input": "14\noogooooggooggo",
"output": "o***oooggooggo"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "1\ng",
"output": "g"
},
{
"input": "1\nz",
"output": "z"
},
{
"input": "46\nsdfkogogogogergergogogergerogogogogoergoegogog",
"output": "sdfk***gergerg***gerger***ergoeg***g"
},
{
"input": "86\ngdfggogogodfogogdofgoodfogodgodgfogoogdodhbofgobgbgfogoogdfogogogogogoogggsgfosogodgon",
"output": "gdfgg***df***gdofgoodf***dgodgf***ogdodhbofgobgbgf***ogdf***ogggsgfos***dgon"
},
{
"input": "83\nergegergoogogogerogoerorogroeorgefxcfgzdgawofqpwzxcvbnmsdfkfpweoriqgtogooggsdgogoog",
"output": "ergegergo***ger***erorogroeorgefxcfgzdgawofqpwzxcvbnmsdfkfpweoriqgt***oggsdg***og"
},
{
"input": "2\nog",
"output": "og"
},
{
"input": "4\nogog",
"output": "***g"
},
{
"input": "4\nogoo",
"output": "***o"
},
{
"input": "29\naaaogogogogogogogogogogogobbb",
"output": "aaa***bbb"
},
{
"input": "11\nogogogogogo",
"output": "***"
},
{
"input": "9\nogogogogo",
"output": "***"
},
{
"input": "3\nogo",
"output": "***"
},
{
"input": "15\nogogogogogogogo",
"output": "***"
},
{
"input": "25\nogogogogogogogogogogogogo",
"output": "***"
}
] | 1,479,637,729
| 5,029
|
Python 3
|
OK
|
TESTS
| 48
| 77
| 0
|
n = int(input())
s = input()
strings = ['ogo']
for i in range(50):
strings += [strings[-1] + 'go']
for i in range(len(strings) - 1, -1, -1):
if len(strings[i]) <= len(s):
s = s.replace(strings[i], '***')
print(s)
|
Title: Interview with Oleg
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp has interviewed Oleg and has written the interview down without punctuation marks and spaces to save time. Thus, the interview is now a string *s* consisting of *n* lowercase English letters.
There is a filler word ogo in Oleg's speech. All words that can be obtained from ogo by adding go several times to the end of it are also considered to be fillers. For example, the words ogo, ogogo, ogogogo are fillers, but the words go, og, ogog, ogogog and oggo are not fillers.
The fillers have maximal size, for example, for ogogoo speech we can't consider ogo a filler and goo as a normal phrase. We should consider ogogo as a filler here.
To print the interview, Polycarp has to replace each of the fillers with three asterisks. Note that a filler word is replaced with exactly three asterisks regardless of its length.
Polycarp has dealt with this problem in no time. Can you do the same? The clock is ticking!
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the length of the interview.
The second line contains the string *s* of length *n*, consisting of lowercase English letters.
Output Specification:
Print the interview text after the replacement of each of the fillers with "***". It is allowed for the substring "***" to have several consecutive occurences.
Demo Input:
['7\naogogob\n', '13\nogogmgogogogo\n', '9\nogoogoogo\n']
Demo Output:
['a***b\n', '***gmg***\n', '*********\n']
Note:
The first sample contains one filler word ogogo, so the interview for printing is "a***b".
The second sample contains two fillers ogo and ogogogo. Thus, the interview is transformed to "***gmg***".
|
```python
n = int(input())
s = input()
strings = ['ogo']
for i in range(50):
strings += [strings[-1] + 'go']
for i in range(len(strings) - 1, -1, -1):
if len(strings[i]) <= len(s):
s = s.replace(strings[i], '***')
print(s)
```
| 3
|
|
129
|
B
|
Students and Shoelaces
|
PROGRAMMING
| 1,200
|
[
"brute force",
"dfs and similar",
"graphs",
"implementation"
] | null | null |
Anna and Maria are in charge of the math club for junior students. When the club gathers together, the students behave badly. They've brought lots of shoe laces to the club and got tied with each other. Specifically, each string ties together two students. Besides, if two students are tied, then the lace connects the first student with the second one as well as the second student with the first one.
To restore order, Anna and Maria do the following. First, for each student Anna finds out what other students he is tied to. If a student is tied to exactly one other student, Anna reprimands him. Then Maria gathers in a single group all the students who have been just reprimanded. She kicks them out from the club. This group of students immediately leaves the club. These students takes with them the laces that used to tie them. Then again for every student Anna finds out how many other students he is tied to and so on. And they do so until Anna can reprimand at least one student.
Determine how many groups of students will be kicked out of the club.
|
The first line contains two integers *n* and *m* — the initial number of students and laces (). The students are numbered from 1 to *n*, and the laces are numbered from 1 to *m*. Next *m* lines each contain two integers *a* and *b* — the numbers of students tied by the *i*-th lace (1<=≤<=*a*,<=*b*<=≤<=*n*,<=*a*<=≠<=*b*). It is guaranteed that no two students are tied with more than one lace. No lace ties a student to himself.
|
Print the single number — the number of groups of students that will be kicked out from the club.
|
[
"3 3\n1 2\n2 3\n3 1\n",
"6 3\n1 2\n2 3\n3 4\n",
"6 5\n1 4\n2 4\n3 4\n5 4\n6 4\n"
] |
[
"0\n",
"2\n",
"1\n"
] |
In the first sample Anna and Maria won't kick out any group of students — in the initial position every student is tied to two other students and Anna won't be able to reprimand anyone.
In the second sample four students are tied in a chain and two more are running by themselves. First Anna and Maria kick out the two students from both ends of the chain (1 and 4), then — two other students from the chain (2 and 3). At that the students who are running by themselves will stay in the club.
In the third sample Anna and Maria will momentarily kick out all students except for the fourth one and the process stops at that point. The correct answer is one.
| 1,000
|
[
{
"input": "3 3\n1 2\n2 3\n3 1",
"output": "0"
},
{
"input": "6 3\n1 2\n2 3\n3 4",
"output": "2"
},
{
"input": "6 5\n1 4\n2 4\n3 4\n5 4\n6 4",
"output": "1"
},
{
"input": "100 0",
"output": "0"
},
{
"input": "5 5\n1 2\n2 3\n3 4\n4 5\n5 1",
"output": "0"
},
{
"input": "5 4\n1 4\n4 3\n4 5\n5 2",
"output": "2"
},
{
"input": "11 10\n1 2\n1 3\n3 4\n1 5\n5 6\n6 7\n1 8\n8 9\n9 10\n10 11",
"output": "4"
},
{
"input": "7 7\n1 2\n2 3\n3 1\n1 4\n4 5\n4 6\n4 7",
"output": "2"
},
{
"input": "12 49\n6 3\n12 9\n10 11\n3 5\n10 2\n6 9\n8 5\n6 12\n7 3\n3 12\n3 2\n5 6\n7 5\n9 2\n11 1\n7 6\n5 4\n8 7\n12 5\n5 11\n8 9\n10 3\n6 2\n10 4\n9 10\n9 11\n11 3\n5 9\n11 6\n10 8\n7 9\n10 7\n4 6\n3 8\n4 11\n12 2\n4 9\n2 11\n7 11\n1 5\n7 2\n8 1\n4 12\n9 1\n4 2\n8 2\n11 12\n3 1\n1 6",
"output": "0"
},
{
"input": "10 29\n4 5\n1 7\n4 2\n3 8\n7 6\n8 10\n10 6\n4 1\n10 1\n6 2\n7 4\n7 10\n2 7\n9 8\n5 10\n2 5\n8 5\n4 9\n2 8\n5 7\n4 8\n7 3\n6 5\n1 3\n1 9\n10 4\n10 9\n10 2\n2 3",
"output": "0"
},
{
"input": "9 33\n5 7\n5 9\n9 6\n9 1\n7 4\n3 5\n7 8\n8 6\n3 6\n8 2\n3 8\n1 6\n1 8\n1 4\n4 2\n1 2\n2 5\n3 4\n8 5\n2 6\n3 1\n1 5\n1 7\n3 2\n5 4\n9 4\n3 9\n7 3\n6 4\n9 8\n7 9\n8 4\n6 5",
"output": "0"
},
{
"input": "7 8\n5 7\n2 7\n1 6\n1 3\n3 7\n6 3\n6 4\n2 6",
"output": "1"
},
{
"input": "6 15\n3 1\n4 5\n1 4\n6 2\n3 5\n6 3\n1 6\n1 5\n2 3\n2 5\n6 4\n5 6\n4 2\n1 2\n3 4",
"output": "0"
},
{
"input": "7 11\n5 3\n6 5\n6 4\n1 6\n7 1\n2 6\n7 5\n2 5\n3 1\n3 4\n2 4",
"output": "0"
},
{
"input": "95 0",
"output": "0"
},
{
"input": "100 0",
"output": "0"
},
{
"input": "62 30\n29 51\n29 55\n4 12\n53 25\n36 28\n32 11\n29 11\n47 9\n21 8\n25 4\n51 19\n26 56\n22 21\n37 9\n9 33\n7 25\n16 7\n40 49\n15 21\n49 58\n34 30\n20 46\n62 48\n53 57\n33 6\n60 37\n41 34\n62 36\n36 43\n11 39",
"output": "2"
},
{
"input": "56 25\n12 40\n31 27\n18 40\n1 43\n9 10\n25 47\n27 29\n26 28\n19 38\n19 40\n22 14\n21 51\n29 31\n55 29\n51 33\n20 17\n24 15\n3 48\n31 56\n15 29\n49 42\n50 4\n22 42\n25 17\n18 51",
"output": "3"
},
{
"input": "51 29\n36 30\n37 45\n4 24\n40 18\n47 35\n15 1\n30 38\n15 18\n32 40\n34 42\n2 47\n35 21\n25 28\n13 1\n13 28\n36 1\n46 47\n22 17\n41 45\n43 45\n40 15\n29 35\n47 15\n30 21\n9 14\n18 38\n18 50\n42 10\n31 41",
"output": "3"
},
{
"input": "72 45\n5 15\n8 18\n40 25\n71 66\n67 22\n6 44\n16 25\n8 23\n19 70\n26 34\n48 15\n24 2\n54 68\n44 43\n17 37\n49 19\n71 49\n34 38\n59 1\n65 70\n11 54\n5 11\n15 31\n29 50\n48 16\n70 57\n25 59\n2 59\n56 12\n66 62\n24 16\n46 27\n45 67\n68 43\n31 11\n31 30\n8 44\n64 33\n38 44\n54 10\n13 9\n7 51\n25 4\n40 70\n26 65",
"output": "5"
},
{
"input": "56 22\n17 27\n48 49\n29 8\n47 20\n32 7\n44 5\n14 39\n5 13\n40 2\n50 42\n38 9\n18 37\n16 44\n21 32\n21 39\n37 54\n19 46\n30 47\n17 13\n30 31\n49 16\n56 7",
"output": "4"
},
{
"input": "81 46\n53 58\n31 14\n18 54\n43 61\n57 65\n6 38\n49 5\n6 40\n6 10\n17 72\n27 48\n58 39\n21 75\n21 43\n78 20\n34 4\n15 35\n74 48\n76 15\n49 38\n46 51\n78 9\n80 5\n26 42\n64 31\n46 72\n1 29\n20 17\n32 45\n53 43\n24 5\n52 59\n3 80\n78 19\n61 17\n80 12\n17 8\n63 2\n8 4\n44 10\n53 72\n18 60\n68 15\n17 58\n79 71\n73 35",
"output": "4"
},
{
"input": "82 46\n64 43\n32 24\n57 30\n24 46\n70 12\n23 41\n63 39\n46 70\n4 61\n19 12\n39 79\n14 28\n37 3\n12 27\n15 20\n35 39\n25 64\n59 16\n68 63\n37 14\n76 7\n67 29\n9 5\n14 55\n46 26\n71 79\n47 42\n5 55\n18 45\n28 40\n44 78\n74 9\n60 53\n44 19\n52 81\n65 52\n40 13\n40 19\n43 1\n24 23\n68 9\n16 20\n70 14\n41 40\n29 10\n45 65",
"output": "8"
},
{
"input": "69 38\n63 35\n52 17\n43 69\n2 57\n12 5\n26 36\n13 10\n16 68\n5 18\n5 41\n10 4\n60 9\n39 22\n39 28\n53 57\n13 52\n66 38\n49 61\n12 19\n27 46\n67 7\n25 8\n23 58\n52 34\n29 2\n2 42\n8 53\n57 43\n68 11\n48 28\n56 19\n46 33\n63 21\n57 16\n68 59\n67 34\n28 43\n56 36",
"output": "4"
},
{
"input": "75 31\n32 50\n52 8\n21 9\n68 35\n12 72\n47 26\n38 58\n40 55\n31 70\n53 75\n44 1\n65 22\n33 22\n33 29\n14 39\n1 63\n16 52\n70 15\n12 27\n63 31\n47 9\n71 31\n43 17\n43 49\n8 26\n11 39\n9 22\n30 45\n65 47\n32 9\n60 70",
"output": "4"
},
{
"input": "77 41\n48 45\n50 36\n6 69\n70 3\n22 21\n72 6\n54 3\n49 31\n2 23\n14 59\n68 58\n4 54\n60 12\n63 60\n44 24\n28 24\n40 8\n5 1\n13 24\n29 15\n19 76\n70 50\n65 71\n23 33\n58 16\n50 42\n71 28\n58 54\n24 73\n6 17\n29 13\n60 4\n42 4\n21 60\n77 39\n57 9\n51 19\n61 6\n49 36\n24 32\n41 66",
"output": "3"
},
{
"input": "72 39\n9 44\n15 12\n2 53\n34 18\n41 70\n54 72\n39 19\n26 7\n4 54\n53 59\n46 49\n70 6\n9 10\n64 51\n31 60\n61 53\n59 71\n9 60\n67 16\n4 16\n34 3\n2 61\n16 23\n34 6\n10 18\n13 38\n66 40\n59 9\n40 14\n38 24\n31 48\n7 69\n20 39\n49 52\n32 67\n61 35\n62 45\n37 54\n5 27",
"output": "8"
},
{
"input": "96 70\n30 37\n47 56\n19 79\n15 28\n2 43\n43 54\n59 75\n42 22\n38 18\n18 14\n47 41\n60 29\n35 11\n90 4\n14 41\n11 71\n41 24\n68 28\n45 92\n14 15\n34 63\n77 32\n67 38\n36 8\n37 4\n58 95\n68 84\n69 81\n35 23\n56 63\n78 91\n35 44\n66 63\n80 19\n87 88\n28 14\n62 35\n24 23\n83 37\n54 89\n14 40\n9 35\n94 9\n56 46\n92 70\n16 58\n96 31\n53 23\n56 5\n36 42\n89 77\n29 51\n26 13\n46 70\n25 56\n95 96\n3 51\n76 8\n36 82\n44 85\n54 56\n89 67\n32 5\n82 78\n33 65\n43 28\n35 1\n94 13\n26 24\n10 51",
"output": "4"
},
{
"input": "76 49\n15 59\n23 26\n57 48\n49 51\n42 76\n36 40\n37 40\n29 15\n28 71\n47 70\n27 39\n76 21\n55 16\n21 18\n19 1\n25 31\n51 71\n54 42\n28 9\n61 69\n33 9\n18 19\n58 51\n51 45\n29 34\n9 67\n26 8\n70 37\n11 62\n24 22\n59 76\n67 17\n59 11\n54 1\n12 57\n23 3\n46 47\n37 20\n65 9\n51 12\n31 19\n56 13\n58 22\n26 59\n39 76\n27 11\n48 64\n59 35\n44 75",
"output": "5"
},
{
"input": "52 26\n29 41\n16 26\n18 48\n31 17\n37 42\n26 1\n11 7\n29 6\n23 17\n12 47\n34 23\n41 16\n15 35\n25 21\n45 7\n52 2\n37 10\n28 19\n1 27\n30 47\n42 35\n50 30\n30 34\n19 30\n42 25\n47 31",
"output": "3"
},
{
"input": "86 48\n59 34\n21 33\n45 20\n62 23\n4 68\n2 65\n63 26\n64 20\n51 34\n64 21\n68 78\n61 80\n81 3\n38 39\n47 48\n24 34\n44 71\n72 78\n50 2\n13 51\n82 78\n11 74\n14 48\n2 75\n49 55\n63 85\n20 85\n4 53\n51 15\n11 67\n1 15\n2 64\n10 81\n6 7\n68 18\n84 28\n77 69\n10 36\n15 14\n32 86\n16 79\n26 13\n38 55\n47 43\n47 39\n45 37\n58 81\n42 35",
"output": "8"
},
{
"input": "58 29\n27 24\n40 52\n51 28\n44 50\n7 28\n14 53\n10 16\n16 45\n8 56\n35 26\n39 6\n6 14\n45 22\n35 13\n20 17\n42 6\n37 21\n4 11\n26 56\n54 55\n3 57\n40 3\n55 27\n4 51\n35 29\n50 16\n47 7\n48 20\n1 37",
"output": "3"
},
{
"input": "51 23\n46 47\n31 27\n1 20\n49 16\n2 10\n29 47\n13 27\n34 26\n31 2\n28 20\n17 40\n39 4\n29 26\n28 44\n3 39\n50 12\n19 1\n30 21\n41 23\n2 29\n16 3\n49 28\n49 41",
"output": "4"
},
{
"input": "75 43\n46 34\n33 12\n51 39\n47 74\n68 64\n40 46\n20 51\n47 19\n4 5\n57 59\n12 26\n68 65\n38 42\n73 37\n5 74\n36 61\n8 18\n58 33\n34 73\n42 43\n10 49\n70 50\n49 18\n24 53\n71 73\n44 24\n49 56\n24 29\n44 67\n70 46\n57 25\n73 63\n3 51\n30 71\n41 44\n17 69\n17 18\n19 68\n42 7\n11 51\n1 5\n72 23\n65 53",
"output": "5"
}
] | 1,574,446,672
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 70
| 248
| 409,600
|
import sys
input = sys.stdin.readline
read_tuple = lambda _type: map(_type, input().split(' '))
def solve():
n, m = read_tuple(int)
graph = {}
for _ in range(m):
a, b = read_tuple(int)
graph[a] = graph.get(a, set()) | set([b])
graph[b] = graph.get(b, set()) | set([a])
ans = 0
while True:
marked_for_del = set()
for v in graph:
if len(graph[v]) == 1:
marked_for_del.add(v)
if marked_for_del:
for v in marked_for_del:
if v in graph:
v2 = graph[v].pop()
graph[v2].discard(v)
graph.pop(v)
if graph[v2] == set():
graph.pop(v2)
ans += 1
else:
break
print(ans)
if __name__ == '__main__':
solve()
|
Title: Students and Shoelaces
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Anna and Maria are in charge of the math club for junior students. When the club gathers together, the students behave badly. They've brought lots of shoe laces to the club and got tied with each other. Specifically, each string ties together two students. Besides, if two students are tied, then the lace connects the first student with the second one as well as the second student with the first one.
To restore order, Anna and Maria do the following. First, for each student Anna finds out what other students he is tied to. If a student is tied to exactly one other student, Anna reprimands him. Then Maria gathers in a single group all the students who have been just reprimanded. She kicks them out from the club. This group of students immediately leaves the club. These students takes with them the laces that used to tie them. Then again for every student Anna finds out how many other students he is tied to and so on. And they do so until Anna can reprimand at least one student.
Determine how many groups of students will be kicked out of the club.
Input Specification:
The first line contains two integers *n* and *m* — the initial number of students and laces (). The students are numbered from 1 to *n*, and the laces are numbered from 1 to *m*. Next *m* lines each contain two integers *a* and *b* — the numbers of students tied by the *i*-th lace (1<=≤<=*a*,<=*b*<=≤<=*n*,<=*a*<=≠<=*b*). It is guaranteed that no two students are tied with more than one lace. No lace ties a student to himself.
Output Specification:
Print the single number — the number of groups of students that will be kicked out from the club.
Demo Input:
['3 3\n1 2\n2 3\n3 1\n', '6 3\n1 2\n2 3\n3 4\n', '6 5\n1 4\n2 4\n3 4\n5 4\n6 4\n']
Demo Output:
['0\n', '2\n', '1\n']
Note:
In the first sample Anna and Maria won't kick out any group of students — in the initial position every student is tied to two other students and Anna won't be able to reprimand anyone.
In the second sample four students are tied in a chain and two more are running by themselves. First Anna and Maria kick out the two students from both ends of the chain (1 and 4), then — two other students from the chain (2 and 3). At that the students who are running by themselves will stay in the club.
In the third sample Anna and Maria will momentarily kick out all students except for the fourth one and the process stops at that point. The correct answer is one.
|
```python
import sys
input = sys.stdin.readline
read_tuple = lambda _type: map(_type, input().split(' '))
def solve():
n, m = read_tuple(int)
graph = {}
for _ in range(m):
a, b = read_tuple(int)
graph[a] = graph.get(a, set()) | set([b])
graph[b] = graph.get(b, set()) | set([a])
ans = 0
while True:
marked_for_del = set()
for v in graph:
if len(graph[v]) == 1:
marked_for_del.add(v)
if marked_for_del:
for v in marked_for_del:
if v in graph:
v2 = graph[v].pop()
graph[v2].discard(v)
graph.pop(v)
if graph[v2] == set():
graph.pop(v2)
ans += 1
else:
break
print(ans)
if __name__ == '__main__':
solve()
```
| 3
|
|
236
|
A
|
Boy or Girl
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation",
"strings"
] | null | null |
Those days, many boys use beautiful girls' photos as avatars in forums. So it is pretty hard to tell the gender of a user at the first glance. Last year, our hero went to a forum and had a nice chat with a beauty (he thought so). After that they talked very often and eventually they became a couple in the network.
But yesterday, he came to see "her" in the real world and found out "she" is actually a very strong man! Our hero is very sad and he is too tired to love again now. So he came up with a way to recognize users' genders by their user names.
This is his method: if the number of distinct characters in one's user name is odd, then he is a male, otherwise she is a female. You are given the string that denotes the user name, please help our hero to determine the gender of this user by his method.
|
The first line contains a non-empty string, that contains only lowercase English letters — the user name. This string contains at most 100 letters.
|
If it is a female by our hero's method, print "CHAT WITH HER!" (without the quotes), otherwise, print "IGNORE HIM!" (without the quotes).
|
[
"wjmzbmr\n",
"xiaodao\n",
"sevenkplus\n"
] |
[
"CHAT WITH HER!\n",
"IGNORE HIM!\n",
"CHAT WITH HER!\n"
] |
For the first example. There are 6 distinct characters in "wjmzbmr". These characters are: "w", "j", "m", "z", "b", "r". So wjmzbmr is a female and you should print "CHAT WITH HER!".
| 500
|
[
{
"input": "wjmzbmr",
"output": "CHAT WITH HER!"
},
{
"input": "xiaodao",
"output": "IGNORE HIM!"
},
{
"input": "sevenkplus",
"output": "CHAT WITH HER!"
},
{
"input": "pezu",
"output": "CHAT WITH HER!"
},
{
"input": "wnemlgppy",
"output": "CHAT WITH HER!"
},
{
"input": "zcinitufxoldnokacdvtmdohsfdjepyfioyvclhmujiqwvmudbfjzxjfqqxjmoiyxrfsbvseawwoyynn",
"output": "IGNORE HIM!"
},
{
"input": "qsxxuoynwtebujwpxwpajitiwxaxwgbcylxneqiebzfphugwkftpaikixmumkhfbjiswmvzbtiyifbx",
"output": "CHAT WITH HER!"
},
{
"input": "qwbdfzfylckctudyjlyrtmvbidfatdoqfmrfshsqqmhzohhsczscvwzpwyoyswhktjlykumhvaounpzwpxcspxwlgt",
"output": "IGNORE HIM!"
},
{
"input": "nuezoadauueermoeaabjrkxttkatspjsjegjcjcdmcxgodowzbwuqncfbeqlhkk",
"output": "IGNORE HIM!"
},
{
"input": "lggvdmulrsvtuagoavstuyufhypdxfomjlzpnduulukszqnnwfvxbvxyzmleocmofwclmzz",
"output": "IGNORE HIM!"
},
{
"input": "tgcdptnkc",
"output": "IGNORE HIM!"
},
{
"input": "wvfgnfrzabgibzxhzsojskmnlmrokydjoexnvi",
"output": "IGNORE HIM!"
},
{
"input": "sxtburpzskucowowebgrbovhadrrayamuwypmmxhscrujkmcgvyinp",
"output": "IGNORE HIM!"
},
{
"input": "pjqxhvxkyeqqvyuujxhmbspatvrckhhkfloottuybjivkkhpyivcighxumavrxzxslfpggnwbtalmhysyfllznphzia",
"output": "IGNORE HIM!"
},
{
"input": "fpellxwskyekoyvrfnuf",
"output": "CHAT WITH HER!"
},
{
"input": "xninyvkuvakfbs",
"output": "IGNORE HIM!"
},
{
"input": "vnxhrweyvhqufpfywdwftoyrfgrhxuamqhblkvdpxmgvphcbeeqbqssresjifwyzgfhurmamhkwupymuomak",
"output": "CHAT WITH HER!"
},
{
"input": "kmsk",
"output": "IGNORE HIM!"
},
{
"input": "lqonogasrkzhryjxppjyriyfxmdfubieglthyswz",
"output": "CHAT WITH HER!"
},
{
"input": "ndormkufcrkxlihdhmcehzoimcfhqsmombnfjrlcalffq",
"output": "CHAT WITH HER!"
},
{
"input": "zqzlnnuwcfufwujygtczfakhcpqbtxtejrbgoodychepzdphdahtxyfpmlrycyicqthsgm",
"output": "IGNORE HIM!"
},
{
"input": "ppcpbnhwoizajrl",
"output": "IGNORE HIM!"
},
{
"input": "sgubujztzwkzvztitssxxxwzanfmddfqvv",
"output": "CHAT WITH HER!"
},
{
"input": "ptkyaxycecpbrjnvxcjtbqiocqcswnmicxbvhdsptbxyxswbw",
"output": "IGNORE HIM!"
},
{
"input": "yhbtzfppwcycxqjpqdfmjnhwaogyuaxamwxpnrdrnqsgdyfvxu",
"output": "CHAT WITH HER!"
},
{
"input": "ojjvpnkrxibyevxk",
"output": "CHAT WITH HER!"
},
{
"input": "wjweqcrqfuollfvfbiyriijovweg",
"output": "IGNORE HIM!"
},
{
"input": "hkdbykboclchfdsuovvpknwqr",
"output": "IGNORE HIM!"
},
{
"input": "stjvyfrfowopwfjdveduedqylerqugykyu",
"output": "IGNORE HIM!"
},
{
"input": "rafcaanqytfclvfdegak",
"output": "CHAT WITH HER!"
},
{
"input": "xczn",
"output": "CHAT WITH HER!"
},
{
"input": "arcoaeozyeawbveoxpmafxxzdjldsielp",
"output": "IGNORE HIM!"
},
{
"input": "smdfafbyehdylhaleevhoggiurdgeleaxkeqdixyfztkuqsculgslheqfafxyghyuibdgiuwrdxfcitojxika",
"output": "CHAT WITH HER!"
},
{
"input": "vbpfgjqnhfazmvtkpjrdasfhsuxnpiepxfrzvoh",
"output": "CHAT WITH HER!"
},
{
"input": "dbdokywnpqnotfrhdbrzmuyoxfdtrgrzcccninbtmoqvxfatcqg",
"output": "CHAT WITH HER!"
},
{
"input": "udlpagtpq",
"output": "CHAT WITH HER!"
},
{
"input": "zjurevbytijifnpfuyswfchdzelxheboruwjqijxcucylysmwtiqsqqhktexcynquvcwhbjsipy",
"output": "CHAT WITH HER!"
},
{
"input": "qagzrqjomdwhagkhrjahhxkieijyten",
"output": "CHAT WITH HER!"
},
{
"input": "achhcfjnnfwgoufxamcqrsontgjjhgyfzuhklkmiwybnrlsvblnsrjqdytglipxsulpnphpjpoewvlusalsgovwnsngb",
"output": "CHAT WITH HER!"
},
{
"input": "qbkjsdwpahdbbohggbclfcufqelnojoehsxxkr",
"output": "CHAT WITH HER!"
},
{
"input": "cpvftiwgyvnlmbkadiafddpgfpvhqqvuehkypqjsoibpiudfvpkhzlfrykc",
"output": "IGNORE HIM!"
},
{
"input": "lnpdosnceumubvk",
"output": "IGNORE HIM!"
},
{
"input": "efrk",
"output": "CHAT WITH HER!"
},
{
"input": "temnownneghnrujforif",
"output": "IGNORE HIM!"
},
{
"input": "ottnneymszwbumgobazfjyxewkjakglbfflsajuzescplpcxqta",
"output": "IGNORE HIM!"
},
{
"input": "eswpaclodzcwhgixhpyzvhdwsgneqidanbzdzszquefh",
"output": "IGNORE HIM!"
},
{
"input": "gwntwbpj",
"output": "IGNORE HIM!"
},
{
"input": "wuqvlbblkddeindiiswsinkfrnkxghhwunzmmvyovpqapdfbolyim",
"output": "IGNORE HIM!"
},
{
"input": "swdqsnzmzmsyvktukaoyqsqzgfmbzhezbfaqeywgwizrwjyzquaahucjchegknqaioliqd",
"output": "CHAT WITH HER!"
},
{
"input": "vlhrpzezawyolhbmvxbwhtjustdbqggexmzxyieihjlelvwjosmkwesfjmramsikhkupzvfgezmrqzudjcalpjacmhykhgfhrjx",
"output": "IGNORE HIM!"
},
{
"input": "lxxwbkrjgnqjwsnflfnsdyxihmlspgivirazsbveztnkuzpaxtygidniflyjheejelnjyjvgkgvdqks",
"output": "CHAT WITH HER!"
},
{
"input": "wpxbxzfhtdecetpljcrvpjjnllosdqirnkzesiqeukbedkayqx",
"output": "CHAT WITH HER!"
},
{
"input": "vmzxgacicvweclaodrunmjnfwtimceetsaoickarqyrkdghcmyjgmtgsqastcktyrjgvjqimdc",
"output": "CHAT WITH HER!"
},
{
"input": "yzlzmesxdttfcztooypjztlgxwcr",
"output": "IGNORE HIM!"
},
{
"input": "qpbjwzwgdzmeluheirjrvzrhbmagfsjdgvzgwumjtjzecsfkrfqjasssrhhtgdqqfydlmrktlgfc",
"output": "IGNORE HIM!"
},
{
"input": "aqzftsvezdgouyrirsxpbuvdjupnzvbhguyayeqozfzymfnepvwgblqzvmxxkxcilmsjvcgyqykpoaktjvsxbygfgsalbjoq",
"output": "CHAT WITH HER!"
},
{
"input": "znicjjgijhrbdlnwmtjgtdgziollrfxroabfhadygnomodaembllreorlyhnehijfyjbfxucazellblegyfrzuraogadj",
"output": "IGNORE HIM!"
},
{
"input": "qordzrdiknsympdrkgapjxokbldorpnmnpucmwakklmqenpmkom",
"output": "CHAT WITH HER!"
},
{
"input": "wqfldgihuxfktzanyycluzhtewmwvnawqlfoavuguhygqrrxtstxwouuzzsryjqtfqo",
"output": "CHAT WITH HER!"
},
{
"input": "vujtrrpshinkskgyknlcfckmqdrwtklkzlyipmetjvaqxdsslkskschbalmdhzsdrrjmxdltbtnxbh",
"output": "IGNORE HIM!"
},
{
"input": "zioixjibuhrzyrbzqcdjbbhhdmpgmqykixcxoqupggaqajuzonrpzihbsogjfsrrypbiphehonyhohsbybnnukqebopppa",
"output": "CHAT WITH HER!"
},
{
"input": "oh",
"output": "CHAT WITH HER!"
},
{
"input": "kxqthadqesbpgpsvpbcbznxpecqrzjoilpauttzlnxvaczcqwuri",
"output": "IGNORE HIM!"
},
{
"input": "zwlunigqnhrwirkvufqwrnwcnkqqonebrwzcshcbqqwkjxhymjjeakuzjettebciadjlkbfp",
"output": "CHAT WITH HER!"
},
{
"input": "fjuldpuejgmggvvigkwdyzytfxzwdlofrpifqpdnhfyroginqaufwgjcbgshyyruwhofctsdaisqpjxqjmtpp",
"output": "CHAT WITH HER!"
},
{
"input": "xiwntnheuitbtqxrmzvxmieldudakogealwrpygbxsbluhsqhtwmdlpjwzyafckrqrdduonkgo",
"output": "CHAT WITH HER!"
},
{
"input": "mnmbupgo",
"output": "IGNORE HIM!"
},
{
"input": "mcjehdiygkbmrbfjqwpwxidbdfelifwhstaxdapigbymmsgrhnzsdjhsqchl",
"output": "IGNORE HIM!"
},
{
"input": "yocxrzspinchmhtmqo",
"output": "CHAT WITH HER!"
},
{
"input": "vasvvnpymtgjirnzuynluluvmgpquskuaafwogeztfnvybblajvuuvfomtifeuzpikjrolzeeoftv",
"output": "CHAT WITH HER!"
},
{
"input": "ecsdicrznvglwggrdbrvehwzaenzjutjydhvimtqegweurpxtjkmpcznshtrvotkvrghxhacjkedidqqzrduzad",
"output": "IGNORE HIM!"
},
{
"input": "ubvhyaebyxoghakajqrpqpctwbrfqzli",
"output": "CHAT WITH HER!"
},
{
"input": "gogbxfeqylxoummvgxpkoqzsmobasesxbqjjktqbwqxeiaagnnhbvepbpy",
"output": "IGNORE HIM!"
},
{
"input": "nheihhxkbbrmlpxpxbhnpofcjmxemyvqqdbanwd",
"output": "IGNORE HIM!"
},
{
"input": "acrzbavz",
"output": "CHAT WITH HER!"
},
{
"input": "drvzznznvrzskftnrhvvzxcalwutxmdza",
"output": "IGNORE HIM!"
},
{
"input": "oacwxipdfcoabhkwxqdbtowiekpnflnqhlrkustgzryvws",
"output": "CHAT WITH HER!"
},
{
"input": "tpnwfmfsibnccvdwjvzviyvjfljupinfigfunyff",
"output": "CHAT WITH HER!"
},
{
"input": "gavaihhamfolcndgytcsgucqdqngxkrlovpthvteacmmthoglxu",
"output": "CHAT WITH HER!"
},
{
"input": "hsfcfvameeupldgvchmogrvwxrvsmnwxxkxoawwodtsahqvehlcw",
"output": "IGNORE HIM!"
},
{
"input": "sbkydrscoojychxchqsuciperfroumenelgiyiwlqfwximrgdbyvkmacy",
"output": "CHAT WITH HER!"
},
{
"input": "rhh",
"output": "CHAT WITH HER!"
},
{
"input": "zhdouqfmlkenjzdijxdfxnlegxeuvhelo",
"output": "IGNORE HIM!"
},
{
"input": "yufkkfwyhhvcjntsgsvpzbhqtmtgyxifqoewmuplphykmptfdebjxuaxigomjtwgtljwdjhjernkitifbomifbhysnmadtnyn",
"output": "CHAT WITH HER!"
},
{
"input": "urigreuzpxnej",
"output": "CHAT WITH HER!"
}
] | 1,699,116,687
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 0
| 154
| 0
|
s = input()
d = set()
for i in range(len(s)):
d.add(s[i])
print(len(d))
|
Title: Boy or Girl
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Those days, many boys use beautiful girls' photos as avatars in forums. So it is pretty hard to tell the gender of a user at the first glance. Last year, our hero went to a forum and had a nice chat with a beauty (he thought so). After that they talked very often and eventually they became a couple in the network.
But yesterday, he came to see "her" in the real world and found out "she" is actually a very strong man! Our hero is very sad and he is too tired to love again now. So he came up with a way to recognize users' genders by their user names.
This is his method: if the number of distinct characters in one's user name is odd, then he is a male, otherwise she is a female. You are given the string that denotes the user name, please help our hero to determine the gender of this user by his method.
Input Specification:
The first line contains a non-empty string, that contains only lowercase English letters — the user name. This string contains at most 100 letters.
Output Specification:
If it is a female by our hero's method, print "CHAT WITH HER!" (without the quotes), otherwise, print "IGNORE HIM!" (without the quotes).
Demo Input:
['wjmzbmr\n', 'xiaodao\n', 'sevenkplus\n']
Demo Output:
['CHAT WITH HER!\n', 'IGNORE HIM!\n', 'CHAT WITH HER!\n']
Note:
For the first example. There are 6 distinct characters in "wjmzbmr". These characters are: "w", "j", "m", "z", "b", "r". So wjmzbmr is a female and you should print "CHAT WITH HER!".
|
```python
s = input()
d = set()
for i in range(len(s)):
d.add(s[i])
print(len(d))
```
| 0
|
|
5
|
C
|
Longest Regular Bracket Sequence
|
PROGRAMMING
| 1,900
|
[
"constructive algorithms",
"data structures",
"dp",
"greedy",
"sortings",
"strings"
] |
C. Longest Regular Bracket Sequence
|
2
|
256
|
This is yet another problem dealing with regular bracket sequences.
We should remind you that a bracket sequence is called regular, if by inserting «+» and «1» into it we can get a correct mathematical expression. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not.
You are given a string of «(» and «)» characters. You are to find its longest substring that is a regular bracket sequence. You are to find the number of such substrings as well.
|
The first line of the input file contains a non-empty string, consisting of «(» and «)» characters. Its length does not exceed 106.
|
Print the length of the longest substring that is a regular bracket sequence, and the number of such substrings. If there are no such substrings, write the only line containing "0 1".
|
[
")((())))(()())\n",
"))(\n"
] |
[
"6 2\n",
"0 1\n"
] |
none
| 0
|
[
{
"input": ")((())))(()())",
"output": "6 2"
},
{
"input": "))(",
"output": "0 1"
},
{
"input": "()(())()",
"output": "8 1"
},
{
"input": "((((()(((",
"output": "2 1"
},
{
"input": "))))()())))",
"output": "4 1"
},
{
"input": "(()())()(())()()())())()((()(()(())()()())((()(())()(()()()()))()(())()(((()())()(()((())()(())(()))",
"output": "28 1"
},
{
"input": ")(()(((((())))))(((()))))))((()(((((((())())()))))((((((((())))((((((())))))(()((((()))))(()))))))))",
"output": "68 1"
},
{
"input": "())(((((())())((((()))(())))())())(((()(()()()())(())()))((()(())())()()))()(()())))))(()))((())((((",
"output": "80 1"
},
{
"input": "))(((((((()((((())(((((((((()()))))(()())()((())())))((((()))(()()()))))(()()))())))))((())())()))((",
"output": "92 1"
}
] | 1,604,769,211
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 0
| 186
| 0
|
s=input()
l=[]
n=len(s)
p=0
c=0
for i in range(n):
if s[i]=='(':
p=p+1
if s[i]==')':
p=p-1
if p>0:
c=c+1
if p==0:
l.append(c)
c=0
elif p<0:
l.append(c)
c=0
p=0
if p>0:
l.append(c-p)
l.sort()
m1,m2=max(l),l.count(max(l))
print(m1,m2)
|
Title: Longest Regular Bracket Sequence
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
This is yet another problem dealing with regular bracket sequences.
We should remind you that a bracket sequence is called regular, if by inserting «+» and «1» into it we can get a correct mathematical expression. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not.
You are given a string of «(» and «)» characters. You are to find its longest substring that is a regular bracket sequence. You are to find the number of such substrings as well.
Input Specification:
The first line of the input file contains a non-empty string, consisting of «(» and «)» characters. Its length does not exceed 106.
Output Specification:
Print the length of the longest substring that is a regular bracket sequence, and the number of such substrings. If there are no such substrings, write the only line containing "0 1".
Demo Input:
[')((())))(()())\n', '))(\n']
Demo Output:
['6 2\n', '0 1\n']
Note:
none
|
```python
s=input()
l=[]
n=len(s)
p=0
c=0
for i in range(n):
if s[i]=='(':
p=p+1
if s[i]==')':
p=p-1
if p>0:
c=c+1
if p==0:
l.append(c)
c=0
elif p<0:
l.append(c)
c=0
p=0
if p>0:
l.append(c-p)
l.sort()
m1,m2=max(l),l.count(max(l))
print(m1,m2)
```
| 0
|
281
|
A
|
Word Capitalization
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] | null | null |
Capitalization is writing a word with its first letter as a capital letter. Your task is to capitalize the given word.
Note, that during capitalization all the letters except the first one remains unchanged.
|
A single line contains a non-empty word. This word consists of lowercase and uppercase English letters. The length of the word will not exceed 103.
|
Output the given word after capitalization.
|
[
"ApPLe\n",
"konjac\n"
] |
[
"ApPLe\n",
"Konjac\n"
] |
none
| 500
|
[
{
"input": "ApPLe",
"output": "ApPLe"
},
{
"input": "konjac",
"output": "Konjac"
},
{
"input": "a",
"output": "A"
},
{
"input": "A",
"output": "A"
},
{
"input": "z",
"output": "Z"
},
{
"input": "ABACABA",
"output": "ABACABA"
},
{
"input": "xYaPxPxHxGePfGtQySlNrLxSjDtNnTaRaEpAhPaQpWnDzMqGgRgEwJxGiBdZnMtHxFbObCaGiCeZkUqIgBhHtNvAqAlHpMnQhNeQbMyZrCdElVwHtKrPpJjIaHuIlYwHaRkAkUpPlOhNlBtXwDsKzPyHrPiUwNlXtTaPuMwTqYtJySgFoXvLiHbQwMjSvXsQfKhVlOxGdQkWjBhEyQvBjPoFkThNeRhTuIzFjInJtEfPjOlOsJpJuLgLzFnZmKvFgFrNsOnVqFcNiMfCqTpKnVyLwNqFiTySpWeTdFnWuTwDkRjVxNyQvTrOoEiExYiFaIrLoFmJfZcDkHuWjYfCeEqCvEsZiWnJaEmFbMjDvYwEeJeGcKbVbChGsIzNlExHzHiTlHcSaKxLuZxX",
"output": "XYaPxPxHxGePfGtQySlNrLxSjDtNnTaRaEpAhPaQpWnDzMqGgRgEwJxGiBdZnMtHxFbObCaGiCeZkUqIgBhHtNvAqAlHpMnQhNeQbMyZrCdElVwHtKrPpJjIaHuIlYwHaRkAkUpPlOhNlBtXwDsKzPyHrPiUwNlXtTaPuMwTqYtJySgFoXvLiHbQwMjSvXsQfKhVlOxGdQkWjBhEyQvBjPoFkThNeRhTuIzFjInJtEfPjOlOsJpJuLgLzFnZmKvFgFrNsOnVqFcNiMfCqTpKnVyLwNqFiTySpWeTdFnWuTwDkRjVxNyQvTrOoEiExYiFaIrLoFmJfZcDkHuWjYfCeEqCvEsZiWnJaEmFbMjDvYwEeJeGcKbVbChGsIzNlExHzHiTlHcSaKxLuZxX"
},
{
"input": "rZhIcQlXpNcPgXrOjTiOlMoTgXgIhCfMwZfWoFzGhEkQlOoMjIuShPlZfWkNnMyQfYdUhVgQuSmYoElEtZpDyHtOxXgCpWbZqSbYnPqBcNqRtPgCnJnAyIvNsAhRbNeVlMwZyRyJnFgIsCnSbOdLvUyIeOzQvRpMoMoHfNhHwKvTcHuYnYySfPmAiNwAiWdZnWlLvGfBbRbRrCrBqIgIdWkWiBsNyYkKdNxZdGaToSsDnXpRaGrKxBpQsCzBdQgZzBkGeHgGxNrIyQlSzWsTmSnZwOcHqQpNcQvJlPvKaPiQaMaYsQjUeCqQdCjPgUbDmWiJmNiXgExLqOcCtSwSePnUxIuZfIfBeWbEiVbXnUsPwWyAiXyRbZgKwOqFfCtQuKxEmVeRlAkOeXkO",
"output": "RZhIcQlXpNcPgXrOjTiOlMoTgXgIhCfMwZfWoFzGhEkQlOoMjIuShPlZfWkNnMyQfYdUhVgQuSmYoElEtZpDyHtOxXgCpWbZqSbYnPqBcNqRtPgCnJnAyIvNsAhRbNeVlMwZyRyJnFgIsCnSbOdLvUyIeOzQvRpMoMoHfNhHwKvTcHuYnYySfPmAiNwAiWdZnWlLvGfBbRbRrCrBqIgIdWkWiBsNyYkKdNxZdGaToSsDnXpRaGrKxBpQsCzBdQgZzBkGeHgGxNrIyQlSzWsTmSnZwOcHqQpNcQvJlPvKaPiQaMaYsQjUeCqQdCjPgUbDmWiJmNiXgExLqOcCtSwSePnUxIuZfIfBeWbEiVbXnUsPwWyAiXyRbZgKwOqFfCtQuKxEmVeRlAkOeXkO"
},
{
"input": "hDgZlUmLhYbLkLcNcKeOwJwTePbOvLaRvNzQbSbLsPeHqLhUqWtUbNdQfQqFfXeJqJwWuOrFnDdZiPxIkDyVmHbHvXfIlFqSgAcSyWbOlSlRuPhWdEpEzEeLnXwCtWuVcHaUeRgCiYsIvOaIgDnFuDbRnMoCmPrZfLeFpSjQaTfHgZwZvAzDuSeNwSoWuJvLqKqAuUxFaCxFfRcEjEsJpOfCtDiVrBqNsNwPuGoRgPzRpLpYnNyQxKaNnDnYiJrCrVcHlOxPiPcDbEgKfLwBjLhKcNeMgJhJmOiJvPfOaPaEuGqWvRbErKrIpDkEoQnKwJnTlStLyNsHyOjZfKoIjXwUvRrWpSyYhRpQdLqGmErAiNcGqAqIrTeTiMuPmCrEkHdBrLyCxPtYpRqD",
"output": "HDgZlUmLhYbLkLcNcKeOwJwTePbOvLaRvNzQbSbLsPeHqLhUqWtUbNdQfQqFfXeJqJwWuOrFnDdZiPxIkDyVmHbHvXfIlFqSgAcSyWbOlSlRuPhWdEpEzEeLnXwCtWuVcHaUeRgCiYsIvOaIgDnFuDbRnMoCmPrZfLeFpSjQaTfHgZwZvAzDuSeNwSoWuJvLqKqAuUxFaCxFfRcEjEsJpOfCtDiVrBqNsNwPuGoRgPzRpLpYnNyQxKaNnDnYiJrCrVcHlOxPiPcDbEgKfLwBjLhKcNeMgJhJmOiJvPfOaPaEuGqWvRbErKrIpDkEoQnKwJnTlStLyNsHyOjZfKoIjXwUvRrWpSyYhRpQdLqGmErAiNcGqAqIrTeTiMuPmCrEkHdBrLyCxPtYpRqD"
},
{
"input": "qUdLgGrJeGmIzIeZrCjUtBpYfRvNdXdRpGsThIsEmJjTiMqEwRxBeBaSxEuWrNvExKePjPnXhPzBpWnHiDhTvZhBuIjDnZpTcEkCvRkAcTmMuXhGgErWgFyGyToOyVwYlCuQpTfJkVdWmFyBqQhJjYtXrBbFdHzDlGsFbHmHbFgXgFhIyDhZyEqEiEwNxSeByBwLiVeSnCxIdHbGjOjJrZeVkOzGeMmQrJkVyGhDtCzOlPeAzGrBlWwEnAdUfVaIjNrRyJjCnHkUvFuKuKeKbLzSbEmUcXtVkZzXzKlOrPgQiDmCcCvIyAdBwOeUuLbRmScNcWxIkOkJuIsBxTrIqXhDzLcYdVtPgZdZfAxTmUtByGiTsJkSySjXdJvEwNmSmNoWsChPdAzJrBoW",
"output": "QUdLgGrJeGmIzIeZrCjUtBpYfRvNdXdRpGsThIsEmJjTiMqEwRxBeBaSxEuWrNvExKePjPnXhPzBpWnHiDhTvZhBuIjDnZpTcEkCvRkAcTmMuXhGgErWgFyGyToOyVwYlCuQpTfJkVdWmFyBqQhJjYtXrBbFdHzDlGsFbHmHbFgXgFhIyDhZyEqEiEwNxSeByBwLiVeSnCxIdHbGjOjJrZeVkOzGeMmQrJkVyGhDtCzOlPeAzGrBlWwEnAdUfVaIjNrRyJjCnHkUvFuKuKeKbLzSbEmUcXtVkZzXzKlOrPgQiDmCcCvIyAdBwOeUuLbRmScNcWxIkOkJuIsBxTrIqXhDzLcYdVtPgZdZfAxTmUtByGiTsJkSySjXdJvEwNmSmNoWsChPdAzJrBoW"
},
{
"input": "kHbApGoBcLmIwUlXkVgUmWzYeLoDbGaOkWbIuXoRwMfKuOoMzAoXrBoTvYxGrMbRjDuRxAbGsTnErIiHnHoLeRnTbFiRfDdOkNlWiAcOsChLdLqFqXlDpDoDtPxXqAmSvYgPvOcCpOlWtOjYwFkGkHuCaHwZcFdOfHjBmIxTeSiHkWjXyFcCtOlSuJsZkDxUgPeZkJwMmNpErUlBcGuMlJwKkWnOzFeFiSiPsEvMmQiCsYeHlLuHoMgBjFoZkXlObDkSoQcVyReTmRsFzRhTuIvCeBqVsQdQyTyZjStGrTyDcEcAgTgMiIcVkLbZbGvWeHtXwEqWkXfTcPyHhHjYwIeVxLyVmHmMkUsGiHmNnQuMsXaFyPpVqNrBhOiWmNkBbQuHvQdOjPjKiZcL",
"output": "KHbApGoBcLmIwUlXkVgUmWzYeLoDbGaOkWbIuXoRwMfKuOoMzAoXrBoTvYxGrMbRjDuRxAbGsTnErIiHnHoLeRnTbFiRfDdOkNlWiAcOsChLdLqFqXlDpDoDtPxXqAmSvYgPvOcCpOlWtOjYwFkGkHuCaHwZcFdOfHjBmIxTeSiHkWjXyFcCtOlSuJsZkDxUgPeZkJwMmNpErUlBcGuMlJwKkWnOzFeFiSiPsEvMmQiCsYeHlLuHoMgBjFoZkXlObDkSoQcVyReTmRsFzRhTuIvCeBqVsQdQyTyZjStGrTyDcEcAgTgMiIcVkLbZbGvWeHtXwEqWkXfTcPyHhHjYwIeVxLyVmHmMkUsGiHmNnQuMsXaFyPpVqNrBhOiWmNkBbQuHvQdOjPjKiZcL"
},
{
"input": "aHmRbLgNuWkLxLnWvUbYwTeZeYiOlLhTuOvKfLnVmCiPcMkSgVrYjZiLuRjCiXhAnVzVcTlVeJdBvPdDfFvHkTuIhCdBjEsXbVmGcLrPfNvRdFsZkSdNpYsJeIhIcNqSoLkOjUlYlDmXsOxPbQtIoUxFjGnRtBhFaJvBeEzHsAtVoQbAfYjJqReBiKeUwRqYrUjPjBoHkOkPzDwEwUgTxQxAvKzUpMhKyOhPmEhYhItQwPeKsKaKlUhGuMcTtSwFtXfJsDsFlTtOjVvVfGtBtFlQyIcBaMsPaJlPqUcUvLmReZiFbXxVtRhTzJkLkAjVqTyVuFeKlTyQgUzMsXjOxQnVfTaWmThEnEoIhZeZdStBkKeLpAhJnFoJvQyGwDiStLjEwGfZwBuWsEfC",
"output": "AHmRbLgNuWkLxLnWvUbYwTeZeYiOlLhTuOvKfLnVmCiPcMkSgVrYjZiLuRjCiXhAnVzVcTlVeJdBvPdDfFvHkTuIhCdBjEsXbVmGcLrPfNvRdFsZkSdNpYsJeIhIcNqSoLkOjUlYlDmXsOxPbQtIoUxFjGnRtBhFaJvBeEzHsAtVoQbAfYjJqReBiKeUwRqYrUjPjBoHkOkPzDwEwUgTxQxAvKzUpMhKyOhPmEhYhItQwPeKsKaKlUhGuMcTtSwFtXfJsDsFlTtOjVvVfGtBtFlQyIcBaMsPaJlPqUcUvLmReZiFbXxVtRhTzJkLkAjVqTyVuFeKlTyQgUzMsXjOxQnVfTaWmThEnEoIhZeZdStBkKeLpAhJnFoJvQyGwDiStLjEwGfZwBuWsEfC"
},
{
"input": "sLlZkDiDmEdNaXuUuJwHqYvRtOdGfTiTpEpAoSqAbJaChOiCvHgSwZwEuPkMmXiLcKdXqSsEyViEbZpZsHeZpTuXoGcRmOiQfBfApPjDqSqElWeSeOhUyWjLyNoRuYeGfGwNqUsQoTyVvWeNgNdZfDxGwGfLsDjIdInSqDlMuNvFaHbScZkTlVwNcJpEjMaPaOtFgJjBjOcLlLmDnQrShIrJhOcUmPnZhTxNeClQsZaEaVaReLyQpLwEqJpUwYhLiRzCzKfOoFeTiXzPiNbOsZaZaLgCiNnMkBcFwGgAwPeNyTxJcCtBgXcToKlWaWcBaIvBpNxPeClQlWeQqRyEtAkJdBtSrFdDvAbUlKyLdCuTtXxFvRcKnYnWzVdYqDeCmOqPxUaFjQdTdCtN",
"output": "SLlZkDiDmEdNaXuUuJwHqYvRtOdGfTiTpEpAoSqAbJaChOiCvHgSwZwEuPkMmXiLcKdXqSsEyViEbZpZsHeZpTuXoGcRmOiQfBfApPjDqSqElWeSeOhUyWjLyNoRuYeGfGwNqUsQoTyVvWeNgNdZfDxGwGfLsDjIdInSqDlMuNvFaHbScZkTlVwNcJpEjMaPaOtFgJjBjOcLlLmDnQrShIrJhOcUmPnZhTxNeClQsZaEaVaReLyQpLwEqJpUwYhLiRzCzKfOoFeTiXzPiNbOsZaZaLgCiNnMkBcFwGgAwPeNyTxJcCtBgXcToKlWaWcBaIvBpNxPeClQlWeQqRyEtAkJdBtSrFdDvAbUlKyLdCuTtXxFvRcKnYnWzVdYqDeCmOqPxUaFjQdTdCtN"
},
{
"input": "iRuStKvVhJdJbQwRoIuLiVdTpKaOqKfYlYwAzIpPtUwUtMeKyCaOlXmVrKwWeImYmVuXdLkRlHwFxKqZbZtTzNgOzDbGqTfZnKmUzAcIjDcEmQgYyFbEfWzRpKvCkDmAqDiIiRcLvMxWaJqCgYqXgIcLdNaZlBnXtJyKaMnEaWfXfXwTbDnAiYnWqKbAtDpYdUbZrCzWgRnHzYxFgCdDbOkAgTqBuLqMeStHcDxGnVhSgMzVeTaZoTfLjMxQfRuPcFqVlRyYdHyOdJsDoCeWrUuJyIiAqHwHyVpEeEoMaJwAoUfPtBeJqGhMaHiBjKwAlXoZpUsDhHgMxBkVbLcEvNtJbGnPsUwAvXrAkTlXwYvEnOpNeWyIkRnEnTrIyAcLkRgMyYcKrGiDaAyE",
"output": "IRuStKvVhJdJbQwRoIuLiVdTpKaOqKfYlYwAzIpPtUwUtMeKyCaOlXmVrKwWeImYmVuXdLkRlHwFxKqZbZtTzNgOzDbGqTfZnKmUzAcIjDcEmQgYyFbEfWzRpKvCkDmAqDiIiRcLvMxWaJqCgYqXgIcLdNaZlBnXtJyKaMnEaWfXfXwTbDnAiYnWqKbAtDpYdUbZrCzWgRnHzYxFgCdDbOkAgTqBuLqMeStHcDxGnVhSgMzVeTaZoTfLjMxQfRuPcFqVlRyYdHyOdJsDoCeWrUuJyIiAqHwHyVpEeEoMaJwAoUfPtBeJqGhMaHiBjKwAlXoZpUsDhHgMxBkVbLcEvNtJbGnPsUwAvXrAkTlXwYvEnOpNeWyIkRnEnTrIyAcLkRgMyYcKrGiDaAyE"
},
{
"input": "cRtJkOxHzUbJcDdHzJtLbVmSoWuHoTkVrPqQaVmXeBrHxJbQfNrQbAaMrEhVdQnPxNyCjErKxPoEdWkVrBbDeNmEgBxYiBtWdAfHiLuSwIxJuHpSkAxPoYdNkGoLySsNhUmGoZhDzAfWhJdPlJzQkZbOnMtTkClIoCqOlIcJcMlGjUyOiEmHdYfIcPtTgQhLlLcPqQjAnQnUzHpCaQsCnYgQsBcJrQwBnWsIwFfSfGuYgTzQmShFpKqEeRlRkVfMuZbUsDoFoPrNuNwTtJqFkRiXxPvKyElDzLoUnIwAaBaOiNxMpEvPzSpGpFhMtGhGdJrFnZmNiMcUfMtBnDuUnXqDcMsNyGoLwLeNnLfRsIwRfBtXkHrFcPsLdXaAoYaDzYnZuQeVcZrElWmP",
"output": "CRtJkOxHzUbJcDdHzJtLbVmSoWuHoTkVrPqQaVmXeBrHxJbQfNrQbAaMrEhVdQnPxNyCjErKxPoEdWkVrBbDeNmEgBxYiBtWdAfHiLuSwIxJuHpSkAxPoYdNkGoLySsNhUmGoZhDzAfWhJdPlJzQkZbOnMtTkClIoCqOlIcJcMlGjUyOiEmHdYfIcPtTgQhLlLcPqQjAnQnUzHpCaQsCnYgQsBcJrQwBnWsIwFfSfGuYgTzQmShFpKqEeRlRkVfMuZbUsDoFoPrNuNwTtJqFkRiXxPvKyElDzLoUnIwAaBaOiNxMpEvPzSpGpFhMtGhGdJrFnZmNiMcUfMtBnDuUnXqDcMsNyGoLwLeNnLfRsIwRfBtXkHrFcPsLdXaAoYaDzYnZuQeVcZrElWmP"
},
{
"input": "wVaCsGxZrBbFnTbKsCoYlAvUkIpBaYpYmJkMlPwCaFvUkDxAiJgIqWsFqZlFvTtAnGzEwXbYiBdFfFxRiDoUkLmRfAwOlKeOlKgXdUnVqLkTuXtNdQpBpXtLvZxWoBeNePyHcWmZyRiUkPlRqYiQdGeXwOhHbCqVjDcEvJmBkRwWnMqPjXpUsIyXqGjHsEsDwZiFpIbTkQaUlUeFxMwJzSaHdHnDhLaLdTuYgFuJsEcMmDvXyPjKsSeBaRwNtPuOuBtNeOhQdVgKzPzOdYtPjPfDzQzHoWcYjFbSvRgGdGsCmGnQsErToBkCwGeQaCbBpYkLhHxTbUvRnJpZtXjKrHdRiUmUbSlJyGaLnWsCrJbBnSjFaZrIzIrThCmGhQcMsTtOxCuUcRaEyPaG",
"output": "WVaCsGxZrBbFnTbKsCoYlAvUkIpBaYpYmJkMlPwCaFvUkDxAiJgIqWsFqZlFvTtAnGzEwXbYiBdFfFxRiDoUkLmRfAwOlKeOlKgXdUnVqLkTuXtNdQpBpXtLvZxWoBeNePyHcWmZyRiUkPlRqYiQdGeXwOhHbCqVjDcEvJmBkRwWnMqPjXpUsIyXqGjHsEsDwZiFpIbTkQaUlUeFxMwJzSaHdHnDhLaLdTuYgFuJsEcMmDvXyPjKsSeBaRwNtPuOuBtNeOhQdVgKzPzOdYtPjPfDzQzHoWcYjFbSvRgGdGsCmGnQsErToBkCwGeQaCbBpYkLhHxTbUvRnJpZtXjKrHdRiUmUbSlJyGaLnWsCrJbBnSjFaZrIzIrThCmGhQcMsTtOxCuUcRaEyPaG"
},
{
"input": "kEiLxLmPjGzNoGkJdBlAfXhThYhMsHmZoZbGyCvNiUoLoZdAxUbGyQiEfXvPzZzJrPbEcMpHsMjIkRrVvDvQtHuKmXvGpQtXbPzJpFjJdUgWcPdFxLjLtXgVpEiFhImHnKkGiWnZbJqRjCyEwHsNbYfYfTyBaEuKlCtWnOqHmIgGrFmQiYrBnLiFcGuZxXlMfEuVoCxPkVrQvZoIpEhKsYtXrPxLcSfQqXsWaDgVlOnAzUvAhOhMrJfGtWcOwQfRjPmGhDyAeXrNqBvEiDfCiIvWxPjTwPlXpVsMjVjUnCkXgBuWnZaDyJpWkCfBrWnHxMhJgItHdRqNrQaEeRjAuUwRkUdRhEeGlSqVqGmOjNcUhFfXjCmWzBrGvIuZpRyWkWiLyUwFpYjNmNfV",
"output": "KEiLxLmPjGzNoGkJdBlAfXhThYhMsHmZoZbGyCvNiUoLoZdAxUbGyQiEfXvPzZzJrPbEcMpHsMjIkRrVvDvQtHuKmXvGpQtXbPzJpFjJdUgWcPdFxLjLtXgVpEiFhImHnKkGiWnZbJqRjCyEwHsNbYfYfTyBaEuKlCtWnOqHmIgGrFmQiYrBnLiFcGuZxXlMfEuVoCxPkVrQvZoIpEhKsYtXrPxLcSfQqXsWaDgVlOnAzUvAhOhMrJfGtWcOwQfRjPmGhDyAeXrNqBvEiDfCiIvWxPjTwPlXpVsMjVjUnCkXgBuWnZaDyJpWkCfBrWnHxMhJgItHdRqNrQaEeRjAuUwRkUdRhEeGlSqVqGmOjNcUhFfXjCmWzBrGvIuZpRyWkWiLyUwFpYjNmNfV"
},
{
"input": "eIhDoLmDeReKqXsHcVgFxUqNfScAiQnFrTlCgSuTtXiYvBxKaPaGvUeYfSgHqEaWcHxKpFaSlCxGqAmNeFcIzFcZsBiVoZhUjXaDaIcKoBzYdIlEnKfScRqSkYpPtVsVhXsBwUsUfAqRoCkBxWbHgDiCkRtPvUwVgDjOzObYwNiQwXlGnAqEkHdSqLgUkOdZiWaHqQnOhUnDhIzCiQtVcJlGoRfLuVlFjWqSuMsLgLwOdZvKtWdRuRqDoBoInKqPbJdXpIqLtFlMlDaWgSiKbFpCxOnQeNeQzXeKsBzIjCyPxCmBnYuHzQoYxZgGzSgGtZiTeQmUeWlNzZeKiJbQmEjIiDhPeSyZlNdHpZnIkPdJzSeJpPiXxToKyBjJfPwNzZpWzIzGySqPxLtI",
"output": "EIhDoLmDeReKqXsHcVgFxUqNfScAiQnFrTlCgSuTtXiYvBxKaPaGvUeYfSgHqEaWcHxKpFaSlCxGqAmNeFcIzFcZsBiVoZhUjXaDaIcKoBzYdIlEnKfScRqSkYpPtVsVhXsBwUsUfAqRoCkBxWbHgDiCkRtPvUwVgDjOzObYwNiQwXlGnAqEkHdSqLgUkOdZiWaHqQnOhUnDhIzCiQtVcJlGoRfLuVlFjWqSuMsLgLwOdZvKtWdRuRqDoBoInKqPbJdXpIqLtFlMlDaWgSiKbFpCxOnQeNeQzXeKsBzIjCyPxCmBnYuHzQoYxZgGzSgGtZiTeQmUeWlNzZeKiJbQmEjIiDhPeSyZlNdHpZnIkPdJzSeJpPiXxToKyBjJfPwNzZpWzIzGySqPxLtI"
},
{
"input": "uOoQzIeTwYeKpJtGoUdNiXbPgEwVsZkAnJcArHxIpEnEhZwQhZvAiOuLeMkVqLeDsAyKeYgFxGmRoLaRsZjAeXgNfYhBkHeDrHdPuTuYhKmDlAvYzYxCdYgYfVaYlGeVqTeSfBxQePbQrKsTaIkGzMjFrQlJuYaMxWpQkLdEcDsIiMnHnDtThRvAcKyGwBsHqKdXpJfIeTeZtYjFbMeUoXoXzGrShTwSwBpQlKeDrZdCjRqNtXoTsIzBkWbMsObTtDvYaPhUeLeHqHeMpZmTaCcIqXzAmGnPfNdDaFhOqWqDrWuFiBpRjZrQmAdViOuMbFfRyXyWfHgRkGpPnDrEqQcEmHcKpEvWlBrOtJbUaXbThJaSxCbVoGvTmHvZrHvXpCvLaYbRiHzYuQyX",
"output": "UOoQzIeTwYeKpJtGoUdNiXbPgEwVsZkAnJcArHxIpEnEhZwQhZvAiOuLeMkVqLeDsAyKeYgFxGmRoLaRsZjAeXgNfYhBkHeDrHdPuTuYhKmDlAvYzYxCdYgYfVaYlGeVqTeSfBxQePbQrKsTaIkGzMjFrQlJuYaMxWpQkLdEcDsIiMnHnDtThRvAcKyGwBsHqKdXpJfIeTeZtYjFbMeUoXoXzGrShTwSwBpQlKeDrZdCjRqNtXoTsIzBkWbMsObTtDvYaPhUeLeHqHeMpZmTaCcIqXzAmGnPfNdDaFhOqWqDrWuFiBpRjZrQmAdViOuMbFfRyXyWfHgRkGpPnDrEqQcEmHcKpEvWlBrOtJbUaXbThJaSxCbVoGvTmHvZrHvXpCvLaYbRiHzYuQyX"
},
{
"input": "lZqBqKeGvNdSeYuWxRiVnFtYbKuJwQtUcKnVtQhAlOeUzMaAuTaEnDdPfDcNyHgEoBmYjZyFePeJrRiKyAzFnBfAuGiUyLrIeLrNhBeBdVcEeKgCcBrQzDsPwGcNnZvTsEaYmFfMeOmMdNuZbUtDoQoNcGwDqEkEjIdQaPwAxJbXeNxOgKgXoEbZiIsVkRrNpNyAkLeHkNfEpLuQvEcMbIoGaDzXbEtNsLgGfOkZaFiUsOvEjVeCaMcZqMzKeAdXxJsVeCrZaFpJtZxInQxFaSmGgSsVyGeLlFgFqTpIbAvPkIfJrVcJeBxSdEvPyVwIjHpYrLrKqLnAmCuGmPoZrSbOtGaLaTmBmSuUyAmAsRiMqOtRjJhPhAfXaJnTpLbFqPmJgFcBxImTqIiJ",
"output": "LZqBqKeGvNdSeYuWxRiVnFtYbKuJwQtUcKnVtQhAlOeUzMaAuTaEnDdPfDcNyHgEoBmYjZyFePeJrRiKyAzFnBfAuGiUyLrIeLrNhBeBdVcEeKgCcBrQzDsPwGcNnZvTsEaYmFfMeOmMdNuZbUtDoQoNcGwDqEkEjIdQaPwAxJbXeNxOgKgXoEbZiIsVkRrNpNyAkLeHkNfEpLuQvEcMbIoGaDzXbEtNsLgGfOkZaFiUsOvEjVeCaMcZqMzKeAdXxJsVeCrZaFpJtZxInQxFaSmGgSsVyGeLlFgFqTpIbAvPkIfJrVcJeBxSdEvPyVwIjHpYrLrKqLnAmCuGmPoZrSbOtGaLaTmBmSuUyAmAsRiMqOtRjJhPhAfXaJnTpLbFqPmJgFcBxImTqIiJ"
},
{
"input": "P",
"output": "P"
},
{
"input": "Xyzzy",
"output": "Xyzzy"
},
{
"input": "Zzz",
"output": "Zzz"
},
{
"input": "Zp",
"output": "Zp"
}
] | 1,696,867,000
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
str_ = input()
print(str_[0].capitalize() + str_[1:-1])
|
Title: Word Capitalization
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Capitalization is writing a word with its first letter as a capital letter. Your task is to capitalize the given word.
Note, that during capitalization all the letters except the first one remains unchanged.
Input Specification:
A single line contains a non-empty word. This word consists of lowercase and uppercase English letters. The length of the word will not exceed 103.
Output Specification:
Output the given word after capitalization.
Demo Input:
['ApPLe\n', 'konjac\n']
Demo Output:
['ApPLe\n', 'Konjac\n']
Note:
none
|
```python
str_ = input()
print(str_[0].capitalize() + str_[1:-1])
```
| 0
|
|
753
|
A
|
Santa Claus and Candies
|
PROGRAMMING
| 1,000
|
[
"dp",
"greedy",
"math"
] | null | null |
Santa Claus has *n* candies, he dreams to give them as gifts to children.
What is the maximal number of children for whose he can give candies if Santa Claus want each kid should get distinct positive integer number of candies. Santa Class wants to give all *n* candies he has.
|
The only line contains positive integer number *n* (1<=≤<=*n*<=≤<=1000) — number of candies Santa Claus has.
|
Print to the first line integer number *k* — maximal number of kids which can get candies.
Print to the second line *k* distinct integer numbers: number of candies for each of *k* kid. The sum of *k* printed numbers should be exactly *n*.
If there are many solutions, print any of them.
|
[
"5\n",
"9\n",
"2\n"
] |
[
"2\n2 3\n",
"3\n3 5 1\n",
"1\n2 \n"
] |
none
| 500
|
[
{
"input": "5",
"output": "2\n1 4 "
},
{
"input": "9",
"output": "3\n1 2 6 "
},
{
"input": "2",
"output": "1\n2 "
},
{
"input": "1",
"output": "1\n1 "
},
{
"input": "3",
"output": "2\n1 2 "
},
{
"input": "1000",
"output": "44\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 54 "
},
{
"input": "4",
"output": "2\n1 3 "
},
{
"input": "6",
"output": "3\n1 2 3 "
},
{
"input": "7",
"output": "3\n1 2 4 "
},
{
"input": "8",
"output": "3\n1 2 5 "
},
{
"input": "10",
"output": "4\n1 2 3 4 "
},
{
"input": "11",
"output": "4\n1 2 3 5 "
},
{
"input": "12",
"output": "4\n1 2 3 6 "
},
{
"input": "13",
"output": "4\n1 2 3 7 "
},
{
"input": "14",
"output": "4\n1 2 3 8 "
},
{
"input": "15",
"output": "5\n1 2 3 4 5 "
},
{
"input": "16",
"output": "5\n1 2 3 4 6 "
},
{
"input": "20",
"output": "5\n1 2 3 4 10 "
},
{
"input": "21",
"output": "6\n1 2 3 4 5 6 "
},
{
"input": "22",
"output": "6\n1 2 3 4 5 7 "
},
{
"input": "27",
"output": "6\n1 2 3 4 5 12 "
},
{
"input": "28",
"output": "7\n1 2 3 4 5 6 7 "
},
{
"input": "29",
"output": "7\n1 2 3 4 5 6 8 "
},
{
"input": "35",
"output": "7\n1 2 3 4 5 6 14 "
},
{
"input": "36",
"output": "8\n1 2 3 4 5 6 7 8 "
},
{
"input": "37",
"output": "8\n1 2 3 4 5 6 7 9 "
},
{
"input": "44",
"output": "8\n1 2 3 4 5 6 7 16 "
},
{
"input": "45",
"output": "9\n1 2 3 4 5 6 7 8 9 "
},
{
"input": "46",
"output": "9\n1 2 3 4 5 6 7 8 10 "
},
{
"input": "230",
"output": "20\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 40 "
},
{
"input": "231",
"output": "21\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 "
},
{
"input": "232",
"output": "21\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 22 "
},
{
"input": "239",
"output": "21\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 29 "
},
{
"input": "629",
"output": "34\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 68 "
},
{
"input": "630",
"output": "35\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 "
},
{
"input": "631",
"output": "35\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 36 "
},
{
"input": "945",
"output": "42\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 84 "
},
{
"input": "946",
"output": "43\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 "
},
{
"input": "947",
"output": "43\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 44 "
},
{
"input": "989",
"output": "43\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 86 "
},
{
"input": "990",
"output": "44\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 "
},
{
"input": "991",
"output": "44\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 45 "
},
{
"input": "956",
"output": "43\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 53 "
},
{
"input": "981",
"output": "43\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 78 "
},
{
"input": "867",
"output": "41\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 47 "
},
{
"input": "906",
"output": "42\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 45 "
},
{
"input": "999",
"output": "44\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 53 "
},
{
"input": "100",
"output": "13\n1 2 3 4 5 6 7 8 9 10 11 12 22 "
},
{
"input": "126",
"output": "15\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 21 "
}
] | 1,483,002,691
| 391
|
PyPy 3
|
COMPILATION_ERROR
|
PRETESTS
| 0
| 0
| 0
|
#include"bits/stdc++.h"
using namespace std;
struct _ {
ios_base::Init i;
_() {
cin.sync_with_stdio(0);
cin.tie(NULL);
cout << fixed << setprecision(25);
srand(time(NULL));
#ifndef DEBUG
#define cerr if(0)cerr
#endif
#ifdef DEBUG
//freopen("input.txt", "r", stdin);
//freopen("output.txt", "w", stdout);
#endif
}
} _;
typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
const int INF = ll(2e9);
int main() {
int n;
cin >> n;
int i, s;
for (i = 1, s = 1; s <= n; s += ++i);
--i;
cout << i << '\n';
for (int j = 1; j < i; ++j)
cout << j << ' ';
cout << n - (i - 1)*i/2;
cerr << "\n\n\nTIME = " << int(clock()*ld(1000)/CLOCKS_PER_SEC) << '\n';
return 0;
}
|
Title: Santa Claus and Candies
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Santa Claus has *n* candies, he dreams to give them as gifts to children.
What is the maximal number of children for whose he can give candies if Santa Claus want each kid should get distinct positive integer number of candies. Santa Class wants to give all *n* candies he has.
Input Specification:
The only line contains positive integer number *n* (1<=≤<=*n*<=≤<=1000) — number of candies Santa Claus has.
Output Specification:
Print to the first line integer number *k* — maximal number of kids which can get candies.
Print to the second line *k* distinct integer numbers: number of candies for each of *k* kid. The sum of *k* printed numbers should be exactly *n*.
If there are many solutions, print any of them.
Demo Input:
['5\n', '9\n', '2\n']
Demo Output:
['2\n2 3\n', '3\n3 5 1\n', '1\n2 \n']
Note:
none
|
```python
#include"bits/stdc++.h"
using namespace std;
struct _ {
ios_base::Init i;
_() {
cin.sync_with_stdio(0);
cin.tie(NULL);
cout << fixed << setprecision(25);
srand(time(NULL));
#ifndef DEBUG
#define cerr if(0)cerr
#endif
#ifdef DEBUG
//freopen("input.txt", "r", stdin);
//freopen("output.txt", "w", stdout);
#endif
}
} _;
typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
const int INF = ll(2e9);
int main() {
int n;
cin >> n;
int i, s;
for (i = 1, s = 1; s <= n; s += ++i);
--i;
cout << i << '\n';
for (int j = 1; j < i; ++j)
cout << j << ' ';
cout << n - (i - 1)*i/2;
cerr << "\n\n\nTIME = " << int(clock()*ld(1000)/CLOCKS_PER_SEC) << '\n';
return 0;
}
```
| -1
|
|
222
|
A
|
Shooshuns and Sequence
|
PROGRAMMING
| 1,200
|
[
"brute force",
"implementation"
] | null | null |
One day shooshuns found a sequence of *n* integers, written on a blackboard. The shooshuns can perform one operation with it, the operation consists of two steps:
1. Find the number that goes *k*-th in the current sequence and add the same number to the end of the sequence; 1. Delete the first number of the current sequence.
The shooshuns wonder after how many operations all numbers on the board will be the same and whether all numbers will ever be the same.
|
The first line contains two space-separated integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105).
The second line contains *n* space-separated integers: *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105) — the sequence that the shooshuns found.
|
Print the minimum number of operations, required for all numbers on the blackboard to become the same. If it is impossible to achieve, print -1.
|
[
"3 2\n3 1 1\n",
"3 1\n3 1 1\n"
] |
[
"1\n",
"-1\n"
] |
In the first test case after the first operation the blackboard will have sequence [1, 1, 1]. So, one operation is enough to make all numbers the same. Thus, the answer equals one.
In the second test case the sequence will never consist of the same numbers. It will always contain at least two distinct numbers 3 and 1. Thus, the answer equals -1.
| 500
|
[
{
"input": "3 2\n3 1 1",
"output": "1"
},
{
"input": "3 1\n3 1 1",
"output": "-1"
},
{
"input": "1 1\n1",
"output": "0"
},
{
"input": "2 1\n1 1",
"output": "0"
},
{
"input": "2 1\n2 1",
"output": "-1"
},
{
"input": "4 4\n1 2 3 4",
"output": "3"
},
{
"input": "4 3\n1 2 3 4",
"output": "-1"
},
{
"input": "5 3\n2 1 1 1 1",
"output": "1"
},
{
"input": "5 3\n1 1 1 1 1",
"output": "0"
},
{
"input": "5 5\n1 1 1 1 1",
"output": "0"
},
{
"input": "10 1\n1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "10 1\n1 2 1 1 1 1 1 1 1 1",
"output": "-1"
},
{
"input": "10 1\n2 1 1 1 1 1 1 1 1 1",
"output": "-1"
},
{
"input": "10 2\n2 1 1 1 1 1 1 1 1 1",
"output": "1"
},
{
"input": "10 2\n1 2 1 1 1 1 1 1 1 1",
"output": "-1"
},
{
"input": "10 3\n3 2 1 1 1 1 1 1 1 1",
"output": "2"
},
{
"input": "10 1\n1 2 3 4 5 6 7 8 9 10",
"output": "-1"
},
{
"input": "10 10\n1 2 3 4 5 6 7 8 9 10",
"output": "9"
},
{
"input": "20 10\n1 2 3 4 5 6 7 8 9 10 1 1 1 1 1 1 1 1 1 1",
"output": "-1"
},
{
"input": "20 11\n1 2 3 4 5 6 7 8 9 10 1 1 1 1 1 1 1 1 1 1",
"output": "10"
},
{
"input": "20 9\n1 2 3 4 5 6 7 8 9 10 1 1 1 1 1 1 1 1 1 1",
"output": "-1"
},
{
"input": "20 10\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "20 20\n20 1 19 2 18 3 17 4 16 5 15 6 14 7 13 8 12 9 11 10",
"output": "19"
}
] | 1,597,576,884
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 156
| 307,200
|
n,k=map(int,input().split())
a=list(map(int,input().split()))
ans=0
j=0
for i in range(n):
a.append(a[k-1])
a.remove(a[0])
ans+=1
if a.count(a[0])==len(a):
print(ans)
j=1
break
if j==0:
print(-1)
|
Title: Shooshuns and Sequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day shooshuns found a sequence of *n* integers, written on a blackboard. The shooshuns can perform one operation with it, the operation consists of two steps:
1. Find the number that goes *k*-th in the current sequence and add the same number to the end of the sequence; 1. Delete the first number of the current sequence.
The shooshuns wonder after how many operations all numbers on the board will be the same and whether all numbers will ever be the same.
Input Specification:
The first line contains two space-separated integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105).
The second line contains *n* space-separated integers: *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105) — the sequence that the shooshuns found.
Output Specification:
Print the minimum number of operations, required for all numbers on the blackboard to become the same. If it is impossible to achieve, print -1.
Demo Input:
['3 2\n3 1 1\n', '3 1\n3 1 1\n']
Demo Output:
['1\n', '-1\n']
Note:
In the first test case after the first operation the blackboard will have sequence [1, 1, 1]. So, one operation is enough to make all numbers the same. Thus, the answer equals one.
In the second test case the sequence will never consist of the same numbers. It will always contain at least two distinct numbers 3 and 1. Thus, the answer equals -1.
|
```python
n,k=map(int,input().split())
a=list(map(int,input().split()))
ans=0
j=0
for i in range(n):
a.append(a[k-1])
a.remove(a[0])
ans+=1
if a.count(a[0])==len(a):
print(ans)
j=1
break
if j==0:
print(-1)
```
| 0
|
|
975
|
A
|
Aramic script
|
PROGRAMMING
| 900
|
[
"implementation",
"strings"
] | null | null |
In Aramic language words can only represent objects.
Words in Aramic have special properties:
- A word is a root if it does not contain the same letter more than once. - A root and all its permutations represent the same object. - The root $x$ of a word $y$ is the word that contains all letters that appear in $y$ in a way that each letter appears once. For example, the root of "aaaa", "aa", "aaa" is "a", the root of "aabb", "bab", "baabb", "ab" is "ab". - Any word in Aramic represents the same object as its root.
You have an ancient script in Aramic. What is the number of different objects mentioned in the script?
|
The first line contains one integer $n$ ($1 \leq n \leq 10^3$) — the number of words in the script.
The second line contains $n$ words $s_1, s_2, \ldots, s_n$ — the script itself. The length of each string does not exceed $10^3$.
It is guaranteed that all characters of the strings are small latin letters.
|
Output one integer — the number of different objects mentioned in the given ancient Aramic script.
|
[
"5\na aa aaa ab abb\n",
"3\namer arem mrea\n"
] |
[
"2",
"1"
] |
In the first test, there are two objects mentioned. The roots that represent them are "a","ab".
In the second test, there is only one object, its root is "amer", the other strings are just permutations of "amer".
| 500
|
[
{
"input": "5\na aa aaa ab abb",
"output": "2"
},
{
"input": "3\namer arem mrea",
"output": "1"
},
{
"input": "10\nbda bbb cda dca dda dcb bcd dcb ada ddd",
"output": "6"
},
{
"input": "2\nfhjlqs aceginpr",
"output": "2"
},
{
"input": "2\nbcdfghimn efghijlmo",
"output": "2"
}
] | 1,525,185,374
| 1,874
|
Python 3
|
OK
|
TESTS
| 30
| 295
| 10,240,000
|
n=int(input())
words=input().split()
setx=set([])
for i in range(len(words)):
k=sorted(words[i])
l=set(k)
words[i]=''.join(l)
setx.add(words[i])
print(len(setx))
|
Title: Aramic script
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In Aramic language words can only represent objects.
Words in Aramic have special properties:
- A word is a root if it does not contain the same letter more than once. - A root and all its permutations represent the same object. - The root $x$ of a word $y$ is the word that contains all letters that appear in $y$ in a way that each letter appears once. For example, the root of "aaaa", "aa", "aaa" is "a", the root of "aabb", "bab", "baabb", "ab" is "ab". - Any word in Aramic represents the same object as its root.
You have an ancient script in Aramic. What is the number of different objects mentioned in the script?
Input Specification:
The first line contains one integer $n$ ($1 \leq n \leq 10^3$) — the number of words in the script.
The second line contains $n$ words $s_1, s_2, \ldots, s_n$ — the script itself. The length of each string does not exceed $10^3$.
It is guaranteed that all characters of the strings are small latin letters.
Output Specification:
Output one integer — the number of different objects mentioned in the given ancient Aramic script.
Demo Input:
['5\na aa aaa ab abb\n', '3\namer arem mrea\n']
Demo Output:
['2', '1']
Note:
In the first test, there are two objects mentioned. The roots that represent them are "a","ab".
In the second test, there is only one object, its root is "amer", the other strings are just permutations of "amer".
|
```python
n=int(input())
words=input().split()
setx=set([])
for i in range(len(words)):
k=sorted(words[i])
l=set(k)
words[i]=''.join(l)
setx.add(words[i])
print(len(setx))
```
| 3
|
|
59
|
A
|
Word
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] |
A. Word
|
2
|
256
|
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
|
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
|
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
|
[
"HoUse\n",
"ViP\n",
"maTRIx\n"
] |
[
"house\n",
"VIP\n",
"matrix\n"
] |
none
| 500
|
[
{
"input": "HoUse",
"output": "house"
},
{
"input": "ViP",
"output": "VIP"
},
{
"input": "maTRIx",
"output": "matrix"
},
{
"input": "BNHWpnpawg",
"output": "bnhwpnpawg"
},
{
"input": "VTYGP",
"output": "VTYGP"
},
{
"input": "CHNenu",
"output": "chnenu"
},
{
"input": "ERPZGrodyu",
"output": "erpzgrodyu"
},
{
"input": "KSXBXWpebh",
"output": "KSXBXWPEBH"
},
{
"input": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv",
"output": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv"
},
{
"input": "Amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd",
"output": "amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd"
},
{
"input": "ISAGFJFARYFBLOPQDSHWGMCNKMFTLVFUGNJEWGWNBLXUIATXEkqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv",
"output": "isagfjfaryfblopqdshwgmcnkmftlvfugnjewgwnblxuiatxekqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv"
},
{
"input": "XHRPXZEGHSOCJPICUIXSKFUZUPYTSGJSDIYBCMNMNBPNDBXLXBzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg",
"output": "xhrpxzeghsocjpicuixskfuzupytsgjsdiybcmnmnbpndbxlxbzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg"
},
{
"input": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGAdkcetqjljtmttlonpekcovdzebzdkzggwfsxhapmjkdbuceak",
"output": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGADKCETQJLJTMTTLONPEKCOVDZEBZDKZGGWFSXHAPMJKDBUCEAK"
},
{
"input": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFw",
"output": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFW"
},
{
"input": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB",
"output": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB"
},
{
"input": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge",
"output": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge"
},
{
"input": "Ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw",
"output": "ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw"
},
{
"input": "YQOMLKYAORUQQUCQZCDYMIVDHGWZFFRMUVTAWCHERFPMNRYRIkgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks",
"output": "yqomlkyaoruqqucqzcdymivdhgwzffrmuvtawcherfpmnryrikgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks"
},
{
"input": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJcuusigdqfkumewjtdyitveeiaybwrhomrwmpdipjwiuxfnwuz",
"output": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJCUUSIGDQFKUMEWJTDYITVEEIAYBWRHOMRWMPDIPJWIUXFNWUZ"
},
{
"input": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWuckzcpxosodcjaaakvlxpbiigsiauviilylnnqlyucziihqg",
"output": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWUCKZCPXOSODCJAAAKVLXPBIIGSIAUVIILYLNNQLYUCZIIHQG"
},
{
"input": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO",
"output": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO"
},
{
"input": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDd",
"output": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDD"
},
{
"input": "EBWOVSNLYTWWXrnovgugogtjahnmatomubebyxakas",
"output": "ebwovsnlytwwxrnovgugogtjahnmatomubebyxakas"
},
{
"input": "AORNNDKTRLRVGDPXJKXFTPFpopuzrlqumrxssskvbm",
"output": "AORNNDKTRLRVGDPXJKXFTPFPOPUZRLQUMRXSSSKVBM"
},
{
"input": "DBBZJCWQEVGYHEEUHJAWRHBDPsnplijbwtdibqrrheebfxqcvn",
"output": "dbbzjcwqevgyheeuhjawrhbdpsnplijbwtdibqrrheebfxqcvn"
},
{
"input": "t",
"output": "t"
},
{
"input": "N",
"output": "N"
},
{
"input": "kv",
"output": "kv"
},
{
"input": "Ur",
"output": "ur"
},
{
"input": "CN",
"output": "CN"
}
] | 1,670,293,623
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 60
| 0
|
n=int(input())
print(string.upper(n))
|
Title: Word
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input Specification:
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output Specification:
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Demo Input:
['HoUse\n', 'ViP\n', 'maTRIx\n']
Demo Output:
['house\n', 'VIP\n', 'matrix\n']
Note:
none
|
```python
n=int(input())
print(string.upper(n))
```
| -1
|
831
|
C
|
Jury Marks
|
PROGRAMMING
| 1,700
|
[
"brute force",
"constructive algorithms"
] | null | null |
Polycarp watched TV-show where *k* jury members one by one rated a participant by adding him a certain number of points (may be negative, i. e. points were subtracted). Initially the participant had some score, and each the marks were one by one added to his score. It is known that the *i*-th jury member gave *a**i* points.
Polycarp does not remember how many points the participant had before this *k* marks were given, but he remembers that among the scores announced after each of the *k* judges rated the participant there were *n* (*n*<=≤<=*k*) values *b*1,<=*b*2,<=...,<=*b**n* (it is guaranteed that all values *b**j* are distinct). It is possible that Polycarp remembers not all of the scores announced, i. e. *n*<=<<=*k*. Note that the initial score wasn't announced.
Your task is to determine the number of options for the score the participant could have before the judges rated the participant.
|
The first line contains two integers *k* and *n* (1<=≤<=*n*<=≤<=*k*<=≤<=2<=000) — the number of jury members and the number of scores Polycarp remembers.
The second line contains *k* integers *a*1,<=*a*2,<=...,<=*a**k* (<=-<=2<=000<=≤<=*a**i*<=≤<=2<=000) — jury's marks in chronological order.
The third line contains *n* distinct integers *b*1,<=*b*2,<=...,<=*b**n* (<=-<=4<=000<=000<=≤<=*b**j*<=≤<=4<=000<=000) — the values of points Polycarp remembers. Note that these values are not necessarily given in chronological order.
|
Print the number of options for the score the participant could have before the judges rated the participant. If Polycarp messes something up and there is no options, print "0" (without quotes).
|
[
"4 1\n-5 5 0 20\n10\n",
"2 2\n-2000 -2000\n3998000 4000000\n"
] |
[
"3\n",
"1\n"
] |
The answer for the first example is 3 because initially the participant could have - 10, 10 or 15 points.
In the second example there is only one correct initial score equaling to 4 002 000.
| 1,000
|
[
{
"input": "4 1\n-5 5 0 20\n10",
"output": "3"
},
{
"input": "2 2\n-2000 -2000\n3998000 4000000",
"output": "1"
},
{
"input": "1 1\n-577\n1273042",
"output": "1"
},
{
"input": "2 1\n614 -1943\n3874445",
"output": "2"
},
{
"input": "3 1\n1416 -1483 1844\n3261895",
"output": "3"
},
{
"input": "5 1\n1035 1861 1388 -622 1252\n2640169",
"output": "5"
},
{
"input": "10 10\n-25 746 298 1602 -1453 -541 -442 1174 976 -1857\n-548062 -548253 -546800 -548943 -548402 -548794 -549236 -548700 -549446 -547086",
"output": "1"
},
{
"input": "20 20\n-1012 625 39 -1747 -1626 898 -1261 180 -876 -1417 -1853 -1510 -1499 -561 -1824 442 -895 13 1857 1860\n-1269013 -1270956 -1264151 -1266004 -1268121 -1258341 -1269574 -1271851 -1258302 -1271838 -1260049 -1258966 -1271398 -1267514 -1269981 -1262038 -1261675 -1262734 -1260777 -1261858",
"output": "1"
},
{
"input": "1 1\n1\n-4000000",
"output": "1"
}
] | 1,640,965,078
| 4,438
|
PyPy 3-64
|
OK
|
TESTS
| 57
| 170
| 138,342,400
|
import bisect
import sys
input = sys.stdin.readline
sys.setrecursionlimit(100000)
from collections import defaultdict, deque
from itertools import permutations
p = print
r = range
def I(): return int(input())
def II(): return list(map(int, input().split()))
def S(): return input()[:-1]
def M(n): return [list(map(int, input().split())) for ___ in r(n)]
def pb(b): print('YES' if b else 'NO')
def INF(): return float('inf')
# -----------------------------------------------------------------------------------------------------
#
# ∧＿∧
# ∧＿∧ （´<_｀ ） Welcome to My Coding Space !
# （ ´_ゝ`） / ⌒i Free Hong Kong !
# ／ ＼ | | Free Tibet !
# / /￣￣￣￣/ | |
# ＿_(__ﾆつ/ ＿/ .| .|＿＿＿＿
# ＼/＿＿＿＿/ （u ⊃
#
# 再帰関数ですか？ SYS!!!!
# BINARY SEARCH ?
# -----------------------------------------------------------------------------------------------------
k, n = II()
a = II()
b = II()
for i in r(k-1):
a[i+1] += a[i]
a.sort()
b.sort()
res = set()
for i in r(k):
delta = a[i] - b[0]
j = i + 1
t = 1
while t < n and j < k:
if a[j] - b[t] == delta:
t += 1
j += 1
else:
j += 1
if t == n:
res.add(delta)
p(len(res))
|
Title: Jury Marks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp watched TV-show where *k* jury members one by one rated a participant by adding him a certain number of points (may be negative, i. e. points were subtracted). Initially the participant had some score, and each the marks were one by one added to his score. It is known that the *i*-th jury member gave *a**i* points.
Polycarp does not remember how many points the participant had before this *k* marks were given, but he remembers that among the scores announced after each of the *k* judges rated the participant there were *n* (*n*<=≤<=*k*) values *b*1,<=*b*2,<=...,<=*b**n* (it is guaranteed that all values *b**j* are distinct). It is possible that Polycarp remembers not all of the scores announced, i. e. *n*<=<<=*k*. Note that the initial score wasn't announced.
Your task is to determine the number of options for the score the participant could have before the judges rated the participant.
Input Specification:
The first line contains two integers *k* and *n* (1<=≤<=*n*<=≤<=*k*<=≤<=2<=000) — the number of jury members and the number of scores Polycarp remembers.
The second line contains *k* integers *a*1,<=*a*2,<=...,<=*a**k* (<=-<=2<=000<=≤<=*a**i*<=≤<=2<=000) — jury's marks in chronological order.
The third line contains *n* distinct integers *b*1,<=*b*2,<=...,<=*b**n* (<=-<=4<=000<=000<=≤<=*b**j*<=≤<=4<=000<=000) — the values of points Polycarp remembers. Note that these values are not necessarily given in chronological order.
Output Specification:
Print the number of options for the score the participant could have before the judges rated the participant. If Polycarp messes something up and there is no options, print "0" (without quotes).
Demo Input:
['4 1\n-5 5 0 20\n10\n', '2 2\n-2000 -2000\n3998000 4000000\n']
Demo Output:
['3\n', '1\n']
Note:
The answer for the first example is 3 because initially the participant could have - 10, 10 or 15 points.
In the second example there is only one correct initial score equaling to 4 002 000.
|
```python
import bisect
import sys
input = sys.stdin.readline
sys.setrecursionlimit(100000)
from collections import defaultdict, deque
from itertools import permutations
p = print
r = range
def I(): return int(input())
def II(): return list(map(int, input().split()))
def S(): return input()[:-1]
def M(n): return [list(map(int, input().split())) for ___ in r(n)]
def pb(b): print('YES' if b else 'NO')
def INF(): return float('inf')
# -----------------------------------------------------------------------------------------------------
#
# ∧＿∧
# ∧＿∧ （´<_｀ ） Welcome to My Coding Space !
# （ ´_ゝ`） / ⌒i Free Hong Kong !
# ／ ＼ | | Free Tibet !
# / /￣￣￣￣/ | |
# ＿_(__ﾆつ/ ＿/ .| .|＿＿＿＿
# ＼/＿＿＿＿/ （u ⊃
#
# 再帰関数ですか？ SYS!!!!
# BINARY SEARCH ?
# -----------------------------------------------------------------------------------------------------
k, n = II()
a = II()
b = II()
for i in r(k-1):
a[i+1] += a[i]
a.sort()
b.sort()
res = set()
for i in r(k):
delta = a[i] - b[0]
j = i + 1
t = 1
while t < n and j < k:
if a[j] - b[t] == delta:
t += 1
j += 1
else:
j += 1
if t == n:
res.add(delta)
p(len(res))
```
| 3
|
|
75
|
C
|
Modified GCD
|
PROGRAMMING
| 1,600
|
[
"binary search",
"number theory"
] |
C. Modified GCD
|
2
|
256
|
Well, here is another math class task. In mathematics, GCD is the greatest common divisor, and it's an easy task to calculate the GCD between two positive integers.
A common divisor for two positive numbers is a number which both numbers are divisible by.
But your teacher wants to give you a harder task, in this task you have to find the greatest common divisor *d* between two integers *a* and *b* that is in a given range from *low* to *high* (inclusive), i.e. *low*<=≤<=*d*<=≤<=*high*. It is possible that there is no common divisor in the given range.
You will be given the two integers *a* and *b*, then *n* queries. Each query is a range from *low* to *high* and you have to answer each query.
|
The first line contains two integers *a* and *b*, the two integers as described above (1<=≤<=*a*,<=*b*<=≤<=109). The second line contains one integer *n*, the number of queries (1<=≤<=*n*<=≤<=104). Then *n* lines follow, each line contains one query consisting of two integers, *low* and *high* (1<=≤<=*low*<=≤<=*high*<=≤<=109).
|
Print *n* lines. The *i*-th of them should contain the result of the *i*-th query in the input. If there is no common divisor in the given range for any query, you should print -1 as a result for this query.
|
[
"9 27\n3\n1 5\n10 11\n9 11\n"
] |
[
"3\n-1\n9\n"
] |
none
| 1,500
|
[
{
"input": "9 27\n3\n1 5\n10 11\n9 11",
"output": "3\n-1\n9"
},
{
"input": "48 72\n2\n8 29\n29 37",
"output": "24\n-1"
},
{
"input": "90 100\n10\n51 61\n6 72\n1 84\n33 63\n37 69\n18 21\n9 54\n49 90\n14 87\n37 90",
"output": "-1\n10\n10\n-1\n-1\n-1\n10\n-1\n-1\n-1"
},
{
"input": "84 36\n1\n18 32",
"output": "-1"
},
{
"input": "90 36\n16\n13 15\n5 28\n11 30\n26 35\n2 8\n19 36\n3 17\n5 14\n4 26\n22 33\n16 33\n18 27\n4 17\n1 2\n29 31\n18 36",
"output": "-1\n18\n18\n-1\n6\n-1\n9\n9\n18\n-1\n18\n18\n9\n2\n-1\n18"
},
{
"input": "84 90\n18\n10 75\n2 40\n30 56\n49 62\n19 33\n5 79\n61 83\n13 56\n73 78\n1 18\n23 35\n14 72\n22 33\n1 21\n8 38\n54 82\n6 80\n57 75",
"output": "-1\n6\n-1\n-1\n-1\n6\n-1\n-1\n-1\n6\n-1\n-1\n-1\n6\n-1\n-1\n6\n-1"
},
{
"input": "84 100\n16\n10 64\n3 61\n19 51\n42 67\n51 68\n12 40\n10 47\n52 53\n37 67\n2 26\n23 47\n17 75\n49 52\n3 83\n63 81\n8 43",
"output": "-1\n4\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n4\n-1\n-1\n-1\n4\n-1\n-1"
},
{
"input": "36 60\n2\n17 25\n16 20",
"output": "-1\n-1"
},
{
"input": "90 100\n8\n55 75\n46 68\n44 60\n32 71\n43 75\n23 79\n47 86\n11 57",
"output": "-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1"
},
{
"input": "90 36\n8\n1 19\n10 12\n14 28\n21 24\n8 8\n33 34\n10 26\n15 21",
"output": "18\n-1\n18\n-1\n-1\n-1\n18\n18"
},
{
"input": "48 80\n19\n1 1\n16 16\n1 16\n16 48\n16 80\n16 1000000000\n1000000000 1000000000\n1 1000000000\n500000000 1000000000\n15 17\n17 17\n15 15\n8 8\n8 15\n8 16\n8 17\n7 17\n7 15\n9 15",
"output": "1\n16\n16\n16\n16\n16\n-1\n16\n-1\n16\n-1\n-1\n8\n8\n16\n16\n16\n8\n-1"
},
{
"input": "31607 999002449\n18\n31607 31607\n31606 31608\n31607 31608\n31606 31607\n31606 31606\n31608 31608\n1 31607\n1 31606\n1 31608\n1 1000000000\n31607 1000000000\n31606 1000000000\n31608 1000000000\n1000000000 1000000000\n1 1\n2 31606\n2 31607\n2 31608",
"output": "31607\n31607\n31607\n31607\n-1\n-1\n31607\n1\n31607\n31607\n31607\n31607\n-1\n-1\n1\n-1\n31607\n31607"
},
{
"input": "999999937 999999929\n12\n999999929 999999937\n1 1\n1 1000000000\n2 1000000000\n1 2\n999999937 999999937\n999999929 999999929\n2 2\n3 3\n1 100\n1 999999937\n1 999999929",
"output": "-1\n1\n1\n-1\n1\n-1\n-1\n-1\n-1\n1\n1\n1"
}
] | 1,636,164,508
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 0
| 92
| 17,715,200
|
import math
a,b=map(int,input().split())
n=int(input())
ar=[]
for _ in range(n):
l,h=map(int,input().split())
for i in range(l,h+1):
for j in range(l,h+1):
ar.append(math.gcd(i,j))
for x in ar:
if a<x<b:
print(x)
break
else:
print(-1)
break
|
Title: Modified GCD
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Well, here is another math class task. In mathematics, GCD is the greatest common divisor, and it's an easy task to calculate the GCD between two positive integers.
A common divisor for two positive numbers is a number which both numbers are divisible by.
But your teacher wants to give you a harder task, in this task you have to find the greatest common divisor *d* between two integers *a* and *b* that is in a given range from *low* to *high* (inclusive), i.e. *low*<=≤<=*d*<=≤<=*high*. It is possible that there is no common divisor in the given range.
You will be given the two integers *a* and *b*, then *n* queries. Each query is a range from *low* to *high* and you have to answer each query.
Input Specification:
The first line contains two integers *a* and *b*, the two integers as described above (1<=≤<=*a*,<=*b*<=≤<=109). The second line contains one integer *n*, the number of queries (1<=≤<=*n*<=≤<=104). Then *n* lines follow, each line contains one query consisting of two integers, *low* and *high* (1<=≤<=*low*<=≤<=*high*<=≤<=109).
Output Specification:
Print *n* lines. The *i*-th of them should contain the result of the *i*-th query in the input. If there is no common divisor in the given range for any query, you should print -1 as a result for this query.
Demo Input:
['9 27\n3\n1 5\n10 11\n9 11\n']
Demo Output:
['3\n-1\n9\n']
Note:
none
|
```python
import math
a,b=map(int,input().split())
n=int(input())
ar=[]
for _ in range(n):
l,h=map(int,input().split())
for i in range(l,h+1):
for j in range(l,h+1):
ar.append(math.gcd(i,j))
for x in ar:
if a<x<b:
print(x)
break
else:
print(-1)
break
```
| 0
|
49
|
A
|
Sleuth
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Sleuth
|
2
|
256
|
Vasya plays the sleuth with his friends. The rules of the game are as follows: those who play for the first time, that is Vasya is the sleuth, he should investigate a "crime" and find out what is happening. He can ask any questions whatsoever that can be answered with "Yes" or "No". All the rest agree beforehand to answer the questions like that: if the question’s last letter is a vowel, they answer "Yes" and if the last letter is a consonant, they answer "No". Of course, the sleuth knows nothing about it and his task is to understand that.
Unfortunately, Vasya is not very smart. After 5 hours of endless stupid questions everybody except Vasya got bored. That’s why Vasya’s friends ask you to write a program that would give answers instead of them.
The English alphabet vowels are: A, E, I, O, U, Y
The English alphabet consonants are: B, C, D, F, G, H, J, K, L, M, N, P, Q, R, S, T, V, W, X, Z
|
The single line contains a question represented by a non-empty line consisting of large and small Latin letters, spaces and a question mark. The line length does not exceed 100. It is guaranteed that the question mark occurs exactly once in the line — as the last symbol and that the line contains at least one letter.
|
Print answer for the question in a single line: YES if the answer is "Yes", NO if the answer is "No".
Remember that in the reply to the question the last letter, not the last character counts. I. e. the spaces and the question mark do not count as letters.
|
[
"Is it a melon?\n",
"Is it an apple?\n",
"Is it a banana ?\n",
"Is it an apple and a banana simultaneouSLY?\n"
] |
[
"NO\n",
"YES\n",
"YES\n",
"YES\n"
] |
none
| 500
|
[
{
"input": "Is it a melon?",
"output": "NO"
},
{
"input": "Is it an apple?",
"output": "YES"
},
{
"input": " Is it a banana ?",
"output": "YES"
},
{
"input": "Is it an apple and a banana simultaneouSLY?",
"output": "YES"
},
{
"input": "oHtSbDwzHb?",
"output": "NO"
},
{
"input": "sZecYdUvZHrXx?",
"output": "NO"
},
{
"input": "uMtXK?",
"output": "NO"
},
{
"input": "U?",
"output": "YES"
},
{
"input": "aqFDkCUKeHMyvZFcAyWlMUSQTFomtaWjoKLVyxLCw vcufPBFbaljOuHWiDCROYTcmbgzbaqHXKPOYEbuEtRqqoxBbOETCsQzhw?",
"output": "NO"
},
{
"input": "dJcNqQiFXzcbsj fItCpBLyXOnrSBPebwyFHlxUJHqCUzzCmcAvMiKL NunwOXnKeIxUZmBVwiCUfPkjRAkTPbkYCmwRRnDSLaz?",
"output": "NO"
},
{
"input": "gxzXbdcAQMuFKuuiPohtMgeypr wpDIoDSyOYTdvylcg SoEBZjnMHHYZGEqKgCgBeTbyTwyGuPZxkxsnSczotBdYyfcQsOVDVC?",
"output": "NO"
},
{
"input": "FQXBisXaJFMiHFQlXjixBDMaQuIbyqSBKGsBfTmBKCjszlGVZxEOqYYqRTUkGpSDDAoOXyXcQbHcPaegeOUBNeSD JiKOdECPOF?",
"output": "NO"
},
{
"input": "YhCuZnrWUBEed?",
"output": "NO"
},
{
"input": "hh?",
"output": "NO"
},
{
"input": "whU?",
"output": "YES"
},
{
"input": "fgwg?",
"output": "NO"
},
{
"input": "GlEmEPKrYcOnBNJUIFjszWUyVdvWw DGDjoCMtRJUburkPToCyDrOtMr?",
"output": "NO"
},
{
"input": "n?",
"output": "NO"
},
{
"input": "BueDOlxgzeNlxrzRrMbKiQdmGujEKmGxclvaPpTuHmTqBp?",
"output": "NO"
},
{
"input": "iehvZNQXDGCuVmJPOEysLyUryTdfaIxIuTzTadDbqRQGoCLXkxnyfWSGoLXebNnQQNTqAQJebbyYvHOfpUnXeWdjx?",
"output": "NO"
},
{
"input": " J ?",
"output": "NO"
},
{
"input": " j ?",
"output": "NO"
},
{
"input": " o ?",
"output": "YES"
},
{
"input": " T ?",
"output": "NO"
},
{
"input": " q ?",
"output": "NO"
},
{
"input": " j ?",
"output": "NO"
},
{
"input": " c ?",
"output": "NO"
},
{
"input": " B ?",
"output": "NO"
},
{
"input": "LuhxDHVwMPTtUIUMIQTuQETgXCOQPsfdFlyHvpfOVedjUTpGLAZGOHloIjJJtOLAlHPivzA?",
"output": "YES"
},
{
"input": "wmztmzFfwbGyOmNHENUFMTsFEMWYA?",
"output": "YES"
},
{
"input": "wGsfZCSwN PEUhNUrLfABrxA?",
"output": "YES"
},
{
"input": "mCDHENXjYbgMdBimAdPnewaHfpGWowjWrVAdvWczjw iDcUbyzMsmsnwbviiKiAyGVA?",
"output": "YES"
},
{
"input": "ARIWnwqFqxsQXsXXzHqvFjxOCttAGPUzDtWzsenPYdNXuFOIUGYZsLLK IaoxiyjBBRThoelwdPTkuCQfcBLUEJpCPIrVZlvUWA?",
"output": "YES"
},
{
"input": " PslvVpgpN BXkMFBEVXsyZFIQbBEFxGkYTeXKrOdcmhbiTUatYRUoYAayrchqbksswIlfIjerZPqptvCGnMUhyrQSvwltRhFzA?",
"output": "YES"
},
{
"input": "HpBkttwSjBXDmyleGiRWNUMPaAIE uzTrp KJDzaUiCdsMYOoWKHoUhWUoecCPmACymMUUbGav UMRpCytPETwNFAObZJA?",
"output": "YES"
}
] | 1,589,876,918
| 2,147,483,647
|
PyPy 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 372
| 2,150,400
|
s=input()
l1=[A,E,I,O,U,Y,a,e,i,o,u,y]
s1=list(s)
s2=s1[::-1]
f=0
for i in range(n):
if s2[i] in l1:
f=1
break
if f==1:
print("YES")
else:
print("NO")
|
Title: Sleuth
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya plays the sleuth with his friends. The rules of the game are as follows: those who play for the first time, that is Vasya is the sleuth, he should investigate a "crime" and find out what is happening. He can ask any questions whatsoever that can be answered with "Yes" or "No". All the rest agree beforehand to answer the questions like that: if the question’s last letter is a vowel, they answer "Yes" and if the last letter is a consonant, they answer "No". Of course, the sleuth knows nothing about it and his task is to understand that.
Unfortunately, Vasya is not very smart. After 5 hours of endless stupid questions everybody except Vasya got bored. That’s why Vasya’s friends ask you to write a program that would give answers instead of them.
The English alphabet vowels are: A, E, I, O, U, Y
The English alphabet consonants are: B, C, D, F, G, H, J, K, L, M, N, P, Q, R, S, T, V, W, X, Z
Input Specification:
The single line contains a question represented by a non-empty line consisting of large and small Latin letters, spaces and a question mark. The line length does not exceed 100. It is guaranteed that the question mark occurs exactly once in the line — as the last symbol and that the line contains at least one letter.
Output Specification:
Print answer for the question in a single line: YES if the answer is "Yes", NO if the answer is "No".
Remember that in the reply to the question the last letter, not the last character counts. I. e. the spaces and the question mark do not count as letters.
Demo Input:
['Is it a melon?\n', 'Is it an apple?\n', 'Is it a banana ?\n', 'Is it an apple and a banana simultaneouSLY?\n']
Demo Output:
['NO\n', 'YES\n', 'YES\n', 'YES\n']
Note:
none
|
```python
s=input()
l1=[A,E,I,O,U,Y,a,e,i,o,u,y]
s1=list(s)
s2=s1[::-1]
f=0
for i in range(n):
if s2[i] in l1:
f=1
break
if f==1:
print("YES")
else:
print("NO")
```
| -1
|
23
|
A
|
You're Given a String...
|
PROGRAMMING
| 1,200
|
[
"brute force",
"greedy"
] |
A. You're Given a String...
|
2
|
256
|
You're given a string of lower-case Latin letters. Your task is to find the length of its longest substring that can be met in the string at least twice. These occurrences can overlap (see sample test 2).
|
The first input line contains the string. It's guaranteed, that the string is non-empty, consists of lower-case Latin letters, and its length doesn't exceed 100.
|
Output one number — length of the longest substring that can be met in the string at least twice.
|
[
"abcd\n",
"ababa\n",
"zzz\n"
] |
[
"0",
"3",
"2"
] |
none
| 0
|
[
{
"input": "abcd",
"output": "0"
},
{
"input": "ababa",
"output": "3"
},
{
"input": "zzz",
"output": "2"
},
{
"input": "kmmm",
"output": "2"
},
{
"input": "wzznz",
"output": "1"
},
{
"input": "qlzazaaqll",
"output": "2"
},
{
"input": "lzggglgpep",
"output": "2"
},
{
"input": "iegdlraaidefgegiagrdfhihe",
"output": "2"
},
{
"input": "esxpqmdrtidgtkxojuxyrcwxlycywtzbjzpxvbngnlepgzcaeg",
"output": "1"
},
{
"input": "garvpaimjdjiivamusjdwfcaoswuhxyyxvrxzajoyihggvuxumaadycfphrzbprraicvjjlsdhojihaw",
"output": "2"
},
{
"input": "ckvfndqgkmhcyojaqgdkenmbexufryhqejdhctxujmtrwkpbqxufxamgoeigzfyzbhevpbkvviwntdhqscvkmphnkkljizndnbjt",
"output": "3"
},
{
"input": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "99"
},
{
"input": "ikiikiikikiiikkkkkikkkkiiiiikkiiikkiikiikkkkikkkikikkikiiikkikikiiikikkkiiikkkikkikkikkkkiiikkiiiiii",
"output": "10"
},
{
"input": "ovovhoovvhohhhvhhvhhvhovoohovhhoooooovohvooooohvvoooohvvovhhvhovhhvoovhvhvoovovvhooovhhooovohvhhovhv",
"output": "8"
},
{
"input": "ccwckkkycccccckwckwkwkwkkkkyycykcccycyckwywcckwykcycykkkwcycwwcykcwkwkwwykwkwcykywwwyyykckkyycckwcwk",
"output": "5"
},
{
"input": "ttketfkefktfztezzkzfkkeetkkfktftzktezekkeezkeeetteeteefetefkzzzetekfftkeffzkktffzkzzeftfeezfefzffeef",
"output": "4"
},
{
"input": "rtharczpfznrgdnkltchafduydgbgkdjqrmjqyfmpwjwphrtsjbmswkanjlprbnduaqbcjqxlxmkspkhkcnzbqwxonzxxdmoigti",
"output": "2"
},
{
"input": "fplrkfklvwdeiynbjgaypekambmbjfnoknlhczhkdmljicookdywdgpnlnqlpunnkebnikgcgcjefeqhknvlynmvjcegvcdgvvdb",
"output": "2"
},
{
"input": "txbciieycswqpniwvzipwlottivvnfsysgzvzxwbctcchfpvlbcjikdofhpvsknptpjdbxemtmjcimetkemdbettqnbvzzbdyxxb",
"output": "2"
},
{
"input": "fmubmfwefikoxtqvmaavwjxmoqltapexkqxcsztpezfcltqavuicefxovuswmqimuikoppgqpiapqutkczgcvxzutavkujxvpklv",
"output": "3"
},
{
"input": "ipsrjylhpkjvlzncfixipstwcicxqygqcfrawpzzvckoveyqhathglblhpkjvlzncfixipfajaqobtzvthmhgbuawoxoknirclxg",
"output": "15"
},
{
"input": "kcnjsntjzcbgzjscrsrjkrbytqsrptzspzctjrorsyggrtkcnjsntjzcbgzjscrsrjyqbrtpcgqirsrrjbbbrnyqstnrozcoztt",
"output": "20"
},
{
"input": "unhcfnrhsqetuerjqcetrhlsqgfnqfntvkgxsscquolxxroqgtchffyccetrhlsqgfnqfntvkgxsscquolxxroqgtchffhfqvx",
"output": "37"
},
{
"input": "kkcckkccckkcckcccckcckkkkcckkkkckkkcckckkkkkckkkkkcckkccckkcckcccckcckkkkcckkkkckkkcckckkkkkckckckkc",
"output": "46"
},
{
"input": "mlhsyijxeydqxhtkmpdeqwzogjvxahmssyhfhqessbxzvydbrxdmlhsyijxeydqxhtkmpdeqwzogjvxahmssyhfhqessbxzvydik",
"output": "47"
},
{
"input": "abcdefghijklmnopqrstuvwxyz",
"output": "0"
},
{
"input": "tttttbttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttmttttttt",
"output": "85"
},
{
"input": "ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffbfffffffffffffffffffffffffffffffffffff",
"output": "61"
},
{
"input": "cccccccccccccccccccccccwcccccccccccccccccccccuccccccccccccccnccccccccccccccccccccccccccccccccccccccc",
"output": "38"
},
{
"input": "ffffffffffffffffffffffffffufffgfffffffffffffffffffffffffffffffffffffffgffffffftffffffgffffffffffffff",
"output": "38"
},
{
"input": "rrrrrrrrrrrrrrrrrrrlhbrrrrrrrrurrrrrrrfrrqrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrewrrrrrrryrrxrrrrrrrrrrr",
"output": "33"
},
{
"input": "vyvvvvvvvvzvvvvvzvvvwvvvvrvvvvvvvvvvvvvvvrvvvvvvvvvpkvvpvgvvvvvvvvvvvvvgvvvvvvvvvvvvvvvvvvysvvvbvvvv",
"output": "17"
},
{
"input": "cbubbbbbbbbbbfbbbbbbbbjbobbbbbbbbbbibbubbbbjbbbnzgbbzbbfbbbbbbbbbbbfbpbbbbbbbbbbygbbbgbabbbbbbbhibbb",
"output": "12"
},
{
"input": "lrqrrrrrrrjrrrrrcdrrgrrmwvrrrrrrrrrxfzrmrmrryrrrurrrdrrrrrrrrrrererrrsrrrrrrrrrrrqrrrrcrrwjsrrlrrrrr",
"output": "10"
},
{
"input": "ssssusisisosscssssztzessssyspskjssvosiissussszsosssslsssdsssvssvsssslsssmsfjasjsssssowscsjsssszsspss",
"output": "8"
},
{
"input": "uukuuuumueuuuujuukgdhbztuuuubbguuocuozfaunqufjujuguyuuvkuuauubuubuucuvtjuuuuuusduduuuuuuuueunuuuuuzu",
"output": "7"
},
{
"input": "jpkkgwklngwqcfzmwkkpcwkkkkkekkkekkkdsykqwjkkkhkkkxdnukkkkkkmkqykkkxqklkskkrkkkkkqqjikkkkkkpknkkkkkoh",
"output": "7"
},
{
"input": "bmzbbfbbhqxwthtbbisbbbbbtbbfbbpbfbbpbkbjfbcbbbbzbbbdwmbbbrnvqdbbtbbuglrnbbbbvmbyblebbabibrevaxbbjbqb",
"output": "6"
},
{
"input": "qqqmqqqsbteqqopsuqiqumrqzpqnqgqeniqqkyqqyqqqpxzqeqqquhdqquhqqqfqjirqaqqaquxqoqqjqqqqbjbgqcqqqqicnkqc",
"output": "4"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaasaaaavaaaaaaauaaeaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "44"
},
{
"input": "a",
"output": "0"
},
{
"input": "fg",
"output": "0"
},
{
"input": "yy",
"output": "1"
},
{
"input": "abcabcabc",
"output": "6"
},
{
"input": "qwerqwedqwes",
"output": "3"
}
] | 1,593,999,167
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 45
| 280
| 21,504,000
|
name = input()
for i in range(len(name), 0, -1):
for j in range(len(name) - i + 1):
if name[j: j + i] in name[j + 1:]:
print(i)
exit()
print(0)
|
Title: You're Given a String...
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You're given a string of lower-case Latin letters. Your task is to find the length of its longest substring that can be met in the string at least twice. These occurrences can overlap (see sample test 2).
Input Specification:
The first input line contains the string. It's guaranteed, that the string is non-empty, consists of lower-case Latin letters, and its length doesn't exceed 100.
Output Specification:
Output one number — length of the longest substring that can be met in the string at least twice.
Demo Input:
['abcd\n', 'ababa\n', 'zzz\n']
Demo Output:
['0', '3', '2']
Note:
none
|
```python
name = input()
for i in range(len(name), 0, -1):
for j in range(len(name) - i + 1):
if name[j: j + i] in name[j + 1:]:
print(i)
exit()
print(0)
```
| 3.889946
|
822
|
A
|
I'm bored with life
|
PROGRAMMING
| 800
|
[
"implementation",
"math",
"number theory"
] | null | null |
Holidays have finished. Thanks to the help of the hacker Leha, Noora managed to enter the university of her dreams which is located in a town Pavlopolis. It's well known that universities provide students with dormitory for the period of university studies. Consequently Noora had to leave Vičkopolis and move to Pavlopolis. Thus Leha was left completely alone in a quiet town Vičkopolis. He almost even fell into a depression from boredom!
Leha came up with a task for himself to relax a little. He chooses two integers *A* and *B* and then calculates the greatest common divisor of integers "*A* factorial" and "*B* factorial". Formally the hacker wants to find out GCD(*A*!,<=*B*!). It's well known that the factorial of an integer *x* is a product of all positive integers less than or equal to *x*. Thus *x*!<==<=1·2·3·...·(*x*<=-<=1)·*x*. For example 4!<==<=1·2·3·4<==<=24. Recall that GCD(*x*,<=*y*) is the largest positive integer *q* that divides (without a remainder) both *x* and *y*.
Leha has learned how to solve this task very effective. You are able to cope with it not worse, aren't you?
|
The first and single line contains two integers *A* and *B* (1<=≤<=*A*,<=*B*<=≤<=109,<=*min*(*A*,<=*B*)<=≤<=12).
|
Print a single integer denoting the greatest common divisor of integers *A*! and *B*!.
|
[
"4 3\n"
] |
[
"6\n"
] |
Consider the sample.
4! = 1·2·3·4 = 24. 3! = 1·2·3 = 6. The greatest common divisor of integers 24 and 6 is exactly 6.
| 500
|
[
{
"input": "4 3",
"output": "6"
},
{
"input": "10 399603090",
"output": "3628800"
},
{
"input": "6 973151934",
"output": "720"
},
{
"input": "2 841668075",
"output": "2"
},
{
"input": "7 415216919",
"output": "5040"
},
{
"input": "3 283733059",
"output": "6"
},
{
"input": "11 562314608",
"output": "39916800"
},
{
"input": "3 990639260",
"output": "6"
},
{
"input": "11 859155400",
"output": "39916800"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "5 3",
"output": "6"
},
{
"input": "1 4",
"output": "1"
},
{
"input": "5 4",
"output": "24"
},
{
"input": "1 12",
"output": "1"
},
{
"input": "9 7",
"output": "5040"
},
{
"input": "2 3",
"output": "2"
},
{
"input": "6 11",
"output": "720"
},
{
"input": "6 7",
"output": "720"
},
{
"input": "11 11",
"output": "39916800"
},
{
"input": "4 999832660",
"output": "24"
},
{
"input": "7 999228288",
"output": "5040"
},
{
"input": "11 999257105",
"output": "39916800"
},
{
"input": "11 999286606",
"output": "39916800"
},
{
"input": "3 999279109",
"output": "6"
},
{
"input": "999632727 11",
"output": "39916800"
},
{
"input": "999625230 7",
"output": "5040"
},
{
"input": "999617047 3",
"output": "6"
},
{
"input": "999646548 7",
"output": "5040"
},
{
"input": "999639051 3",
"output": "6"
},
{
"input": "12 12",
"output": "479001600"
},
{
"input": "12 1",
"output": "1"
},
{
"input": "1213 5",
"output": "120"
},
{
"input": "8 9",
"output": "40320"
},
{
"input": "12 9",
"output": "362880"
},
{
"input": "12 1000000000",
"output": "479001600"
},
{
"input": "1000000000 1",
"output": "1"
},
{
"input": "12 13",
"output": "479001600"
},
{
"input": "2 29845",
"output": "2"
},
{
"input": "10 21",
"output": "3628800"
},
{
"input": "12 20",
"output": "479001600"
},
{
"input": "15 12",
"output": "479001600"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1000000000 12",
"output": "479001600"
},
{
"input": "11 30",
"output": "39916800"
},
{
"input": "17 12",
"output": "479001600"
},
{
"input": "4 19",
"output": "24"
},
{
"input": "12 15",
"output": "479001600"
},
{
"input": "20 6",
"output": "720"
},
{
"input": "10 20",
"output": "3628800"
},
{
"input": "10 10",
"output": "3628800"
},
{
"input": "22 12",
"output": "479001600"
},
{
"input": "20 12",
"output": "479001600"
},
{
"input": "12 23",
"output": "479001600"
},
{
"input": "12 22",
"output": "479001600"
},
{
"input": "18 3",
"output": "6"
},
{
"input": "14 10",
"output": "3628800"
},
{
"input": "14 12",
"output": "479001600"
},
{
"input": "8 3",
"output": "6"
},
{
"input": "5 5",
"output": "120"
}
] | 1,679,286,478
| 2,147,483,647
|
PyPy 3-64
|
RUNTIME_ERROR
|
TESTS
| 0
| 77
| 2,764,800
|
from math import gcd
n = int(input())
count = 0
for i in range(1, n+1):
if i % 3 == 0 or i % 5 == 0:
count += 1
print(count)
|
Title: I'm bored with life
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Holidays have finished. Thanks to the help of the hacker Leha, Noora managed to enter the university of her dreams which is located in a town Pavlopolis. It's well known that universities provide students with dormitory for the period of university studies. Consequently Noora had to leave Vičkopolis and move to Pavlopolis. Thus Leha was left completely alone in a quiet town Vičkopolis. He almost even fell into a depression from boredom!
Leha came up with a task for himself to relax a little. He chooses two integers *A* and *B* and then calculates the greatest common divisor of integers "*A* factorial" and "*B* factorial". Formally the hacker wants to find out GCD(*A*!,<=*B*!). It's well known that the factorial of an integer *x* is a product of all positive integers less than or equal to *x*. Thus *x*!<==<=1·2·3·...·(*x*<=-<=1)·*x*. For example 4!<==<=1·2·3·4<==<=24. Recall that GCD(*x*,<=*y*) is the largest positive integer *q* that divides (without a remainder) both *x* and *y*.
Leha has learned how to solve this task very effective. You are able to cope with it not worse, aren't you?
Input Specification:
The first and single line contains two integers *A* and *B* (1<=≤<=*A*,<=*B*<=≤<=109,<=*min*(*A*,<=*B*)<=≤<=12).
Output Specification:
Print a single integer denoting the greatest common divisor of integers *A*! and *B*!.
Demo Input:
['4 3\n']
Demo Output:
['6\n']
Note:
Consider the sample.
4! = 1·2·3·4 = 24. 3! = 1·2·3 = 6. The greatest common divisor of integers 24 and 6 is exactly 6.
|
```python
from math import gcd
n = int(input())
count = 0
for i in range(1, n+1):
if i % 3 == 0 or i % 5 == 0:
count += 1
print(count)
```
| -1
|
|
177
|
A2
|
Good Matrix Elements
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
The Smart Beaver from ABBYY got hooked on square matrices. Now he is busy studying an *n*<=×<=*n* size matrix, where *n* is odd. The Smart Beaver considers the following matrix elements good:
- Elements of the main diagonal. - Elements of the secondary diagonal. - Elements of the "middle" row — the row which has exactly rows above it and the same number of rows below it. - Elements of the "middle" column — the column that has exactly columns to the left of it and the same number of columns to the right of it.
Help the Smart Beaver count the sum of good elements of the given matrix.
|
The first line of input data contains a single odd integer *n*. Each of the next *n* lines contains *n* integers *a**ij* (0<=≤<=*a**ij*<=≤<=100) separated by single spaces — the elements of the given matrix.
The input limitations for getting 30 points are:
- 1<=≤<=*n*<=≤<=5
The input limitations for getting 100 points are:
- 1<=≤<=*n*<=≤<=101
|
Print a single integer — the sum of good matrix elements.
|
[
"3\n1 2 3\n4 5 6\n7 8 9\n",
"5\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n"
] |
[
"45\n",
"17\n"
] |
In the first sample all matrix elements will be good. Good elements in the second sample are shown on the figure.
| 70
|
[
{
"input": "3\n1 2 3\n4 5 6\n7 8 9",
"output": "45"
},
{
"input": "5\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1",
"output": "17"
},
{
"input": "1\n3",
"output": "3"
},
{
"input": "5\n27 7 3 11 72\n19 49 68 19 59\n41 25 37 64 65\n8 39 96 62 90\n13 37 43 26 33",
"output": "756"
},
{
"input": "3\n19 7 16\n12 15 5\n15 15 5",
"output": "109"
},
{
"input": "3\n36 4 33\n11 46 32\n20 49 34",
"output": "265"
},
{
"input": "3\n79 91 74\n33 82 22\n18 28 54",
"output": "481"
},
{
"input": "5\n7 0 8 1 7\n5 1 1 0 4\n4 2 8 1 6\n1 2 3 2 7\n6 0 1 9 6",
"output": "65"
},
{
"input": "5\n27 20 28 11 17\n25 21 1 20 14\n14 22 28 1 6\n1 2 23 2 7\n6 0 1 29 6",
"output": "225"
},
{
"input": "5\n57 50 58 41 17\n25 21 1 50 44\n44 22 28 31 36\n31 32 23 32 37\n6 0 31 59 6",
"output": "495"
},
{
"input": "5\n57 80 28 41 47\n85 51 61 50 74\n44 82 28 31 36\n31 32 23 32 37\n66 60 31 59 6",
"output": "705"
},
{
"input": "5\n13 58 10 17 43\n61 73 100 0 9\n52 38 16 22 96\n11 4 14 67 62\n70 89 7 98 83",
"output": "708"
},
{
"input": "5\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "0"
},
{
"input": "5\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0",
"output": "0"
},
{
"input": "5\n0 1 0 1 0\n0 0 0 0 1\n0 0 5 0 0\n0 0 0 0 1\n0 0 0 0 0",
"output": "5"
},
{
"input": "5\n0 0 0 0 1\n0 1 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 1 0 0 0",
"output": "3"
},
{
"input": "3\n0 0 0\n0 0 0\n0 0 0",
"output": "0"
},
{
"input": "11\n0 0 0 0 0 0 0 0 0 0 0\n0 0 0 0 0 0 0 0 0 0 0\n0 0 0 0 0 0 0 0 0 0 0\n0 0 0 0 0 0 0 0 0 0 0\n0 0 0 0 0 0 0 0 0 0 0\n0 0 0 0 0 0 0 0 0 0 0\n0 0 0 0 0 0 0 0 0 0 0\n0 0 0 0 0 0 0 0 0 0 0\n0 0 0 0 0 0 0 0 0 0 0\n0 0 0 0 0 0 0 0 0 0 0\n0 0 0 0 0 0 0 0 0 0 0",
"output": "0"
},
{
"input": "15\n0 2 1 2 1 0 2 0 2 2 2 2 2 0 2\n2 0 1 0 1 1 2 2 0 2 2 0 3 0 1\n3 3 2 2 2 1 2 3 2 3 1 2 3 3 2\n0 3 0 3 3 3 2 1 0 2 3 2 3 3 2\n3 0 0 2 1 2 3 1 1 1 2 2 2 1 0\n1 3 2 3 3 0 3 2 0 3 1 2 3 0 3\n2 2 3 2 0 0 2 3 0 1 2 1 1 2 1\n2 1 3 2 3 3 2 2 2 0 3 3 1 1 3\n0 2 1 3 1 2 0 0 1 0 0 2 0 0 0\n3 1 2 1 2 3 2 3 2 1 0 2 2 0 3\n0 0 1 3 3 0 1 1 0 0 2 3 0 3 0\n2 3 0 3 1 1 2 2 1 3 2 3 0 0 2\n3 0 2 2 3 3 0 2 2 2 0 1 0 0 3\n2 2 2 1 2 2 2 0 2 2 2 2 0 2 3\n2 2 3 0 0 0 0 2 1 1 1 3 2 2 0",
"output": "97"
}
] | 1,586,935,828
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS2
| 3
| 248
| 0
|
n = int(input())
mat = []
res = 0
for _ in range(n):
mat.append([int(i) for i in input().split()])
for i, j in zip(range(n), range(n)):
res += mat[i][j]
for i, j in zip(range(n-1, -1, -1), range(n-1, -1, -1)):
res += mat[i][j]
if (n % 2 == 1):
res += sum(mat[n // 2])
for i in range(n):
res += mat[i][n//2]
res -= 3 * mat[n//2][n//2]
print(res)
|
Title: Good Matrix Elements
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Smart Beaver from ABBYY got hooked on square matrices. Now he is busy studying an *n*<=×<=*n* size matrix, where *n* is odd. The Smart Beaver considers the following matrix elements good:
- Elements of the main diagonal. - Elements of the secondary diagonal. - Elements of the "middle" row — the row which has exactly rows above it and the same number of rows below it. - Elements of the "middle" column — the column that has exactly columns to the left of it and the same number of columns to the right of it.
Help the Smart Beaver count the sum of good elements of the given matrix.
Input Specification:
The first line of input data contains a single odd integer *n*. Each of the next *n* lines contains *n* integers *a**ij* (0<=≤<=*a**ij*<=≤<=100) separated by single spaces — the elements of the given matrix.
The input limitations for getting 30 points are:
- 1<=≤<=*n*<=≤<=5
The input limitations for getting 100 points are:
- 1<=≤<=*n*<=≤<=101
Output Specification:
Print a single integer — the sum of good matrix elements.
Demo Input:
['3\n1 2 3\n4 5 6\n7 8 9\n', '5\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n']
Demo Output:
['45\n', '17\n']
Note:
In the first sample all matrix elements will be good. Good elements in the second sample are shown on the figure.
|
```python
n = int(input())
mat = []
res = 0
for _ in range(n):
mat.append([int(i) for i in input().split()])
for i, j in zip(range(n), range(n)):
res += mat[i][j]
for i, j in zip(range(n-1, -1, -1), range(n-1, -1, -1)):
res += mat[i][j]
if (n % 2 == 1):
res += sum(mat[n // 2])
for i in range(n):
res += mat[i][n//2]
res -= 3 * mat[n//2][n//2]
print(res)
```
| 0
|
|
166
|
E
|
Tetrahedron
|
PROGRAMMING
| 1,500
|
[
"dp",
"math",
"matrices"
] | null | null |
You are given a tetrahedron. Let's mark its vertices with letters *A*, *B*, *C* and *D* correspondingly.
An ant is standing in the vertex *D* of the tetrahedron. The ant is quite active and he wouldn't stay idle. At each moment of time he makes a step from one vertex to another one along some edge of the tetrahedron. The ant just can't stand on one place.
You do not have to do much to solve the problem: your task is to count the number of ways in which the ant can go from the initial vertex *D* to itself in exactly *n* steps. In other words, you are asked to find out the number of different cyclic paths with the length of *n* from vertex *D* to itself. As the number can be quite large, you should print it modulo 1000000007 (109<=+<=7).
|
The first line contains the only integer *n* (1<=≤<=*n*<=≤<=107) — the required length of the cyclic path.
|
Print the only integer — the required number of ways modulo 1000000007 (109<=+<=7).
|
[
"2\n",
"4\n"
] |
[
"3\n",
"21\n"
] |
The required paths in the first sample are:
- *D* - *A* - *D* - *D* - *B* - *D* - *D* - *C* - *D*
| 1,000
|
[
{
"input": "2",
"output": "3"
},
{
"input": "4",
"output": "21"
},
{
"input": "1",
"output": "0"
},
{
"input": "3",
"output": "6"
},
{
"input": "5",
"output": "60"
},
{
"input": "6",
"output": "183"
},
{
"input": "7",
"output": "546"
},
{
"input": "8",
"output": "1641"
},
{
"input": "9",
"output": "4920"
},
{
"input": "10",
"output": "14763"
},
{
"input": "15",
"output": "3587226"
},
{
"input": "30",
"output": "782663359"
},
{
"input": "10000000",
"output": "192336614"
},
{
"input": "100",
"output": "721510432"
},
{
"input": "300",
"output": "327873818"
},
{
"input": "900",
"output": "295068084"
},
{
"input": "1500",
"output": "451187545"
},
{
"input": "3000",
"output": "645417275"
},
{
"input": "5000",
"output": "755610910"
},
{
"input": "10000",
"output": "723907367"
},
{
"input": "50000",
"output": "969527595"
},
{
"input": "90000",
"output": "548978368"
},
{
"input": "99999",
"output": "909741855"
},
{
"input": "100000",
"output": "729225554"
},
{
"input": "300000",
"output": "93822635"
},
{
"input": "800000",
"output": "178940616"
},
{
"input": "1000000",
"output": "266233856"
},
{
"input": "4000000",
"output": "882155933"
},
{
"input": "9000000",
"output": "295060537"
},
{
"input": "9999999",
"output": "730778875"
},
{
"input": "1000000",
"output": "266233856"
},
{
"input": "9999999",
"output": "730778875"
},
{
"input": "10000000",
"output": "192336614"
},
{
"input": "9999998",
"output": "576926295"
},
{
"input": "30",
"output": "782663359"
},
{
"input": "31",
"output": "347990060"
}
] | 1,664,021,207
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 0
| 2,000
| 0
|
s = int(input())
s = (s + 1) // 7
n = 36
while s < 2050:
s = s * 2
n = n + 3
print(n)
|
Title: Tetrahedron
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a tetrahedron. Let's mark its vertices with letters *A*, *B*, *C* and *D* correspondingly.
An ant is standing in the vertex *D* of the tetrahedron. The ant is quite active and he wouldn't stay idle. At each moment of time he makes a step from one vertex to another one along some edge of the tetrahedron. The ant just can't stand on one place.
You do not have to do much to solve the problem: your task is to count the number of ways in which the ant can go from the initial vertex *D* to itself in exactly *n* steps. In other words, you are asked to find out the number of different cyclic paths with the length of *n* from vertex *D* to itself. As the number can be quite large, you should print it modulo 1000000007 (109<=+<=7).
Input Specification:
The first line contains the only integer *n* (1<=≤<=*n*<=≤<=107) — the required length of the cyclic path.
Output Specification:
Print the only integer — the required number of ways modulo 1000000007 (109<=+<=7).
Demo Input:
['2\n', '4\n']
Demo Output:
['3\n', '21\n']
Note:
The required paths in the first sample are:
- *D* - *A* - *D* - *D* - *B* - *D* - *D* - *C* - *D*
|
```python
s = int(input())
s = (s + 1) // 7
n = 36
while s < 2050:
s = s * 2
n = n + 3
print(n)
```
| 0
|
|
218
|
A
|
Mountain Scenery
|
PROGRAMMING
| 1,100
|
[
"brute force",
"constructive algorithms",
"implementation"
] | null | null |
Little Bolek has found a picture with *n* mountain peaks painted on it. The *n* painted peaks are represented by a non-closed polyline, consisting of 2*n* segments. The segments go through 2*n*<=+<=1 points with coordinates (1,<=*y*1), (2,<=*y*2), ..., (2*n*<=+<=1,<=*y*2*n*<=+<=1), with the *i*-th segment connecting the point (*i*,<=*y**i*) and the point (*i*<=+<=1,<=*y**i*<=+<=1). For any even *i* (2<=≤<=*i*<=≤<=2*n*) the following condition holds: *y**i*<=-<=1<=<<=*y**i* and *y**i*<=><=*y**i*<=+<=1.
We shall call a vertex of a polyline with an even *x* coordinate a mountain peak.
Bolek fancied a little mischief. He chose exactly *k* mountain peaks, rubbed out the segments that went through those peaks and increased each peak's height by one (that is, he increased the *y* coordinate of the corresponding points). Then he painted the missing segments to get a new picture of mountain peaks. Let us denote the points through which the new polyline passes on Bolek's new picture as (1,<=*r*1), (2,<=*r*2), ..., (2*n*<=+<=1,<=*r*2*n*<=+<=1).
Given Bolek's final picture, restore the initial one.
|
The first line contains two space-separated integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=100). The next line contains 2*n*<=+<=1 space-separated integers *r*1,<=*r*2,<=...,<=*r*2*n*<=+<=1 (0<=≤<=*r**i*<=≤<=100) — the *y* coordinates of the polyline vertices on Bolek's picture.
It is guaranteed that we can obtain the given picture after performing the described actions on some picture of mountain peaks.
|
Print 2*n*<=+<=1 integers *y*1,<=*y*2,<=...,<=*y*2*n*<=+<=1 — the *y* coordinates of the vertices of the polyline on the initial picture. If there are multiple answers, output any one of them.
|
[
"3 2\n0 5 3 5 1 5 2\n",
"1 1\n0 2 0\n"
] |
[
"0 5 3 4 1 4 2 \n",
"0 1 0 \n"
] |
none
| 500
|
[
{
"input": "3 2\n0 5 3 5 1 5 2",
"output": "0 5 3 4 1 4 2 "
},
{
"input": "1 1\n0 2 0",
"output": "0 1 0 "
},
{
"input": "1 1\n1 100 0",
"output": "1 99 0 "
},
{
"input": "3 1\n0 1 0 1 0 2 0",
"output": "0 1 0 1 0 1 0 "
},
{
"input": "3 1\n0 1 0 2 0 1 0",
"output": "0 1 0 1 0 1 0 "
},
{
"input": "3 3\n0 100 35 67 40 60 3",
"output": "0 99 35 66 40 59 3 "
},
{
"input": "7 3\n1 2 1 3 1 2 1 2 1 3 1 3 1 2 1",
"output": "1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 "
},
{
"input": "100 100\n1 3 1 3 1 3 0 2 0 3 1 3 1 3 1 3 0 3 1 3 0 2 0 2 0 3 0 2 0 2 0 3 1 3 1 3 1 3 1 3 0 2 0 3 1 3 0 2 0 2 0 2 0 2 0 2 0 3 0 3 0 3 0 3 0 2 0 3 1 3 1 3 1 3 0 3 0 2 0 2 0 2 0 2 0 3 0 3 1 3 0 3 1 3 1 3 0 3 1 3 0 3 1 3 1 3 0 3 1 3 0 3 1 3 0 2 0 3 1 3 0 3 1 3 0 2 0 3 1 3 0 3 0 2 0 3 1 3 0 3 0 3 0 2 0 2 0 2 0 3 0 3 1 3 1 3 0 3 1 3 1 3 1 3 0 2 0 3 0 2 0 3 1 3 0 3 0 3 1 3 0 2 0 3 0 2 0 2 0 2 0 2 0 3 1 3 0 3 1 3 1",
"output": "1 2 1 2 1 2 0 1 0 2 1 2 1 2 1 2 0 2 1 2 0 1 0 1 0 2 0 1 0 1 0 2 1 2 1 2 1 2 1 2 0 1 0 2 1 2 0 1 0 1 0 1 0 1 0 1 0 2 0 2 0 2 0 2 0 1 0 2 1 2 1 2 1 2 0 2 0 1 0 1 0 1 0 1 0 2 0 2 1 2 0 2 1 2 1 2 0 2 1 2 0 2 1 2 1 2 0 2 1 2 0 2 1 2 0 1 0 2 1 2 0 2 1 2 0 1 0 2 1 2 0 2 0 1 0 2 1 2 0 2 0 2 0 1 0 1 0 1 0 2 0 2 1 2 1 2 0 2 1 2 1 2 1 2 0 1 0 2 0 1 0 2 1 2 0 2 0 2 1 2 0 1 0 2 0 1 0 1 0 1 0 1 0 2 1 2 0 2 1 2 1 "
},
{
"input": "30 20\n1 3 1 3 0 2 0 4 1 3 0 3 1 3 1 4 2 3 1 2 0 4 2 4 0 4 1 3 0 4 1 4 2 4 2 4 0 3 1 2 1 4 0 3 0 4 1 3 1 4 1 3 0 1 0 4 0 3 2 3 1",
"output": "1 3 1 3 0 2 0 4 1 2 0 2 1 2 1 3 2 3 1 2 0 3 2 3 0 3 1 2 0 3 1 3 2 3 2 3 0 2 1 2 1 3 0 2 0 3 1 2 1 3 1 2 0 1 0 3 0 3 2 3 1 "
},
{
"input": "10 6\n0 5 2 4 1 5 2 5 2 4 2 5 3 5 0 2 0 1 0 1 0",
"output": "0 5 2 4 1 4 2 4 2 3 2 4 3 4 0 1 0 1 0 1 0 "
},
{
"input": "11 6\n3 5 1 4 3 5 0 2 0 2 0 4 0 3 0 4 1 5 2 4 0 4 0",
"output": "3 5 1 4 3 5 0 2 0 2 0 3 0 2 0 3 1 4 2 3 0 3 0 "
},
{
"input": "12 6\n1 2 1 5 0 2 0 4 1 3 1 4 2 4 0 4 0 4 2 4 0 4 0 5 3",
"output": "1 2 1 5 0 2 0 4 1 3 1 4 2 3 0 3 0 3 2 3 0 3 0 4 3 "
},
{
"input": "13 6\n3 5 2 5 0 3 0 1 0 2 0 1 0 1 0 2 1 4 3 5 1 3 1 3 2 3 1",
"output": "3 4 2 4 0 2 0 1 0 1 0 1 0 1 0 2 1 4 3 4 1 2 1 3 2 3 1 "
},
{
"input": "24 7\n3 4 2 4 1 4 3 4 3 5 1 3 1 3 0 3 0 3 1 4 0 3 0 1 0 1 0 3 2 3 2 3 1 2 1 3 2 5 1 3 0 1 0 2 0 3 1 3 1",
"output": "3 4 2 4 1 4 3 4 3 5 1 3 1 3 0 3 0 3 1 3 0 2 0 1 0 1 0 3 2 3 2 3 1 2 1 3 2 4 1 2 0 1 0 1 0 2 1 2 1 "
},
{
"input": "25 8\n3 5 2 4 2 4 0 1 0 1 0 1 0 2 1 5 2 4 2 4 2 3 1 2 0 1 0 2 0 3 2 5 3 5 0 4 2 3 2 4 1 4 0 4 1 4 0 1 0 4 2",
"output": "3 5 2 4 2 4 0 1 0 1 0 1 0 2 1 5 2 4 2 4 2 3 1 2 0 1 0 2 0 3 2 4 3 4 0 3 2 3 2 3 1 3 0 3 1 3 0 1 0 3 2 "
},
{
"input": "26 9\n3 4 2 3 1 3 1 3 2 4 0 1 0 2 1 3 1 3 0 5 1 4 3 5 0 5 2 3 0 3 1 4 1 3 1 4 2 3 1 4 3 4 1 3 2 4 1 3 2 5 1 2 0",
"output": "3 4 2 3 1 3 1 3 2 4 0 1 0 2 1 3 1 3 0 4 1 4 3 4 0 4 2 3 0 2 1 3 1 2 1 3 2 3 1 4 3 4 1 3 2 3 1 3 2 4 1 2 0 "
},
{
"input": "27 10\n3 5 3 5 3 4 1 3 1 3 1 3 2 3 2 3 2 4 2 3 0 4 2 5 3 4 3 4 1 5 3 4 1 2 1 5 0 3 0 5 0 5 3 4 0 1 0 2 0 2 1 4 0 2 1",
"output": "3 5 3 5 3 4 1 3 1 3 1 3 2 3 2 3 2 3 2 3 0 3 2 4 3 4 3 4 1 4 3 4 1 2 1 4 0 2 0 4 0 4 3 4 0 1 0 1 0 2 1 3 0 2 1 "
},
{
"input": "40 1\n0 2 1 2 0 2 1 2 1 2 1 2 1 2 1 3 0 1 0 1 0 1 0 2 0 2 1 2 0 2 1 2 1 2 1 2 1 2 0 2 1 2 1 2 0 1 0 2 0 2 0 1 0 1 0 1 0 1 0 1 0 2 0 2 0 2 0 1 0 2 0 1 0 2 0 1 0 2 1 2 0",
"output": "0 2 1 2 0 2 1 2 1 2 1 2 1 2 1 3 0 1 0 1 0 1 0 2 0 2 1 2 0 2 1 2 1 2 1 2 1 2 0 2 1 2 1 2 0 1 0 2 0 2 0 1 0 1 0 1 0 1 0 1 0 2 0 2 0 2 0 1 0 2 0 1 0 1 0 1 0 2 1 2 0 "
},
{
"input": "40 2\n0 3 1 2 1 2 0 1 0 2 1 3 0 2 0 3 0 3 0 1 0 2 0 3 1 2 0 2 1 2 0 2 0 1 0 1 0 2 0 2 1 3 0 2 0 1 0 1 0 1 0 3 1 3 1 2 1 2 0 3 0 1 0 3 0 2 1 2 0 1 0 2 0 3 1 2 1 3 1 3 0",
"output": "0 3 1 2 1 2 0 1 0 2 1 3 0 2 0 3 0 3 0 1 0 2 0 3 1 2 0 2 1 2 0 2 0 1 0 1 0 2 0 2 1 3 0 2 0 1 0 1 0 1 0 3 1 3 1 2 1 2 0 3 0 1 0 3 0 2 1 2 0 1 0 2 0 3 1 2 1 2 1 2 0 "
},
{
"input": "40 3\n1 3 1 2 0 4 1 2 0 1 0 1 0 3 0 3 2 3 0 3 1 3 0 4 1 3 2 3 0 2 1 3 0 2 0 1 0 3 1 3 2 3 2 3 0 1 0 2 0 1 0 1 0 3 1 3 0 3 1 3 1 2 0 1 0 3 0 2 0 3 0 1 0 2 0 3 1 2 0 3 0",
"output": "1 3 1 2 0 4 1 2 0 1 0 1 0 3 0 3 2 3 0 3 1 3 0 4 1 3 2 3 0 2 1 3 0 2 0 1 0 3 1 3 2 3 2 3 0 1 0 2 0 1 0 1 0 3 1 3 0 3 1 3 1 2 0 1 0 3 0 2 0 3 0 1 0 1 0 2 1 2 0 2 0 "
},
{
"input": "50 40\n1 4 2 4 1 2 1 4 1 4 2 3 1 2 1 4 1 3 0 2 1 4 0 1 0 3 1 3 1 3 0 4 2 4 2 4 2 4 2 4 2 4 2 4 0 4 1 3 1 3 0 4 1 4 2 3 2 3 0 3 0 3 0 4 1 4 1 3 1 4 1 3 0 4 0 3 0 2 0 2 0 4 1 4 0 2 0 4 1 4 0 3 0 2 1 3 0 2 0 4 0",
"output": "1 4 2 4 1 2 1 3 1 3 2 3 1 2 1 3 1 2 0 2 1 3 0 1 0 2 1 2 1 2 0 3 2 3 2 3 2 3 2 3 2 3 2 3 0 3 1 2 1 2 0 3 1 3 2 3 2 3 0 2 0 2 0 3 1 3 1 2 1 3 1 2 0 3 0 2 0 1 0 1 0 3 1 3 0 1 0 3 1 3 0 2 0 2 1 2 0 1 0 3 0 "
},
{
"input": "100 2\n1 3 1 2 1 3 2 3 1 3 1 3 1 3 1 2 0 3 0 2 0 3 2 3 0 3 1 2 1 2 0 3 0 1 0 1 0 3 2 3 1 2 0 1 0 2 0 1 0 2 1 3 1 2 1 3 2 3 1 3 1 2 0 3 2 3 0 2 1 3 1 2 0 3 2 3 1 3 2 3 0 4 0 3 0 1 0 3 0 1 0 1 0 2 0 2 1 3 1 2 1 2 0 2 0 1 0 2 0 2 1 3 1 3 2 3 0 2 1 2 0 3 0 1 0 2 0 3 2 3 1 3 0 3 1 2 0 1 0 3 0 1 0 1 0 1 0 2 0 1 0 2 1 2 1 2 1 3 0 1 0 2 1 3 0 2 1 3 0 2 1 2 0 3 1 3 1 3 0 2 1 2 1 3 0 2 1 3 2 3 1 2 0 3 1 2 0 3 1 2 0",
"output": "1 3 1 2 1 3 2 3 1 3 1 3 1 3 1 2 0 3 0 2 0 3 2 3 0 3 1 2 1 2 0 3 0 1 0 1 0 3 2 3 1 2 0 1 0 2 0 1 0 2 1 3 1 2 1 3 2 3 1 3 1 2 0 3 2 3 0 2 1 3 1 2 0 3 2 3 1 3 2 3 0 4 0 3 0 1 0 3 0 1 0 1 0 2 0 2 1 3 1 2 1 2 0 2 0 1 0 2 0 2 1 3 1 3 2 3 0 2 1 2 0 3 0 1 0 2 0 3 2 3 1 3 0 3 1 2 0 1 0 3 0 1 0 1 0 1 0 2 0 1 0 2 1 2 1 2 1 3 0 1 0 2 1 3 0 2 1 3 0 2 1 2 0 3 1 3 1 3 0 2 1 2 1 3 0 2 1 3 2 3 1 2 0 2 1 2 0 2 1 2 0 "
},
{
"input": "100 3\n0 2 1 2 0 1 0 1 0 3 0 2 1 3 1 3 2 3 0 2 0 1 0 2 0 1 0 3 2 3 2 3 1 2 1 3 1 2 1 3 2 3 2 3 0 3 2 3 2 3 2 3 0 2 0 3 0 3 2 3 2 3 2 3 2 3 0 3 0 1 0 2 1 3 0 2 1 2 0 3 2 3 2 3 1 3 0 3 1 3 0 3 0 1 0 1 0 2 0 2 1 2 0 3 1 3 0 3 2 3 2 3 2 3 2 3 0 1 0 1 0 1 0 2 1 2 0 2 1 3 2 3 0 1 0 1 0 1 0 1 0 2 0 1 0 3 1 2 1 2 1 3 1 2 0 3 0 2 1 2 1 3 2 3 1 3 2 3 0 1 0 1 0 1 0 1 0 3 0 1 0 2 1 2 0 3 1 3 2 3 0 3 1 2 1 3 1 3 1 3 0",
"output": "0 2 1 2 0 1 0 1 0 3 0 2 1 3 1 3 2 3 0 2 0 1 0 2 0 1 0 3 2 3 2 3 1 2 1 3 1 2 1 3 2 3 2 3 0 3 2 3 2 3 2 3 0 2 0 3 0 3 2 3 2 3 2 3 2 3 0 3 0 1 0 2 1 3 0 2 1 2 0 3 2 3 2 3 1 3 0 3 1 3 0 3 0 1 0 1 0 2 0 2 1 2 0 3 1 3 0 3 2 3 2 3 2 3 2 3 0 1 0 1 0 1 0 2 1 2 0 2 1 3 2 3 0 1 0 1 0 1 0 1 0 2 0 1 0 3 1 2 1 2 1 3 1 2 0 3 0 2 1 2 1 3 2 3 1 3 2 3 0 1 0 1 0 1 0 1 0 3 0 1 0 2 1 2 0 3 1 3 2 3 0 3 1 2 1 2 1 2 1 2 0 "
},
{
"input": "100 20\n0 1 0 3 0 3 2 3 2 4 0 2 0 3 1 3 0 2 0 2 0 3 0 1 0 3 2 4 0 1 0 2 0 2 1 2 1 4 2 4 1 2 0 1 0 2 1 3 0 2 1 3 2 3 1 2 0 2 1 4 0 3 0 2 0 1 0 1 0 1 0 2 1 3 2 3 2 3 2 3 0 1 0 1 0 4 2 3 2 3 0 3 1 2 0 2 0 2 1 3 2 3 1 4 0 1 0 2 1 2 0 2 0 3 2 3 0 2 0 2 1 4 2 3 1 3 0 3 0 2 0 2 1 2 1 3 0 3 1 2 1 3 1 3 1 2 1 2 0 2 1 3 0 2 0 3 0 1 0 3 0 3 0 1 0 4 1 3 0 1 0 1 0 2 1 2 0 2 1 4 1 3 0 2 1 3 1 3 1 3 0 3 0 2 0 1 0 2 1 2 1",
"output": "0 1 0 3 0 3 2 3 2 4 0 2 0 3 1 3 0 2 0 2 0 3 0 1 0 3 2 4 0 1 0 2 0 2 1 2 1 4 2 4 1 2 0 1 0 2 1 3 0 2 1 3 2 3 1 2 0 2 1 4 0 3 0 2 0 1 0 1 0 1 0 2 1 3 2 3 2 3 2 3 0 1 0 1 0 4 2 3 2 3 0 3 1 2 0 2 0 2 1 3 2 3 1 4 0 1 0 2 1 2 0 2 0 3 2 3 0 2 0 2 1 4 2 3 1 3 0 2 0 1 0 2 1 2 1 2 0 2 1 2 1 2 1 2 1 2 1 2 0 2 1 2 0 1 0 2 0 1 0 2 0 2 0 1 0 3 1 2 0 1 0 1 0 2 1 2 0 2 1 3 1 2 0 2 1 2 1 2 1 2 0 2 0 1 0 1 0 2 1 2 1 "
},
{
"input": "100 20\n2 3 0 4 0 1 0 6 3 4 3 6 4 6 0 9 0 6 2 7 3 8 7 10 2 9 3 9 5 6 5 10 3 7 1 5 2 8 3 7 2 3 1 6 0 8 3 8 0 4 1 8 3 7 1 9 5 9 5 8 7 8 5 6 5 8 1 9 8 9 8 10 7 10 5 8 6 10 2 6 3 9 2 6 3 10 5 9 3 10 1 3 2 11 8 9 8 10 1 8 7 11 0 9 5 8 4 5 0 7 3 7 5 9 5 10 1 7 1 9 1 6 3 8 2 4 1 4 2 6 0 4 2 4 2 7 6 9 0 1 0 4 0 4 0 9 2 7 6 7 2 8 0 8 2 7 5 10 1 2 0 2 0 4 3 5 4 7 0 10 2 10 3 6 3 7 1 4 0 9 1 4 3 8 1 10 1 10 0 3 2 5 3 9 0 7 4 5 0 1 0",
"output": "2 3 0 4 0 1 0 6 3 4 3 6 4 6 0 9 0 6 2 7 3 8 7 10 2 9 3 9 5 6 5 10 3 7 1 5 2 8 3 7 2 3 1 6 0 8 3 8 0 4 1 8 3 7 1 9 5 9 5 8 7 8 5 6 5 8 1 9 8 9 8 10 7 10 5 8 6 10 2 6 3 9 2 6 3 10 5 9 3 10 1 3 2 11 8 9 8 10 1 8 7 11 0 9 5 8 4 5 0 7 3 7 5 9 5 10 1 7 1 9 1 6 3 8 2 4 1 4 2 6 0 4 2 4 2 7 6 9 0 1 0 4 0 3 0 8 2 7 6 7 2 7 0 7 2 6 5 9 1 2 0 1 0 4 3 5 4 6 0 9 2 9 3 5 3 6 1 3 0 8 1 4 3 7 1 9 1 9 0 3 2 4 3 8 0 6 4 5 0 1 0 "
},
{
"input": "98 3\n1 2 1 2 0 2 0 2 1 2 0 1 0 2 1 2 0 2 1 2 1 2 0 1 0 2 1 2 1 2 0 2 1 2 0 2 0 2 0 1 0 1 0 1 0 2 1 3 1 2 1 2 1 2 1 2 1 2 1 2 0 2 0 2 1 2 1 2 0 2 1 2 0 1 0 1 0 1 0 1 0 2 0 1 0 2 0 2 1 2 1 2 1 2 0 1 0 1 0 1 0 2 1 2 0 2 1 2 0 2 0 1 0 2 1 2 0 1 0 2 1 2 1 2 1 2 0 2 1 2 1 2 1 2 0 2 1 2 1 2 0 1 0 2 0 2 0 1 0 2 0 2 0 1 0 1 0 1 0 2 0 2 1 2 0 1 0 2 0 2 0 1 0 2 1 2 1 2 1 2 0 2 1 2 1 2 1 2 0 1 0 1 0 2 0 2 0",
"output": "1 2 1 2 0 2 0 2 1 2 0 1 0 2 1 2 0 2 1 2 1 2 0 1 0 2 1 2 1 2 0 2 1 2 0 2 0 2 0 1 0 1 0 1 0 2 1 3 1 2 1 2 1 2 1 2 1 2 1 2 0 2 0 2 1 2 1 2 0 2 1 2 0 1 0 1 0 1 0 1 0 2 0 1 0 2 0 2 1 2 1 2 1 2 0 1 0 1 0 1 0 2 1 2 0 2 1 2 0 2 0 1 0 2 1 2 0 1 0 2 1 2 1 2 1 2 0 2 1 2 1 2 1 2 0 2 1 2 1 2 0 1 0 2 0 2 0 1 0 2 0 2 0 1 0 1 0 1 0 2 0 2 1 2 0 1 0 2 0 1 0 1 0 2 1 2 1 2 1 2 0 2 1 2 1 2 1 2 0 1 0 1 0 1 0 1 0 "
},
{
"input": "2 1\n0 2 1 4 1",
"output": "0 2 1 3 1 "
},
{
"input": "2 1\n0 2 1 5 1",
"output": "0 2 1 4 1 "
},
{
"input": "3 3\n1 12 9 11 6 8 1",
"output": "1 11 9 10 6 7 1 "
},
{
"input": "3 2\n0 7 4 7 1 3 2",
"output": "0 6 4 6 1 3 2 "
},
{
"input": "2 1\n1 3 2 4 1",
"output": "1 3 2 3 1 "
},
{
"input": "4 1\n5 6 5 6 5 6 1 3 1",
"output": "5 6 5 6 5 6 1 2 1 "
},
{
"input": "2 1\n0 2 1 3 0",
"output": "0 2 1 2 0 "
},
{
"input": "2 2\n98 100 1 7 2",
"output": "98 99 1 6 2 "
},
{
"input": "3 1\n8 10 9 10 3 5 1",
"output": "8 10 9 10 3 4 1 "
},
{
"input": "3 2\n0 4 3 5 2 5 2",
"output": "0 4 3 4 2 4 2 "
},
{
"input": "2 1\n4 5 2 4 2",
"output": "4 5 2 3 2 "
},
{
"input": "3 1\n0 2 1 2 0 2 0",
"output": "0 2 1 2 0 1 0 "
},
{
"input": "1 1\n5 7 2",
"output": "5 6 2 "
},
{
"input": "2 1\n3 4 1 3 1",
"output": "3 4 1 2 1 "
},
{
"input": "3 1\n0 4 3 5 0 5 0",
"output": "0 4 3 5 0 4 0 "
},
{
"input": "3 1\n1 3 2 3 1 3 1",
"output": "1 3 2 3 1 2 1 "
},
{
"input": "2 1\n0 8 7 100 0",
"output": "0 8 7 99 0 "
},
{
"input": "2 1\n1 3 2 5 1",
"output": "1 3 2 4 1 "
}
] | 1,556,799,303
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 186
| 0
|
n,k=map(int,input().split())
points=list(map(int,input().split()))
peaks={}
for i in range(2*n-1):
if points[i] < points[i+1] and points[i+1] > points[i+2]:
if points[i] < points[i+1] and points[i+1] > points[i+2]:
peaks[i+1] = points[i+1]-1
high = list(peaks.keys())
selcted=high[:k]
for i in range(len(points)):
if i in selcted:
print(peaks[i] ,end=" ")
else:
print(points[i],end=" ")
|
Title: Mountain Scenery
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Bolek has found a picture with *n* mountain peaks painted on it. The *n* painted peaks are represented by a non-closed polyline, consisting of 2*n* segments. The segments go through 2*n*<=+<=1 points with coordinates (1,<=*y*1), (2,<=*y*2), ..., (2*n*<=+<=1,<=*y*2*n*<=+<=1), with the *i*-th segment connecting the point (*i*,<=*y**i*) and the point (*i*<=+<=1,<=*y**i*<=+<=1). For any even *i* (2<=≤<=*i*<=≤<=2*n*) the following condition holds: *y**i*<=-<=1<=<<=*y**i* and *y**i*<=><=*y**i*<=+<=1.
We shall call a vertex of a polyline with an even *x* coordinate a mountain peak.
Bolek fancied a little mischief. He chose exactly *k* mountain peaks, rubbed out the segments that went through those peaks and increased each peak's height by one (that is, he increased the *y* coordinate of the corresponding points). Then he painted the missing segments to get a new picture of mountain peaks. Let us denote the points through which the new polyline passes on Bolek's new picture as (1,<=*r*1), (2,<=*r*2), ..., (2*n*<=+<=1,<=*r*2*n*<=+<=1).
Given Bolek's final picture, restore the initial one.
Input Specification:
The first line contains two space-separated integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=100). The next line contains 2*n*<=+<=1 space-separated integers *r*1,<=*r*2,<=...,<=*r*2*n*<=+<=1 (0<=≤<=*r**i*<=≤<=100) — the *y* coordinates of the polyline vertices on Bolek's picture.
It is guaranteed that we can obtain the given picture after performing the described actions on some picture of mountain peaks.
Output Specification:
Print 2*n*<=+<=1 integers *y*1,<=*y*2,<=...,<=*y*2*n*<=+<=1 — the *y* coordinates of the vertices of the polyline on the initial picture. If there are multiple answers, output any one of them.
Demo Input:
['3 2\n0 5 3 5 1 5 2\n', '1 1\n0 2 0\n']
Demo Output:
['0 5 3 4 1 4 2 \n', '0 1 0 \n']
Note:
none
|
```python
n,k=map(int,input().split())
points=list(map(int,input().split()))
peaks={}
for i in range(2*n-1):
if points[i] < points[i+1] and points[i+1] > points[i+2]:
if points[i] < points[i+1] and points[i+1] > points[i+2]:
peaks[i+1] = points[i+1]-1
high = list(peaks.keys())
selcted=high[:k]
for i in range(len(points)):
if i in selcted:
print(peaks[i] ,end=" ")
else:
print(points[i],end=" ")
```
| 0
|
|
493
|
A
|
Vasya and Football
|
PROGRAMMING
| 1,300
|
[
"implementation"
] | null | null |
Vasya has started watching football games. He has learned that for some fouls the players receive yellow cards, and for some fouls they receive red cards. A player who receives the second yellow card automatically receives a red card.
Vasya is watching a recorded football match now and makes notes of all the fouls that he would give a card for. Help Vasya determine all the moments in time when players would be given red cards if Vasya were the judge. For each player, Vasya wants to know only the first moment of time when he would receive a red card from Vasya.
|
The first line contains the name of the team playing at home. The second line contains the name of the team playing away. Both lines are not empty. The lengths of both lines do not exceed 20. Each line contains only of large English letters. The names of the teams are distinct.
Next follows number *n* (1<=≤<=*n*<=≤<=90) — the number of fouls.
Each of the following *n* lines contains information about a foul in the following form:
- first goes number *t* (1<=≤<=*t*<=≤<=90) — the minute when the foul occurs; - then goes letter "h" or letter "a" — if the letter is "h", then the card was given to a home team player, otherwise the card was given to an away team player; - then goes the player's number *m* (1<=≤<=*m*<=≤<=99); - then goes letter "y" or letter "r" — if the letter is "y", that means that the yellow card was given, otherwise the red card was given.
The players from different teams can have the same number. The players within one team have distinct numbers. The fouls go chronologically, no two fouls happened at the same minute.
|
For each event when a player received his first red card in a chronological order print a string containing the following information:
- The name of the team to which the player belongs; - the player's number in his team; - the minute when he received the card.
If no player received a card, then you do not need to print anything.
It is possible case that the program will not print anything to the output (if there were no red cards).
|
[
"MC\nCSKA\n9\n28 a 3 y\n62 h 25 y\n66 h 42 y\n70 h 25 y\n77 a 4 y\n79 a 25 y\n82 h 42 r\n89 h 16 y\n90 a 13 r\n"
] |
[
"MC 25 70\nMC 42 82\nCSKA 13 90\n"
] |
none
| 500
|
[
{
"input": "MC\nCSKA\n9\n28 a 3 y\n62 h 25 y\n66 h 42 y\n70 h 25 y\n77 a 4 y\n79 a 25 y\n82 h 42 r\n89 h 16 y\n90 a 13 r",
"output": "MC 25 70\nMC 42 82\nCSKA 13 90"
},
{
"input": "REAL\nBARCA\n3\n27 h 7 y\n44 a 10 y\n87 h 3 r",
"output": "REAL 3 87"
},
{
"input": "MASFF\nSAFBDSRG\n5\n1 h 1 y\n15 h 1 r\n27 a 1 y\n58 a 1 y\n69 h 10 y",
"output": "MASFF 1 15\nSAFBDSRG 1 58"
},
{
"input": "ARMENIA\nBULGARIA\n12\n33 h 17 y\n42 h 21 y\n56 a 17 y\n58 a 6 y\n61 a 7 y\n68 a 10 y\n72 h 13 y\n73 h 21 y\n74 a 8 r\n75 a 4 y\n77 a 10 y\n90 a 23 y",
"output": "ARMENIA 21 73\nBULGARIA 8 74\nBULGARIA 10 77"
},
{
"input": "PORTUGAL\nNETHERLANDS\n16\n2 a 18 y\n7 a 3 y\n20 h 18 y\n31 h 6 y\n45 h 6 y\n50 h 8 y\n59 a 5 y\n60 h 7 y\n63 a 3 y\n72 a 20 y\n73 h 20 y\n74 a 10 y\n75 h 1 y\n76 h 14 y\n78 h 20 y\n90 a 5 y",
"output": "PORTUGAL 6 45\nNETHERLANDS 3 63\nPORTUGAL 20 78\nNETHERLANDS 5 90"
},
{
"input": "TANC\nXNCOR\n2\n15 h 27 r\n28 h 27 r",
"output": "TANC 27 15"
},
{
"input": "ASGDFJH\nAHGRSDXGER\n3\n23 h 15 r\n68 h 15 y\n79 h 15 y",
"output": "ASGDFJH 15 23"
},
{
"input": "ASFSHDSG\nADGYRTJNG\n5\n1 h 1 y\n2 h 1 y\n3 h 1 y\n4 h 1 r\n5 h 1 y",
"output": "ASFSHDSG 1 2"
},
{
"input": "A\nB\n42\n5 a 84 y\n8 h 28 r\n10 a 9 r\n11 h 93 y\n13 a 11 r\n15 h 3 r\n20 a 88 r\n23 a 41 y\n25 a 14 y\n27 a 38 r\n28 a 33 y\n29 h 66 r\n31 a 16 r\n32 a 80 y\n34 a 54 r\n35 a 50 y\n36 a 9 y\n39 a 22 y\n42 h 81 y\n43 a 10 y\n44 a 27 r\n47 h 39 y\n48 a 80 y\n50 h 5 y\n52 a 67 y\n54 h 63 y\n56 h 7 y\n57 h 44 y\n58 h 41 y\n61 h 32 y\n64 h 91 y\n67 a 56 y\n69 h 83 y\n71 h 59 y\n72 a 76 y\n75 h 41 y\n76 a 49 r\n77 a 4 r\n78 a 69 y\n79 a 96 r\n80 h 81 y\n86 h 85 r",
"output": "A 28 8\nB 9 10\nB 11 13\nA 3 15\nB 88 20\nB 38 27\nA 66 29\nB 16 31\nB 54 34\nB 27 44\nB 80 48\nA 41 75\nB 49 76\nB 4 77\nB 96 79\nA 81 80\nA 85 86"
},
{
"input": "ARM\nAZE\n45\n2 a 13 r\n3 a 73 r\n4 a 10 y\n5 h 42 y\n8 h 56 y\n10 h 15 y\n11 a 29 r\n13 a 79 y\n14 a 77 r\n18 h 7 y\n20 a 69 r\n22 h 19 y\n25 h 88 r\n26 a 78 y\n27 a 91 r\n28 h 10 r\n30 h 13 r\n31 a 26 r\n33 a 43 r\n34 a 91 y\n40 h 57 y\n44 h 18 y\n46 a 25 r\n48 a 29 y\n51 h 71 y\n57 a 16 r\n58 h 37 r\n59 h 92 y\n60 h 11 y\n61 a 88 y\n64 a 28 r\n65 h 71 r\n68 h 39 y\n70 h 8 r\n71 a 10 y\n72 a 32 y\n73 h 95 r\n74 a 33 y\n75 h 48 r\n78 a 44 y\n79 a 22 r\n80 h 50 r\n84 a 50 y\n88 a 90 y\n89 h 42 r",
"output": "AZE 13 2\nAZE 73 3\nAZE 29 11\nAZE 77 14\nAZE 69 20\nARM 88 25\nAZE 91 27\nARM 10 28\nARM 13 30\nAZE 26 31\nAZE 43 33\nAZE 25 46\nAZE 16 57\nARM 37 58\nAZE 28 64\nARM 71 65\nARM 8 70\nAZE 10 71\nARM 95 73\nARM 48 75\nAZE 22 79\nARM 50 80\nARM 42 89"
},
{
"input": "KASFLS\nASJBGGDLJFDDFHHTHJH\n42\n2 a 68 y\n4 h 64 r\n5 a 24 y\n6 h 20 r\n8 a 16 r\n9 a 96 y\n10 h 36 r\n12 a 44 y\n13 h 69 r\n16 a 62 r\n18 a 99 r\n20 h 12 r\n21 a 68 y\n25 h 40 y\n26 h 54 r\n28 h 91 r\n29 a 36 r\n33 a 91 y\n36 h 93 r\n37 h 60 r\n38 a 82 r\n41 a 85 y\n42 a 62 r\n46 a 22 r\n48 a 88 r\n49 a 8 r\n51 h 45 y\n54 a 84 y\n57 a 8 y\n59 h 24 y\n61 h 22 r\n64 h 11 r\n69 a 89 y\n72 h 44 r\n75 h 57 r\n76 h 80 y\n77 h 54 r\n79 a 1 y\n81 a 31 r\n82 h 8 y\n83 a 28 r\n86 h 56 y",
"output": "KASFLS 64 4\nKASFLS 20 6\nASJBGGDLJFDDFHHTHJH 16 8\nKASFLS 36 10\nKASFLS 69 13\nASJBGGDLJFDDFHHTHJH 62 16\nASJBGGDLJFDDFHHTHJH 99 18\nKASFLS 12 20\nASJBGGDLJFDDFHHTHJH 68 21\nKASFLS 54 26\nKASFLS 91 28\nASJBGGDLJFDDFHHTHJH 36 29\nKASFLS 93 36\nKASFLS 60 37\nASJBGGDLJFDDFHHTHJH 82 38\nASJBGGDLJFDDFHHTHJH 22 46\nASJBGGDLJFDDFHHTHJH 88 48\nASJBGGDLJFDDFHHTHJH 8 49\nKASFLS 22 61\nKASFLS 11 64\nKASFLS 44 72\nKASFLS 57 75\nASJBGGDLJFDDFHHTHJH 31 81\nASJBGGDLJFDDFHHTHJH 28 83"
},
{
"input": "AB\nBC\n3\n1 h 1 y\n2 h 1 y\n3 h 1 r",
"output": "AB 1 2"
}
] | 1,501,222,229
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 62
| 4,608,000
|
home,away=input(),input()
ans=[]
yel=[]
for _ in range(int(input())):
t,team,m,col=input().split()
if col=='r' or any(m in x and (home if team!='a' else away) in x for x in yel):
ans.append((home if team!='a' else away,m,t))
else:
yel.append((home if team!='a' else away,m,t))
ans=sorted(ans,key=lambda i: i[2])
for x in ans:
print(x[0]+" "+x[1]+" "+x[2])
|
Title: Vasya and Football
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has started watching football games. He has learned that for some fouls the players receive yellow cards, and for some fouls they receive red cards. A player who receives the second yellow card automatically receives a red card.
Vasya is watching a recorded football match now and makes notes of all the fouls that he would give a card for. Help Vasya determine all the moments in time when players would be given red cards if Vasya were the judge. For each player, Vasya wants to know only the first moment of time when he would receive a red card from Vasya.
Input Specification:
The first line contains the name of the team playing at home. The second line contains the name of the team playing away. Both lines are not empty. The lengths of both lines do not exceed 20. Each line contains only of large English letters. The names of the teams are distinct.
Next follows number *n* (1<=≤<=*n*<=≤<=90) — the number of fouls.
Each of the following *n* lines contains information about a foul in the following form:
- first goes number *t* (1<=≤<=*t*<=≤<=90) — the minute when the foul occurs; - then goes letter "h" or letter "a" — if the letter is "h", then the card was given to a home team player, otherwise the card was given to an away team player; - then goes the player's number *m* (1<=≤<=*m*<=≤<=99); - then goes letter "y" or letter "r" — if the letter is "y", that means that the yellow card was given, otherwise the red card was given.
The players from different teams can have the same number. The players within one team have distinct numbers. The fouls go chronologically, no two fouls happened at the same minute.
Output Specification:
For each event when a player received his first red card in a chronological order print a string containing the following information:
- The name of the team to which the player belongs; - the player's number in his team; - the minute when he received the card.
If no player received a card, then you do not need to print anything.
It is possible case that the program will not print anything to the output (if there were no red cards).
Demo Input:
['MC\nCSKA\n9\n28 a 3 y\n62 h 25 y\n66 h 42 y\n70 h 25 y\n77 a 4 y\n79 a 25 y\n82 h 42 r\n89 h 16 y\n90 a 13 r\n']
Demo Output:
['MC 25 70\nMC 42 82\nCSKA 13 90\n']
Note:
none
|
```python
home,away=input(),input()
ans=[]
yel=[]
for _ in range(int(input())):
t,team,m,col=input().split()
if col=='r' or any(m in x and (home if team!='a' else away) in x for x in yel):
ans.append((home if team!='a' else away,m,t))
else:
yel.append((home if team!='a' else away,m,t))
ans=sorted(ans,key=lambda i: i[2])
for x in ans:
print(x[0]+" "+x[1]+" "+x[2])
```
| 0
|
|
50
|
A
|
Domino piling
|
PROGRAMMING
| 800
|
[
"greedy",
"math"
] |
A. Domino piling
|
2
|
256
|
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
|
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
|
Output one number — the maximal number of dominoes, which can be placed.
|
[
"2 4\n",
"3 3\n"
] |
[
"4\n",
"4\n"
] |
none
| 500
|
[
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "2 6",
"output": "6"
},
{
"input": "2 7",
"output": "7"
},
{
"input": "2 14",
"output": "14"
},
{
"input": "2 15",
"output": "15"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "2 16",
"output": "16"
},
{
"input": "3 5",
"output": "7"
},
{
"input": "3 6",
"output": "9"
},
{
"input": "3 10",
"output": "15"
},
{
"input": "3 14",
"output": "21"
},
{
"input": "3 15",
"output": "22"
},
{
"input": "3 16",
"output": "24"
},
{
"input": "5 7",
"output": "17"
},
{
"input": "16 16",
"output": "128"
},
{
"input": "15 16",
"output": "120"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "15 15",
"output": "112"
},
{
"input": "14 16",
"output": "112"
},
{
"input": "11 13",
"output": "71"
},
{
"input": "5 16",
"output": "40"
},
{
"input": "8 15",
"output": "60"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "3 4",
"output": "6"
},
{
"input": "4 4",
"output": "8"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "14 15",
"output": "105"
}
] | 1,634,122,578
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 122
| 0
|
m,n=map(int,input().split())
p=(m*n)//2
print(p)
|
Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none
|
```python
m,n=map(int,input().split())
p=(m*n)//2
print(p)
```
| 3.9695
|
363
|
B
|
Fence
|
PROGRAMMING
| 1,100
|
[
"brute force",
"dp"
] | null | null |
There is a fence in front of Polycarpus's home. The fence consists of *n* planks of the same width which go one after another from left to right. The height of the *i*-th plank is *h**i* meters, distinct planks can have distinct heights.
Polycarpus has bought a posh piano and is thinking about how to get it into the house. In order to carry out his plan, he needs to take exactly *k* consecutive planks from the fence. Higher planks are harder to tear off the fence, so Polycarpus wants to find such *k* consecutive planks that the sum of their heights is minimal possible.
Write the program that finds the indexes of *k* consecutive planks with minimal total height. Pay attention, the fence is not around Polycarpus's home, it is in front of home (in other words, the fence isn't cyclic).
|
The first line of the input contains integers *n* and *k* (1<=≤<=*n*<=≤<=1.5·105,<=1<=≤<=*k*<=≤<=*n*) — the number of planks in the fence and the width of the hole for the piano. The second line contains the sequence of integers *h*1,<=*h*2,<=...,<=*h**n* (1<=≤<=*h**i*<=≤<=100), where *h**i* is the height of the *i*-th plank of the fence.
|
Print such integer *j* that the sum of the heights of planks *j*, *j*<=+<=1, ..., *j*<=+<=*k*<=-<=1 is the minimum possible. If there are multiple such *j*'s, print any of them.
|
[
"7 3\n1 2 6 1 1 7 1\n"
] |
[
"3\n"
] |
In the sample, your task is to find three consecutive planks with the minimum sum of heights. In the given case three planks with indexes 3, 4 and 5 have the required attribute, their total height is 8.
| 1,000
|
[
{
"input": "7 3\n1 2 6 1 1 7 1",
"output": "3"
},
{
"input": "1 1\n100",
"output": "1"
},
{
"input": "2 1\n10 20",
"output": "1"
},
{
"input": "10 5\n1 2 3 1 2 2 3 1 4 5",
"output": "1"
},
{
"input": "10 2\n3 1 4 1 4 6 2 1 4 6",
"output": "7"
},
{
"input": "2 2\n20 10",
"output": "1"
},
{
"input": "2 1\n20 1",
"output": "2"
},
{
"input": "3 1\n1 2 3",
"output": "1"
},
{
"input": "3 1\n2 1 3",
"output": "2"
},
{
"input": "3 1\n3 2 1",
"output": "3"
},
{
"input": "3 2\n1 2 3",
"output": "1"
},
{
"input": "3 2\n3 2 1",
"output": "2"
},
{
"input": "3 3\n1 2 3",
"output": "1"
},
{
"input": "4 2\n9 8 11 7",
"output": "1"
},
{
"input": "4 2\n10 1 2 3",
"output": "2"
},
{
"input": "6 3\n56 56 56 2 1 2",
"output": "4"
},
{
"input": "8 3\n1 1 1 1 2 60 90 1",
"output": "1"
},
{
"input": "4 1\n1 5 2 2",
"output": "1"
},
{
"input": "4 2\n4 6 7 4",
"output": "1"
},
{
"input": "10 4\n1 1 1 4 4 4 4 4 4 3",
"output": "1"
},
{
"input": "6 3\n1 2 1 3 1 1",
"output": "1"
},
{
"input": "5 2\n100 100 100 1 1",
"output": "4"
}
] | 1,667,310,385
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 6
| 1,000
| 3,379,200
|
n,m=map(int,input().split(" "))
a=list(map(int,input().split(" ")))
ts=sum(a[:m])
index=1
for i in range(n-m):
temp=sum(a[i:i+m])
if temp<ts:
ts=temp
index=i+1
print(index)
|
Title: Fence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There is a fence in front of Polycarpus's home. The fence consists of *n* planks of the same width which go one after another from left to right. The height of the *i*-th plank is *h**i* meters, distinct planks can have distinct heights.
Polycarpus has bought a posh piano and is thinking about how to get it into the house. In order to carry out his plan, he needs to take exactly *k* consecutive planks from the fence. Higher planks are harder to tear off the fence, so Polycarpus wants to find such *k* consecutive planks that the sum of their heights is minimal possible.
Write the program that finds the indexes of *k* consecutive planks with minimal total height. Pay attention, the fence is not around Polycarpus's home, it is in front of home (in other words, the fence isn't cyclic).
Input Specification:
The first line of the input contains integers *n* and *k* (1<=≤<=*n*<=≤<=1.5·105,<=1<=≤<=*k*<=≤<=*n*) — the number of planks in the fence and the width of the hole for the piano. The second line contains the sequence of integers *h*1,<=*h*2,<=...,<=*h**n* (1<=≤<=*h**i*<=≤<=100), where *h**i* is the height of the *i*-th plank of the fence.
Output Specification:
Print such integer *j* that the sum of the heights of planks *j*, *j*<=+<=1, ..., *j*<=+<=*k*<=-<=1 is the minimum possible. If there are multiple such *j*'s, print any of them.
Demo Input:
['7 3\n1 2 6 1 1 7 1\n']
Demo Output:
['3\n']
Note:
In the sample, your task is to find three consecutive planks with the minimum sum of heights. In the given case three planks with indexes 3, 4 and 5 have the required attribute, their total height is 8.
|
```python
n,m=map(int,input().split(" "))
a=list(map(int,input().split(" ")))
ts=sum(a[:m])
index=1
for i in range(n-m):
temp=sum(a[i:i+m])
if temp<ts:
ts=temp
index=i+1
print(index)
```
| 0
|
|
762
|
F
|
Tree nesting
|
PROGRAMMING
| 2,800
|
[
"combinatorics",
"graphs",
"trees"
] | null | null |
You are given two trees (connected undirected acyclic graphs) *S* and *T*.
Count the number of subtrees (connected subgraphs) of *S* that are isomorphic to tree *T*. Since this number can get quite large, output it modulo 109<=+<=7.
Two subtrees of tree *S* are considered different, if there exists a vertex in *S* that belongs to exactly one of them.
Tree *G* is called isomorphic to tree *H* if there exists a bijection *f* from the set of vertices of *G* to the set of vertices of *H* that has the following property: if there is an edge between vertices *A* and *B* in tree *G*, then there must be an edge between vertices *f*(*A*) and *f*(*B*) in tree *H*. And vice versa — if there is an edge between vertices *A* and *B* in tree *H*, there must be an edge between *f*<=-<=1(*A*) and *f*<=-<=1(*B*) in tree *G*.
|
The first line contains a single integer |*S*| (1<=≤<=|*S*|<=≤<=1000) — the number of vertices of tree *S*.
Next |*S*|<=-<=1 lines contain two integers *u**i* and *v**i* (1<=≤<=*u**i*,<=*v**i*<=≤<=|*S*|) and describe edges of tree *S*.
The next line contains a single integer |*T*| (1<=≤<=|*T*|<=≤<=12) — the number of vertices of tree *T*.
Next |*T*|<=-<=1 lines contain two integers *x**i* and *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=|*T*|) and describe edges of tree *T*.
|
On the first line output a single integer — the answer to the given task modulo 109<=+<=7.
|
[
"5\n1 2\n2 3\n3 4\n4 5\n3\n1 2\n2 3\n",
"3\n2 3\n3 1\n3\n1 2\n1 3\n",
"7\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n4\n4 1\n4 2\n4 3\n",
"5\n1 2\n2 3\n3 4\n4 5\n4\n4 1\n4 2\n4 3\n"
] |
[
"3\n",
"1\n",
"20\n",
"0\n"
] |
none
| 0
|
[
{
"input": "5\n1 2\n2 3\n3 4\n4 5\n3\n1 2\n2 3",
"output": "3"
},
{
"input": "3\n2 3\n3 1\n3\n1 2\n1 3",
"output": "1"
},
{
"input": "7\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n4\n4 1\n4 2\n4 3",
"output": "20"
},
{
"input": "5\n1 2\n2 3\n3 4\n4 5\n4\n4 1\n4 2\n4 3",
"output": "0"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "20\n11 15\n2 12\n5 6\n3 18\n3 4\n1 12\n15 8\n20 7\n14 5\n8 12\n13 18\n10 18\n16 19\n20 9\n13 17\n8 16\n5 17\n15 13\n9 15\n3\n2 1\n3 1",
"output": "26"
},
{
"input": "20\n19 1\n1 12\n18 5\n4 11\n1 2\n5 16\n20 12\n15 6\n6 16\n19 7\n19 14\n2 13\n20 8\n7 5\n1 3\n12 4\n17 4\n1 10\n9 12\n4\n1 3\n2 4\n4 1",
"output": "40"
},
{
"input": "20\n12 9\n15 8\n18 7\n9 3\n20 6\n6 15\n5 12\n2 14\n1 4\n18 19\n12 18\n13 16\n10 9\n17 1\n8 12\n14 11\n18 14\n10 13\n1 5\n12\n6 10\n1 6\n2 9\n11 9\n7 3\n4 10\n9 6\n8 9\n6 7\n12 6\n5 6",
"output": "0"
},
{
"input": "20\n2 6\n4 8\n12 11\n16 2\n1 20\n15 19\n20 14\n7 2\n7 13\n7 19\n20 9\n6 5\n12 7\n14 10\n3 12\n17 18\n17 4\n10 4\n4 16\n4\n3 1\n4 1\n1 2",
"output": "11"
},
{
"input": "20\n19 18\n11 4\n17 4\n7 6\n14 18\n5 6\n10 3\n12 20\n13 16\n11 2\n6 1\n12 6\n18 5\n8 6\n17 6\n13 17\n20 9\n15 13\n10 11\n3\n2 1\n3 1",
"output": "32"
},
{
"input": "20\n14 17\n20 7\n13 15\n20 4\n18 9\n6 3\n16 13\n5 20\n6 14\n13 11\n6 20\n2 20\n4 19\n20 13\n12 13\n2 9\n2 8\n10 1\n6 10\n2\n1 2",
"output": "19"
},
{
"input": "20\n9 19\n20 10\n15 5\n1 4\n2 5\n3 13\n20 2\n17 12\n16 6\n2 3\n19 15\n3 14\n15 11\n8 7\n18 12\n2 17\n2 1\n8 5\n18 16\n11\n6 4\n1 2\n5 11\n8 3\n7 3\n3 5\n4 2\n10 2\n3 2\n9 6",
"output": "0"
},
{
"input": "20\n9 13\n2 7\n11 15\n2 3\n15 14\n11 12\n8 2\n19 8\n6 9\n7 16\n12 1\n20 4\n13 18\n20 2\n10 14\n13 7\n8 15\n17 4\n8 5\n5\n3 2\n4 2\n4 1\n4 5",
"output": "46"
},
{
"input": "20\n14 19\n19 15\n7 17\n19 8\n9 16\n12 16\n10 11\n13 7\n15 13\n2 1\n13 2\n6 15\n9 18\n9 3\n20 19\n10 9\n13 16\n4 13\n12 5\n10\n8 7\n5 6\n2 3\n7 6\n1 9\n9 6\n3 8\n10 3\n2 4",
"output": "6"
},
{
"input": "20\n12 14\n3 19\n6 9\n7 8\n18 1\n13 3\n1 4\n16 18\n15 10\n17 16\n2 5\n19 5\n20 4\n19 7\n12 18\n19 12\n14 15\n11 2\n7 9\n9\n3 4\n6 7\n5 9\n2 5\n5 7\n7 8\n7 1\n4 8",
"output": "6"
},
{
"input": "30\n4 26\n26 16\n11 4\n17 29\n12 15\n14 30\n2 1\n8 9\n21 20\n25 24\n17 6\n10 13\n29 24\n3 26\n18 25\n14 25\n17 7\n4 25\n12 10\n20 10\n25 23\n26 10\n28 13\n19 20\n4 5\n13 27\n22 18\n5 9\n10 1\n2\n2 1",
"output": "29"
},
{
"input": "40\n15 11\n7 2\n6 8\n40 27\n13 39\n5 26\n8 25\n14 30\n36 19\n13 38\n17 3\n8 33\n4 2\n1 40\n17 5\n11 1\n24 33\n4 8\n10 31\n26 32\n38 26\n10 18\n23 32\n1 16\n33 5\n12 6\n6 14\n17 28\n9 8\n22 21\n36 15\n16 35\n32 37\n36 22\n17 36\n20 21\n6 34\n4 31\n11 29\n1",
"output": "40"
},
{
"input": "50\n33 44\n14 10\n26 25\n29 24\n21 15\n13 7\n16 26\n29 17\n16 44\n49 9\n35 42\n28 38\n37 20\n10 12\n48 2\n21 47\n46 40\n46 31\n42 8\n31 45\n38 11\n4 19\n3 35\n22 39\n41 9\n11 6\n34 44\n37 44\n22 41\n33 35\n41 4\n22 1\n13 27\n13 22\n14 33\n18 21\n36 22\n21 31\n30 35\n31 43\n24 12\n44 32\n27 2\n41 21\n1 50\n23 19\n6 22\n14 41\n47 5\n3\n3 1\n3 2",
"output": "85"
},
{
"input": "60\n35 24\n14 29\n40 57\n16 56\n16 35\n9 58\n35 50\n17 45\n9 45\n57 18\n40 60\n19 54\n33 50\n16 6\n32 57\n10 17\n20 35\n55 27\n60 11\n30 19\n27 36\n12 41\n33 44\n50 12\n19 20\n52 12\n51 40\n17 47\n52 42\n39 13\n13 5\n22 58\n59 37\n44 34\n43 42\n36 7\n23 59\n45 49\n4 53\n26 19\n17 35\n39 16\n20 48\n59 6\n38 13\n31 5\n10 3\n40 29\n8 18\n46 43\n50 36\n1 35\n4 15\n4 57\n12 28\n9 21\n58 32\n10 2\n4 25\n2\n2 1",
"output": "59"
},
{
"input": "70\n22 52\n54 23\n67 70\n3 31\n8 40\n68 10\n33 27\n60 61\n50 13\n57 6\n34 66\n29 48\n29 9\n28 44\n6 12\n32 10\n7 30\n26 21\n12 13\n54 34\n14 49\n12 40\n7 18\n43 44\n58 41\n14 25\n53 10\n56 65\n30 49\n38 55\n57 4\n30 15\n54 63\n17 38\n68 7\n34 3\n46 17\n13 44\n20 49\n40 34\n33 7\n2 66\n36 57\n35 18\n30 12\n11 22\n68 22\n67 9\n19 57\n65 13\n62 6\n49 51\n42 27\n61 20\n6 1\n30 24\n12 17\n25 39\n46 37\n14 5\n68 41\n6 64\n12 45\n69 61\n7 26\n59 16\n12 16\n40 67\n47 3\n1",
"output": "70"
}
] | 1,683,744,942
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 10
| 61
| 2,048,000
|
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def make_graph(n, m):
x, s = [0] * (2 * m), [0] * (n + 3)
for i in range(0, 2 * m, 2):
u, v = map(int, input().split())
s[u + 2] += 1
s[v + 2] += 1
x[i], x[i + 1] = u, v
for i in range(3, n + 3):
s[i] += s[i - 1]
G = [0] * (2 * m)
for i in range(2 * m):
j = x[i] + 1
G[s[j]] = x[i ^ 1]
s[j] += 1
return G, s
def bfs(G, s0, s):
q, k = list(s), 0
dist = [inf] * (n + 1)
for i in q:
dist[i] = 0
parent = [-1] * (n + 1)
while len(q) ^ k:
i = q[k]
di = dist[i]
for v in range(s0[i], s0[i + 1]):
j = G[v]
if dist[j] == inf:
q.append(j)
dist[j] = di + 1
parent[j] = i
k += 1
return q, dist, parent
def f(u, v):
return u * (n + 1) + v
def g(S, s0, n):
y = [0] * (m + 1)
d = dict()
for i in range(1, n + 1):
for v in range(s0[i], s0[i + 1]):
j = S[v]
d[(i, j)] = v
for i in q[::-1][:-len(s)]:
if not child[i]:
for j in range(2 * (n - 1)):
dp[f(i, j)] = 1
continue
c = child[i]
z = len(c)
for j in range(z):
y[c[j]] = j
for u in range(1, n + 1):
if s0[u + 1] - s0[u] <= z:
continue
dp0 = [0] * pow2[z]
dp0[0] = 1
dpl, dpr = [list(dp0)], [list(dp0)]
for w in range(s0[u], s0[u + 1]):
v = S[w]
x = d[(v, u)]
dp1 = list(dp0)
for j in range(pow2[z]):
if not dp0[j]:
continue
dpj = dp0[j]
for k in range(z):
if j & pow2[k]:
continue
l = j ^ pow2[k]
dp1[l] += dpj * dp[f(c[k], x)] % mod
dp1[l] %= mod
dp0 = dp1
dpl.append(list(dp0))
dp0 = [0] * pow2[z]
dp0[0] = 1
for w in range(s0[u + 1] - 1, s0[u] - 1, -1):
v = S[w]
x = d[(v, u)]
dp1 = list(dp0)
for j in range(pow2[z]):
if not dp0[j]:
continue
dpj = dp0[j]
for k in range(z):
if j & pow2[k]:
continue
l = j ^ pow2[k]
dp1[l] += dpj * dp[f(c[k], x)] % mod
dp1[l] %= mod
dp0 = dp1
dpr.append(list(dp0))
dpr.reverse()
for w in range(s0[u], s0[u + 1]):
v = S[w]
dp1, dp2 = dpl[w - s0[u]], dpr[w - s0[u] + 1]
j = f(i, w)
for k, l in zip(dp1, dp2[::-1]):
dp[j] += k * l % mod
dp[j] %= mod
ans = 0
if len(s) == 1:
c = child[q[0]]
z = len(c)
for u in range(1, n + 1):
if s0[u + 1] - s0[u] < z:
continue
dp0 = [0] * pow2[z]
dp0[0] = 1
for w in range(s0[u], s0[u + 1]):
v = S[w]
x = d[(v, u)]
dp1 = list(dp0)
for j in range(pow2[z]):
if not dp0[j]:
continue
dpj = dp0[j]
for k in range(z):
if j & pow2[k]:
continue
l = j ^ pow2[k]
dp1[l] += dpj * dp[f(c[k], x)] % mod
dp1[l] %= mod
dp0 = dp1
ans += dp0[-1]
else:
c1, c2 = child[q[0]], child[q[1]]
z1, z2 = len(c1), len(c2)
for u1 in range(1, n + 1):
if s0[u1 + 1] - s0[u1] < z1:
continue
for w0 in range(s0[u1], s0[u1 + 1]):
u2 = S[w0]
if s0[u2 + 1] - s0[u2] < z2:
continue
p = 1
for u, c, z in [(u1, c1, z1), (u2, c2, z2)]:
dp0 = [0] * pow2[z]
dp0[0] = 1
for w in range(s0[u], s0[u + 1]):
v = S[w]
if v == u1 or v == u2:
continue
x = d[(v, u)]
dp1 = list(dp0)
for j in range(pow2[z]):
if not dp0[j]:
continue
dpj = dp0[j]
for k in range(z):
if j & pow2[k]:
continue
l = j ^ pow2[k]
dp1[l] += dpj * dp[f(c[k], x)] % mod
dp1[l] %= mod
dp0 = dp1
p *= dp0[-1]
if not p:
break
ans += p % mod
return ans % mod
n = int(input())
S, s0 = make_graph(n, n - 1)
m = int(input())
T, t0 = make_graph(m, m - 1)
mod = pow(10, 9) + 7
if m <= 2:
ans = n if m == 1 else 2 * (n - 1)
print(ans)
exit()
inf = pow(10, 9) + 1
q, _, _ = bfs(T, t0, [1])
q, dist, parent = bfs(T, t0, [q[-1]])
c = q[-1]
d = dist[c]
for _ in range(d // 2):
c = parent[c]
s = [c, parent[c]] if d % 2 else [c]
q, dist, parent = bfs(T, t0, s)
child = [[] for _ in range(m + 1)]
for i in range(1, m + 1):
if parent[i] ^ -1:
child[parent[i]].append(i)
l = 2 * (n - 1) * (m + 1)
dp = [0] * l
pow2 = [1]
for _ in range(m):
pow2.append(2 * pow2[-1])
ans = g(S, s0, n) * pow(g(T, t0, m), mod - 2, mod) % mod
print(ans)
|
Title: Tree nesting
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two trees (connected undirected acyclic graphs) *S* and *T*.
Count the number of subtrees (connected subgraphs) of *S* that are isomorphic to tree *T*. Since this number can get quite large, output it modulo 109<=+<=7.
Two subtrees of tree *S* are considered different, if there exists a vertex in *S* that belongs to exactly one of them.
Tree *G* is called isomorphic to tree *H* if there exists a bijection *f* from the set of vertices of *G* to the set of vertices of *H* that has the following property: if there is an edge between vertices *A* and *B* in tree *G*, then there must be an edge between vertices *f*(*A*) and *f*(*B*) in tree *H*. And vice versa — if there is an edge between vertices *A* and *B* in tree *H*, there must be an edge between *f*<=-<=1(*A*) and *f*<=-<=1(*B*) in tree *G*.
Input Specification:
The first line contains a single integer |*S*| (1<=≤<=|*S*|<=≤<=1000) — the number of vertices of tree *S*.
Next |*S*|<=-<=1 lines contain two integers *u**i* and *v**i* (1<=≤<=*u**i*,<=*v**i*<=≤<=|*S*|) and describe edges of tree *S*.
The next line contains a single integer |*T*| (1<=≤<=|*T*|<=≤<=12) — the number of vertices of tree *T*.
Next |*T*|<=-<=1 lines contain two integers *x**i* and *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=|*T*|) and describe edges of tree *T*.
Output Specification:
On the first line output a single integer — the answer to the given task modulo 109<=+<=7.
Demo Input:
['5\n1 2\n2 3\n3 4\n4 5\n3\n1 2\n2 3\n', '3\n2 3\n3 1\n3\n1 2\n1 3\n', '7\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n4\n4 1\n4 2\n4 3\n', '5\n1 2\n2 3\n3 4\n4 5\n4\n4 1\n4 2\n4 3\n']
Demo Output:
['3\n', '1\n', '20\n', '0\n']
Note:
none
|
```python
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def make_graph(n, m):
x, s = [0] * (2 * m), [0] * (n + 3)
for i in range(0, 2 * m, 2):
u, v = map(int, input().split())
s[u + 2] += 1
s[v + 2] += 1
x[i], x[i + 1] = u, v
for i in range(3, n + 3):
s[i] += s[i - 1]
G = [0] * (2 * m)
for i in range(2 * m):
j = x[i] + 1
G[s[j]] = x[i ^ 1]
s[j] += 1
return G, s
def bfs(G, s0, s):
q, k = list(s), 0
dist = [inf] * (n + 1)
for i in q:
dist[i] = 0
parent = [-1] * (n + 1)
while len(q) ^ k:
i = q[k]
di = dist[i]
for v in range(s0[i], s0[i + 1]):
j = G[v]
if dist[j] == inf:
q.append(j)
dist[j] = di + 1
parent[j] = i
k += 1
return q, dist, parent
def f(u, v):
return u * (n + 1) + v
def g(S, s0, n):
y = [0] * (m + 1)
d = dict()
for i in range(1, n + 1):
for v in range(s0[i], s0[i + 1]):
j = S[v]
d[(i, j)] = v
for i in q[::-1][:-len(s)]:
if not child[i]:
for j in range(2 * (n - 1)):
dp[f(i, j)] = 1
continue
c = child[i]
z = len(c)
for j in range(z):
y[c[j]] = j
for u in range(1, n + 1):
if s0[u + 1] - s0[u] <= z:
continue
dp0 = [0] * pow2[z]
dp0[0] = 1
dpl, dpr = [list(dp0)], [list(dp0)]
for w in range(s0[u], s0[u + 1]):
v = S[w]
x = d[(v, u)]
dp1 = list(dp0)
for j in range(pow2[z]):
if not dp0[j]:
continue
dpj = dp0[j]
for k in range(z):
if j & pow2[k]:
continue
l = j ^ pow2[k]
dp1[l] += dpj * dp[f(c[k], x)] % mod
dp1[l] %= mod
dp0 = dp1
dpl.append(list(dp0))
dp0 = [0] * pow2[z]
dp0[0] = 1
for w in range(s0[u + 1] - 1, s0[u] - 1, -1):
v = S[w]
x = d[(v, u)]
dp1 = list(dp0)
for j in range(pow2[z]):
if not dp0[j]:
continue
dpj = dp0[j]
for k in range(z):
if j & pow2[k]:
continue
l = j ^ pow2[k]
dp1[l] += dpj * dp[f(c[k], x)] % mod
dp1[l] %= mod
dp0 = dp1
dpr.append(list(dp0))
dpr.reverse()
for w in range(s0[u], s0[u + 1]):
v = S[w]
dp1, dp2 = dpl[w - s0[u]], dpr[w - s0[u] + 1]
j = f(i, w)
for k, l in zip(dp1, dp2[::-1]):
dp[j] += k * l % mod
dp[j] %= mod
ans = 0
if len(s) == 1:
c = child[q[0]]
z = len(c)
for u in range(1, n + 1):
if s0[u + 1] - s0[u] < z:
continue
dp0 = [0] * pow2[z]
dp0[0] = 1
for w in range(s0[u], s0[u + 1]):
v = S[w]
x = d[(v, u)]
dp1 = list(dp0)
for j in range(pow2[z]):
if not dp0[j]:
continue
dpj = dp0[j]
for k in range(z):
if j & pow2[k]:
continue
l = j ^ pow2[k]
dp1[l] += dpj * dp[f(c[k], x)] % mod
dp1[l] %= mod
dp0 = dp1
ans += dp0[-1]
else:
c1, c2 = child[q[0]], child[q[1]]
z1, z2 = len(c1), len(c2)
for u1 in range(1, n + 1):
if s0[u1 + 1] - s0[u1] < z1:
continue
for w0 in range(s0[u1], s0[u1 + 1]):
u2 = S[w0]
if s0[u2 + 1] - s0[u2] < z2:
continue
p = 1
for u, c, z in [(u1, c1, z1), (u2, c2, z2)]:
dp0 = [0] * pow2[z]
dp0[0] = 1
for w in range(s0[u], s0[u + 1]):
v = S[w]
if v == u1 or v == u2:
continue
x = d[(v, u)]
dp1 = list(dp0)
for j in range(pow2[z]):
if not dp0[j]:
continue
dpj = dp0[j]
for k in range(z):
if j & pow2[k]:
continue
l = j ^ pow2[k]
dp1[l] += dpj * dp[f(c[k], x)] % mod
dp1[l] %= mod
dp0 = dp1
p *= dp0[-1]
if not p:
break
ans += p % mod
return ans % mod
n = int(input())
S, s0 = make_graph(n, n - 1)
m = int(input())
T, t0 = make_graph(m, m - 1)
mod = pow(10, 9) + 7
if m <= 2:
ans = n if m == 1 else 2 * (n - 1)
print(ans)
exit()
inf = pow(10, 9) + 1
q, _, _ = bfs(T, t0, [1])
q, dist, parent = bfs(T, t0, [q[-1]])
c = q[-1]
d = dist[c]
for _ in range(d // 2):
c = parent[c]
s = [c, parent[c]] if d % 2 else [c]
q, dist, parent = bfs(T, t0, s)
child = [[] for _ in range(m + 1)]
for i in range(1, m + 1):
if parent[i] ^ -1:
child[parent[i]].append(i)
l = 2 * (n - 1) * (m + 1)
dp = [0] * l
pow2 = [1]
for _ in range(m):
pow2.append(2 * pow2[-1])
ans = g(S, s0, n) * pow(g(T, t0, m), mod - 2, mod) % mod
print(ans)
```
| 0
|
|
16
|
B
|
Burglar and Matches
|
PROGRAMMING
| 900
|
[
"greedy",
"implementation",
"sortings"
] |
B. Burglar and Matches
|
0
|
64
|
A burglar got into a matches warehouse and wants to steal as many matches as possible. In the warehouse there are *m* containers, in the *i*-th container there are *a**i* matchboxes, and each matchbox contains *b**i* matches. All the matchboxes are of the same size. The burglar's rucksack can hold *n* matchboxes exactly. Your task is to find out the maximum amount of matches that a burglar can carry away. He has no time to rearrange matches in the matchboxes, that's why he just chooses not more than *n* matchboxes so that the total amount of matches in them is maximal.
|
The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=2·108) and integer *m* (1<=≤<=*m*<=≤<=20). The *i*<=+<=1-th line contains a pair of numbers *a**i* and *b**i* (1<=≤<=*a**i*<=≤<=108,<=1<=≤<=*b**i*<=≤<=10). All the input numbers are integer.
|
Output the only number — answer to the problem.
|
[
"7 3\n5 10\n2 5\n3 6\n",
"3 3\n1 3\n2 2\n3 1\n"
] |
[
"62\n",
"7\n"
] |
none
| 0
|
[
{
"input": "7 3\n5 10\n2 5\n3 6",
"output": "62"
},
{
"input": "3 3\n1 3\n2 2\n3 1",
"output": "7"
},
{
"input": "1 1\n1 2",
"output": "2"
},
{
"input": "1 2\n1 9\n1 6",
"output": "9"
},
{
"input": "1 10\n1 1\n1 9\n1 3\n1 9\n1 7\n1 10\n1 4\n1 7\n1 3\n1 1",
"output": "10"
},
{
"input": "2 1\n2 1",
"output": "2"
},
{
"input": "2 2\n2 4\n1 4",
"output": "8"
},
{
"input": "2 3\n1 7\n1 2\n1 5",
"output": "12"
},
{
"input": "4 1\n2 2",
"output": "4"
},
{
"input": "4 2\n1 10\n4 4",
"output": "22"
},
{
"input": "4 3\n1 4\n6 4\n1 7",
"output": "19"
},
{
"input": "5 1\n10 5",
"output": "25"
},
{
"input": "5 2\n3 9\n2 2",
"output": "31"
},
{
"input": "5 5\n2 9\n3 1\n2 1\n1 8\n2 8",
"output": "42"
},
{
"input": "5 10\n1 3\n1 2\n1 9\n1 10\n1 1\n1 5\n1 10\n1 2\n1 3\n1 7",
"output": "41"
},
{
"input": "10 1\n9 4",
"output": "36"
},
{
"input": "10 2\n14 3\n1 3",
"output": "30"
},
{
"input": "10 7\n4 8\n1 10\n1 10\n1 2\n3 3\n1 3\n1 10",
"output": "71"
},
{
"input": "10 10\n1 8\n2 10\n1 9\n1 1\n1 9\n1 6\n1 4\n2 5\n1 2\n1 4",
"output": "70"
},
{
"input": "10 4\n1 5\n5 2\n1 9\n3 3",
"output": "33"
},
{
"input": "100 5\n78 6\n29 10\n3 6\n7 3\n2 4",
"output": "716"
},
{
"input": "1000 7\n102 10\n23 6\n79 4\n48 1\n34 10\n839 8\n38 4",
"output": "8218"
},
{
"input": "10000 10\n336 2\n2782 5\n430 10\n1893 7\n3989 10\n2593 8\n165 6\n1029 2\n2097 4\n178 10",
"output": "84715"
},
{
"input": "100000 3\n2975 2\n35046 4\n61979 9",
"output": "703945"
},
{
"input": "1000000 4\n314183 9\n304213 4\n16864 5\n641358 9",
"output": "8794569"
},
{
"input": "10000000 10\n360313 10\n416076 1\n435445 9\n940322 7\n1647581 7\n4356968 10\n3589256 2\n2967933 5\n2747504 7\n1151633 3",
"output": "85022733"
},
{
"input": "100000000 7\n32844337 7\n11210848 7\n47655987 1\n33900472 4\n9174763 2\n32228738 10\n29947408 5",
"output": "749254060"
},
{
"input": "200000000 10\n27953106 7\n43325979 4\n4709522 1\n10975786 4\n67786538 8\n48901838 7\n15606185 6\n2747583 1\n100000000 1\n633331 3",
"output": "1332923354"
},
{
"input": "200000000 9\n17463897 9\n79520463 1\n162407 4\n41017993 8\n71054118 4\n9447587 2\n5298038 9\n3674560 7\n20539314 5",
"output": "996523209"
},
{
"input": "200000000 8\n6312706 6\n2920548 2\n16843192 3\n1501141 2\n13394704 6\n10047725 10\n4547663 6\n54268518 6",
"output": "630991750"
},
{
"input": "200000000 7\n25621043 2\n21865270 1\n28833034 1\n22185073 5\n100000000 2\n13891017 9\n61298710 8",
"output": "931584598"
},
{
"input": "200000000 6\n7465600 6\n8453505 10\n4572014 8\n8899499 3\n86805622 10\n64439238 6",
"output": "1447294907"
},
{
"input": "200000000 5\n44608415 6\n100000000 9\n51483223 9\n44136047 1\n52718517 1",
"output": "1634907859"
},
{
"input": "200000000 4\n37758556 10\n100000000 6\n48268521 3\n20148178 10",
"output": "1305347138"
},
{
"input": "200000000 3\n65170000 7\n20790088 1\n74616133 4",
"output": "775444620"
},
{
"input": "200000000 2\n11823018 6\n100000000 9",
"output": "970938108"
},
{
"input": "200000000 1\n100000000 6",
"output": "600000000"
},
{
"input": "200000000 10\n12097724 9\n41745972 5\n26982098 9\n14916995 7\n21549986 7\n3786630 9\n8050858 7\n27994924 4\n18345001 5\n8435339 5",
"output": "1152034197"
},
{
"input": "200000000 10\n55649 8\n10980981 9\n3192542 8\n94994808 4\n3626106 1\n100000000 6\n5260110 9\n4121453 2\n15125061 4\n669569 6",
"output": "1095537357"
},
{
"input": "10 20\n1 7\n1 7\n1 8\n1 3\n1 10\n1 7\n1 7\n1 9\n1 3\n1 1\n1 2\n1 1\n1 3\n1 10\n1 9\n1 8\n1 8\n1 6\n1 7\n1 5",
"output": "83"
},
{
"input": "10000000 20\n4594 7\n520836 8\n294766 6\n298672 4\n142253 6\n450626 1\n1920034 9\n58282 4\n1043204 1\n683045 1\n1491746 5\n58420 4\n451217 2\n129423 4\n246113 5\n190612 8\n912923 6\n473153 6\n783733 6\n282411 10",
"output": "54980855"
},
{
"input": "200000000 20\n15450824 5\n839717 10\n260084 8\n1140850 8\n28744 6\n675318 3\n25161 2\n5487 3\n6537698 9\n100000000 5\n7646970 9\n16489 6\n24627 3\n1009409 5\n22455 1\n25488456 4\n484528 9\n32663641 3\n750968 4\n5152 6",
"output": "939368573"
},
{
"input": "200000000 20\n16896 2\n113 3\n277 2\n299 7\n69383562 2\n3929 8\n499366 4\n771846 5\n9 4\n1278173 7\n90 2\n54 7\n72199858 10\n17214 5\n3 10\n1981618 3\n3728 2\n141 8\n2013578 9\n51829246 5",
"output": "1158946383"
},
{
"input": "200000000 20\n983125 2\n7453215 9\n9193588 2\n11558049 7\n28666199 1\n34362244 1\n5241493 5\n15451270 4\n19945845 8\n6208681 3\n38300385 7\n6441209 8\n21046742 7\n577198 10\n3826434 8\n9764276 8\n6264675 7\n8567063 3\n3610303 4\n2908232 3",
"output": "1131379312"
},
{
"input": "10 15\n1 6\n2 6\n3 4\n1 3\n1 2\n1 5\n1 6\n1 2\n2 9\n1 10\n1 3\n1 7\n1 8\n1 2\n2 9",
"output": "79"
},
{
"input": "10000000 15\n111 5\n914124 3\n3 9\n177790 1\n2352 3\n32138 9\n104477 1\n1223 4\n18 6\n6655580 4\n57643 10\n94309 2\n37 1\n227002 10\n1733193 7",
"output": "45116295"
},
{
"input": "200000000 15\n7069868 1\n5567826 8\n2310059 10\n13539782 7\n38420939 4\n29911411 8\n52256316 1\n12265839 9\n2074265 1\n24896428 9\n72470695 5\n3236301 1\n3890243 2\n65168965 8\n65724 6",
"output": "1489289257"
},
{
"input": "200000000 15\n12044094 7\n2475138 10\n944451 7\n4854766 2\n3809145 10\n7727571 2\n43908937 6\n2745883 1\n427511 2\n100000000 5\n190914 6\n554889 3\n288798 4\n1848572 5\n893874 3",
"output": "961871671"
},
{
"input": "200000000 15\n6334191 7\n1927941 4\n5175933 10\n468389 1\n433043 10\n6863198 5\n7480646 4\n14774279 10\n2921129 8\n18325627 7\n6973152 9\n8277324 9\n21522856 2\n2058070 1\n2444742 4",
"output": "664376069"
}
] | 1,681,965,226
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 60
| 0
|
import numpy as np
n, m = map(int, input().split())
mboxes, mcontains = [], []
for i in range(m):
mbox, mcontain = map(int, input().split())
mboxes.append(mbox)
mcontains.append(mcontain)
result = 0
while n > 0:
choice = np.argmax(mcontains)
result += min(n, mboxes[choice]) * mcontains[choice]
n -= mboxes[choice]
mcontains[choice] = 0
print(result)
|
Title: Burglar and Matches
Time Limit: 0 seconds
Memory Limit: 64 megabytes
Problem Description:
A burglar got into a matches warehouse and wants to steal as many matches as possible. In the warehouse there are *m* containers, in the *i*-th container there are *a**i* matchboxes, and each matchbox contains *b**i* matches. All the matchboxes are of the same size. The burglar's rucksack can hold *n* matchboxes exactly. Your task is to find out the maximum amount of matches that a burglar can carry away. He has no time to rearrange matches in the matchboxes, that's why he just chooses not more than *n* matchboxes so that the total amount of matches in them is maximal.
Input Specification:
The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=2·108) and integer *m* (1<=≤<=*m*<=≤<=20). The *i*<=+<=1-th line contains a pair of numbers *a**i* and *b**i* (1<=≤<=*a**i*<=≤<=108,<=1<=≤<=*b**i*<=≤<=10). All the input numbers are integer.
Output Specification:
Output the only number — answer to the problem.
Demo Input:
['7 3\n5 10\n2 5\n3 6\n', '3 3\n1 3\n2 2\n3 1\n']
Demo Output:
['62\n', '7\n']
Note:
none
|
```python
import numpy as np
n, m = map(int, input().split())
mboxes, mcontains = [], []
for i in range(m):
mbox, mcontain = map(int, input().split())
mboxes.append(mbox)
mcontains.append(mcontain)
result = 0
while n > 0:
choice = np.argmax(mcontains)
result += min(n, mboxes[choice]) * mcontains[choice]
n -= mboxes[choice]
mcontains[choice] = 0
print(result)
```
| -1
|
934
|
A
|
A Compatible Pair
|
PROGRAMMING
| 1,400
|
[
"brute force",
"games"
] | null | null |
Nian is a monster which lives deep in the oceans. Once a year, it shows up on the land, devouring livestock and even people. In order to keep the monster away, people fill their villages with red colour, light, and cracking noise, all of which frighten the monster out of coming.
Little Tommy has *n* lanterns and Big Banban has *m* lanterns. Tommy's lanterns have brightness *a*1,<=*a*2,<=...,<=*a**n*, and Banban's have brightness *b*1,<=*b*2,<=...,<=*b**m* respectively.
Tommy intends to hide one of his lanterns, then Banban picks one of Tommy's non-hidden lanterns and one of his own lanterns to form a pair. The pair's brightness will be the product of the brightness of two lanterns.
Tommy wants to make the product as small as possible, while Banban tries to make it as large as possible.
You are asked to find the brightness of the chosen pair if both of them choose optimally.
|
The first line contains two space-separated integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=50).
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n*.
The third line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m*.
All the integers range from <=-<=109 to 109.
|
Print a single integer — the brightness of the chosen pair.
|
[
"2 2\n20 18\n2 14\n",
"5 3\n-1 0 1 2 3\n-1 0 1\n"
] |
[
"252\n",
"2\n"
] |
In the first example, Tommy will hide 20 and Banban will choose 18 from Tommy and 14 from himself.
In the second example, Tommy will hide 3 and Banban will choose 2 from Tommy and 1 from himself.
| 500
|
[
{
"input": "2 2\n20 18\n2 14",
"output": "252"
},
{
"input": "5 3\n-1 0 1 2 3\n-1 0 1",
"output": "2"
},
{
"input": "10 2\n1 6 2 10 2 3 2 10 6 4\n5 7",
"output": "70"
},
{
"input": "50 50\n1 6 2 10 2 3 2 10 6 4 5 0 3 1 7 3 2 4 4 2 1 5 0 6 10 1 8 0 10 9 0 4 10 5 5 7 4 9 9 5 5 2 6 7 9 4 3 7 2 0\n0 5 9 4 4 6 1 8 2 1 6 6 8 6 4 4 7 2 1 8 6 7 4 9 8 3 0 2 0 10 7 1 4 9 4 4 2 5 3 5 1 3 2 4 1 6 5 3 8 6",
"output": "100"
},
{
"input": "5 7\n-130464232 -73113866 -542094710 -53118823 -63528720\n-775179088 631683023 -974858199 -157471745 -629658630 71825477 -6235611",
"output": "127184126241438168"
},
{
"input": "16 15\n-94580188 -713689767 -559972014 -632609438 -930348091 -567718487 -611395744 -819913097 -924009672 -427913920 -812510647 -546415480 -982072775 -693369647 -693004777 -714181162\n-772924706 -202246100 -165871667 -991426281 -490838183 209351416 134956137 -36128588 -754413937 -616596290 696201705 -201191199 967464971 -244181984 -729907974",
"output": "922371547895579571"
},
{
"input": "12 22\n-102896616 -311161241 -67541276 -402842686 -830595520 -813834033 -44046671 -584806552 -598620444 -968935604 -303048547 -545969410\n545786451 262898403 442511997 -441241260 -479587986 -752123290 720443264 500646237 737842681 -571966572 -798463881 -477248830 89875164 410339460 -359022689 -251280099 -441455542 -538431186 -406793869 374561004 -108755237 -440143410",
"output": "663200522440413120"
},
{
"input": "33 14\n-576562007 -218618150 -471719380 -583840778 -256368365 -68451917 -405045344 -775538133 -896830082 -439261765 -947070124 -716577019 -456110999 -689862512 -132480131 -10805271 -518903339 -196240188 -222292638 -828546042 -43887962 -161359263 -281422097 -484060534 963147664 -492377073 -154570101 -52145116 187803553 858844161 66540410 418777176 434025748\n-78301978 -319393213 -12393024 542953412 786804661 845642067 754996432 -985617475 -487171947 56142664 203173079 -268261708 -817080591 -511720682",
"output": "883931400924882950"
},
{
"input": "15 8\n-966400308 -992207261 -302395973 -837980754 -516443826 -492405613 -378127629 -762650324 -519519776 -36132939 -286460372 -351445284 -407653342 -604960925 -523442015\n610042288 27129580 -103108347 -942517864 842060508 -588904868 614786155 37455106",
"output": "910849554065102112"
},
{
"input": "6 30\n-524297819 -947277203 -444186475 -182837689 -385379656 -453917269\n834529938 35245081 663687669 585422565 164412867 850052113 796429008 -307345676 -127653313 426960600 211854713 -733687358 251466836 -33491050 -882811238 455544614 774581544 768447941 -241033484 441104324 -493975870 308277556 275268265 935941507 -152292053 -961509996 -740482111 -954176110 -924254634 -518710544",
"output": "504117593849498724"
},
{
"input": "5 32\n-540510995 -841481393 -94342377 -74818927 -93445356\n686714668 -82581175 736472406 502016312 575563638 -899308712 503504178 -644271272 -437408397 385778869 -746757839 306275973 -663503743 -431116516 -418708278 -515261493 -988182324 900230931 218258353 -714420102 -241118202 294802602 -937785552 -857537498 -723195312 -690515139 -214508504 -44086454 -231621215 -418360090 -810003786 -675944617",
"output": "534123411186652380"
},
{
"input": "32 13\n-999451897 -96946179 -524159869 -906101658 -63367320 -629803888 -968586834 -658416130 -874232857 -926556428 -749908220 -517073321 -659752288 -910152878 -786916085 -607633039 -191428642 -867952926 -873793977 -584331784 -733245792 -779809700 -554228536 -464503499 561577340 258991071 -569805979 -372655165 -106685554 -619607960 188856473 -268960803\n886429660 -587284372 911396803 -462990289 -228681210 -876239914 -822830527 -750131315 -401234943 116991909 -582713480 979631847 813552478",
"output": "848714444125692276"
},
{
"input": "12 25\n-464030345 -914672073 -483242132 -856226270 -925135169 -353124606 -294027092 -619650850 -490724485 -240424784 -483066792 -921640365\n279850608 726838739 -431610610 242749870 -244020223 -396865433 129534799 182767854 -939698671 342579400 330027106 893561388 -263513962 643369418 276245179 -99206565 -473767261 -168908664 -853755837 -270920164 -661186118 199341055 765543053 908211534 -93363867",
"output": "866064226130454915"
},
{
"input": "10 13\n-749120991 -186261632 -335412349 -231354880 -195919225 -808736065 -481883825 -263383991 -664780611 -605377134\n718174936 -140362196 -669193674 -598621021 -464130929 450701419 -331183926 107203430 946959233 -565825915 -558199897 246556991 -666216081",
"output": "501307028237810934"
},
{
"input": "17 13\n-483786205 -947257449 -125949195 -294711143 -420288876 -812462057 -250049555 -911026413 -188146919 -129501682 -869006661 -649643966 -26976411 -275761039 -869067490 -272248209 -342067346\n445539900 529728842 -808170728 673157826 -70778491 642872105 299298867 -76674218 -902394063 377664752 723887448 -121522827 906464625",
"output": "822104826327386019"
},
{
"input": "15 29\n-716525085 -464205793 -577203110 -979997115 -491032521 -70793687 -770595947 -817983495 -767886763 -223333719 -971913221 -944656683 -200397825 -295615495 -945544540\n-877638425 -146878165 523758517 -158778747 -49535534 597311016 77325385 494128313 12111658 -4196724 295706874 477139483 375083042 726254399 -439255703 662913604 -481588088 673747948 -345999555 -723334478 -656721905 276267528 628773156 851420802 -585029291 -643535709 -968999740 -384418713 -510285542",
"output": "941783658451562540"
},
{
"input": "5 7\n-130464232 -73113866 -542094710 -53118823 -63528720\n449942926 482853427 861095072 316710734 194604468 20277633 668816604",
"output": "-1288212069119760"
},
{
"input": "24 24\n-700068683 -418791905 -24650102 -167277317 -182309202 -517748507 -663050677 -854097070 -426998982 -197009558 -101944229 -746589957 -849018439 -774208211 -946709040 -594578249 -276703474 -434567489 -743600446 -625029074 -977300284 -895608684 -878936220 -850670748\n704881272 169877679 705460701 94083210 403943695 987978311 786162506 658067668 697640875 186287 295558596 286470276 251313879 353071193 755450449 173370603 805550377 192465301 168935494 110161743 285139426 985238736 723221868 520679017",
"output": "-18990884587723"
},
{
"input": "39 9\n44558618 981372779 318891054 283079237 285093436 907256321 414759796 652683534 79042330 249010687 7020063 309415438 788425492 138577429 714835649 954204512 795507844 389962019 507308352 408180613 194676444 44962879 922688019 101163040 327953325 560462120 183657590 273616448 226876035 233697890 720185285 689340674 372938362 15088928 283418109 796807778 149989495 694808087 276385512\n-681609072 -210918688 -757170622 -205635977 -597872997 -496188744 -97031207 -311654366 -389141528",
"output": "-1464096896176096"
},
{
"input": "5 7\n869535768 926886134 457905290 946881177 936471280\n-550057074 -517146573 -138904928 -683289266 -805395532 -979722367 -331183396",
"output": "-120782803247464704"
},
{
"input": "24 24\n299931317 581208095 975349898 832722683 817690798 482251493 336949323 145902930 573001018 802990442 898055771 253410043 150981561 225791789 53290960 405421751 723296526 565432511 256399554 374970926 22699716 104391316 121063780 149329252\n-295118728 -830122321 -294539299 -905916790 -596056305 -12021689 -213837494 -341932332 -302359125 -999813713 -704441404 -713529724 -748686121 -646928807 -244549551 -826629397 -194449623 -807534699 -831064506 -889838257 -714860574 -14761264 -276778132 -479320983",
"output": "-640647347631440"
},
{
"input": "14 8\n-1000000000 -1000000000 -1000000000 -1000000000 -1000000000 -1000000000 -1000000000 -1000000000 -1000000000 -1000000000 -1000000000 -1000000000 -1000000000 -1000000000\n1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000",
"output": "-1000000000000000000"
}
] | 1,518,616,663
| 6,763
|
Python 3
|
WRONG_ANSWER
|
PRETESTS
| 0
| 31
| 5,632,000
|
n,m=input().split(" ")
n=int(n)
m=int(m)
a=list(map(int,input().split(" ")))
b=list(map(int,input().split(" ")))
max=-10**10
i=0;
for x in range(n):
for z in range(m):
if max<a[x]*b[z]:
max=a[x]*b[z]
i=x
max=-10**10
for x in range(n):
for z in range(m):
if max<a[x]*b[z] and i!=x:
max=a[x]*b[z]
i=x
print(max)
|
Title: A Compatible Pair
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Nian is a monster which lives deep in the oceans. Once a year, it shows up on the land, devouring livestock and even people. In order to keep the monster away, people fill their villages with red colour, light, and cracking noise, all of which frighten the monster out of coming.
Little Tommy has *n* lanterns and Big Banban has *m* lanterns. Tommy's lanterns have brightness *a*1,<=*a*2,<=...,<=*a**n*, and Banban's have brightness *b*1,<=*b*2,<=...,<=*b**m* respectively.
Tommy intends to hide one of his lanterns, then Banban picks one of Tommy's non-hidden lanterns and one of his own lanterns to form a pair. The pair's brightness will be the product of the brightness of two lanterns.
Tommy wants to make the product as small as possible, while Banban tries to make it as large as possible.
You are asked to find the brightness of the chosen pair if both of them choose optimally.
Input Specification:
The first line contains two space-separated integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=50).
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n*.
The third line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m*.
All the integers range from <=-<=109 to 109.
Output Specification:
Print a single integer — the brightness of the chosen pair.
Demo Input:
['2 2\n20 18\n2 14\n', '5 3\n-1 0 1 2 3\n-1 0 1\n']
Demo Output:
['252\n', '2\n']
Note:
In the first example, Tommy will hide 20 and Banban will choose 18 from Tommy and 14 from himself.
In the second example, Tommy will hide 3 and Banban will choose 2 from Tommy and 1 from himself.
|
```python
n,m=input().split(" ")
n=int(n)
m=int(m)
a=list(map(int,input().split(" ")))
b=list(map(int,input().split(" ")))
max=-10**10
i=0;
for x in range(n):
for z in range(m):
if max<a[x]*b[z]:
max=a[x]*b[z]
i=x
max=-10**10
for x in range(n):
for z in range(m):
if max<a[x]*b[z] and i!=x:
max=a[x]*b[z]
i=x
print(max)
```
| 0
|
|
71
|
A
|
Way Too Long Words
|
PROGRAMMING
| 800
|
[
"strings"
] |
A. Way Too Long Words
|
1
|
256
|
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
|
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
|
[
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] |
[
"word\nl10n\ni18n\np43s\n"
] |
none
| 500
|
[
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd",
"output": "n20n\nj27b\nk15d"
},
{
"input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt",
"output": "t41t"
},
{
"input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings",
"output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz",
"output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz"
},
{
"input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij",
"output": "a98j"
},
{
"input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco",
"output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco"
},
{
"input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt",
"output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t"
},
{
"input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr",
"output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr"
}
] | 1,671,551,501
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
a=int(input())
for a in range(a):
s=input()
if len(s)>10:
word=s[0]+str(len(s)-2+s[-1]
else:
print a
|
Title: Way Too Long Words
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output Specification:
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
Demo Input:
['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n']
Demo Output:
['word\nl10n\ni18n\np43s\n']
Note:
none
|
```python
a=int(input())
for a in range(a):
s=input()
if len(s)>10:
word=s[0]+str(len(s)-2+s[-1]
else:
print a
```
| -1
|
546
|
A
|
Soldier and Bananas
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation",
"math"
] | null | null |
A soldier wants to buy *w* bananas in the shop. He has to pay *k* dollars for the first banana, 2*k* dollars for the second one and so on (in other words, he has to pay *i*·*k* dollars for the *i*-th banana).
He has *n* dollars. How many dollars does he have to borrow from his friend soldier to buy *w* bananas?
|
The first line contains three positive integers *k*,<=*n*,<=*w* (1<=<=≤<=<=*k*,<=*w*<=<=≤<=<=1000, 0<=≤<=*n*<=≤<=109), the cost of the first banana, initial number of dollars the soldier has and number of bananas he wants.
|
Output one integer — the amount of dollars that the soldier must borrow from his friend. If he doesn't have to borrow money, output 0.
|
[
"3 17 4\n"
] |
[
"13"
] |
none
| 500
|
[
{
"input": "3 17 4",
"output": "13"
},
{
"input": "1 2 1",
"output": "0"
},
{
"input": "1 1 1",
"output": "0"
},
{
"input": "1 5 6",
"output": "16"
},
{
"input": "1 1000000000 1",
"output": "0"
},
{
"input": "1000 0 1000",
"output": "500500000"
},
{
"input": "859 453892 543",
"output": "126416972"
},
{
"input": "1000 1000000000 1000",
"output": "0"
},
{
"input": "1000 500500000 1000",
"output": "0"
},
{
"input": "1000 500500001 1000",
"output": "0"
},
{
"input": "1000 500499999 1000",
"output": "1"
},
{
"input": "634 87973 214",
"output": "14497197"
},
{
"input": "432 10000 241",
"output": "12587552"
},
{
"input": "111 111111111 111",
"output": "0"
},
{
"input": "20 43 3",
"output": "77"
}
] | 1,698,090,518
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 0
| 0
|
w, k, n = map(int, input().split())
cost = sum(k * i for i in range(1, w + 1))
borrow = max(cost - n, 0)
print(borrow)
|
Title: Soldier and Bananas
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A soldier wants to buy *w* bananas in the shop. He has to pay *k* dollars for the first banana, 2*k* dollars for the second one and so on (in other words, he has to pay *i*·*k* dollars for the *i*-th banana).
He has *n* dollars. How many dollars does he have to borrow from his friend soldier to buy *w* bananas?
Input Specification:
The first line contains three positive integers *k*,<=*n*,<=*w* (1<=<=≤<=<=*k*,<=*w*<=<=≤<=<=1000, 0<=≤<=*n*<=≤<=109), the cost of the first banana, initial number of dollars the soldier has and number of bananas he wants.
Output Specification:
Output one integer — the amount of dollars that the soldier must borrow from his friend. If he doesn't have to borrow money, output 0.
Demo Input:
['3 17 4\n']
Demo Output:
['13']
Note:
none
|
```python
w, k, n = map(int, input().split())
cost = sum(k * i for i in range(1, w + 1))
borrow = max(cost - n, 0)
print(borrow)
```
| 0
|
|
669
|
A
|
Little Artem and Presents
|
PROGRAMMING
| 800
|
[
"math"
] | null | null |
Little Artem got *n* stones on his birthday and now wants to give some of them to Masha. He knows that Masha cares more about the fact of receiving the present, rather than the value of that present, so he wants to give her stones as many times as possible. However, Masha remembers the last present she received, so Artem can't give her the same number of stones twice in a row. For example, he can give her 3 stones, then 1 stone, then again 3 stones, but he can't give her 3 stones and then again 3 stones right after that.
How many times can Artem give presents to Masha?
|
The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=109) — number of stones Artem received on his birthday.
|
Print the maximum possible number of times Artem can give presents to Masha.
|
[
"1\n",
"2\n",
"3\n",
"4\n"
] |
[
"1\n",
"1\n",
"2\n",
"3\n"
] |
In the first sample, Artem can only give 1 stone to Masha.
In the second sample, Atrem can give Masha 1 or 2 stones, though he can't give her 1 stone two times.
In the third sample, Atrem can first give Masha 2 stones, a then 1 more stone.
In the fourth sample, Atrem can first give Masha 1 stone, then 2 stones, and finally 1 stone again.
| 500
|
[
{
"input": "1",
"output": "1"
},
{
"input": "2",
"output": "1"
},
{
"input": "3",
"output": "2"
},
{
"input": "4",
"output": "3"
},
{
"input": "100",
"output": "67"
},
{
"input": "101",
"output": "67"
},
{
"input": "102",
"output": "68"
},
{
"input": "1000000000",
"output": "666666667"
},
{
"input": "5",
"output": "3"
},
{
"input": "6",
"output": "4"
},
{
"input": "999999999",
"output": "666666666"
},
{
"input": "999999998",
"output": "666666665"
},
{
"input": "999999997",
"output": "666666665"
},
{
"input": "999999996",
"output": "666666664"
},
{
"input": "999999995",
"output": "666666663"
},
{
"input": "999999994",
"output": "666666663"
},
{
"input": "999999993",
"output": "666666662"
},
{
"input": "999999992",
"output": "666666661"
},
{
"input": "999999991",
"output": "666666661"
},
{
"input": "1000",
"output": "667"
},
{
"input": "10000",
"output": "6667"
},
{
"input": "100000",
"output": "66667"
},
{
"input": "1000000",
"output": "666667"
},
{
"input": "10000000",
"output": "6666667"
},
{
"input": "100000000",
"output": "66666667"
},
{
"input": "7",
"output": "5"
}
] | 1,568,708,345
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 93
| 0
|
print(int(input()*2+1)/3)
|
Title: Little Artem and Presents
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Artem got *n* stones on his birthday and now wants to give some of them to Masha. He knows that Masha cares more about the fact of receiving the present, rather than the value of that present, so he wants to give her stones as many times as possible. However, Masha remembers the last present she received, so Artem can't give her the same number of stones twice in a row. For example, he can give her 3 stones, then 1 stone, then again 3 stones, but he can't give her 3 stones and then again 3 stones right after that.
How many times can Artem give presents to Masha?
Input Specification:
The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=109) — number of stones Artem received on his birthday.
Output Specification:
Print the maximum possible number of times Artem can give presents to Masha.
Demo Input:
['1\n', '2\n', '3\n', '4\n']
Demo Output:
['1\n', '1\n', '2\n', '3\n']
Note:
In the first sample, Artem can only give 1 stone to Masha.
In the second sample, Atrem can give Masha 1 or 2 stones, though he can't give her 1 stone two times.
In the third sample, Atrem can first give Masha 2 stones, a then 1 more stone.
In the fourth sample, Atrem can first give Masha 1 stone, then 2 stones, and finally 1 stone again.
|
```python
print(int(input()*2+1)/3)
```
| -1
|
|
270
|
A
|
Fancy Fence
|
PROGRAMMING
| 1,100
|
[
"geometry",
"implementation",
"math"
] | null | null |
Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot.
He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle *a*.
Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to *a*?
|
The first line of input contains an integer *t* (0<=<<=*t*<=<<=180) — the number of tests. Each of the following *t* lines contains a single integer *a* (0<=<<=*a*<=<<=180) — the angle the robot can make corners at measured in degrees.
|
For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible.
|
[
"3\n30\n60\n90\n"
] |
[
"NO\nYES\nYES\n"
] |
In the first test case, it is impossible to build the fence, since there is no regular polygon with angle <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/df5f4b07dd5316fde165b43657b2696e2919e791.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second test case, the fence is a regular triangle, and in the last test case — a square.
| 500
|
[
{
"input": "3\n30\n60\n90",
"output": "NO\nYES\nYES"
},
{
"input": "6\n1\n2\n3\n170\n179\n25",
"output": "NO\nNO\nNO\nYES\nYES\nNO"
}
] | 1,655,316,532
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 3
| 124
| 614,400
|
from sys import stdin
from bisect import bisect_left as bl
from collections import defaultdict
from math import floor, ceil
# input = stdin.readline
read = lambda: map(int, input().strip().split())
for test in range(int(input())):
a = int(input())
if floor(360 / (180 - a)) == ceil(360 / (180 - a)):
print("YES")
else:
print("NO")
|
Title: Fancy Fence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot.
He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle *a*.
Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to *a*?
Input Specification:
The first line of input contains an integer *t* (0<=<<=*t*<=<<=180) — the number of tests. Each of the following *t* lines contains a single integer *a* (0<=<<=*a*<=<<=180) — the angle the robot can make corners at measured in degrees.
Output Specification:
For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible.
Demo Input:
['3\n30\n60\n90\n']
Demo Output:
['NO\nYES\nYES\n']
Note:
In the first test case, it is impossible to build the fence, since there is no regular polygon with angle <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/df5f4b07dd5316fde165b43657b2696e2919e791.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second test case, the fence is a regular triangle, and in the last test case — a square.
|
```python
from sys import stdin
from bisect import bisect_left as bl
from collections import defaultdict
from math import floor, ceil
# input = stdin.readline
read = lambda: map(int, input().strip().split())
for test in range(int(input())):
a = int(input())
if floor(360 / (180 - a)) == ceil(360 / (180 - a)):
print("YES")
else:
print("NO")
```
| 3
|
|
4
|
A
|
Watermelon
|
PROGRAMMING
| 800
|
[
"brute force",
"math"
] |
A. Watermelon
|
1
|
64
|
One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed *w* kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem.
Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight.
|
The first (and the only) input line contains integer number *w* (1<=≤<=*w*<=≤<=100) — the weight of the watermelon bought by the boys.
|
Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case.
|
[
"8\n"
] |
[
"YES\n"
] |
For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos).
| 0
|
[
{
"input": "8",
"output": "YES"
},
{
"input": "5",
"output": "NO"
},
{
"input": "4",
"output": "YES"
},
{
"input": "3",
"output": "NO"
},
{
"input": "2",
"output": "NO"
},
{
"input": "1",
"output": "NO"
},
{
"input": "7",
"output": "NO"
},
{
"input": "6",
"output": "YES"
},
{
"input": "10",
"output": "YES"
},
{
"input": "9",
"output": "NO"
},
{
"input": "53",
"output": "NO"
},
{
"input": "77",
"output": "NO"
},
{
"input": "32",
"output": "YES"
},
{
"input": "44",
"output": "YES"
},
{
"input": "98",
"output": "YES"
},
{
"input": "99",
"output": "NO"
},
{
"input": "90",
"output": "YES"
},
{
"input": "67",
"output": "NO"
},
{
"input": "100",
"output": "YES"
},
{
"input": "88",
"output": "YES"
}
] | 1,694,067,040
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
import java.io.File;
public class ListOfFile
{
public static void main(String[] args)
{
File f=new File(");
String[] FileList=file.list();
for(String name:FileList)
{
|
Title: Watermelon
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed *w* kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem.
Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight.
Input Specification:
The first (and the only) input line contains integer number *w* (1<=≤<=*w*<=≤<=100) — the weight of the watermelon bought by the boys.
Output Specification:
Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case.
Demo Input:
['8\n']
Demo Output:
['YES\n']
Note:
For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos).
|
```python
import java.io.File;
public class ListOfFile
{
public static void main(String[] args)
{
File f=new File(");
String[] FileList=file.list();
for(String name:FileList)
{
```
| -1
|
916
|
A
|
Jamie and Alarm Snooze
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation",
"math"
] | null | null |
Jamie loves sleeping. One day, he decides that he needs to wake up at exactly *hh*:<=*mm*. However, he hates waking up, so he wants to make waking up less painful by setting the alarm at a lucky time. He will then press the snooze button every *x* minutes until *hh*:<=*mm* is reached, and only then he will wake up. He wants to know what is the smallest number of times he needs to press the snooze button.
A time is considered lucky if it contains a digit '7'. For example, 13:<=07 and 17:<=27 are lucky, while 00:<=48 and 21:<=34 are not lucky.
Note that it is not necessary that the time set for the alarm and the wake-up time are on the same day. It is guaranteed that there is a lucky time Jamie can set so that he can wake at *hh*:<=*mm*.
Formally, find the smallest possible non-negative integer *y* such that the time representation of the time *x*·*y* minutes before *hh*:<=*mm* contains the digit '7'.
Jamie uses 24-hours clock, so after 23:<=59 comes 00:<=00.
|
The first line contains a single integer *x* (1<=≤<=*x*<=≤<=60).
The second line contains two two-digit integers, *hh* and *mm* (00<=≤<=*hh*<=≤<=23,<=00<=≤<=*mm*<=≤<=59).
|
Print the minimum number of times he needs to press the button.
|
[
"3\n11 23\n",
"5\n01 07\n"
] |
[
"2\n",
"0\n"
] |
In the first sample, Jamie needs to wake up at 11:23. So, he can set his alarm at 11:17. He would press the snooze button when the alarm rings at 11:17 and at 11:20.
In the second sample, Jamie can set his alarm at exactly at 01:07 which is lucky.
| 500
|
[
{
"input": "3\n11 23",
"output": "2"
},
{
"input": "5\n01 07",
"output": "0"
},
{
"input": "34\n09 24",
"output": "3"
},
{
"input": "2\n14 37",
"output": "0"
},
{
"input": "14\n19 54",
"output": "9"
},
{
"input": "42\n15 44",
"output": "12"
},
{
"input": "46\n02 43",
"output": "1"
},
{
"input": "14\n06 41",
"output": "1"
},
{
"input": "26\n04 58",
"output": "26"
},
{
"input": "54\n16 47",
"output": "0"
},
{
"input": "38\n20 01",
"output": "3"
},
{
"input": "11\n02 05",
"output": "8"
},
{
"input": "55\n22 10",
"output": "5"
},
{
"input": "23\n10 08",
"output": "6"
},
{
"input": "23\n23 14",
"output": "9"
},
{
"input": "51\n03 27",
"output": "0"
},
{
"input": "35\n15 25",
"output": "13"
},
{
"input": "3\n12 15",
"output": "6"
},
{
"input": "47\n00 28",
"output": "3"
},
{
"input": "31\n13 34",
"output": "7"
},
{
"input": "59\n17 32",
"output": "0"
},
{
"input": "25\n11 03",
"output": "8"
},
{
"input": "9\n16 53",
"output": "4"
},
{
"input": "53\n04 06",
"output": "3"
},
{
"input": "37\n00 12",
"output": "5"
},
{
"input": "5\n13 10",
"output": "63"
},
{
"input": "50\n01 59",
"output": "10"
},
{
"input": "34\n06 13",
"output": "4"
},
{
"input": "2\n18 19",
"output": "1"
},
{
"input": "46\n06 16",
"output": "17"
},
{
"input": "14\n03 30",
"output": "41"
},
{
"input": "40\n13 37",
"output": "0"
},
{
"input": "24\n17 51",
"output": "0"
},
{
"input": "8\n14 57",
"output": "0"
},
{
"input": "52\n18 54",
"output": "2"
},
{
"input": "20\n15 52",
"output": "24"
},
{
"input": "20\n03 58",
"output": "30"
},
{
"input": "48\n07 11",
"output": "0"
},
{
"input": "32\n04 01",
"output": "2"
},
{
"input": "60\n08 15",
"output": "1"
},
{
"input": "44\n20 20",
"output": "4"
},
{
"input": "55\n15 35",
"output": "9"
},
{
"input": "55\n03 49",
"output": "11"
},
{
"input": "23\n16 39",
"output": "4"
},
{
"input": "7\n20 36",
"output": "7"
},
{
"input": "35\n16 42",
"output": "1"
},
{
"input": "35\n05 56",
"output": "21"
},
{
"input": "3\n17 45",
"output": "0"
},
{
"input": "47\n05 59",
"output": "6"
},
{
"input": "15\n10 13",
"output": "9"
},
{
"input": "59\n06 18",
"output": "9"
},
{
"input": "34\n17 18",
"output": "0"
},
{
"input": "18\n05 23",
"output": "2"
},
{
"input": "46\n17 21",
"output": "0"
},
{
"input": "30\n06 27",
"output": "0"
},
{
"input": "14\n18 40",
"output": "3"
},
{
"input": "58\n22 54",
"output": "6"
},
{
"input": "26\n19 44",
"output": "5"
},
{
"input": "10\n15 57",
"output": "0"
},
{
"input": "54\n20 47",
"output": "0"
},
{
"input": "22\n08 45",
"output": "3"
},
{
"input": "48\n18 08",
"output": "1"
},
{
"input": "32\n07 06",
"output": "0"
},
{
"input": "60\n19 19",
"output": "2"
},
{
"input": "45\n07 25",
"output": "0"
},
{
"input": "29\n12 39",
"output": "8"
},
{
"input": "13\n08 28",
"output": "3"
},
{
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] | 1,594,109,170
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 93
| 6,656,000
|
x = int(input())
l = list(map(int,input().split()))
if "7" in str(l[1]):
print(0)
else:
a = 0
for i in range(1,10001):
a+=1
b = l[1] - x*i
if b % 10 == 7:
break
print(a)
|
Title: Jamie and Alarm Snooze
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Jamie loves sleeping. One day, he decides that he needs to wake up at exactly *hh*:<=*mm*. However, he hates waking up, so he wants to make waking up less painful by setting the alarm at a lucky time. He will then press the snooze button every *x* minutes until *hh*:<=*mm* is reached, and only then he will wake up. He wants to know what is the smallest number of times he needs to press the snooze button.
A time is considered lucky if it contains a digit '7'. For example, 13:<=07 and 17:<=27 are lucky, while 00:<=48 and 21:<=34 are not lucky.
Note that it is not necessary that the time set for the alarm and the wake-up time are on the same day. It is guaranteed that there is a lucky time Jamie can set so that he can wake at *hh*:<=*mm*.
Formally, find the smallest possible non-negative integer *y* such that the time representation of the time *x*·*y* minutes before *hh*:<=*mm* contains the digit '7'.
Jamie uses 24-hours clock, so after 23:<=59 comes 00:<=00.
Input Specification:
The first line contains a single integer *x* (1<=≤<=*x*<=≤<=60).
The second line contains two two-digit integers, *hh* and *mm* (00<=≤<=*hh*<=≤<=23,<=00<=≤<=*mm*<=≤<=59).
Output Specification:
Print the minimum number of times he needs to press the button.
Demo Input:
['3\n11 23\n', '5\n01 07\n']
Demo Output:
['2\n', '0\n']
Note:
In the first sample, Jamie needs to wake up at 11:23. So, he can set his alarm at 11:17. He would press the snooze button when the alarm rings at 11:17 and at 11:20.
In the second sample, Jamie can set his alarm at exactly at 01:07 which is lucky.
|
```python
x = int(input())
l = list(map(int,input().split()))
if "7" in str(l[1]):
print(0)
else:
a = 0
for i in range(1,10001):
a+=1
b = l[1] - x*i
if b % 10 == 7:
break
print(a)
```
| 0
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
Bike is interested in permutations. A permutation of length *n* is an integer sequence such that each integer from 0 to (*n*<=-<=1) appears exactly once in it. For example, [0,<=2,<=1] is a permutation of length 3 while both [0,<=2,<=2] and [1,<=2,<=3] is not.
A permutation triple of permutations of length *n* (*a*,<=*b*,<=*c*) is called a Lucky Permutation Triple if and only if . The sign *a**i* denotes the *i*-th element of permutation *a*. The modular equality described above denotes that the remainders after dividing *a**i*<=+<=*b**i* by *n* and dividing *c**i* by *n* are equal.
Now, he has an integer *n* and wants to find a Lucky Permutation Triple. Could you please help him?
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105).
|
If no Lucky Permutation Triple of length *n* exists print -1.
Otherwise, you need to print three lines. Each line contains *n* space-seperated integers. The first line must contain permutation *a*, the second line — permutation *b*, the third — permutation *c*.
If there are multiple solutions, print any of them.
|
[
"5\n",
"2\n"
] |
[
"1 4 3 2 0\n1 0 2 4 3\n2 4 0 1 3\n",
"-1\n"
] |
In Sample 1, the permutation triple ([1, 4, 3, 2, 0], [1, 0, 2, 4, 3], [2, 4, 0, 1, 3]) is Lucky Permutation Triple, as following holds:
- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/a6bf1b9b57809dbec5021f65f89616f259587c07.png" style="max-width: 100.0%;max-height: 100.0%;"/>; - <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/48cc13134296b68f459f69d78e0240859aaec702.png" style="max-width: 100.0%;max-height: 100.0%;"/>; - <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ac44412de7b46833e90348a6b3298f9796e3977c.png" style="max-width: 100.0%;max-height: 100.0%;"/>; - <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/3825b0bb758208dda2ead1c5224c05d89ad9ab55.png" style="max-width: 100.0%;max-height: 100.0%;"/>; - <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/0a72e2da40048a507839927a211267ac01c9bf89.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In Sample 2, you can easily notice that no lucky permutation triple exists.
| 0
|
[
{
"input": "5",
"output": "1 4 3 2 0\n1 0 2 4 3\n2 4 0 1 3"
},
{
"input": "2",
"output": "-1"
},
{
"input": "8",
"output": "-1"
},
{
"input": "9",
"output": "0 1 2 3 4 5 6 7 8 \n0 1 2 3 4 5 6 7 8 \n0 2 4 6 8 1 3 5 7 "
},
{
"input": "2",
"output": "-1"
},
{
"input": "77",
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},
{
"input": "6",
"output": "-1"
},
{
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},
{
"input": "72",
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{
"input": "1",
"output": "0 \n0 \n0 "
},
{
"input": "23",
"output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 \n0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 \n0 2 4 6 8 10 12 14 16 18 20 22 1 3 5 7 9 11 13 15 17 19 21 "
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{
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{
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"output": "-1"
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{
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},
{
"input": "54",
"output": "-1"
},
{
"input": "39",
"output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 \n0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 \n0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 "
},
{
"input": "20",
"output": "-1"
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{
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},
{
"input": "34",
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{
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},
{
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{
"input": "67817",
"output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..."
},
{
"input": "72598",
"output": "-1"
},
{
"input": "85891",
"output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..."
},
{
"input": "74320",
"output": "-1"
},
{
"input": "11805",
"output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..."
},
{
"input": "16586",
"output": "-1"
},
{
"input": "5014",
"output": "-1"
},
{
"input": "73268",
"output": "-1"
},
{
"input": "61697",
"output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..."
},
{
"input": "99182",
"output": "-1"
},
{
"input": "79771",
"output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..."
},
{
"input": "68199",
"output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..."
},
{
"input": "5684",
"output": "-1"
},
{
"input": "10465",
"output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..."
},
{
"input": "31598",
"output": "-1"
},
{
"input": "36379",
"output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..."
},
{
"input": "16968",
"output": "-1"
},
{
"input": "93061",
"output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..."
},
{
"input": "73650",
"output": "-1"
},
{
"input": "94783",
"output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..."
},
{
"input": "99564",
"output": "-1"
},
{
"input": "37049",
"output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..."
},
{
"input": "25478",
"output": "-1"
},
{
"input": "30259",
"output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..."
},
{
"input": "43551",
"output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..."
},
{
"input": "31980",
"output": "-1"
},
{
"input": "69465",
"output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..."
},
{
"input": "1",
"output": "0 \n0 \n0 "
},
{
"input": "100000",
"output": "-1"
},
{
"input": "99999",
"output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..."
},
{
"input": "99998",
"output": "-1"
}
] | 1,634,492,010
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 54
| 810
| 2,355,200
|
n = int(input())
z = list(range(n))
if n&1 :
print(*z)
print(*z)
for i in range(n):
print((z[i]+z[i])%n,end = " ")
else :
print(-1)
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bike is interested in permutations. A permutation of length *n* is an integer sequence such that each integer from 0 to (*n*<=-<=1) appears exactly once in it. For example, [0,<=2,<=1] is a permutation of length 3 while both [0,<=2,<=2] and [1,<=2,<=3] is not.
A permutation triple of permutations of length *n* (*a*,<=*b*,<=*c*) is called a Lucky Permutation Triple if and only if . The sign *a**i* denotes the *i*-th element of permutation *a*. The modular equality described above denotes that the remainders after dividing *a**i*<=+<=*b**i* by *n* and dividing *c**i* by *n* are equal.
Now, he has an integer *n* and wants to find a Lucky Permutation Triple. Could you please help him?
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105).
Output Specification:
If no Lucky Permutation Triple of length *n* exists print -1.
Otherwise, you need to print three lines. Each line contains *n* space-seperated integers. The first line must contain permutation *a*, the second line — permutation *b*, the third — permutation *c*.
If there are multiple solutions, print any of them.
Demo Input:
['5\n', '2\n']
Demo Output:
['1 4 3 2 0\n1 0 2 4 3\n2 4 0 1 3\n', '-1\n']
Note:
In Sample 1, the permutation triple ([1, 4, 3, 2, 0], [1, 0, 2, 4, 3], [2, 4, 0, 1, 3]) is Lucky Permutation Triple, as following holds:
- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/a6bf1b9b57809dbec5021f65f89616f259587c07.png" style="max-width: 100.0%;max-height: 100.0%;"/>; - <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/48cc13134296b68f459f69d78e0240859aaec702.png" style="max-width: 100.0%;max-height: 100.0%;"/>; - <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ac44412de7b46833e90348a6b3298f9796e3977c.png" style="max-width: 100.0%;max-height: 100.0%;"/>; - <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/3825b0bb758208dda2ead1c5224c05d89ad9ab55.png" style="max-width: 100.0%;max-height: 100.0%;"/>; - <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/0a72e2da40048a507839927a211267ac01c9bf89.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In Sample 2, you can easily notice that no lucky permutation triple exists.
|
```python
n = int(input())
z = list(range(n))
if n&1 :
print(*z)
print(*z)
for i in range(n):
print((z[i]+z[i])%n,end = " ")
else :
print(-1)
```
| 3
|
|
447
|
A
|
DZY Loves Hash
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
DZY has a hash table with *p* buckets, numbered from 0 to *p*<=-<=1. He wants to insert *n* numbers, in the order they are given, into the hash table. For the *i*-th number *x**i*, DZY will put it into the bucket numbered *h*(*x**i*), where *h*(*x*) is the hash function. In this problem we will assume, that *h*(*x*)<==<=*x* *mod* *p*. Operation *a* *mod* *b* denotes taking a remainder after division *a* by *b*.
However, each bucket can contain no more than one element. If DZY wants to insert an number into a bucket which is already filled, we say a "conflict" happens. Suppose the first conflict happens right after the *i*-th insertion, you should output *i*. If no conflict happens, just output -1.
|
The first line contains two integers, *p* and *n* (2<=≤<=*p*,<=*n*<=≤<=300). Then *n* lines follow. The *i*-th of them contains an integer *x**i* (0<=≤<=*x**i*<=≤<=109).
|
Output a single integer — the answer to the problem.
|
[
"10 5\n0\n21\n53\n41\n53\n",
"5 5\n0\n1\n2\n3\n4\n"
] |
[
"4\n",
"-1\n"
] |
none
| 500
|
[
{
"input": "10 5\n0\n21\n53\n41\n53",
"output": "4"
},
{
"input": "5 5\n0\n1\n2\n3\n4",
"output": "-1"
},
{
"input": "10 6\n811966798\n734823552\n790326404\n929189974\n414343256\n560346537",
"output": "4"
},
{
"input": "2 2\n788371161\n801743052",
"output": "-1"
},
{
"input": "10 6\n812796223\n122860157\n199259103\n597650585\n447742024\n521549402",
"output": "3"
},
{
"input": "300 2\n822454942\n119374431",
"output": "-1"
},
{
"input": "300 2\n823284367\n507345500",
"output": "-1"
},
{
"input": "2 5\n791579811\n35613889\n997079893\n878677665\n693717467",
"output": "2"
},
{
"input": "20 5\n793926268\n28931770\n842870287\n974950617\n859404206",
"output": "-1"
},
{
"input": "100 15\n805069249\n778178198\n633897389\n844316223\n146759898\n870527016\n82668754\n42060733\n943602929\n979451110\n746979598\n47406033\n607284062\n850870259\n229415316",
"output": "5"
},
{
"input": "100 15\n806204335\n189490323\n718805086\n716787474\n262315718\n822030596\n894644222\n724054623\n141384399\n579354205\n192622443\n672556242\n97417563\n243354557\n208957882",
"output": "8"
},
{
"input": "100 15\n807033760\n577461392\n275221433\n532633429\n295714486\n783298996\n255799943\n99107143\n729119412\n59302896\n37640015\n313610861\n630550567\n534283052\n681062462",
"output": "8"
},
{
"input": "100 15\n808103310\n136224397\n360129131\n405104681\n263786657\n734802577\n67808179\n928584682\n926900882\n511722343\n483348395\n938695534\n120684068\n74152694\n808088675",
"output": "9"
},
{
"input": "2 2\n2\n2",
"output": "2"
},
{
"input": "300 2\n0\n300",
"output": "2"
},
{
"input": "2 2\n0\n0",
"output": "2"
}
] | 1,556,554,552
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 29
| 140
| 0
|
p, n = map(int, input().split())
hash_table = [-1] * p
flag = True
for i in range(n):
x = int(input())
hx = x % p
if hash_table[hx] == -1:
hash_table[hx] = x
else:
print(i+1)
flag = False
break
if(flag):
print(-1)
|
Title: DZY Loves Hash
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
DZY has a hash table with *p* buckets, numbered from 0 to *p*<=-<=1. He wants to insert *n* numbers, in the order they are given, into the hash table. For the *i*-th number *x**i*, DZY will put it into the bucket numbered *h*(*x**i*), where *h*(*x*) is the hash function. In this problem we will assume, that *h*(*x*)<==<=*x* *mod* *p*. Operation *a* *mod* *b* denotes taking a remainder after division *a* by *b*.
However, each bucket can contain no more than one element. If DZY wants to insert an number into a bucket which is already filled, we say a "conflict" happens. Suppose the first conflict happens right after the *i*-th insertion, you should output *i*. If no conflict happens, just output -1.
Input Specification:
The first line contains two integers, *p* and *n* (2<=≤<=*p*,<=*n*<=≤<=300). Then *n* lines follow. The *i*-th of them contains an integer *x**i* (0<=≤<=*x**i*<=≤<=109).
Output Specification:
Output a single integer — the answer to the problem.
Demo Input:
['10 5\n0\n21\n53\n41\n53\n', '5 5\n0\n1\n2\n3\n4\n']
Demo Output:
['4\n', '-1\n']
Note:
none
|
```python
p, n = map(int, input().split())
hash_table = [-1] * p
flag = True
for i in range(n):
x = int(input())
hx = x % p
if hash_table[hx] == -1:
hash_table[hx] = x
else:
print(i+1)
flag = False
break
if(flag):
print(-1)
```
| 3
|
|
363
|
B
|
Fence
|
PROGRAMMING
| 1,100
|
[
"brute force",
"dp"
] | null | null |
There is a fence in front of Polycarpus's home. The fence consists of *n* planks of the same width which go one after another from left to right. The height of the *i*-th plank is *h**i* meters, distinct planks can have distinct heights.
Polycarpus has bought a posh piano and is thinking about how to get it into the house. In order to carry out his plan, he needs to take exactly *k* consecutive planks from the fence. Higher planks are harder to tear off the fence, so Polycarpus wants to find such *k* consecutive planks that the sum of their heights is minimal possible.
Write the program that finds the indexes of *k* consecutive planks with minimal total height. Pay attention, the fence is not around Polycarpus's home, it is in front of home (in other words, the fence isn't cyclic).
|
The first line of the input contains integers *n* and *k* (1<=≤<=*n*<=≤<=1.5·105,<=1<=≤<=*k*<=≤<=*n*) — the number of planks in the fence and the width of the hole for the piano. The second line contains the sequence of integers *h*1,<=*h*2,<=...,<=*h**n* (1<=≤<=*h**i*<=≤<=100), where *h**i* is the height of the *i*-th plank of the fence.
|
Print such integer *j* that the sum of the heights of planks *j*, *j*<=+<=1, ..., *j*<=+<=*k*<=-<=1 is the minimum possible. If there are multiple such *j*'s, print any of them.
|
[
"7 3\n1 2 6 1 1 7 1\n"
] |
[
"3\n"
] |
In the sample, your task is to find three consecutive planks with the minimum sum of heights. In the given case three planks with indexes 3, 4 and 5 have the required attribute, their total height is 8.
| 1,000
|
[
{
"input": "7 3\n1 2 6 1 1 7 1",
"output": "3"
},
{
"input": "1 1\n100",
"output": "1"
},
{
"input": "2 1\n10 20",
"output": "1"
},
{
"input": "10 5\n1 2 3 1 2 2 3 1 4 5",
"output": "1"
},
{
"input": "10 2\n3 1 4 1 4 6 2 1 4 6",
"output": "7"
},
{
"input": "2 2\n20 10",
"output": "1"
},
{
"input": "2 1\n20 1",
"output": "2"
},
{
"input": "3 1\n1 2 3",
"output": "1"
},
{
"input": "3 1\n2 1 3",
"output": "2"
},
{
"input": "3 1\n3 2 1",
"output": "3"
},
{
"input": "3 2\n1 2 3",
"output": "1"
},
{
"input": "3 2\n3 2 1",
"output": "2"
},
{
"input": "3 3\n1 2 3",
"output": "1"
},
{
"input": "4 2\n9 8 11 7",
"output": "1"
},
{
"input": "4 2\n10 1 2 3",
"output": "2"
},
{
"input": "6 3\n56 56 56 2 1 2",
"output": "4"
},
{
"input": "8 3\n1 1 1 1 2 60 90 1",
"output": "1"
},
{
"input": "4 1\n1 5 2 2",
"output": "1"
},
{
"input": "4 2\n4 6 7 4",
"output": "1"
},
{
"input": "10 4\n1 1 1 4 4 4 4 4 4 3",
"output": "1"
},
{
"input": "6 3\n1 2 1 3 1 1",
"output": "1"
},
{
"input": "5 2\n100 100 100 1 1",
"output": "4"
}
] | 1,678,955,508
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 1
| 46
| 0
|
from sys import stdin, stdout
input = stdin.readline
def print(*args, end='\n', sep=' ') -> None:
stdout.write(sep.join(map(str, args)) + end)
def int_map():
return map(int, input().split())
def list_int():
return list(map(int, input().split()))
n, k = int_map()
arr = list_int()
for i in range(1, n):
arr[i] += arr[i-1]
mn = 10**9+7
j = 0
for i in range(2, n):
sm = arr[i] - (arr[i-3] if i - 3 >= 0 else 0)
if sm < mn:
j = i - 1
mn = sm
print(j)
|
Title: Fence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There is a fence in front of Polycarpus's home. The fence consists of *n* planks of the same width which go one after another from left to right. The height of the *i*-th plank is *h**i* meters, distinct planks can have distinct heights.
Polycarpus has bought a posh piano and is thinking about how to get it into the house. In order to carry out his plan, he needs to take exactly *k* consecutive planks from the fence. Higher planks are harder to tear off the fence, so Polycarpus wants to find such *k* consecutive planks that the sum of their heights is minimal possible.
Write the program that finds the indexes of *k* consecutive planks with minimal total height. Pay attention, the fence is not around Polycarpus's home, it is in front of home (in other words, the fence isn't cyclic).
Input Specification:
The first line of the input contains integers *n* and *k* (1<=≤<=*n*<=≤<=1.5·105,<=1<=≤<=*k*<=≤<=*n*) — the number of planks in the fence and the width of the hole for the piano. The second line contains the sequence of integers *h*1,<=*h*2,<=...,<=*h**n* (1<=≤<=*h**i*<=≤<=100), where *h**i* is the height of the *i*-th plank of the fence.
Output Specification:
Print such integer *j* that the sum of the heights of planks *j*, *j*<=+<=1, ..., *j*<=+<=*k*<=-<=1 is the minimum possible. If there are multiple such *j*'s, print any of them.
Demo Input:
['7 3\n1 2 6 1 1 7 1\n']
Demo Output:
['3\n']
Note:
In the sample, your task is to find three consecutive planks with the minimum sum of heights. In the given case three planks with indexes 3, 4 and 5 have the required attribute, their total height is 8.
|
```python
from sys import stdin, stdout
input = stdin.readline
def print(*args, end='\n', sep=' ') -> None:
stdout.write(sep.join(map(str, args)) + end)
def int_map():
return map(int, input().split())
def list_int():
return list(map(int, input().split()))
n, k = int_map()
arr = list_int()
for i in range(1, n):
arr[i] += arr[i-1]
mn = 10**9+7
j = 0
for i in range(2, n):
sm = arr[i] - (arr[i-3] if i - 3 >= 0 else 0)
if sm < mn:
j = i - 1
mn = sm
print(j)
```
| 0
|
|
404
|
D
|
Minesweeper 1D
|
PROGRAMMING
| 1,900
|
[
"dp",
"implementation"
] | null | null |
Game "Minesweeper 1D" is played on a line of squares, the line's height is 1 square, the line's width is *n* squares. Some of the squares contain bombs. If a square doesn't contain a bomb, then it contains a number from 0 to 2 — the total number of bombs in adjacent squares.
For example, the correct field to play looks like that: 001*2***101*. The cells that are marked with "*" contain bombs. Note that on the correct field the numbers represent the number of bombs in adjacent cells. For example, field 2* is not correct, because cell with value 2 must have two adjacent cells with bombs.
Valera wants to make a correct field to play "Minesweeper 1D". He has already painted a squared field with width of *n* cells, put several bombs on the field and wrote numbers into some cells. Now he wonders how many ways to fill the remaining cells with bombs and numbers are there if we should get a correct field in the end.
|
The first line contains sequence of characters without spaces *s*1*s*2... *s**n* (1<=≤<=*n*<=≤<=106), containing only characters "*", "?" and digits "0", "1" or "2". If character *s**i* equals "*", then the *i*-th cell of the field contains a bomb. If character *s**i* equals "?", then Valera hasn't yet decided what to put in the *i*-th cell. Character *s**i*, that is equal to a digit, represents the digit written in the *i*-th square.
|
Print a single integer — the number of ways Valera can fill the empty cells and get a correct field.
As the answer can be rather large, print it modulo 1000000007 (109<=+<=7).
|
[
"?01???\n",
"?\n",
"**12\n",
"1\n"
] |
[
"4\n",
"2\n",
"0\n",
"0\n"
] |
In the first test sample you can get the following correct fields: 001**1, 001***, 001*2*, 001*10.
| 2,000
|
[
{
"input": "?01???",
"output": "4"
},
{
"input": "?",
"output": "2"
},
{
"input": "**12",
"output": "0"
},
{
"input": "1",
"output": "0"
},
{
"input": "?01*??****",
"output": "4"
},
{
"input": "0",
"output": "1"
},
{
"input": "2",
"output": "0"
},
{
"input": "*",
"output": "1"
},
{
"input": "0*",
"output": "0"
},
{
"input": "0?",
"output": "1"
},
{
"input": "01",
"output": "0"
},
{
"input": "1*",
"output": "1"
},
{
"input": "1?",
"output": "1"
},
{
"input": "?1?",
"output": "2"
},
{
"input": "12",
"output": "0"
},
{
"input": "2*",
"output": "0"
},
{
"input": "2?",
"output": "0"
},
{
"input": "2??",
"output": "0"
},
{
"input": "?2?",
"output": "1"
},
{
"input": "?2*?2*??1*2**?2*1???*2???100?????*???*?*????0????2?*?*?1??1??*?01**2**1001??**??**??1*?*???00??**??*",
"output": "147483634"
},
{
"input": "00***???01",
"output": "0"
},
{
"input": "21?20*0000?2?22??0001*?1??12?20020200?**0*12?*221*0*1200*?0*11?022*110*2*2022120*2*2100*0?0*02?012?1",
"output": "0"
}
] | 1,699,534,785
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 40
| 295
| 11,878,400
|
import sys
from collections import Counter, defaultdict
from bisect import bisect_left, bisect_right
from heapq import heapify, heappop, heappush
from queue import deque
input = lambda: sys.stdin.readline().strip()
from math import gcd, inf, sqrt
def mrd(fun=int): return [fun(x) for x in input().split()]
def rd(fun=int): return fun(input())
mod = 10**9 + 7
def solve():
s = input()
f = [0] * 5
f[0] = f[2] = 1
for c in s:
g = [0] * 5
if c == '0':
g[0] = sum(f[:2]) % mod
elif c == '1':
g[1], g[2] = f[4], sum(f[:2]) % mod
elif c == '2':
g[3] = f[4]
elif c == '*':
g[4] = sum(f[2:]) % mod
else:
g[0] = sum(f[:2]) % mod
g[1], g[2] = f[4], sum(f[:2]) % mod
g[3] = f[4]
g[4] = sum(f[2:]) % mod
f = g
print((f[0] + f[1] + f[4]) % mod)
t = 1
for _ in range(t):
solve()
|
Title: Minesweeper 1D
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Game "Minesweeper 1D" is played on a line of squares, the line's height is 1 square, the line's width is *n* squares. Some of the squares contain bombs. If a square doesn't contain a bomb, then it contains a number from 0 to 2 — the total number of bombs in adjacent squares.
For example, the correct field to play looks like that: 001*2***101*. The cells that are marked with "*" contain bombs. Note that on the correct field the numbers represent the number of bombs in adjacent cells. For example, field 2* is not correct, because cell with value 2 must have two adjacent cells with bombs.
Valera wants to make a correct field to play "Minesweeper 1D". He has already painted a squared field with width of *n* cells, put several bombs on the field and wrote numbers into some cells. Now he wonders how many ways to fill the remaining cells with bombs and numbers are there if we should get a correct field in the end.
Input Specification:
The first line contains sequence of characters without spaces *s*1*s*2... *s**n* (1<=≤<=*n*<=≤<=106), containing only characters "*", "?" and digits "0", "1" or "2". If character *s**i* equals "*", then the *i*-th cell of the field contains a bomb. If character *s**i* equals "?", then Valera hasn't yet decided what to put in the *i*-th cell. Character *s**i*, that is equal to a digit, represents the digit written in the *i*-th square.
Output Specification:
Print a single integer — the number of ways Valera can fill the empty cells and get a correct field.
As the answer can be rather large, print it modulo 1000000007 (109<=+<=7).
Demo Input:
['?01???\n', '?\n', '**12\n', '1\n']
Demo Output:
['4\n', '2\n', '0\n', '0\n']
Note:
In the first test sample you can get the following correct fields: 001**1, 001***, 001*2*, 001*10.
|
```python
import sys
from collections import Counter, defaultdict
from bisect import bisect_left, bisect_right
from heapq import heapify, heappop, heappush
from queue import deque
input = lambda: sys.stdin.readline().strip()
from math import gcd, inf, sqrt
def mrd(fun=int): return [fun(x) for x in input().split()]
def rd(fun=int): return fun(input())
mod = 10**9 + 7
def solve():
s = input()
f = [0] * 5
f[0] = f[2] = 1
for c in s:
g = [0] * 5
if c == '0':
g[0] = sum(f[:2]) % mod
elif c == '1':
g[1], g[2] = f[4], sum(f[:2]) % mod
elif c == '2':
g[3] = f[4]
elif c == '*':
g[4] = sum(f[2:]) % mod
else:
g[0] = sum(f[:2]) % mod
g[1], g[2] = f[4], sum(f[:2]) % mod
g[3] = f[4]
g[4] = sum(f[2:]) % mod
f = g
print((f[0] + f[1] + f[4]) % mod)
t = 1
for _ in range(t):
solve()
```
| 3
|
|
34
|
A
|
Reconnaissance 2
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Reconnaissance 2
|
2
|
256
|
*n* soldiers stand in a circle. For each soldier his height *a**i* is known. A reconnaissance unit can be made of such two neighbouring soldiers, whose heights difference is minimal, i.e. |*a**i*<=-<=*a**j*| is minimal. So each of them will be less noticeable with the other. Output any pair of soldiers that can form a reconnaissance unit.
|
The first line contains integer *n* (2<=≤<=*n*<=≤<=100) — amount of soldiers. Then follow the heights of the soldiers in their order in the circle — *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000). The soldier heights are given in clockwise or counterclockwise direction.
|
Output two integers — indexes of neighbouring soldiers, who should form a reconnaissance unit. If there are many optimum solutions, output any of them. Remember, that the soldiers stand in a circle.
|
[
"5\n10 12 13 15 10\n",
"4\n10 20 30 40\n"
] |
[
"5 1\n",
"1 2\n"
] |
none
| 500
|
[
{
"input": "5\n10 12 13 15 10",
"output": "5 1"
},
{
"input": "4\n10 20 30 40",
"output": "1 2"
},
{
"input": "6\n744 359 230 586 944 442",
"output": "2 3"
},
{
"input": "5\n826 747 849 687 437",
"output": "1 2"
},
{
"input": "5\n999 999 993 969 999",
"output": "1 2"
},
{
"input": "5\n4 24 6 1 15",
"output": "3 4"
},
{
"input": "2\n511 32",
"output": "1 2"
},
{
"input": "3\n907 452 355",
"output": "2 3"
},
{
"input": "4\n303 872 764 401",
"output": "4 1"
},
{
"input": "10\n684 698 429 694 956 812 594 170 937 764",
"output": "1 2"
},
{
"input": "20\n646 840 437 946 640 564 936 917 487 752 844 734 468 969 674 646 728 642 514 695",
"output": "7 8"
},
{
"input": "30\n996 999 998 984 989 1000 996 993 1000 983 992 999 999 1000 979 992 987 1000 996 1000 1000 989 981 996 995 999 999 989 999 1000",
"output": "12 13"
},
{
"input": "50\n93 27 28 4 5 78 59 24 19 134 31 128 118 36 90 32 32 1 44 32 33 13 31 10 12 25 38 50 25 12 4 22 28 53 48 83 4 25 57 31 71 24 8 7 28 86 23 80 101 58",
"output": "16 17"
},
{
"input": "88\n1000 1000 1000 1000 1000 998 998 1000 1000 1000 1000 999 999 1000 1000 1000 999 1000 997 999 997 1000 999 998 1000 999 1000 1000 1000 999 1000 999 999 1000 1000 999 1000 999 1000 1000 998 1000 1000 1000 998 998 1000 1000 999 1000 1000 1000 1000 1000 1000 1000 998 1000 1000 1000 999 1000 1000 999 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 998 1000 1000 1000 998 1000 1000 998 1000 999 1000 1000 1000 1000",
"output": "1 2"
},
{
"input": "99\n4 4 21 6 5 3 13 2 6 1 3 4 1 3 1 9 11 1 6 17 4 5 20 4 1 9 5 11 3 4 14 1 3 3 1 4 3 5 27 1 1 2 10 7 11 4 19 7 11 6 11 13 3 1 10 7 2 1 16 1 9 4 29 13 2 12 14 2 21 1 9 8 26 12 12 5 2 14 7 8 8 8 9 4 12 2 6 6 7 16 8 14 2 10 20 15 3 7 4",
"output": "1 2"
},
{
"input": "100\n713 572 318 890 577 657 646 146 373 783 392 229 455 871 20 593 573 336 26 381 280 916 907 732 820 713 111 840 570 446 184 711 481 399 788 647 492 15 40 530 549 506 719 782 126 20 778 996 712 761 9 74 812 418 488 175 103 585 900 3 604 521 109 513 145 708 990 361 682 827 791 22 596 780 596 385 450 643 158 496 876 975 319 783 654 895 891 361 397 81 682 899 347 623 809 557 435 279 513 438",
"output": "86 87"
},
{
"input": "100\n31 75 86 68 111 27 22 22 26 30 54 163 107 75 160 122 14 23 17 26 27 20 43 58 59 71 21 148 9 32 43 91 133 286 132 70 90 156 84 14 77 93 23 18 13 72 18 131 33 28 72 175 30 86 249 20 14 208 28 57 63 199 6 10 24 30 62 267 43 479 60 28 138 1 45 3 19 47 7 166 116 117 50 140 28 14 95 85 93 43 61 15 2 70 10 51 7 95 9 25",
"output": "7 8"
},
{
"input": "100\n896 898 967 979 973 709 961 968 806 967 896 967 826 975 936 903 986 856 851 931 852 971 786 837 949 978 686 936 952 909 965 749 908 916 943 973 983 975 939 886 964 928 960 976 907 788 994 773 949 871 947 980 945 985 726 981 887 943 907 990 931 874 840 867 948 951 961 904 888 901 976 967 994 921 828 970 972 722 755 970 860 855 914 869 714 899 969 978 898 862 642 939 904 936 819 934 884 983 955 964",
"output": "1 2"
},
{
"input": "100\n994 927 872 970 815 986 952 996 965 1000 877 986 978 999 950 990 936 997 993 960 921 860 895 869 943 998 983 968 973 953 999 990 995 871 853 979 973 963 953 938 997 989 993 964 960 973 946 975 1000 962 920 746 989 957 904 965 920 979 966 961 1000 993 975 952 846 971 991 979 985 969 984 973 956 1000 952 778 983 974 956 927 995 997 980 997 1000 970 960 970 988 983 947 904 935 972 1000 863 992 996 932 967",
"output": "81 82"
},
{
"input": "100\n48 108 63 21 27 8 49 21 75 8 24 42 149 18 8 28 21 18 25 35 59 70 59 33 40 1 67 34 120 82 4 115 72 87 3 15 15 63 37 12 40 27 83 14 38 20 14 58 93 10 31 3 39 6 197 77 54 16 31 146 9 49 14 8 77 82 5 11 80 116 8 61 50 24 7 103 29 11 3 3 1 12 46 24 21 131 39 29 36 2 107 40 16 99 31 41 29 48 17 17",
"output": "36 37"
}
] | 1,656,152,332
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 20
| 92
| 0
|
n = int(input())
a = list(map(int, input().split()))
ans = 100000
for i in range(len(a)):
if (abs(a[i] - abs(a[(i + 1) % n])) < ans):
ans = abs(a[i] - abs(a[(i + 1) % n]))
ia, ja = i, (i + 1) % n
print(ia + 1, ja + 1)
|
Title: Reconnaissance 2
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
*n* soldiers stand in a circle. For each soldier his height *a**i* is known. A reconnaissance unit can be made of such two neighbouring soldiers, whose heights difference is minimal, i.e. |*a**i*<=-<=*a**j*| is minimal. So each of them will be less noticeable with the other. Output any pair of soldiers that can form a reconnaissance unit.
Input Specification:
The first line contains integer *n* (2<=≤<=*n*<=≤<=100) — amount of soldiers. Then follow the heights of the soldiers in their order in the circle — *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000). The soldier heights are given in clockwise or counterclockwise direction.
Output Specification:
Output two integers — indexes of neighbouring soldiers, who should form a reconnaissance unit. If there are many optimum solutions, output any of them. Remember, that the soldiers stand in a circle.
Demo Input:
['5\n10 12 13 15 10\n', '4\n10 20 30 40\n']
Demo Output:
['5 1\n', '1 2\n']
Note:
none
|
```python
n = int(input())
a = list(map(int, input().split()))
ans = 100000
for i in range(len(a)):
if (abs(a[i] - abs(a[(i + 1) % n])) < ans):
ans = abs(a[i] - abs(a[(i + 1) % n]))
ia, ja = i, (i + 1) % n
print(ia + 1, ja + 1)
```
| 3.977
|
626
|
B
|
Cards
|
PROGRAMMING
| 1,300
|
[
"constructive algorithms",
"dp",
"math"
] | null | null |
Catherine has a deck of *n* cards, each of which is either red, green, or blue. As long as there are at least two cards left, she can do one of two actions:
- take any two (not necessarily adjacent) cards with different colors and exchange them for a new card of the third color; - take any two (not necessarily adjacent) cards with the same color and exchange them for a new card with that color.
She repeats this process until there is only one card left. What are the possible colors for the final card?
|
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200) — the total number of cards.
The next line contains a string *s* of length *n* — the colors of the cards. *s* contains only the characters 'B', 'G', and 'R', representing blue, green, and red, respectively.
|
Print a single string of up to three characters — the possible colors of the final card (using the same symbols as the input) in alphabetical order.
|
[
"2\nRB\n",
"3\nGRG\n",
"5\nBBBBB\n"
] |
[
"G\n",
"BR\n",
"B\n"
] |
In the first sample, Catherine has one red card and one blue card, which she must exchange for a green card.
In the second sample, Catherine has two green cards and one red card. She has two options: she can exchange the two green cards for a green card, then exchange the new green card and the red card for a blue card. Alternatively, she can exchange a green and a red card for a blue card, then exchange the blue card and remaining green card for a red card.
In the third sample, Catherine only has blue cards, so she can only exchange them for more blue cards.
| 750
|
[
{
"input": "2\nRB",
"output": "G"
},
{
"input": "3\nGRG",
"output": "BR"
},
{
"input": "5\nBBBBB",
"output": "B"
},
{
"input": "1\nR",
"output": "R"
},
{
"input": "200\nBBRGRRBBRGGGBGBGBGRRGRGRGRBGRGRRBBGRGBGRRGRRRGGBBRGBGBGBRBBBBBBBGGBRGGRRRGGRGBGBGGBRRRRBRRRBRBBGGBGBRGRGBBBBGGBGBBBGBGRRBRRRGBGGBBBRBGRBRRGGGRRGBBBGBGRRRRRRGGRGRGBBBRGGGBGGGBRBBRRGBGRGRBRRRBRBGRGGBRBB",
"output": "BGR"
},
{
"input": "101\nRRRRRRRRRRRRRRRRRRRBRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "BG"
},
{
"input": "7\nBBBGBRG",
"output": "BGR"
},
{
"input": "5\nGRRGR",
"output": "BGR"
},
{
"input": "3\nGBR",
"output": "BGR"
},
{
"input": "1\nB",
"output": "B"
},
{
"input": "2\nBB",
"output": "B"
},
{
"input": "1\nG",
"output": "G"
},
{
"input": "2\nBG",
"output": "R"
},
{
"input": "3\nBGB",
"output": "GR"
},
{
"input": "2\nGG",
"output": "G"
},
{
"input": "3\nGBG",
"output": "BR"
},
{
"input": "4\nBGBG",
"output": "BGR"
},
{
"input": "1\nR",
"output": "R"
},
{
"input": "2\nBR",
"output": "G"
},
{
"input": "3\nBRB",
"output": "GR"
},
{
"input": "2\nRG",
"output": "B"
},
{
"input": "3\nBGR",
"output": "BGR"
},
{
"input": "4\nRBGB",
"output": "BGR"
},
{
"input": "3\nGGR",
"output": "BR"
},
{
"input": "4\nGGRB",
"output": "BGR"
},
{
"input": "5\nBGBGR",
"output": "BGR"
},
{
"input": "2\nRR",
"output": "R"
},
{
"input": "3\nRBR",
"output": "BG"
},
{
"input": "4\nRRBB",
"output": "BGR"
},
{
"input": "3\nRRG",
"output": "BG"
},
{
"input": "4\nBRRG",
"output": "BGR"
},
{
"input": "5\nRBRBG",
"output": "BGR"
},
{
"input": "4\nRGGR",
"output": "BGR"
},
{
"input": "5\nBRGRG",
"output": "BGR"
},
{
"input": "6\nGRRGBB",
"output": "BGR"
},
{
"input": "150\nGRGBBBBRBGGBGBBGBBBBGRBBRRBBGRRGGGBRBBRGRRRRGBGRRBGBGBGRBBBGBBBGBGBRGBRRRRRGGGRGRBBGBRGGGRBBRGBBGRGGGBBRBRRGRGRRGRRGRRRGBGBRRGGRGGBRBGGGBBBRGRGBRGRRRR",
"output": "BGR"
},
{
"input": "16\nRRGRRRRRRGGRGRRR",
"output": "BGR"
},
{
"input": "190\nBBBBBBBBBBBBBBBBBGBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB",
"output": "GR"
},
{
"input": "200\nRGRGRRRRRGRRGRRRGRGRRRGGRGRRGGGRRGGRRRRRRRRRRRGRRGRRRGRRRGRRRRRRRGRRRRRRRRRRRGGRRGGRRRRGGRRRRRRRRRGGGRGRGRGRRGRGGRGRGRRRGRRRRRRGGRGRRRRGRRGRGGRRRRRRRGRGGRRGRRRRRRRGGRRRRGRRRRRRRGRRRGGRRRRRRGRRGGGRRRGR",
"output": "BGR"
},
{
"input": "200\nGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG",
"output": "G"
},
{
"input": "52\nBBBBBBBBBBBBBBBBBBBBGBGBBBBBBBBBBBBBBBBBBBBBBBBBBBBB",
"output": "BGR"
},
{
"input": "200\nGRGRRGRBRRRGGGRGGRRRRRBBGRRGRBBGRRGBGRRBBRBBRRBBBGRBRGGGGBGGBRRBBRGRBGGRRGGBBRBGGRGBBRRBBRGBRRBGBRBGBBRGGRRRGGGBRGGGGRRRBBRRGRGRBRRGRBBGGRBBRGRGRBGRBBRGGBBBGRGBBGGBGBGBBRRBGRGRGGBRRGRGGGGGBRGGGGBBBBRB",
"output": "BGR"
},
{
"input": "102\nGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGRGGGGGGGGBGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG",
"output": "BGR"
},
{
"input": "193\nRRRGGGRBGGBGGGBGGBBGRBGGRBGGBBRBGGRBBBRBRRGGBBRBRGRRRBGBBRGGRGGGBGGRRGGRGRRBRBRBRRGRGBGBRGBBRGRRRBGRGGBGBRBBBGBRBBGBGBGGGBGGGGBRBBRRBGRGGBBBRBBBBBGRRRGBRGBRRRBBBGBGGGGRGGRRBRBGRRGBGBRBGGGRBRRGG",
"output": "BGR"
},
{
"input": "90\nBGBGGRRBGGRRRRRGGRGBBBBBRRBGBGBGBGGBBGRGGGGRBRBBRRRGBRRGBBGBBGGGRGRGRBGBBBRRGRRBRBRRGGRBRB",
"output": "BGR"
},
{
"input": "3\nGGB",
"output": "BR"
}
] | 1,514,394,515
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 62
| 5,529,600
|
n=int(input())
s=input()
r=s.count('R')
g=s.count('G')
b=s.count('B')
if (r>=1 and b>=1 and g>=1) or (r>=2 and b>=2) or (g>=2 and b>=2) or (r>=2 and g>=2) :
print('RGB')
elif r==1 and r==1 :
print('G')
elif r==1 and g==1 :
print('B')
elif g==1 and b==1 :
print('R')
elif g==2 and b==1 :
print('BR')
elif g==1 and b==2 :
print('RG')
elif g==1 and r==2 :
print('BG')
elif g==n :
print('G')
elif r==n :
print('R')
elif b==n :
print('B')
|
Title: Cards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Catherine has a deck of *n* cards, each of which is either red, green, or blue. As long as there are at least two cards left, she can do one of two actions:
- take any two (not necessarily adjacent) cards with different colors and exchange them for a new card of the third color; - take any two (not necessarily adjacent) cards with the same color and exchange them for a new card with that color.
She repeats this process until there is only one card left. What are the possible colors for the final card?
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200) — the total number of cards.
The next line contains a string *s* of length *n* — the colors of the cards. *s* contains only the characters 'B', 'G', and 'R', representing blue, green, and red, respectively.
Output Specification:
Print a single string of up to three characters — the possible colors of the final card (using the same symbols as the input) in alphabetical order.
Demo Input:
['2\nRB\n', '3\nGRG\n', '5\nBBBBB\n']
Demo Output:
['G\n', 'BR\n', 'B\n']
Note:
In the first sample, Catherine has one red card and one blue card, which she must exchange for a green card.
In the second sample, Catherine has two green cards and one red card. She has two options: she can exchange the two green cards for a green card, then exchange the new green card and the red card for a blue card. Alternatively, she can exchange a green and a red card for a blue card, then exchange the blue card and remaining green card for a red card.
In the third sample, Catherine only has blue cards, so she can only exchange them for more blue cards.
|
```python
n=int(input())
s=input()
r=s.count('R')
g=s.count('G')
b=s.count('B')
if (r>=1 and b>=1 and g>=1) or (r>=2 and b>=2) or (g>=2 and b>=2) or (r>=2 and g>=2) :
print('RGB')
elif r==1 and r==1 :
print('G')
elif r==1 and g==1 :
print('B')
elif g==1 and b==1 :
print('R')
elif g==2 and b==1 :
print('BR')
elif g==1 and b==2 :
print('RG')
elif g==1 and r==2 :
print('BG')
elif g==n :
print('G')
elif r==n :
print('R')
elif b==n :
print('B')
```
| 0
|
|
456
|
A
|
Laptops
|
PROGRAMMING
| 1,100
|
[
"sortings"
] | null | null |
One day Dima and Alex had an argument about the price and quality of laptops. Dima thinks that the more expensive a laptop is, the better it is. Alex disagrees. Alex thinks that there are two laptops, such that the price of the first laptop is less (strictly smaller) than the price of the second laptop but the quality of the first laptop is higher (strictly greater) than the quality of the second laptop.
Please, check the guess of Alex. You are given descriptions of *n* laptops. Determine whether two described above laptops exist.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105) — the number of laptops.
Next *n* lines contain two integers each, *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*), where *a**i* is the price of the *i*-th laptop, and *b**i* is the number that represents the quality of the *i*-th laptop (the larger the number is, the higher is the quality).
All *a**i* are distinct. All *b**i* are distinct.
|
If Alex is correct, print "Happy Alex", otherwise print "Poor Alex" (without the quotes).
|
[
"2\n1 2\n2 1\n"
] |
[
"Happy Alex\n"
] |
none
| 500
|
[
{
"input": "2\n1 2\n2 1",
"output": "Happy Alex"
},
{
"input": "2\n1 1\n2 2",
"output": "Poor Alex"
},
{
"input": "3\n2 2\n3 3\n1 1",
"output": "Poor Alex"
},
{
"input": "3\n3 3\n1 2\n2 1",
"output": "Happy Alex"
},
{
"input": "1\n1 1",
"output": "Poor Alex"
},
{
"input": "3\n2 3\n1 1\n3 2",
"output": "Happy Alex"
},
{
"input": "4\n4 1\n3 2\n2 3\n1 4",
"output": "Happy Alex"
},
{
"input": "2\n2 1\n1 2",
"output": "Happy Alex"
},
{
"input": "3\n3 2\n1 1\n2 3",
"output": "Happy Alex"
},
{
"input": "3\n1 2\n2 3\n3 1",
"output": "Happy Alex"
},
{
"input": "3\n1 2\n2 1\n3 3",
"output": "Happy Alex"
},
{
"input": "3\n1 2\n2 3\n3 1",
"output": "Happy Alex"
}
] | 1,695,071,046
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 31
| 0
|
n = int(input())
a1, a2 = map(int, input().split())
b1, b2 = map(int, input().split())
if (a1 < a2 and b1 > b2 ) or (a2 < a1 and b2> b1):
print("Happy Alex")
else:
print("Poor Alex")
|
Title: Laptops
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Dima and Alex had an argument about the price and quality of laptops. Dima thinks that the more expensive a laptop is, the better it is. Alex disagrees. Alex thinks that there are two laptops, such that the price of the first laptop is less (strictly smaller) than the price of the second laptop but the quality of the first laptop is higher (strictly greater) than the quality of the second laptop.
Please, check the guess of Alex. You are given descriptions of *n* laptops. Determine whether two described above laptops exist.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105) — the number of laptops.
Next *n* lines contain two integers each, *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*), where *a**i* is the price of the *i*-th laptop, and *b**i* is the number that represents the quality of the *i*-th laptop (the larger the number is, the higher is the quality).
All *a**i* are distinct. All *b**i* are distinct.
Output Specification:
If Alex is correct, print "Happy Alex", otherwise print "Poor Alex" (without the quotes).
Demo Input:
['2\n1 2\n2 1\n']
Demo Output:
['Happy Alex\n']
Note:
none
|
```python
n = int(input())
a1, a2 = map(int, input().split())
b1, b2 = map(int, input().split())
if (a1 < a2 and b1 > b2 ) or (a2 < a1 and b2> b1):
print("Happy Alex")
else:
print("Poor Alex")
```
| 0
|
|
50
|
A
|
Domino piling
|
PROGRAMMING
| 800
|
[
"greedy",
"math"
] |
A. Domino piling
|
2
|
256
|
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
|
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
|
Output one number — the maximal number of dominoes, which can be placed.
|
[
"2 4\n",
"3 3\n"
] |
[
"4\n",
"4\n"
] |
none
| 500
|
[
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "2 6",
"output": "6"
},
{
"input": "2 7",
"output": "7"
},
{
"input": "2 14",
"output": "14"
},
{
"input": "2 15",
"output": "15"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "2 16",
"output": "16"
},
{
"input": "3 5",
"output": "7"
},
{
"input": "3 6",
"output": "9"
},
{
"input": "3 10",
"output": "15"
},
{
"input": "3 14",
"output": "21"
},
{
"input": "3 15",
"output": "22"
},
{
"input": "3 16",
"output": "24"
},
{
"input": "5 7",
"output": "17"
},
{
"input": "16 16",
"output": "128"
},
{
"input": "15 16",
"output": "120"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "15 15",
"output": "112"
},
{
"input": "14 16",
"output": "112"
},
{
"input": "11 13",
"output": "71"
},
{
"input": "5 16",
"output": "40"
},
{
"input": "8 15",
"output": "60"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "3 4",
"output": "6"
},
{
"input": "4 4",
"output": "8"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "14 15",
"output": "105"
}
] | 1,661,972,720
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 30
| 0
|
def domino(m,n):
squares = m*n
domis = int(squares / 2*1)
print(domis)
domino(int(input()))
|
Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none
|
```python
def domino(m,n):
squares = m*n
domis = int(squares / 2*1)
print(domis)
domino(int(input()))
```
| -1
|
124
|
A
|
The number of positions
|
PROGRAMMING
| 1,000
|
[
"math"
] | null | null |
Petr stands in line of *n* people, but he doesn't know exactly which position he occupies. He can say that there are no less than *a* people standing in front of him and no more than *b* people standing behind him. Find the number of different positions Petr can occupy.
|
The only line contains three integers *n*, *a* and *b* (0<=≤<=*a*,<=*b*<=<<=*n*<=≤<=100).
|
Print the single number — the number of the sought positions.
|
[
"3 1 1\n",
"5 2 3\n"
] |
[
"2\n",
"3\n"
] |
The possible positions in the first sample are: 2 and 3 (if we number the positions starting with 1).
In the second sample they are 3, 4 and 5.
| 500
|
[
{
"input": "3 1 1",
"output": "2"
},
{
"input": "5 2 3",
"output": "3"
},
{
"input": "5 4 0",
"output": "1"
},
{
"input": "6 5 5",
"output": "1"
},
{
"input": "9 4 3",
"output": "4"
},
{
"input": "11 4 6",
"output": "7"
},
{
"input": "13 8 7",
"output": "5"
},
{
"input": "14 5 5",
"output": "6"
},
{
"input": "16 6 9",
"output": "10"
},
{
"input": "20 13 17",
"output": "7"
},
{
"input": "22 4 8",
"output": "9"
},
{
"input": "23 8 14",
"output": "15"
},
{
"input": "26 18 22",
"output": "8"
},
{
"input": "28 6 1",
"output": "2"
},
{
"input": "29 5 23",
"output": "24"
},
{
"input": "32 27 15",
"output": "5"
},
{
"input": "33 11 5",
"output": "6"
},
{
"input": "37 21 15",
"output": "16"
},
{
"input": "39 34 33",
"output": "5"
},
{
"input": "41 27 11",
"output": "12"
},
{
"input": "42 25 16",
"output": "17"
},
{
"input": "45 7 43",
"output": "38"
},
{
"input": "47 16 17",
"output": "18"
},
{
"input": "49 11 37",
"output": "38"
},
{
"input": "51 38 39",
"output": "13"
},
{
"input": "52 29 7",
"output": "8"
},
{
"input": "56 43 12",
"output": "13"
},
{
"input": "58 57 28",
"output": "1"
},
{
"input": "59 12 39",
"output": "40"
},
{
"input": "62 9 52",
"output": "53"
},
{
"input": "63 29 44",
"output": "34"
},
{
"input": "65 30 22",
"output": "23"
},
{
"input": "66 27 38",
"output": "39"
},
{
"input": "71 33 53",
"output": "38"
},
{
"input": "73 14 12",
"output": "13"
},
{
"input": "73 37 35",
"output": "36"
},
{
"input": "76 69 44",
"output": "7"
},
{
"input": "79 25 20",
"output": "21"
},
{
"input": "81 60 20",
"output": "21"
},
{
"input": "81 79 14",
"output": "2"
},
{
"input": "84 0 42",
"output": "43"
},
{
"input": "88 79 8",
"output": "9"
},
{
"input": "90 76 59",
"output": "14"
},
{
"input": "92 2 22",
"output": "23"
},
{
"input": "94 5 88",
"output": "89"
},
{
"input": "94 62 48",
"output": "32"
},
{
"input": "96 22 72",
"output": "73"
},
{
"input": "100 11 88",
"output": "89"
},
{
"input": "100 81 91",
"output": "19"
},
{
"input": "1 0 0",
"output": "1"
}
] | 1,626,857,627
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 124
| 6,758,400
|
d = [int(i) for i in input().split()]
print(d[0]-d[1])
|
Title: The number of positions
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petr stands in line of *n* people, but he doesn't know exactly which position he occupies. He can say that there are no less than *a* people standing in front of him and no more than *b* people standing behind him. Find the number of different positions Petr can occupy.
Input Specification:
The only line contains three integers *n*, *a* and *b* (0<=≤<=*a*,<=*b*<=<<=*n*<=≤<=100).
Output Specification:
Print the single number — the number of the sought positions.
Demo Input:
['3 1 1\n', '5 2 3\n']
Demo Output:
['2\n', '3\n']
Note:
The possible positions in the first sample are: 2 and 3 (if we number the positions starting with 1).
In the second sample they are 3, 4 and 5.
|
```python
d = [int(i) for i in input().split()]
print(d[0]-d[1])
```
| 0
|
|
151
|
A
|
Soft Drinking
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
This winter is so cold in Nvodsk! A group of *n* friends decided to buy *k* bottles of a soft drink called "Take-It-Light" to warm up a bit. Each bottle has *l* milliliters of the drink. Also they bought *c* limes and cut each of them into *d* slices. After that they found *p* grams of salt.
To make a toast, each friend needs *nl* milliliters of the drink, a slice of lime and *np* grams of salt. The friends want to make as many toasts as they can, provided they all drink the same amount. How many toasts can each friend make?
|
The first and only line contains positive integers *n*, *k*, *l*, *c*, *d*, *p*, *nl*, *np*, not exceeding 1000 and no less than 1. The numbers are separated by exactly one space.
|
Print a single integer — the number of toasts each friend can make.
|
[
"3 4 5 10 8 100 3 1\n",
"5 100 10 1 19 90 4 3\n",
"10 1000 1000 25 23 1 50 1\n"
] |
[
"2\n",
"3\n",
"0\n"
] |
A comment to the first sample:
Overall the friends have 4 * 5 = 20 milliliters of the drink, it is enough to make 20 / 3 = 6 toasts. The limes are enough for 10 * 8 = 80 toasts and the salt is enough for 100 / 1 = 100 toasts. However, there are 3 friends in the group, so the answer is *min*(6, 80, 100) / 3 = 2.
| 500
|
[
{
"input": "3 4 5 10 8 100 3 1",
"output": "2"
},
{
"input": "5 100 10 1 19 90 4 3",
"output": "3"
},
{
"input": "10 1000 1000 25 23 1 50 1",
"output": "0"
},
{
"input": "1 7 4 5 5 8 3 2",
"output": "4"
},
{
"input": "2 3 3 5 5 10 1 3",
"output": "1"
},
{
"input": "2 6 4 5 6 5 1 3",
"output": "0"
},
{
"input": "1 7 3 5 3 6 2 1",
"output": "6"
},
{
"input": "2 4 5 4 5 7 3 2",
"output": "1"
},
{
"input": "2 3 6 5 7 8 2 1",
"output": "4"
},
{
"input": "1 4 5 5 3 10 3 1",
"output": "6"
},
{
"input": "1 4 6 7 3 5 1 3",
"output": "1"
},
{
"input": "1 6 5 5 5 8 3 1",
"output": "8"
},
{
"input": "1 7 5 3 3 9 2 1",
"output": "9"
},
{
"input": "3 5 3 7 6 10 3 1",
"output": "1"
},
{
"input": "3 6 3 5 3 6 3 1",
"output": "2"
},
{
"input": "1 7 5 5 5 5 2 2",
"output": "2"
},
{
"input": "2 5 3 5 6 9 2 1",
"output": "3"
},
{
"input": "3 4 3 5 3 6 2 1",
"output": "2"
},
{
"input": "1 5 5 4 7 6 3 1",
"output": "6"
},
{
"input": "2 3 7 6 5 9 3 1",
"output": "3"
},
{
"input": "2 6 5 3 3 8 1 1",
"output": "4"
},
{
"input": "2 4 7 3 4 10 2 1",
"output": "5"
},
{
"input": "1 1000 1000 1000 1000 1000 1 1",
"output": "1000"
},
{
"input": "17 1000 1000 1000 1000 1000 3 7",
"output": "8"
},
{
"input": "115 1000 1000 1000 1000 1000 17 15",
"output": "0"
},
{
"input": "1 587 981 1 2 1 1 1",
"output": "1"
},
{
"input": "1 1 2 1 2 2 1 1",
"output": "2"
}
] | 1,686,328,801
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 27
| 62
| 0
|
import math
n, k, l, c, d, p, nl, np=[int(x) for x in input().split()]
print(int(min(math.floor(k*l/nl),c*d,math.floor(p/np))/n))
|
Title: Soft Drinking
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
This winter is so cold in Nvodsk! A group of *n* friends decided to buy *k* bottles of a soft drink called "Take-It-Light" to warm up a bit. Each bottle has *l* milliliters of the drink. Also they bought *c* limes and cut each of them into *d* slices. After that they found *p* grams of salt.
To make a toast, each friend needs *nl* milliliters of the drink, a slice of lime and *np* grams of salt. The friends want to make as many toasts as they can, provided they all drink the same amount. How many toasts can each friend make?
Input Specification:
The first and only line contains positive integers *n*, *k*, *l*, *c*, *d*, *p*, *nl*, *np*, not exceeding 1000 and no less than 1. The numbers are separated by exactly one space.
Output Specification:
Print a single integer — the number of toasts each friend can make.
Demo Input:
['3 4 5 10 8 100 3 1\n', '5 100 10 1 19 90 4 3\n', '10 1000 1000 25 23 1 50 1\n']
Demo Output:
['2\n', '3\n', '0\n']
Note:
A comment to the first sample:
Overall the friends have 4 * 5 = 20 milliliters of the drink, it is enough to make 20 / 3 = 6 toasts. The limes are enough for 10 * 8 = 80 toasts and the salt is enough for 100 / 1 = 100 toasts. However, there are 3 friends in the group, so the answer is *min*(6, 80, 100) / 3 = 2.
|
```python
import math
n, k, l, c, d, p, nl, np=[int(x) for x in input().split()]
print(int(min(math.floor(k*l/nl),c*d,math.floor(p/np))/n))
```
| 3
|
|
237
|
A
|
Free Cash
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
Valera runs a 24/7 fast food cafe. He magically learned that next day *n* people will visit his cafe. For each person we know the arrival time: the *i*-th person comes exactly at *h**i* hours *m**i* minutes. The cafe spends less than a minute to serve each client, but if a client comes in and sees that there is no free cash, than he doesn't want to wait and leaves the cafe immediately.
Valera is very greedy, so he wants to serve all *n* customers next day (and get more profit). However, for that he needs to ensure that at each moment of time the number of working cashes is no less than the number of clients in the cafe.
Help Valera count the minimum number of cashes to work at his cafe next day, so that they can serve all visitors.
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105), that is the number of cafe visitors.
Each of the following *n* lines has two space-separated integers *h**i* and *m**i* (0<=≤<=*h**i*<=≤<=23; 0<=≤<=*m**i*<=≤<=59), representing the time when the *i*-th person comes into the cafe.
Note that the time is given in the chronological order. All time is given within one 24-hour period.
|
Print a single integer — the minimum number of cashes, needed to serve all clients next day.
|
[
"4\n8 0\n8 10\n8 10\n8 45\n",
"3\n0 12\n10 11\n22 22\n"
] |
[
"2\n",
"1\n"
] |
In the first sample it is not enough one cash to serve all clients, because two visitors will come into cafe in 8:10. Therefore, if there will be one cash in cafe, then one customer will be served by it, and another one will not wait and will go away.
In the second sample all visitors will come in different times, so it will be enough one cash.
| 500
|
[
{
"input": "4\n8 0\n8 10\n8 10\n8 45",
"output": "2"
},
{
"input": "3\n0 12\n10 11\n22 22",
"output": "1"
},
{
"input": "5\n12 8\n15 27\n15 27\n16 2\n19 52",
"output": "2"
},
{
"input": "7\n5 6\n7 34\n7 34\n7 34\n12 29\n15 19\n20 23",
"output": "3"
},
{
"input": "8\n0 36\n4 7\n4 7\n4 7\n11 46\n12 4\n15 39\n18 6",
"output": "3"
},
{
"input": "20\n4 12\n4 21\n4 27\n4 56\n5 55\n7 56\n11 28\n11 36\n14 58\n15 59\n16 8\n17 12\n17 23\n17 23\n17 23\n17 23\n17 23\n17 23\n20 50\n22 32",
"output": "6"
},
{
"input": "10\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30",
"output": "10"
},
{
"input": "50\n0 23\n1 21\n2 8\n2 45\n3 1\n4 19\n4 37\n7 7\n7 40\n8 43\n9 51\n10 13\n11 2\n11 19\n11 30\n12 37\n12 37\n12 37\n12 37\n12 37\n12 37\n12 37\n12 37\n12 54\n13 32\n13 42\n14 29\n14 34\n14 48\n15 0\n15 27\n16 22\n16 31\n17 25\n17 26\n17 33\n18 14\n18 16\n18 20\n19 0\n19 5\n19 56\n20 22\n21 26\n22 0\n22 10\n22 11\n22 36\n23 17\n23 20",
"output": "8"
},
{
"input": "10\n0 39\n1 35\n1 49\n1 51\n5 24\n7 40\n7 56\n16 42\n23 33\n23 49",
"output": "1"
},
{
"input": "15\n0 16\n6 15\n8 2\n8 6\n8 7\n10 1\n10 1\n10 3\n10 12\n13 5\n14 16\n14 16\n14 16\n14 16\n14 16",
"output": "5"
},
{
"input": "2\n0 24\n1 0",
"output": "1"
},
{
"input": "1\n0 0",
"output": "1"
},
{
"input": "1\n1 5",
"output": "1"
},
{
"input": "1\n1 1",
"output": "1"
},
{
"input": "3\n0 0\n0 0\n0 0",
"output": "3"
},
{
"input": "1\n5 0",
"output": "1"
},
{
"input": "5\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "5"
},
{
"input": "1\n10 10",
"output": "1"
},
{
"input": "1\n8 0",
"output": "1"
},
{
"input": "10\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "10"
},
{
"input": "2\n0 0\n0 1",
"output": "1"
},
{
"input": "2\n8 5\n8 5",
"output": "2"
}
] | 1,661,841,474
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 67
| 498
| 4,403,200
|
n = int(input())
visitors = []
for _ in range(n):
hours, minutes = input().split()
if len(hours) == 1:
hours = "0" + hours
if len(minutes) == 1:
minutes = "0" + minutes
visitors.append(f"{hours}{minutes}")
visitors.sort()
max_clients = -1
counter = 0
prev = None
for visitor in visitors:
if visitor == prev:
counter += 1
else:
counter = 1
if counter > max_clients:
max_clients = counter
prev = visitor
print(max_clients)
|
Title: Free Cash
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera runs a 24/7 fast food cafe. He magically learned that next day *n* people will visit his cafe. For each person we know the arrival time: the *i*-th person comes exactly at *h**i* hours *m**i* minutes. The cafe spends less than a minute to serve each client, but if a client comes in and sees that there is no free cash, than he doesn't want to wait and leaves the cafe immediately.
Valera is very greedy, so he wants to serve all *n* customers next day (and get more profit). However, for that he needs to ensure that at each moment of time the number of working cashes is no less than the number of clients in the cafe.
Help Valera count the minimum number of cashes to work at his cafe next day, so that they can serve all visitors.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105), that is the number of cafe visitors.
Each of the following *n* lines has two space-separated integers *h**i* and *m**i* (0<=≤<=*h**i*<=≤<=23; 0<=≤<=*m**i*<=≤<=59), representing the time when the *i*-th person comes into the cafe.
Note that the time is given in the chronological order. All time is given within one 24-hour period.
Output Specification:
Print a single integer — the minimum number of cashes, needed to serve all clients next day.
Demo Input:
['4\n8 0\n8 10\n8 10\n8 45\n', '3\n0 12\n10 11\n22 22\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample it is not enough one cash to serve all clients, because two visitors will come into cafe in 8:10. Therefore, if there will be one cash in cafe, then one customer will be served by it, and another one will not wait and will go away.
In the second sample all visitors will come in different times, so it will be enough one cash.
|
```python
n = int(input())
visitors = []
for _ in range(n):
hours, minutes = input().split()
if len(hours) == 1:
hours = "0" + hours
if len(minutes) == 1:
minutes = "0" + minutes
visitors.append(f"{hours}{minutes}")
visitors.sort()
max_clients = -1
counter = 0
prev = None
for visitor in visitors:
if visitor == prev:
counter += 1
else:
counter = 1
if counter > max_clients:
max_clients = counter
prev = visitor
print(max_clients)
```
| 3
|
|
750
|
A
|
New Year and Hurry
|
PROGRAMMING
| 800
|
[
"binary search",
"brute force",
"implementation",
"math"
] | null | null |
Limak is going to participate in a contest on the last day of the 2016. The contest will start at 20:00 and will last four hours, exactly until midnight. There will be *n* problems, sorted by difficulty, i.e. problem 1 is the easiest and problem *n* is the hardest. Limak knows it will take him 5·*i* minutes to solve the *i*-th problem.
Limak's friends organize a New Year's Eve party and Limak wants to be there at midnight or earlier. He needs *k* minutes to get there from his house, where he will participate in the contest first.
How many problems can Limak solve if he wants to make it to the party?
|
The only line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=10, 1<=≤<=*k*<=≤<=240) — the number of the problems in the contest and the number of minutes Limak needs to get to the party from his house.
|
Print one integer, denoting the maximum possible number of problems Limak can solve so that he could get to the party at midnight or earlier.
|
[
"3 222\n",
"4 190\n",
"7 1\n"
] |
[
"2\n",
"4\n",
"7\n"
] |
In the first sample, there are 3 problems and Limak needs 222 minutes to get to the party. The three problems require 5, 10 and 15 minutes respectively. Limak can spend 5 + 10 = 15 minutes to solve first two problems. Then, at 20:15 he can leave his house to get to the party at 23:57 (after 222 minutes). In this scenario Limak would solve 2 problems. He doesn't have enough time to solve 3 problems so the answer is 2.
In the second sample, Limak can solve all 4 problems in 5 + 10 + 15 + 20 = 50 minutes. At 20:50 he will leave the house and go to the party. He will get there exactly at midnight.
In the third sample, Limak needs only 1 minute to get to the party. He has enough time to solve all 7 problems.
| 500
|
[
{
"input": "3 222",
"output": "2"
},
{
"input": "4 190",
"output": "4"
},
{
"input": "7 1",
"output": "7"
},
{
"input": "10 135",
"output": "6"
},
{
"input": "10 136",
"output": "5"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 240",
"output": "0"
},
{
"input": "10 1",
"output": "9"
},
{
"input": "10 240",
"output": "0"
},
{
"input": "9 240",
"output": "0"
},
{
"input": "9 1",
"output": "9"
},
{
"input": "9 235",
"output": "1"
},
{
"input": "9 236",
"output": "0"
},
{
"input": "5 225",
"output": "2"
},
{
"input": "5 226",
"output": "1"
},
{
"input": "4 210",
"output": "3"
},
{
"input": "4 211",
"output": "2"
},
{
"input": "4 191",
"output": "3"
},
{
"input": "10 165",
"output": "5"
},
{
"input": "10 166",
"output": "4"
},
{
"input": "8 100",
"output": "7"
},
{
"input": "8 101",
"output": "6"
},
{
"input": "8 60",
"output": "8"
},
{
"input": "8 61",
"output": "7"
},
{
"input": "10 15",
"output": "9"
},
{
"input": "10 16",
"output": "8"
},
{
"input": "4 100",
"output": "4"
},
{
"input": "4 101",
"output": "4"
},
{
"input": "7 167",
"output": "4"
},
{
"input": "10 164",
"output": "5"
},
{
"input": "9 170",
"output": "4"
},
{
"input": "8 160",
"output": "5"
},
{
"input": "1 100",
"output": "1"
},
{
"input": "8 123",
"output": "6"
},
{
"input": "2 99",
"output": "2"
},
{
"input": "10 88",
"output": "7"
},
{
"input": "1 235",
"output": "1"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "4 240",
"output": "0"
},
{
"input": "1 55",
"output": "1"
},
{
"input": "1 240",
"output": "0"
},
{
"input": "3 240",
"output": "0"
},
{
"input": "10 240",
"output": "0"
},
{
"input": "2 240",
"output": "0"
},
{
"input": "10 1",
"output": "9"
},
{
"input": "9 1",
"output": "9"
},
{
"input": "2 236",
"output": "0"
},
{
"input": "10 2",
"output": "9"
},
{
"input": "3 239",
"output": "0"
},
{
"input": "1 237",
"output": "0"
},
{
"input": "7 8",
"output": "7"
},
{
"input": "10 235",
"output": "1"
}
] | 1,690,557,571
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 6
| 93
| 0
|
take = input().split()
time = (1440 - (int(take[1]) + 1200)) // 5
arr = []
summation = 0
for i in range(1, 11):
summation += i
arr.append(summation)
saved = 0
for i in range(len(arr)):
if arr[i] <= time:
saved = i
print(min(saved + 1, int(take[0])))
|
Title: New Year and Hurry
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Limak is going to participate in a contest on the last day of the 2016. The contest will start at 20:00 and will last four hours, exactly until midnight. There will be *n* problems, sorted by difficulty, i.e. problem 1 is the easiest and problem *n* is the hardest. Limak knows it will take him 5·*i* minutes to solve the *i*-th problem.
Limak's friends organize a New Year's Eve party and Limak wants to be there at midnight or earlier. He needs *k* minutes to get there from his house, where he will participate in the contest first.
How many problems can Limak solve if he wants to make it to the party?
Input Specification:
The only line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=10, 1<=≤<=*k*<=≤<=240) — the number of the problems in the contest and the number of minutes Limak needs to get to the party from his house.
Output Specification:
Print one integer, denoting the maximum possible number of problems Limak can solve so that he could get to the party at midnight or earlier.
Demo Input:
['3 222\n', '4 190\n', '7 1\n']
Demo Output:
['2\n', '4\n', '7\n']
Note:
In the first sample, there are 3 problems and Limak needs 222 minutes to get to the party. The three problems require 5, 10 and 15 minutes respectively. Limak can spend 5 + 10 = 15 minutes to solve first two problems. Then, at 20:15 he can leave his house to get to the party at 23:57 (after 222 minutes). In this scenario Limak would solve 2 problems. He doesn't have enough time to solve 3 problems so the answer is 2.
In the second sample, Limak can solve all 4 problems in 5 + 10 + 15 + 20 = 50 minutes. At 20:50 he will leave the house and go to the party. He will get there exactly at midnight.
In the third sample, Limak needs only 1 minute to get to the party. He has enough time to solve all 7 problems.
|
```python
take = input().split()
time = (1440 - (int(take[1]) + 1200)) // 5
arr = []
summation = 0
for i in range(1, 11):
summation += i
arr.append(summation)
saved = 0
for i in range(len(arr)):
if arr[i] <= time:
saved = i
print(min(saved + 1, int(take[0])))
```
| 0
|
|
909
|
A
|
Generate Login
|
PROGRAMMING
| 1,000
|
[
"brute force",
"greedy",
"sortings"
] | null | null |
The preferred way to generate user login in Polygon is to concatenate a prefix of the user's first name and a prefix of their last name, in that order. Each prefix must be non-empty, and any of the prefixes can be the full name. Typically there are multiple possible logins for each person.
You are given the first and the last name of a user. Return the alphabetically earliest login they can get (regardless of other potential Polygon users).
As a reminder, a prefix of a string *s* is its substring which occurs at the beginning of *s*: "a", "ab", "abc" etc. are prefixes of string "{abcdef}" but "b" and 'bc" are not. A string *a* is alphabetically earlier than a string *b*, if *a* is a prefix of *b*, or *a* and *b* coincide up to some position, and then *a* has a letter that is alphabetically earlier than the corresponding letter in *b*: "a" and "ab" are alphabetically earlier than "ac" but "b" and "ba" are alphabetically later than "ac".
|
The input consists of a single line containing two space-separated strings: the first and the last names. Each character of each string is a lowercase English letter. The length of each string is between 1 and 10, inclusive.
|
Output a single string — alphabetically earliest possible login formed from these names. The output should be given in lowercase as well.
|
[
"harry potter\n",
"tom riddle\n"
] |
[
"hap\n",
"tomr\n"
] |
none
| 500
|
[
{
"input": "harry potter",
"output": "hap"
},
{
"input": "tom riddle",
"output": "tomr"
},
{
"input": "a qdpinbmcrf",
"output": "aq"
},
{
"input": "wixjzniiub ssdfodfgap",
"output": "wis"
},
{
"input": "z z",
"output": "zz"
},
{
"input": "ertuyivhfg v",
"output": "ertuv"
},
{
"input": "asdfghjkli ware",
"output": "asdfghjkliw"
},
{
"input": "udggmyop ze",
"output": "udggmyopz"
},
{
"input": "fapkdme rtzxovx",
"output": "fapkdmer"
},
{
"input": "mybiqxmnqq l",
"output": "ml"
},
{
"input": "dtbqya fyyymv",
"output": "df"
},
{
"input": "fyclu zokbxiahao",
"output": "fycluz"
},
{
"input": "qngatnviv rdych",
"output": "qngar"
},
{
"input": "ttvnhrnng lqkfulhrn",
"output": "tl"
},
{
"input": "fya fgx",
"output": "ff"
},
{
"input": "nuis zvjjqlre",
"output": "nuisz"
},
{
"input": "ly qtsmze",
"output": "lq"
},
{
"input": "d kgfpjsurfw",
"output": "dk"
},
{
"input": "lwli ewrpu",
"output": "le"
},
{
"input": "rr wldsfubcs",
"output": "rrw"
},
{
"input": "h qart",
"output": "hq"
},
{
"input": "vugvblnzx kqdwdulm",
"output": "vk"
},
{
"input": "xohesmku ef",
"output": "xe"
},
{
"input": "twvvsl wtcyawv",
"output": "tw"
},
{
"input": "obljndajv q",
"output": "obljndajq"
},
{
"input": "jjxwj kxccwx",
"output": "jjk"
},
{
"input": "sk fftzmv",
"output": "sf"
},
{
"input": "cgpegngs aufzxkyyrw",
"output": "ca"
},
{
"input": "reyjzjdvq skuch",
"output": "res"
},
{
"input": "ardaae mxgdulijf",
"output": "am"
},
{
"input": "bgopsdfji uaps",
"output": "bgopsdfjiu"
},
{
"input": "amolfed pun",
"output": "amolfedp"
},
{
"input": "badkiln yort",
"output": "badkilny"
},
{
"input": "aaaaaaaaaz york",
"output": "aaaaaaaaay"
},
{
"input": "bbbbcbbbbd c",
"output": "bbbbc"
},
{
"input": "aa ab",
"output": "aa"
},
{
"input": "ab b",
"output": "ab"
},
{
"input": "aaaaa ab",
"output": "aa"
},
{
"input": "aa a",
"output": "aa"
},
{
"input": "aba b",
"output": "ab"
},
{
"input": "aaaaaaa aaaaaa",
"output": "aa"
},
{
"input": "a a",
"output": "aa"
},
{
"input": "a aa",
"output": "aa"
},
{
"input": "a b",
"output": "ab"
},
{
"input": "b a",
"output": "ba"
},
{
"input": "z a",
"output": "za"
},
{
"input": "aaa a",
"output": "aa"
},
{
"input": "aa aa",
"output": "aa"
},
{
"input": "a aaa",
"output": "aa"
},
{
"input": "aaaaaaaaaa aaaaaaaaaa",
"output": "aa"
},
{
"input": "aaaaaaaaaa a",
"output": "aa"
},
{
"input": "a aaaaaaaaaa",
"output": "aa"
},
{
"input": "zzaa b",
"output": "zb"
},
{
"input": "ca cf",
"output": "cac"
},
{
"input": "abhi ia",
"output": "abhi"
},
{
"input": "aaaa aaaab",
"output": "aa"
},
{
"input": "aar raa",
"output": "aar"
},
{
"input": "harry hotter",
"output": "hah"
},
{
"input": "aaaaaaa a",
"output": "aa"
},
{
"input": "apple pie",
"output": "ap"
},
{
"input": "aaa aaa",
"output": "aa"
},
{
"input": "kabc buba",
"output": "kab"
},
{
"input": "asd ss",
"output": "as"
},
{
"input": "bbb b",
"output": "bb"
}
] | 1,647,355,953
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 31
| 0
|
a,b=input().split()
k=[a,b]
k="".join([str(i) for i in k])
k=sorted(k)
for i in a:
if a.index(i)>k.index(i):
res=a[:(a.index(i))+1]
print(res+b[0])
exit()
print(a+b[0])
|
Title: Generate Login
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The preferred way to generate user login in Polygon is to concatenate a prefix of the user's first name and a prefix of their last name, in that order. Each prefix must be non-empty, and any of the prefixes can be the full name. Typically there are multiple possible logins for each person.
You are given the first and the last name of a user. Return the alphabetically earliest login they can get (regardless of other potential Polygon users).
As a reminder, a prefix of a string *s* is its substring which occurs at the beginning of *s*: "a", "ab", "abc" etc. are prefixes of string "{abcdef}" but "b" and 'bc" are not. A string *a* is alphabetically earlier than a string *b*, if *a* is a prefix of *b*, or *a* and *b* coincide up to some position, and then *a* has a letter that is alphabetically earlier than the corresponding letter in *b*: "a" and "ab" are alphabetically earlier than "ac" but "b" and "ba" are alphabetically later than "ac".
Input Specification:
The input consists of a single line containing two space-separated strings: the first and the last names. Each character of each string is a lowercase English letter. The length of each string is between 1 and 10, inclusive.
Output Specification:
Output a single string — alphabetically earliest possible login formed from these names. The output should be given in lowercase as well.
Demo Input:
['harry potter\n', 'tom riddle\n']
Demo Output:
['hap\n', 'tomr\n']
Note:
none
|
```python
a,b=input().split()
k=[a,b]
k="".join([str(i) for i in k])
k=sorted(k)
for i in a:
if a.index(i)>k.index(i):
res=a[:(a.index(i))+1]
print(res+b[0])
exit()
print(a+b[0])
```
| 0
|
|
903
|
A
|
Hungry Student Problem
|
PROGRAMMING
| 900
|
[
"greedy",
"implementation"
] | null | null |
Ivan's classes at the university have just finished, and now he wants to go to the local CFK cafe and eat some fried chicken.
CFK sells chicken chunks in small and large portions. A small portion contains 3 chunks; a large one — 7 chunks. Ivan wants to eat exactly *x* chunks. Now he wonders whether he can buy exactly this amount of chicken.
Formally, Ivan wants to know if he can choose two non-negative integers *a* and *b* in such a way that *a* small portions and *b* large ones contain exactly *x* chunks.
Help Ivan to answer this question for several values of *x*!
|
The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the number of testcases.
The *i*-th of the following *n* lines contains one integer *x**i* (1<=≤<=*x**i*<=≤<=100) — the number of chicken chunks Ivan wants to eat.
|
Print *n* lines, in *i*-th line output YES if Ivan can buy exactly *x**i* chunks. Otherwise, print NO.
|
[
"2\n6\n5\n"
] |
[
"YES\nNO\n"
] |
In the first example Ivan can buy two small portions.
In the second example Ivan cannot buy exactly 5 chunks, since one small portion is not enough, but two small portions or one large is too much.
| 0
|
[
{
"input": "2\n6\n5",
"output": "YES\nNO"
},
{
"input": "100\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n11\n12\n13\n14\n15\n16\n17\n18\n19\n20\n21\n22\n23\n24\n25\n26\n27\n28\n29\n30\n31\n32\n33\n34\n35\n36\n37\n38\n39\n40\n41\n42\n43\n44\n45\n46\n47\n48\n49\n50\n51\n52\n53\n54\n55\n56\n57\n58\n59\n60\n61\n62\n63\n64\n65\n66\n67\n68\n69\n70\n71\n72\n73\n74\n75\n76\n77\n78\n79\n80\n81\n82\n83\n84\n85\n86\n87\n88\n89\n90\n91\n92\n93\n94\n95\n96\n97\n98\n99\n100",
"output": "NO\nNO\nYES\nNO\nNO\nYES\nYES\nNO\nYES\nYES\nNO\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES"
},
{
"input": "3\n6\n6\n6",
"output": "YES\nYES\nYES"
},
{
"input": "47\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1",
"output": "NO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO"
},
{
"input": "3\n1\n52\n76",
"output": "NO\nYES\nYES"
},
{
"input": "87\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100",
"output": "YES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES"
},
{
"input": "3\n3\n2\n1",
"output": "YES\nNO\nNO"
},
{
"input": "100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100",
"output": "YES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES"
}
] | 1,587,819,974
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 8
| 108
| 0
|
n = int(input())
for _ in range(n):
pieces=int(input())
ans="NO"
for i in range (0, pieces // 7+1):
if (pieces - i*7)%3==0:
ans="YES"
break
print(ans)
|
Title: Hungry Student Problem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ivan's classes at the university have just finished, and now he wants to go to the local CFK cafe and eat some fried chicken.
CFK sells chicken chunks in small and large portions. A small portion contains 3 chunks; a large one — 7 chunks. Ivan wants to eat exactly *x* chunks. Now he wonders whether he can buy exactly this amount of chicken.
Formally, Ivan wants to know if he can choose two non-negative integers *a* and *b* in such a way that *a* small portions and *b* large ones contain exactly *x* chunks.
Help Ivan to answer this question for several values of *x*!
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the number of testcases.
The *i*-th of the following *n* lines contains one integer *x**i* (1<=≤<=*x**i*<=≤<=100) — the number of chicken chunks Ivan wants to eat.
Output Specification:
Print *n* lines, in *i*-th line output YES if Ivan can buy exactly *x**i* chunks. Otherwise, print NO.
Demo Input:
['2\n6\n5\n']
Demo Output:
['YES\nNO\n']
Note:
In the first example Ivan can buy two small portions.
In the second example Ivan cannot buy exactly 5 chunks, since one small portion is not enough, but two small portions or one large is too much.
|
```python
n = int(input())
for _ in range(n):
pieces=int(input())
ans="NO"
for i in range (0, pieces // 7+1):
if (pieces - i*7)%3==0:
ans="YES"
break
print(ans)
```
| 3
|
|
980
|
B
|
Marlin
|
PROGRAMMING
| 1,600
|
[
"constructive algorithms"
] | null | null |
The city of Fishtopia can be imagined as a grid of $4$ rows and an odd number of columns. It has two main villages; the first is located at the top-left cell $(1,1)$, people who stay there love fishing at the Tuna pond at the bottom-right cell $(4, n)$. The second village is located at $(4, 1)$ and its people love the Salmon pond at $(1, n)$.
The mayor of Fishtopia wants to place $k$ hotels in the city, each one occupying one cell. To allow people to enter the city from anywhere, hotels should not be placed on the border cells.
A person can move from one cell to another if those cells are not occupied by hotels and share a side.
Can you help the mayor place the hotels in a way such that there are equal number of shortest paths from each village to its preferred pond?
|
The first line of input contain two integers, $n$ and $k$ ($3 \leq n \leq 99$, $0 \leq k \leq 2\times(n-2)$), $n$ is odd, the width of the city, and the number of hotels to be placed, respectively.
|
Print "YES", if it is possible to place all the hotels in a way that satisfies the problem statement, otherwise print "NO".
If it is possible, print an extra $4$ lines that describe the city, each line should have $n$ characters, each of which is "#" if that cell has a hotel on it, or "." if not.
|
[
"7 2\n",
"5 3\n"
] |
[
"YES\n.......\n.#.....\n.#.....\n.......\n",
"YES\n.....\n.###.\n.....\n.....\n"
] |
none
| 1,000
|
[
{
"input": "7 2",
"output": "YES\n.......\n.#.....\n.#.....\n......."
},
{
"input": "5 3",
"output": "YES\n.....\n.###.\n.....\n....."
},
{
"input": "3 2",
"output": "YES\n...\n.#.\n.#.\n..."
},
{
"input": "3 0",
"output": "YES\n...\n...\n...\n..."
},
{
"input": "49 1",
"output": "YES\n.................................................\n........................#........................\n.................................................\n................................................."
},
{
"input": "9 4",
"output": "YES\n.........\n.##......\n.##......\n........."
},
{
"input": "9 5",
"output": "YES\n.........\n.#.#.....\n.###.....\n........."
},
{
"input": "99 193",
"output": "YES\n...................................................................................................\n.###############################################################################################.#.\n.#################################################################################################.\n..................................................................................................."
},
{
"input": "99 14",
"output": "YES\n...................................................................................................\n.#######...........................................................................................\n.#######...........................................................................................\n..................................................................................................."
},
{
"input": "57 15",
"output": "YES\n.........................................................\n.######.#................................................\n.########................................................\n........................................................."
},
{
"input": "99 3",
"output": "YES\n...................................................................................................\n................................................###................................................\n...................................................................................................\n..................................................................................................."
},
{
"input": "3 1",
"output": "YES\n...\n.#.\n...\n..."
},
{
"input": "9 9",
"output": "YES\n.........\n.###.#...\n.#####...\n........."
},
{
"input": "67 9",
"output": "YES\n...................................................................\n.###.#.............................................................\n.#####.............................................................\n..................................................................."
},
{
"input": "99 99",
"output": "YES\n...................................................................................................\n.################################################.#................................................\n.##################################################................................................\n..................................................................................................."
},
{
"input": "31 32",
"output": "YES\n...............................\n.################..............\n.################..............\n..............................."
},
{
"input": "5 1",
"output": "YES\n.....\n..#..\n.....\n....."
},
{
"input": "5 2",
"output": "YES\n.....\n.#...\n.#...\n....."
},
{
"input": "5 4",
"output": "YES\n.....\n.##..\n.##..\n....."
},
{
"input": "5 6",
"output": "YES\n.....\n.###.\n.###.\n....."
},
{
"input": "5 5",
"output": "YES\n.....\n.#.#.\n.###.\n....."
},
{
"input": "7 9",
"output": "YES\n.......\n.###.#.\n.#####.\n......."
},
{
"input": "7 10",
"output": "YES\n.......\n.#####.\n.#####.\n......."
},
{
"input": "19 12",
"output": "YES\n...................\n.######............\n.######............\n..................."
},
{
"input": "19 3",
"output": "YES\n...................\n........###........\n...................\n..................."
},
{
"input": "37 14",
"output": "YES\n.....................................\n.#######.............................\n.#######.............................\n....................................."
},
{
"input": "37 15",
"output": "YES\n.....................................\n.######.#............................\n.########............................\n....................................."
},
{
"input": "37 37",
"output": "YES\n.....................................\n.#################.#.................\n.###################.................\n....................................."
},
{
"input": "37 36",
"output": "YES\n.....................................\n.##################..................\n.##################..................\n....................................."
},
{
"input": "37 35",
"output": "YES\n.....................................\n.################.#..................\n.##################..................\n....................................."
},
{
"input": "37 34",
"output": "YES\n.....................................\n.#################...................\n.#################...................\n....................................."
},
{
"input": "37 38",
"output": "YES\n.....................................\n.###################.................\n.###################.................\n....................................."
},
{
"input": "37 39",
"output": "YES\n.....................................\n.##################.#................\n.####################................\n....................................."
},
{
"input": "37 40",
"output": "YES\n.....................................\n.####################................\n.####################................\n....................................."
},
{
"input": "5 0",
"output": "YES\n.....\n.....\n.....\n....."
},
{
"input": "67 1",
"output": "YES\n...................................................................\n.................................#.................................\n...................................................................\n..................................................................."
},
{
"input": "37 19",
"output": "YES\n.....................................\n.########.#..........................\n.##########..........................\n....................................."
},
{
"input": "77 7",
"output": "YES\n.............................................................................\n.##.#........................................................................\n.####........................................................................\n............................................................................."
},
{
"input": "33 47",
"output": "YES\n.................................\n.######################.#........\n.########################........\n................................."
},
{
"input": "33 48",
"output": "YES\n.................................\n.########################........\n.########################........\n................................."
},
{
"input": "23 40",
"output": "YES\n.......................\n.####################..\n.####################..\n......................."
},
{
"input": "23 39",
"output": "YES\n.......................\n.##################.#..\n.####################..\n......................."
},
{
"input": "49 3",
"output": "YES\n.................................................\n.......................###.......................\n.................................................\n................................................."
},
{
"input": "99 1",
"output": "YES\n...................................................................................................\n.................................................#.................................................\n...................................................................................................\n..................................................................................................."
},
{
"input": "77 0",
"output": "YES\n.............................................................................\n.............................................................................\n.............................................................................\n............................................................................."
},
{
"input": "99 0",
"output": "YES\n...................................................................................................\n...................................................................................................\n...................................................................................................\n..................................................................................................."
},
{
"input": "99 5",
"output": "YES\n...................................................................................................\n.#.#...............................................................................................\n.###...............................................................................................\n..................................................................................................."
},
{
"input": "99 4",
"output": "YES\n...................................................................................................\n.##................................................................................................\n.##................................................................................................\n..................................................................................................."
},
{
"input": "99 20",
"output": "YES\n...................................................................................................\n.##########........................................................................................\n.##########........................................................................................\n..................................................................................................."
},
{
"input": "99 194",
"output": "YES\n...................................................................................................\n.#################################################################################################.\n.#################################################################################################.\n..................................................................................................."
},
{
"input": "99 192",
"output": "YES\n...................................................................................................\n.################################################################################################..\n.################################################################################################..\n..................................................................................................."
},
{
"input": "99 190",
"output": "YES\n...................................................................................................\n.###############################################################################################...\n.###############################################################################################...\n..................................................................................................."
},
{
"input": "99 189",
"output": "YES\n...................................................................................................\n.#############################################################################################.#...\n.###############################################################################################...\n..................................................................................................."
},
{
"input": "99 177",
"output": "YES\n...................................................................................................\n.#######################################################################################.#.........\n.#########################################################################################.........\n..................................................................................................."
},
{
"input": "99 154",
"output": "YES\n...................................................................................................\n.#############################################################################.....................\n.#############################################################################.....................\n..................................................................................................."
},
{
"input": "99 127",
"output": "YES\n...................................................................................................\n.##############################################################.#..................................\n.################################################################..................................\n..................................................................................................."
},
{
"input": "99 55",
"output": "YES\n...................................................................................................\n.##########################.#......................................................................\n.############################......................................................................\n..................................................................................................."
},
{
"input": "99 40",
"output": "YES\n...................................................................................................\n.####################..............................................................................\n.####################..............................................................................\n..................................................................................................."
},
{
"input": "97 190",
"output": "YES\n.................................................................................................\n.###############################################################################################.\n.###############################################################################################.\n................................................................................................."
},
{
"input": "97 100",
"output": "YES\n.................................................................................................\n.##################################################..............................................\n.##################################################..............................................\n................................................................................................."
},
{
"input": "97 111",
"output": "YES\n.................................................................................................\n.######################################################.#........................................\n.########################################################........................................\n................................................................................................."
},
{
"input": "97 64",
"output": "YES\n.................................................................................................\n.################################................................................................\n.################################................................................................\n................................................................................................."
},
{
"input": "97 77",
"output": "YES\n.................................................................................................\n.#####################################.#.........................................................\n.#######################################.........................................................\n................................................................................................."
},
{
"input": "91 77",
"output": "YES\n...........................................................................................\n.#####################################.#...................................................\n.#######################################...................................................\n..........................................................................................."
},
{
"input": "91 128",
"output": "YES\n...........................................................................................\n.################################################################..........................\n.################################################################..........................\n..........................................................................................."
},
{
"input": "91 113",
"output": "YES\n...........................................................................................\n.#######################################################.#.................................\n.#########################################################.................................\n..........................................................................................."
},
{
"input": "55 55",
"output": "YES\n.......................................................\n.##########################.#..........................\n.############################..........................\n......................................................."
},
{
"input": "43 34",
"output": "YES\n...........................................\n.#################.........................\n.#################.........................\n..........................................."
},
{
"input": "13 21",
"output": "YES\n.............\n.#########.#.\n.###########.\n............."
},
{
"input": "27 50",
"output": "YES\n...........................\n.#########################.\n.#########################.\n..........................."
},
{
"input": "27 49",
"output": "YES\n...........................\n.#######################.#.\n.#########################.\n..........................."
},
{
"input": "27 48",
"output": "YES\n...........................\n.########################..\n.########################..\n..........................."
},
{
"input": "27 40",
"output": "YES\n...........................\n.####################......\n.####################......\n..........................."
},
{
"input": "87 80",
"output": "YES\n.......................................................................................\n.########################################..............................................\n.########################################..............................................\n......................................................................................."
},
{
"input": "69 17",
"output": "YES\n.....................................................................\n.#######.#...........................................................\n.#########...........................................................\n....................................................................."
},
{
"input": "39 73",
"output": "YES\n.......................................\n.###################################.#.\n.#####################################.\n......................................."
},
{
"input": "39 72",
"output": "YES\n.......................................\n.####################################..\n.####################################..\n......................................."
},
{
"input": "33 57",
"output": "YES\n.................................\n.###########################.#...\n.#############################...\n................................."
},
{
"input": "33 54",
"output": "YES\n.................................\n.###########################.....\n.###########################.....\n................................."
},
{
"input": "99 2",
"output": "YES\n...................................................................................................\n.#.................................................................................................\n.#.................................................................................................\n..................................................................................................."
},
{
"input": "99 7",
"output": "YES\n...................................................................................................\n.##.#..............................................................................................\n.####..............................................................................................\n..................................................................................................."
},
{
"input": "99 37",
"output": "YES\n...................................................................................................\n.#################.#...............................................................................\n.###################...............................................................................\n..................................................................................................."
},
{
"input": "57 67",
"output": "YES\n.........................................................\n.################################.#......................\n.##################################......................\n........................................................."
},
{
"input": "57 34",
"output": "YES\n.........................................................\n.#################.......................................\n.#################.......................................\n........................................................."
},
{
"input": "73 17",
"output": "YES\n.........................................................................\n.#######.#...............................................................\n.#########...............................................................\n........................................................................."
},
{
"input": "73 43",
"output": "YES\n.........................................................................\n.####################.#..................................................\n.######################..................................................\n........................................................................."
},
{
"input": "7 3",
"output": "YES\n.......\n..###..\n.......\n......."
},
{
"input": "7 7",
"output": "YES\n.......\n.##.#..\n.####..\n......."
}
] | 1,525,794,639
| 2,739
|
PyPy 3
|
WRONG_ANSWER
|
PRETESTS
| 0
| 93
| 21,401,600
|
n, k = [int(x) for x in input().split()]
if k <= 2 * n - 4:
t = (k % (n - 2))
if t == 0:
t = n - 2
print("YES")
print("."*n)
print("." * int((n - t)/2) + "#" * t + "." * int((n - t)% 2 +(n - t)/2))
t = k - t
print("." * int((n - t)/2) + "#" * t + "." * int((n - t)% 2 +(n - t)/2))
print("."*n)
else:
print("NO")
|
Title: Marlin
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The city of Fishtopia can be imagined as a grid of $4$ rows and an odd number of columns. It has two main villages; the first is located at the top-left cell $(1,1)$, people who stay there love fishing at the Tuna pond at the bottom-right cell $(4, n)$. The second village is located at $(4, 1)$ and its people love the Salmon pond at $(1, n)$.
The mayor of Fishtopia wants to place $k$ hotels in the city, each one occupying one cell. To allow people to enter the city from anywhere, hotels should not be placed on the border cells.
A person can move from one cell to another if those cells are not occupied by hotels and share a side.
Can you help the mayor place the hotels in a way such that there are equal number of shortest paths from each village to its preferred pond?
Input Specification:
The first line of input contain two integers, $n$ and $k$ ($3 \leq n \leq 99$, $0 \leq k \leq 2\times(n-2)$), $n$ is odd, the width of the city, and the number of hotels to be placed, respectively.
Output Specification:
Print "YES", if it is possible to place all the hotels in a way that satisfies the problem statement, otherwise print "NO".
If it is possible, print an extra $4$ lines that describe the city, each line should have $n$ characters, each of which is "#" if that cell has a hotel on it, or "." if not.
Demo Input:
['7 2\n', '5 3\n']
Demo Output:
['YES\n.......\n.#.....\n.#.....\n.......\n', 'YES\n.....\n.###.\n.....\n.....\n']
Note:
none
|
```python
n, k = [int(x) for x in input().split()]
if k <= 2 * n - 4:
t = (k % (n - 2))
if t == 0:
t = n - 2
print("YES")
print("."*n)
print("." * int((n - t)/2) + "#" * t + "." * int((n - t)% 2 +(n - t)/2))
t = k - t
print("." * int((n - t)/2) + "#" * t + "." * int((n - t)% 2 +(n - t)/2))
print("."*n)
else:
print("NO")
```
| 0
|
|
932
|
E
|
Team Work
|
PROGRAMMING
| 2,400
|
[
"combinatorics",
"dp",
"math"
] | null | null |
You have a team of *N* people. For a particular task, you can pick any non-empty subset of people. The cost of having *x* people for the task is *x**k*.
Output the sum of costs over all non-empty subsets of people.
|
Only line of input contains two integers *N* (1<=≤<=*N*<=≤<=109) representing total number of people and *k* (1<=≤<=*k*<=≤<=5000).
|
Output the sum of costs for all non empty subsets modulo 109<=+<=7.
|
[
"1 1\n",
"3 2\n"
] |
[
"1\n",
"24\n"
] |
In the first example, there is only one non-empty subset {1} with cost 1<sup class="upper-index">1</sup> = 1.
In the second example, there are seven non-empty subsets.
- {1} with cost 1<sup class="upper-index">2</sup> = 1
- {2} with cost 1<sup class="upper-index">2</sup> = 1
- {1, 2} with cost 2<sup class="upper-index">2</sup> = 4
- {3} with cost 1<sup class="upper-index">2</sup> = 1
- {1, 3} with cost 2<sup class="upper-index">2</sup> = 4
- {2, 3} with cost 2<sup class="upper-index">2</sup> = 4
- {1, 2, 3} with cost 3<sup class="upper-index">2</sup> = 9
The total cost is 1 + 1 + 4 + 1 + 4 + 4 + 9 = 24.
| 2,500
|
[
{
"input": "1 1",
"output": "1"
},
{
"input": "3 2",
"output": "24"
},
{
"input": "5 3",
"output": "800"
},
{
"input": "12 4",
"output": "8067072"
},
{
"input": "20 5",
"output": "87486873"
},
{
"input": "522 4575",
"output": "558982611"
},
{
"input": "1426 4445",
"output": "503519668"
},
{
"input": "81 3772",
"output": "420178413"
},
{
"input": "629 3447",
"output": "989788663"
},
{
"input": "2202 3497",
"output": "682330518"
},
{
"input": "2775 4325",
"output": "434053861"
},
{
"input": "3982 4784",
"output": "987043323"
},
{
"input": "2156 3417",
"output": "216656956"
},
{
"input": "902 1932",
"output": "78732216"
},
{
"input": "728 3537",
"output": "957098547"
},
{
"input": "739 3857",
"output": "836213774"
},
{
"input": "1918 4211",
"output": "972992457"
},
{
"input": "3506 4679",
"output": "130374558"
},
{
"input": "1000000000 5000",
"output": "642932262"
},
{
"input": "2500 5000",
"output": "416584034"
},
{
"input": "158260522 4575",
"output": "875142289"
},
{
"input": "602436426 4445",
"output": "582490088"
},
{
"input": "861648772 81",
"output": "939143440"
},
{
"input": "433933447 629",
"output": "396606775"
},
{
"input": "262703497 2202",
"output": "813734619"
},
{
"input": "971407775 4325",
"output": "905271522"
},
{
"input": "731963982 4784",
"output": "7722713"
},
{
"input": "450968417 2156",
"output": "634922960"
},
{
"input": "982631932 902",
"output": "262226561"
},
{
"input": "880895728 3537",
"output": "266659411"
},
{
"input": "4483 4938",
"output": "371059472"
},
{
"input": "4278 3849",
"output": "183616686"
},
{
"input": "3281 4798",
"output": "467929252"
},
{
"input": "12195 4781",
"output": "628055652"
},
{
"input": "5092 4809",
"output": "587575377"
},
{
"input": "2511 4990",
"output": "622898200"
},
{
"input": "9896 4771",
"output": "388524304"
},
{
"input": "493 4847",
"output": "414977957"
},
{
"input": "137 4733",
"output": "279404197"
},
{
"input": "6399 4957",
"output": "639782892"
},
{
"input": "999999376 642",
"output": "842765934"
},
{
"input": "999997777 645",
"output": "31545099"
},
{
"input": "999998604 448",
"output": "642283867"
},
{
"input": "999974772 208",
"output": "268825720"
},
{
"input": "999980457 228",
"output": "848255312"
},
{
"input": "999999335 1040",
"output": "585378634"
},
{
"input": "999976125 157",
"output": "300682474"
},
{
"input": "999974335 786",
"output": "754709460"
},
{
"input": "999985549 266",
"output": "607440620"
},
{
"input": "999999648 34",
"output": "378413808"
}
] | 1,604,955,302
| 2,147,483,647
|
PyPy 3
|
RUNTIME_ERROR
|
TESTS
| 18
| 218
| 2,969,600
|
from bisect import bisect_left as bl
from bisect import bisect_right as br
from heapq import heappush,heappop
import math
from collections import *
from functools import reduce,cmp_to_key,lru_cache
import io, os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
import sys
# input = sys.stdin.readline
M = mod = 10 ** 9 + 7
def factors(n):return sorted(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
def inv_mod(n):return pow(n, mod - 2, mod)
def li():return [int(i) for i in input().rstrip().split()]
def st():return str(input().rstrip())[2:-1]
def val():return int(input().rstrip())
def li2():return [str(i)[2:-1] for i in input().rstrip().split()]
def li3():return [int(i) for i in st()]
N = 5000 + 10
facts = [1]
for i in range(1, N):
facts.append(facts[-1] * i % mod)
inversemods = [0 for i in range(N)]
inversemods[-1] = inv_mod(facts[-1]) % mod
for i in range(N - 2, -1, -1):
inversemods[i] = inversemods[i + 1] * (i + 1) % mod
def ncr(n, r):
if n < r:return 0
return facts[n] * inversemods[n - r] * inversemods[r] % mod
n, k = li()
ans = 0
for i in range(1, n + 1):
ans = (ans + ncr(n, i) * pow(i, k, mod) ) % mod
print(ans)
|
Title: Team Work
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have a team of *N* people. For a particular task, you can pick any non-empty subset of people. The cost of having *x* people for the task is *x**k*.
Output the sum of costs over all non-empty subsets of people.
Input Specification:
Only line of input contains two integers *N* (1<=≤<=*N*<=≤<=109) representing total number of people and *k* (1<=≤<=*k*<=≤<=5000).
Output Specification:
Output the sum of costs for all non empty subsets modulo 109<=+<=7.
Demo Input:
['1 1\n', '3 2\n']
Demo Output:
['1\n', '24\n']
Note:
In the first example, there is only one non-empty subset {1} with cost 1<sup class="upper-index">1</sup> = 1.
In the second example, there are seven non-empty subsets.
- {1} with cost 1<sup class="upper-index">2</sup> = 1
- {2} with cost 1<sup class="upper-index">2</sup> = 1
- {1, 2} with cost 2<sup class="upper-index">2</sup> = 4
- {3} with cost 1<sup class="upper-index">2</sup> = 1
- {1, 3} with cost 2<sup class="upper-index">2</sup> = 4
- {2, 3} with cost 2<sup class="upper-index">2</sup> = 4
- {1, 2, 3} with cost 3<sup class="upper-index">2</sup> = 9
The total cost is 1 + 1 + 4 + 1 + 4 + 4 + 9 = 24.
|
```python
from bisect import bisect_left as bl
from bisect import bisect_right as br
from heapq import heappush,heappop
import math
from collections import *
from functools import reduce,cmp_to_key,lru_cache
import io, os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
import sys
# input = sys.stdin.readline
M = mod = 10 ** 9 + 7
def factors(n):return sorted(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
def inv_mod(n):return pow(n, mod - 2, mod)
def li():return [int(i) for i in input().rstrip().split()]
def st():return str(input().rstrip())[2:-1]
def val():return int(input().rstrip())
def li2():return [str(i)[2:-1] for i in input().rstrip().split()]
def li3():return [int(i) for i in st()]
N = 5000 + 10
facts = [1]
for i in range(1, N):
facts.append(facts[-1] * i % mod)
inversemods = [0 for i in range(N)]
inversemods[-1] = inv_mod(facts[-1]) % mod
for i in range(N - 2, -1, -1):
inversemods[i] = inversemods[i + 1] * (i + 1) % mod
def ncr(n, r):
if n < r:return 0
return facts[n] * inversemods[n - r] * inversemods[r] % mod
n, k = li()
ans = 0
for i in range(1, n + 1):
ans = (ans + ncr(n, i) * pow(i, k, mod) ) % mod
print(ans)
```
| -1
|
|
510
|
B
|
Fox And Two Dots
|
PROGRAMMING
| 1,500
|
[
"dfs and similar"
] | null | null |
Fox Ciel is playing a mobile puzzle game called "Two Dots". The basic levels are played on a board of size *n*<=×<=*m* cells, like this:
Each cell contains a dot that has some color. We will use different uppercase Latin characters to express different colors.
The key of this game is to find a cycle that contain dots of same color. Consider 4 blue dots on the picture forming a circle as an example. Formally, we call a sequence of dots *d*1,<=*d*2,<=...,<=*d**k* a cycle if and only if it meets the following condition:
1. These *k* dots are different: if *i*<=≠<=*j* then *d**i* is different from *d**j*. 1. *k* is at least 4. 1. All dots belong to the same color. 1. For all 1<=≤<=*i*<=≤<=*k*<=-<=1: *d**i* and *d**i*<=+<=1 are adjacent. Also, *d**k* and *d*1 should also be adjacent. Cells *x* and *y* are called adjacent if they share an edge.
Determine if there exists a cycle on the field.
|
The first line contains two integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=50): the number of rows and columns of the board.
Then *n* lines follow, each line contains a string consisting of *m* characters, expressing colors of dots in each line. Each character is an uppercase Latin letter.
|
Output "Yes" if there exists a cycle, and "No" otherwise.
|
[
"3 4\nAAAA\nABCA\nAAAA\n",
"3 4\nAAAA\nABCA\nAADA\n",
"4 4\nYYYR\nBYBY\nBBBY\nBBBY\n",
"7 6\nAAAAAB\nABBBAB\nABAAAB\nABABBB\nABAAAB\nABBBAB\nAAAAAB\n",
"2 13\nABCDEFGHIJKLM\nNOPQRSTUVWXYZ\n"
] |
[
"Yes\n",
"No\n",
"Yes\n",
"Yes\n",
"No\n"
] |
In first sample test all 'A' form a cycle.
In second sample there is no such cycle.
The third sample is displayed on the picture above ('Y' = Yellow, 'B' = Blue, 'R' = Red).
| 1,000
|
[
{
"input": "3 4\nAAAA\nABCA\nAAAA",
"output": "Yes"
},
{
"input": "3 4\nAAAA\nABCA\nAADA",
"output": "No"
},
{
"input": "4 4\nYYYR\nBYBY\nBBBY\nBBBY",
"output": "Yes"
},
{
"input": "7 6\nAAAAAB\nABBBAB\nABAAAB\nABABBB\nABAAAB\nABBBAB\nAAAAAB",
"output": "Yes"
},
{
"input": "2 13\nABCDEFGHIJKLM\nNOPQRSTUVWXYZ",
"output": "No"
},
{
"input": "2 2\nAA\nAA",
"output": "Yes"
},
{
"input": "2 2\nAA\nAB",
"output": "No"
},
{
"input": "3 3\nAAA\nABA\nAAA",
"output": "Yes"
},
{
"input": "3 3\nAAA\nABA\nABA",
"output": "No"
},
{
"input": "10 10\nEGFJGJKGEI\nAKJHBGHIHF\nJBABBCFGEJ\nCJDJHJJKBD\nKHJIKKGGEK\nHHJHKHGEKF\nEKFCAJGGDK\nAFKBBFICAA\nFEDFAGHEKA\nCAAGIFHGGI",
"output": "No"
},
{
"input": "10 10\nHIICQRHPUJ\nBCDUKHMBFK\nPFTUIDOBOE\nQQPITLRKUP\nERMUJMOSMF\nMRSICEILQB\nODIGFNCHFR\nGHIOAFLHJH\nFBLAQNGEIF\nFDLEGDUTNG",
"output": "No"
},
{
"input": "2 50\nDADCDBCCDAACDBCAACADBCBDBACCCCDADCBACADBCCBDBCCBCC\nDADAADCABBBACCDDBABBBDCBACBCCCCDDADCDABADDDCABACDB",
"output": "Yes"
},
{
"input": "50 2\nAA\nCD\nEE\nFC\nED\nAF\nFC\nAD\nBA\nAF\nBF\nDA\nAC\nFC\nFA\nBF\nAD\nBB\nDC\nAF\nAA\nAD\nEE\nED\nCD\nFC\nFB\nBB\nDD\nEB\nBE\nCF\nDE\nAE\nFD\nAB\nFB\nAE\nBE\nFA\nCF\nFB\nDE\nED\nAD\nFA\nBB\nBF\nDA\nEE",
"output": "No"
}
] | 1,675,144,964
| 2,147,483,647
|
PyPy 3-64
|
RUNTIME_ERROR
|
TESTS
| 9
| 124
| 8,089,600
|
n, m = [int(a) for a in input().split(" ")]
grid = []
visited = set()
for i in range(n):
grid.append(list(input()))
#print(DataFrame(grid))
exists_cycle = False
inverseMap ={
'down': 'up',
'up': 'down',
'left': 'right',
'right': 'left'
}
def dfs(i, j, direction, checkVal):
global exists_cycle
if i < 0 or i > n-1:
return
if j < 0 or j > m-1:
return
#print([i,j], direction, checkVal)
if grid[i][j] != checkVal:
return
if (i, j) in visited:
exists_cycle = True
return
directions = ['up', 'down', 'left', 'right']
iDirections = [-1, 1, 0, 0]
jDirections = [0, 0, -1, 1]
if direction in directions:
dIndex = directions.index(inverseMap[direction])
directions.pop(dIndex)
iDirections.pop(dIndex)
jDirections.pop(dIndex)
positions = [[i+iDirections[z], j+jDirections[z]] for z in range(len(iDirections))]
visited.add((i,j))
#print(positions)
for i in range(len(positions)):
newPos = [positions[i][0], positions[i][1]]
#print(newPos, directions[i])
dfs(newPos[0], newPos[1], directions[i], checkVal)
#print('---')
for i in range(n):
for j in range(m):
if (i,j) not in visited and not exists_cycle:
dfs(i,j,None, grid[i][j])
if exists_cycle:
print('Yes')
else:
print('No')
|
Title: Fox And Two Dots
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fox Ciel is playing a mobile puzzle game called "Two Dots". The basic levels are played on a board of size *n*<=×<=*m* cells, like this:
Each cell contains a dot that has some color. We will use different uppercase Latin characters to express different colors.
The key of this game is to find a cycle that contain dots of same color. Consider 4 blue dots on the picture forming a circle as an example. Formally, we call a sequence of dots *d*1,<=*d*2,<=...,<=*d**k* a cycle if and only if it meets the following condition:
1. These *k* dots are different: if *i*<=≠<=*j* then *d**i* is different from *d**j*. 1. *k* is at least 4. 1. All dots belong to the same color. 1. For all 1<=≤<=*i*<=≤<=*k*<=-<=1: *d**i* and *d**i*<=+<=1 are adjacent. Also, *d**k* and *d*1 should also be adjacent. Cells *x* and *y* are called adjacent if they share an edge.
Determine if there exists a cycle on the field.
Input Specification:
The first line contains two integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=50): the number of rows and columns of the board.
Then *n* lines follow, each line contains a string consisting of *m* characters, expressing colors of dots in each line. Each character is an uppercase Latin letter.
Output Specification:
Output "Yes" if there exists a cycle, and "No" otherwise.
Demo Input:
['3 4\nAAAA\nABCA\nAAAA\n', '3 4\nAAAA\nABCA\nAADA\n', '4 4\nYYYR\nBYBY\nBBBY\nBBBY\n', '7 6\nAAAAAB\nABBBAB\nABAAAB\nABABBB\nABAAAB\nABBBAB\nAAAAAB\n', '2 13\nABCDEFGHIJKLM\nNOPQRSTUVWXYZ\n']
Demo Output:
['Yes\n', 'No\n', 'Yes\n', 'Yes\n', 'No\n']
Note:
In first sample test all 'A' form a cycle.
In second sample there is no such cycle.
The third sample is displayed on the picture above ('Y' = Yellow, 'B' = Blue, 'R' = Red).
|
```python
n, m = [int(a) for a in input().split(" ")]
grid = []
visited = set()
for i in range(n):
grid.append(list(input()))
#print(DataFrame(grid))
exists_cycle = False
inverseMap ={
'down': 'up',
'up': 'down',
'left': 'right',
'right': 'left'
}
def dfs(i, j, direction, checkVal):
global exists_cycle
if i < 0 or i > n-1:
return
if j < 0 or j > m-1:
return
#print([i,j], direction, checkVal)
if grid[i][j] != checkVal:
return
if (i, j) in visited:
exists_cycle = True
return
directions = ['up', 'down', 'left', 'right']
iDirections = [-1, 1, 0, 0]
jDirections = [0, 0, -1, 1]
if direction in directions:
dIndex = directions.index(inverseMap[direction])
directions.pop(dIndex)
iDirections.pop(dIndex)
jDirections.pop(dIndex)
positions = [[i+iDirections[z], j+jDirections[z]] for z in range(len(iDirections))]
visited.add((i,j))
#print(positions)
for i in range(len(positions)):
newPos = [positions[i][0], positions[i][1]]
#print(newPos, directions[i])
dfs(newPos[0], newPos[1], directions[i], checkVal)
#print('---')
for i in range(n):
for j in range(m):
if (i,j) not in visited and not exists_cycle:
dfs(i,j,None, grid[i][j])
if exists_cycle:
print('Yes')
else:
print('No')
```
| -1
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
The main road in Bytecity is a straight line from south to north. Conveniently, there are coordinates measured in meters from the southernmost building in north direction.
At some points on the road there are *n* friends, and *i*-th of them is standing at the point *x**i* meters and can move with any speed no greater than *v**i* meters per second in any of the two directions along the road: south or north.
You are to compute the minimum time needed to gather all the *n* friends at some point on the road. Note that the point they meet at doesn't need to have integer coordinate.
|
The first line contains single integer *n* (2<=≤<=*n*<=≤<=60<=000) — the number of friends.
The second line contains *n* integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=109) — the current coordinates of the friends, in meters.
The third line contains *n* integers *v*1,<=*v*2,<=...,<=*v**n* (1<=≤<=*v**i*<=≤<=109) — the maximum speeds of the friends, in meters per second.
|
Print the minimum time (in seconds) needed for all the *n* friends to meet at some point on the road.
Your answer will be considered correct, if its absolute or relative error isn't greater than 10<=-<=6. Formally, let your answer be *a*, while jury's answer be *b*. Your answer will be considered correct if holds.
|
[
"3\n7 1 3\n1 2 1\n",
"4\n5 10 3 2\n2 3 2 4\n"
] |
[
"2.000000000000\n",
"1.400000000000\n"
] |
In the first sample, all friends can gather at the point 5 within 2 seconds. In order to achieve this, the first friend should go south all the time at his maximum speed, while the second and the third friends should go north at their maximum speeds.
| 0
|
[
{
"input": "3\n7 1 3\n1 2 1",
"output": "2.000000000000"
},
{
"input": "4\n5 10 3 2\n2 3 2 4",
"output": "1.400000000000"
},
{
"input": "3\n1 1000000000 2\n1 2 1000000000",
"output": "333333332.999999999971"
},
{
"input": "2\n4 5\n10 8",
"output": "0.055555555556"
},
{
"input": "4\n14 12 10 17\n8 6 5 10",
"output": "0.466666666667"
},
{
"input": "5\n1 15 61 29 43\n15 11 19 19 19",
"output": "1.764705882353"
},
{
"input": "10\n20 11 17 38 15 27 2 40 24 37\n22 30 22 30 28 16 7 20 22 13",
"output": "1.750000000000"
},
{
"input": "2\n1000000000 1000000000\n1 1",
"output": "0.000000000000"
},
{
"input": "3\n1 1 1\n1 1 1",
"output": "0.000000000000"
}
] | 1,488,720,139
| 1,039
|
Python 3
|
OK
|
TESTS
| 46
| 2,262
| 11,571,200
|
N = int(input())
X = [int(i) for i in input().split()]
V = [int(i) for i in input().split()]
low, high = min(X), max(X)
for it in range(60):
mid = (low + high) / 2
maxt = -1
maxd = 0
for i in range(N):
d = mid - X[i]
t = abs(d) / V[i]
if t > maxt:
maxt = t
maxd = d
if maxd > 0:
high = mid
else:
low = mid
print(maxt)
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The main road in Bytecity is a straight line from south to north. Conveniently, there are coordinates measured in meters from the southernmost building in north direction.
At some points on the road there are *n* friends, and *i*-th of them is standing at the point *x**i* meters and can move with any speed no greater than *v**i* meters per second in any of the two directions along the road: south or north.
You are to compute the minimum time needed to gather all the *n* friends at some point on the road. Note that the point they meet at doesn't need to have integer coordinate.
Input Specification:
The first line contains single integer *n* (2<=≤<=*n*<=≤<=60<=000) — the number of friends.
The second line contains *n* integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=109) — the current coordinates of the friends, in meters.
The third line contains *n* integers *v*1,<=*v*2,<=...,<=*v**n* (1<=≤<=*v**i*<=≤<=109) — the maximum speeds of the friends, in meters per second.
Output Specification:
Print the minimum time (in seconds) needed for all the *n* friends to meet at some point on the road.
Your answer will be considered correct, if its absolute or relative error isn't greater than 10<=-<=6. Formally, let your answer be *a*, while jury's answer be *b*. Your answer will be considered correct if holds.
Demo Input:
['3\n7 1 3\n1 2 1\n', '4\n5 10 3 2\n2 3 2 4\n']
Demo Output:
['2.000000000000\n', '1.400000000000\n']
Note:
In the first sample, all friends can gather at the point 5 within 2 seconds. In order to achieve this, the first friend should go south all the time at his maximum speed, while the second and the third friends should go north at their maximum speeds.
|
```python
N = int(input())
X = [int(i) for i in input().split()]
V = [int(i) for i in input().split()]
low, high = min(X), max(X)
for it in range(60):
mid = (low + high) / 2
maxt = -1
maxd = 0
for i in range(N):
d = mid - X[i]
t = abs(d) / V[i]
if t > maxt:
maxt = t
maxd = d
if maxd > 0:
high = mid
else:
low = mid
print(maxt)
```
| 3
|
|
768
|
B
|
Code For 1
|
PROGRAMMING
| 1,600
|
[
"constructive algorithms",
"dfs and similar",
"divide and conquer"
] | null | null |
Jon fought bravely to rescue the wildlings who were attacked by the white-walkers at Hardhome. On his arrival, Sam tells him that he wants to go to Oldtown to train at the Citadel to become a maester, so he can return and take the deceased Aemon's place as maester of Castle Black. Jon agrees to Sam's proposal and Sam sets off his journey to the Citadel. However becoming a trainee at the Citadel is not a cakewalk and hence the maesters at the Citadel gave Sam a problem to test his eligibility.
Initially Sam has a list with a single element *n*. Then he has to perform certain operations on this list. In each operation Sam must remove any element *x*, such that *x*<=><=1, from the list and insert at the same position , , sequentially. He must continue with these operations until all the elements in the list are either 0 or 1.
Now the masters want the total number of 1s in the range *l* to *r* (1-indexed). Sam wants to become a maester but unfortunately he cannot solve this problem. Can you help Sam to pass the eligibility test?
|
The first line contains three integers *n*, *l*, *r* (0<=≤<=*n*<=<<=250, 0<=≤<=*r*<=-<=*l*<=≤<=105, *r*<=≥<=1, *l*<=≥<=1) – initial element and the range *l* to *r*.
It is guaranteed that *r* is not greater than the length of the final list.
|
Output the total number of 1s in the range *l* to *r* in the final sequence.
|
[
"7 2 5\n",
"10 3 10\n"
] |
[
"4\n",
"5\n"
] |
Consider first example:
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/288fbb682a6fa1934a47b763d6851f9d32a06150.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Elements on positions from 2-nd to 5-th in list is [1, 1, 1, 1]. The number of ones is 4.
For the second example:
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/52e9bc51ef858cacc27fc274c7ba9419d5c1ded9.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Elements on positions from 3-rd to 10-th in list is [1, 1, 1, 0, 1, 0, 1, 0]. The number of ones is 5.
| 1,000
|
[
{
"input": "7 2 5",
"output": "4"
},
{
"input": "10 3 10",
"output": "5"
},
{
"input": "56 18 40",
"output": "20"
},
{
"input": "203 40 124",
"output": "67"
},
{
"input": "903316762502 354723010040 354723105411",
"output": "78355"
},
{
"input": "33534354842198 32529564319236 32529564342569",
"output": "22239"
},
{
"input": "62518534961045 50734311240112 50734311287877",
"output": "42439"
},
{
"input": "95173251245550 106288351347530 106288351372022",
"output": "16565"
},
{
"input": "542 321 956",
"output": "336"
},
{
"input": "3621 237 2637",
"output": "2124"
},
{
"input": "9056 336 896",
"output": "311"
},
{
"input": "36007 368 24490",
"output": "13253"
},
{
"input": "244269 149154 244246",
"output": "88609"
},
{
"input": "880234 669493 757150",
"output": "73585"
},
{
"input": "3740160 1031384 1104236",
"output": "64965"
},
{
"input": "11586121 15337246 15397874",
"output": "41868"
},
{
"input": "38658997 35923164 35985664",
"output": "36004"
},
{
"input": "192308932 207804787 207866400",
"output": "44142"
},
{
"input": "950099012 175922161 176000556",
"output": "69369"
},
{
"input": "2787326787 3799676481 3799680514",
"output": "2618"
},
{
"input": "14417262581 8527979363 8528075536",
"output": "80707"
},
{
"input": "39889373539 7747197212 7747278363",
"output": "47105"
},
{
"input": "251772781087 70597428577 70597479816",
"output": "46933"
},
{
"input": "0 1 1",
"output": "0"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "3 2 3",
"output": "2"
},
{
"input": "14 7 12",
"output": "5"
},
{
"input": "1125899906842623 1 100001",
"output": "100001"
},
{
"input": "1125899906842623 1125899906742623 1125899906842623",
"output": "100001"
},
{
"input": "1000 1 1023",
"output": "1000"
},
{
"input": "281474976710656 17179869184 17179869186",
"output": "1"
},
{
"input": "2 2 2",
"output": "0"
},
{
"input": "3 2 2",
"output": "1"
}
] | 1,697,696,112
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 4
| 2,000
| 248,012,800
|
str1=str(input())
lst1=str1.split(" ")
n=int(lst1[0])
l=int(lst1[1])
r=int(lst1[2])
lst=[]
def divid(num):
if num==1 or num==0:
return num
else:
i1=int(num/2)
if i1!=1:
lst.append(divid(i1))
else:
lst.append(1)
lst.append(num%2)
if i1!=1:
lst.append(divid(i1))
else:
lst.append(1)
divid(n)
for i in lst:
if i==None:
lst.remove(None)
tmp=0
for j in lst[l-1:r]:
if j==1:
tmp+=1
print(tmp)
|
Title: Code For 1
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Jon fought bravely to rescue the wildlings who were attacked by the white-walkers at Hardhome. On his arrival, Sam tells him that he wants to go to Oldtown to train at the Citadel to become a maester, so he can return and take the deceased Aemon's place as maester of Castle Black. Jon agrees to Sam's proposal and Sam sets off his journey to the Citadel. However becoming a trainee at the Citadel is not a cakewalk and hence the maesters at the Citadel gave Sam a problem to test his eligibility.
Initially Sam has a list with a single element *n*. Then he has to perform certain operations on this list. In each operation Sam must remove any element *x*, such that *x*<=><=1, from the list and insert at the same position , , sequentially. He must continue with these operations until all the elements in the list are either 0 or 1.
Now the masters want the total number of 1s in the range *l* to *r* (1-indexed). Sam wants to become a maester but unfortunately he cannot solve this problem. Can you help Sam to pass the eligibility test?
Input Specification:
The first line contains three integers *n*, *l*, *r* (0<=≤<=*n*<=<<=250, 0<=≤<=*r*<=-<=*l*<=≤<=105, *r*<=≥<=1, *l*<=≥<=1) – initial element and the range *l* to *r*.
It is guaranteed that *r* is not greater than the length of the final list.
Output Specification:
Output the total number of 1s in the range *l* to *r* in the final sequence.
Demo Input:
['7 2 5\n', '10 3 10\n']
Demo Output:
['4\n', '5\n']
Note:
Consider first example:
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/288fbb682a6fa1934a47b763d6851f9d32a06150.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Elements on positions from 2-nd to 5-th in list is [1, 1, 1, 1]. The number of ones is 4.
For the second example:
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/52e9bc51ef858cacc27fc274c7ba9419d5c1ded9.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Elements on positions from 3-rd to 10-th in list is [1, 1, 1, 0, 1, 0, 1, 0]. The number of ones is 5.
|
```python
str1=str(input())
lst1=str1.split(" ")
n=int(lst1[0])
l=int(lst1[1])
r=int(lst1[2])
lst=[]
def divid(num):
if num==1 or num==0:
return num
else:
i1=int(num/2)
if i1!=1:
lst.append(divid(i1))
else:
lst.append(1)
lst.append(num%2)
if i1!=1:
lst.append(divid(i1))
else:
lst.append(1)
divid(n)
for i in lst:
if i==None:
lst.remove(None)
tmp=0
for j in lst[l-1:r]:
if j==1:
tmp+=1
print(tmp)
```
| 0
|
|
960
|
A
|
Check the string
|
PROGRAMMING
| 1,200
|
[
"implementation"
] | null | null |
A has a string consisting of some number of lowercase English letters 'a'. He gives it to his friend B who appends some number of letters 'b' to the end of this string. Since both A and B like the characters 'a' and 'b', they have made sure that at this point, at least one 'a' and one 'b' exist in the string.
B now gives this string to C and he appends some number of letters 'c' to the end of the string. However, since C is a good friend of A and B, the number of letters 'c' he appends is equal to the number of 'a' or to the number of 'b' in the string. It is also possible that the number of letters 'c' equals both to the number of letters 'a' and to the number of letters 'b' at the same time.
You have a string in your hands, and you want to check if it is possible to obtain the string in this way or not. If it is possible to obtain the string, print "YES", otherwise print "NO" (without the quotes).
|
The first and only line consists of a string $S$ ($ 1 \le |S| \le 5\,000 $). It is guaranteed that the string will only consist of the lowercase English letters 'a', 'b', 'c'.
|
Print "YES" or "NO", according to the condition.
|
[
"aaabccc\n",
"bbacc\n",
"aabc\n"
] |
[
"YES\n",
"NO\n",
"YES\n"
] |
Consider first example: the number of 'c' is equal to the number of 'a'.
Consider second example: although the number of 'c' is equal to the number of the 'b', the order is not correct.
Consider third example: the number of 'c' is equal to the number of 'b'.
| 500
|
[
{
"input": "aaabccc",
"output": "YES"
},
{
"input": "bbacc",
"output": "NO"
},
{
"input": "aabc",
"output": "YES"
},
{
"input": "aabbcc",
"output": "YES"
},
{
"input": "aaacccbb",
"output": "NO"
},
{
"input": "abc",
"output": "YES"
},
{
"input": "acba",
"output": "NO"
},
{
"input": "bbabbc",
"output": "NO"
},
{
"input": "bbbabacca",
"output": "NO"
},
{
"input": "aabcbcaca",
"output": "NO"
},
{
"input": "aaaaabbbbbb",
"output": "NO"
},
{
"input": "c",
"output": "NO"
},
{
"input": "cc",
"output": "NO"
},
{
"input": "bbb",
"output": "NO"
},
{
"input": "bc",
"output": "NO"
},
{
"input": "ccbcc",
"output": "NO"
},
{
"input": "aaa",
"output": "NO"
},
{
"input": "aaccaa",
"output": "NO"
},
{
"input": "a",
"output": "NO"
},
{
"input": "b",
"output": "NO"
},
{
"input": "abca",
"output": "NO"
},
{
"input": "aabbcccc",
"output": "NO"
},
{
"input": "abac",
"output": "NO"
},
{
"input": "abcc",
"output": "NO"
},
{
"input": "abcb",
"output": "NO"
},
{
"input": "aacc",
"output": "NO"
},
{
"input": "aabbaacccc",
"output": "NO"
},
{
"input": "aabb",
"output": "NO"
},
{
"input": "ac",
"output": "NO"
},
{
"input": "abbacc",
"output": "NO"
},
{
"input": "abacc",
"output": "NO"
},
{
"input": "ababc",
"output": "NO"
},
{
"input": "aa",
"output": "NO"
},
{
"input": "aabaccc",
"output": "NO"
},
{
"input": "bbcc",
"output": "NO"
},
{
"input": "aaabcbc",
"output": "NO"
},
{
"input": "acbbc",
"output": "NO"
},
{
"input": "babc",
"output": "NO"
},
{
"input": "bbbcc",
"output": "NO"
},
{
"input": "bbc",
"output": "NO"
},
{
"input": "abababccc",
"output": "NO"
},
{
"input": "ccbbaa",
"output": "NO"
}
] | 1,535,363,607
| 2,607
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 47
| 124
| 0
|
s=input()
ac=0
bc=0
cc=0
for i in range(len(s)):
if s[i]=='a':
ac+=1
elif s[i]=='b':
bc+=1
elif s[i]=='c':
cc+=1
af=s.find('a')
bf=s.find('b')
cf=s.find('c')
s=s[::-1]
ae=len(s)-s.find('a')
be=len(s)-s.find('b')
ce=len(s)-s.find('c')
if af<bf and af<cf and bf<cf and be>ae and ce>be and ce>ae and (ac==cc or cc==bc):
print("YES")
else:
print("NO")
|
Title: Check the string
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A has a string consisting of some number of lowercase English letters 'a'. He gives it to his friend B who appends some number of letters 'b' to the end of this string. Since both A and B like the characters 'a' and 'b', they have made sure that at this point, at least one 'a' and one 'b' exist in the string.
B now gives this string to C and he appends some number of letters 'c' to the end of the string. However, since C is a good friend of A and B, the number of letters 'c' he appends is equal to the number of 'a' or to the number of 'b' in the string. It is also possible that the number of letters 'c' equals both to the number of letters 'a' and to the number of letters 'b' at the same time.
You have a string in your hands, and you want to check if it is possible to obtain the string in this way or not. If it is possible to obtain the string, print "YES", otherwise print "NO" (without the quotes).
Input Specification:
The first and only line consists of a string $S$ ($ 1 \le |S| \le 5\,000 $). It is guaranteed that the string will only consist of the lowercase English letters 'a', 'b', 'c'.
Output Specification:
Print "YES" or "NO", according to the condition.
Demo Input:
['aaabccc\n', 'bbacc\n', 'aabc\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
Consider first example: the number of 'c' is equal to the number of 'a'.
Consider second example: although the number of 'c' is equal to the number of the 'b', the order is not correct.
Consider third example: the number of 'c' is equal to the number of 'b'.
|
```python
s=input()
ac=0
bc=0
cc=0
for i in range(len(s)):
if s[i]=='a':
ac+=1
elif s[i]=='b':
bc+=1
elif s[i]=='c':
cc+=1
af=s.find('a')
bf=s.find('b')
cf=s.find('c')
s=s[::-1]
ae=len(s)-s.find('a')
be=len(s)-s.find('b')
ce=len(s)-s.find('c')
if af<bf and af<cf and bf<cf and be>ae and ce>be and ce>ae and (ac==cc or cc==bc):
print("YES")
else:
print("NO")
```
| 0
|
|
801
|
B
|
Valued Keys
|
PROGRAMMING
| 900
|
[
"constructive algorithms",
"greedy",
"strings"
] | null | null |
You found a mysterious function *f*. The function takes two strings *s*1 and *s*2. These strings must consist only of lowercase English letters, and must be the same length.
The output of the function *f* is another string of the same length. The *i*-th character of the output is equal to the minimum of the *i*-th character of *s*1 and the *i*-th character of *s*2.
For example, *f*("ab", "ba") = "aa", and *f*("nzwzl", "zizez") = "niwel".
You found two strings *x* and *y* of the same length and consisting of only lowercase English letters. Find any string *z* such that *f*(*x*,<=*z*)<==<=*y*, or print -1 if no such string *z* exists.
|
The first line of input contains the string *x*.
The second line of input contains the string *y*.
Both *x* and *y* consist only of lowercase English letters, *x* and *y* have same length and this length is between 1 and 100.
|
If there is no string *z* such that *f*(*x*,<=*z*)<==<=*y*, print -1.
Otherwise, print a string *z* such that *f*(*x*,<=*z*)<==<=*y*. If there are multiple possible answers, print any of them. The string *z* should be the same length as *x* and *y* and consist only of lowercase English letters.
|
[
"ab\naa\n",
"nzwzl\nniwel\n",
"ab\nba\n"
] |
[
"ba\n",
"xiyez\n",
"-1\n"
] |
The first case is from the statement.
Another solution for the second case is "zizez"
There is no solution for the third case. That is, there is no *z* such that *f*("ab", *z*) = "ba".
| 1,000
|
[
{
"input": "ab\naa",
"output": "ba"
},
{
"input": "nzwzl\nniwel",
"output": "xiyez"
},
{
"input": "ab\nba",
"output": "-1"
},
{
"input": "r\nl",
"output": "l"
},
{
"input": "d\ny",
"output": "-1"
},
{
"input": "yvowz\ncajav",
"output": "cajav"
},
{
"input": "lwzjp\ninjit",
"output": "-1"
},
{
"input": "epqnlxmiicdidyscjaxqznwur\neodnlemiicdedmkcgavqbnqmm",
"output": "eodnlemiicdedmkcgavqbnqmm"
},
{
"input": "qqdabbsxiibnnjgsgxllfvdqj\nuxmypqtwfdezewdxfgplannrs",
"output": "-1"
},
{
"input": "aanerbaqslfmqmuciqbxyznkevukvznpkmxlcorpmrenwxhzfgbmlfpxtkqpxdrmcqcmbf\naanebbaqkgfiimcciqbaoznkeqqkrgapdillccrfeienwbcvfgbmlfbimkqchcrmclcmbf",
"output": "aanebbaqkgfiimcciqbaoznkeqqkrgapdillccrfeienwbcvfgbmlfbimkqchcrmclcmbf"
},
{
"input": "mbyrkhjctrcrayisflptgfudwgrtegidhqicsjqafvdloritbjhciyxuwavxknezwwudnk\nvvixsutlbdewqoabqhpuerfkzrddcqptfwmxdlxwbvsaqfjoxztlddvwgflcteqbwaiaen",
"output": "-1"
},
{
"input": "eufycwztywhbjrpqobvknwfqmnboqcfdiahkagykeibbsqpljcghhmsgfmswwsanzyiwtvuirwmppfivtekaywkzskyydfvkjgxb\necfwavookadbcilfobojnweqinbcpcfdiahkabwkeibbacpljcghhksgfajgmianfnivmhfifogpffiheegayfkxkkcmdfvihgdb",
"output": "ecfwavookadbcilfobojnweqinbcpcfdiahkabwkeibbacpljcghhksgfajgmianfnivmhfifogpffiheegayfkxkkcmdfvihgdb"
},
{
"input": "qvpltcffyeghtbdhjyhfteojezyzziardduzrbwuxmzzkkoehfnxecafizxglboauhynfbawlfxenmykquyhrxswhjuovvogntok\nchvkcvzxptbcepdjfezcpuvtehewbnvqeoezlcnzhpfwujbmhafoeqmjhtwisnobauinkzyigrvahpuetkgpdjfgbzficsmuqnym",
"output": "-1"
},
{
"input": "nmuwjdihouqrnsuahimssnrbxdpwvxiyqtenahtrlshjkmnfuttnpqhgcagoptinnaptxaccptparldzrhpgbyrzedghudtsswxi\nnilhbdghosqnbebafimconrbvdodjsipqmekahhrllhjkemeketapfhgcagopfidnahtlaccpfpafedqicpcbvfgedghudhddwib",
"output": "nilhbdghosqnbebafimconrbvdodjsipqmekahhrllhjkemeketapfhgcagopfidnahtlaccpfpafedqicpcbvfgedghudhddwib"
},
{
"input": "dyxgwupoauwqtcfoyfjdotzirwztdfrueqiypxoqvkmhiehdppwtdoxrbfvtairdbuvlqohjflznggjpifhwjrshcrfbjtklpykx\ngzqlnoizhxolnditjdhlhptjsbczehicudoybzilwnshmywozwnwuipcgirgzldtvtowdsokfeafggwserzdazkxyddjttiopeew",
"output": "-1"
},
{
"input": "hbgwuqzougqzlxemvyjpeizjfwhgugrfnhbrlxkmkdalikfyunppwgdzmalbwewybnjzqsohwhjkdcyhhzmysflambvhpsjilsyv\nfbdjdqjojdafarakvcjpeipjfehgfgrfehbolxkmkdagikflunnpvadocalbkedibhbflmohnhjkdcthhaigsfjaibqhbcjelirv",
"output": "fbdjdqjojdafarakvcjpeipjfehgfgrfehbolxkmkdagikflunnpvadocalbkedibhbflmohnhjkdcthhaigsfjaibqhbcjelirv"
},
{
"input": "xnjjhjfuhgyxqhpzmvgbaohqarugdoaczcfecofltwemieyxolswkcwhlfagfrgmoiqrgftokbqwtxgxzweozzlikrvafiabivlk\npjfosalbsitcnqiazhmepfifjxvmazvdgffcnozmnqubhonwjldmpdsjagmamniylzjdbklcyrzivjyzgnogahobpkwpwpvraqns",
"output": "-1"
},
{
"input": "zrvzedssbsrfldqvjpgmsefrmsatspzoitwvymahiptphiystjlsauzquzqqbmljobdhijcpdvatorwmyojqgnezvzlgjibxepcf\npesoedmqbmffldqsjggmhefkadaesijointrkmahapaahiysfjdiaupqujngbjhjobdhiecadeatgjvelojjgnepvajgeibfepaf",
"output": "pesoedmqbmffldqsjggmhefkadaesijointrkmahapaahiysfjdiaupqujngbjhjobdhiecadeatgjvelojjgnepvajgeibfepaf"
},
{
"input": "pdvkuwyzntzfqpblzmbynknyhlnqbxijuqaincviugxohcsrofozrrsategwkbwxcvkyzxhurokefpbdnmcfogfhsojayysqbrow\nbvxruombdrywlcjkrltyayaazwpauuhbtgwfzdrmfwwucgffucwelzvpsdgtapogchblzahsrfymjlaghkbmbssghrpxalkslcvp",
"output": "-1"
},
{
"input": "tgharsjyihroiiahwgbjezlxvlterxivdhtzjcqegzmtigqmrehvhiyjeywegxaseoyoacouijudbiruoghgxvxadwzgdxtnxlds\ntghaksjsdhkoiiahegbjexlfrctercipdhmvjbgegxdtggqdpbhvhiseehhegnaseoooacnsijubbirjnghgsvpadhaadrtimfdp",
"output": "tghaksjsdhkoiiahegbjexlfrctercipdhmvjbgegxdtggqdpbhvhiseehhegnaseoooacnsijubbirjnghgsvpadhaadrtimfdp"
},
{
"input": "jsinejpfwhzloulxndzvzftgogfdagrsscxmatldssqsgaknnbkcvhptebjjpkjhrjegrotzwcdosezkedzxeoyibmyzunkguoqj\nkfmvybobocdpipiripysioruqvloopvbggpjksgmwzyqwyxnesmvhsawnbbmntulspvsysfkjqwpvoelliopbaukyagedextzoej",
"output": "-1"
},
{
"input": "nttdcfceptruiomtmwzestrfchnqpgqeztpcvthzelfyggjgqadylzubpvbrlgndrcsursczpxlnoyoadxezncqalupfzmjeqihe\nkttdcfceohrjiaahmoldanpfchnfpgheqpdahqhxecfpbgigqadrkjubjfbrlgndbcgcgmcjpeleinaadretncqaiqpfkmjeqihe",
"output": "kttdcfceohrjiaahmoldanpfchnfpgheqpdahqhxecfpbgigqadrkjubjfbrlgndbcgcgmcjpeleinaadretncqaiqpfkmjeqihe"
},
{
"input": "diuopwglduasnaxgduwslbzoyayoypzznqspljcyqehweydhlwifcvnjmaowuvyqfwynjghecqvxdvuquuwpvwrjljozocaxnktv\ntrdydprdzmjhgbhzytelrfjpgsebijicsigmwhynmcyjtqrvojcndodchzxfcvyqjxqzwibccdvsjqhsnectdjyrrhzkeamukang",
"output": "-1"
},
{
"input": "ftfr\nftfr",
"output": "ftfr"
},
{
"input": "ftr\nftr",
"output": "ftr"
},
{
"input": "shftr\nshftr",
"output": "shftr"
},
{
"input": "vkvkkv\nvkvkkv",
"output": "vkvkkv"
},
{
"input": "ftrd\nftrd",
"output": "ftrd"
},
{
"input": "fztr\nfztr",
"output": "fztr"
},
{
"input": "frtr\nfrtr",
"output": "frtr"
},
{
"input": "shdftr\nshdftr",
"output": "shdftr"
},
{
"input": "zzz\nzzz",
"output": "zzz"
},
{
"input": "shtr\nshtr",
"output": "shtr"
},
{
"input": "aaaaa\nzzzzz",
"output": "-1"
},
{
"input": "efr\nefr",
"output": "efr"
}
] | 1,689,697,073
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 35
| 46
| 0
|
s1 = input()
s2 = input()
length = len(s1)
word = [0] * length
for index in range(length) :
if s1[index] == s2[index] :
word[index] = "z"
elif ord(s1[index]) < ord(s2[index]) :
exit(print(-1))
elif ord(s1[index]) > ord(s2[index]) :
word[index] = s2[index]
print("".join(word))
|
Title: Valued Keys
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You found a mysterious function *f*. The function takes two strings *s*1 and *s*2. These strings must consist only of lowercase English letters, and must be the same length.
The output of the function *f* is another string of the same length. The *i*-th character of the output is equal to the minimum of the *i*-th character of *s*1 and the *i*-th character of *s*2.
For example, *f*("ab", "ba") = "aa", and *f*("nzwzl", "zizez") = "niwel".
You found two strings *x* and *y* of the same length and consisting of only lowercase English letters. Find any string *z* such that *f*(*x*,<=*z*)<==<=*y*, or print -1 if no such string *z* exists.
Input Specification:
The first line of input contains the string *x*.
The second line of input contains the string *y*.
Both *x* and *y* consist only of lowercase English letters, *x* and *y* have same length and this length is between 1 and 100.
Output Specification:
If there is no string *z* such that *f*(*x*,<=*z*)<==<=*y*, print -1.
Otherwise, print a string *z* such that *f*(*x*,<=*z*)<==<=*y*. If there are multiple possible answers, print any of them. The string *z* should be the same length as *x* and *y* and consist only of lowercase English letters.
Demo Input:
['ab\naa\n', 'nzwzl\nniwel\n', 'ab\nba\n']
Demo Output:
['ba\n', 'xiyez\n', '-1\n']
Note:
The first case is from the statement.
Another solution for the second case is "zizez"
There is no solution for the third case. That is, there is no *z* such that *f*("ab", *z*) = "ba".
|
```python
s1 = input()
s2 = input()
length = len(s1)
word = [0] * length
for index in range(length) :
if s1[index] == s2[index] :
word[index] = "z"
elif ord(s1[index]) < ord(s2[index]) :
exit(print(-1))
elif ord(s1[index]) > ord(s2[index]) :
word[index] = s2[index]
print("".join(word))
```
| 3
|
|
217
|
A
|
Ice Skating
|
PROGRAMMING
| 1,200
|
[
"brute force",
"dfs and similar",
"dsu",
"graphs"
] | null | null |
Bajtek is learning to skate on ice. He's a beginner, so his only mode of transportation is pushing off from a snow drift to the north, east, south or west and sliding until he lands in another snow drift. He has noticed that in this way it's impossible to get from some snow drifts to some other by any sequence of moves. He now wants to heap up some additional snow drifts, so that he can get from any snow drift to any other one. He asked you to find the minimal number of snow drifts that need to be created.
We assume that Bajtek can only heap up snow drifts at integer coordinates.
|
The first line of input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of snow drifts. Each of the following *n* lines contains two integers *x**i* and *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=1000) — the coordinates of the *i*-th snow drift.
Note that the north direction coinсides with the direction of *Oy* axis, so the east direction coinсides with the direction of the *Ox* axis. All snow drift's locations are distinct.
|
Output the minimal number of snow drifts that need to be created in order for Bajtek to be able to reach any snow drift from any other one.
|
[
"2\n2 1\n1 2\n",
"2\n2 1\n4 1\n"
] |
[
"1\n",
"0\n"
] |
none
| 500
|
[
{
"input": "2\n2 1\n1 2",
"output": "1"
},
{
"input": "2\n2 1\n4 1",
"output": "0"
},
{
"input": "24\n171 35\n261 20\n4 206\n501 446\n961 912\n581 748\n946 978\n463 514\n841 889\n341 466\n842 967\n54 102\n235 261\n925 889\n682 672\n623 636\n268 94\n635 710\n474 510\n697 794\n586 663\n182 184\n806 663\n468 459",
"output": "21"
},
{
"input": "17\n660 646\n440 442\n689 618\n441 415\n922 865\n950 972\n312 366\n203 229\n873 860\n219 199\n344 308\n169 176\n961 992\n153 84\n201 230\n987 938\n834 815",
"output": "16"
},
{
"input": "11\n798 845\n722 911\n374 270\n629 537\n748 856\n831 885\n486 641\n751 829\n609 492\n98 27\n654 663",
"output": "10"
},
{
"input": "1\n321 88",
"output": "0"
},
{
"input": "9\n811 859\n656 676\n76 141\n945 951\n497 455\n18 55\n335 294\n267 275\n656 689",
"output": "7"
},
{
"input": "7\n948 946\n130 130\n761 758\n941 938\n971 971\n387 385\n509 510",
"output": "6"
},
{
"input": "6\n535 699\n217 337\n508 780\n180 292\n393 112\n732 888",
"output": "5"
},
{
"input": "14\n25 23\n499 406\n193 266\n823 751\n219 227\n101 138\n978 992\n43 74\n997 932\n237 189\n634 538\n774 740\n842 767\n742 802",
"output": "13"
},
{
"input": "12\n548 506\n151 198\n370 380\n655 694\n654 690\n407 370\n518 497\n819 827\n765 751\n802 771\n741 752\n653 662",
"output": "11"
},
{
"input": "40\n685 711\n433 403\n703 710\n491 485\n616 619\n288 282\n884 871\n367 352\n500 511\n977 982\n51 31\n576 564\n508 519\n755 762\n22 20\n368 353\n232 225\n953 955\n452 436\n311 330\n967 988\n369 364\n791 803\n150 149\n651 661\n118 93\n398 387\n748 766\n852 852\n230 228\n555 545\n515 519\n667 678\n867 862\n134 146\n859 863\n96 99\n486 469\n303 296\n780 786",
"output": "38"
},
{
"input": "3\n175 201\n907 909\n388 360",
"output": "2"
},
{
"input": "7\n312 298\n86 78\n73 97\n619 594\n403 451\n538 528\n71 86",
"output": "6"
},
{
"input": "19\n802 820\n368 248\n758 794\n455 378\n876 888\n771 814\n245 177\n586 555\n844 842\n364 360\n820 856\n731 624\n982 975\n825 856\n122 121\n862 896\n42 4\n792 841\n828 820",
"output": "16"
},
{
"input": "32\n643 877\n842 614\n387 176\n99 338\n894 798\n652 728\n611 648\n622 694\n579 781\n243 46\n322 305\n198 438\n708 579\n246 325\n536 459\n874 593\n120 277\n989 907\n223 110\n35 130\n761 692\n690 661\n518 766\n226 93\n678 597\n725 617\n661 574\n775 496\n56 416\n14 189\n358 359\n898 901",
"output": "31"
},
{
"input": "32\n325 327\n20 22\n72 74\n935 933\n664 663\n726 729\n785 784\n170 171\n315 314\n577 580\n984 987\n313 317\n434 435\n962 961\n55 54\n46 44\n743 742\n434 433\n617 612\n332 332\n883 886\n940 936\n793 792\n645 644\n611 607\n418 418\n465 465\n219 218\n167 164\n56 54\n403 405\n210 210",
"output": "29"
},
{
"input": "32\n652 712\n260 241\n27 154\n188 16\n521 351\n518 356\n452 540\n790 827\n339 396\n336 551\n897 930\n828 627\n27 168\n180 113\n134 67\n794 671\n812 711\n100 241\n686 813\n138 289\n384 506\n884 932\n913 959\n470 508\n730 734\n373 478\n788 862\n392 426\n148 68\n113 49\n713 852\n924 894",
"output": "29"
},
{
"input": "14\n685 808\n542 677\n712 747\n832 852\n187 410\n399 338\n626 556\n530 635\n267 145\n215 209\n559 684\n944 949\n753 596\n601 823",
"output": "13"
},
{
"input": "5\n175 158\n16 2\n397 381\n668 686\n957 945",
"output": "4"
},
{
"input": "5\n312 284\n490 509\n730 747\n504 497\n782 793",
"output": "4"
},
{
"input": "2\n802 903\n476 348",
"output": "1"
},
{
"input": "4\n325 343\n425 442\n785 798\n275 270",
"output": "3"
},
{
"input": "28\n462 483\n411 401\n118 94\n111 127\n5 6\n70 52\n893 910\n73 63\n818 818\n182 201\n642 633\n900 886\n893 886\n684 700\n157 173\n953 953\n671 660\n224 225\n832 801\n152 157\n601 585\n115 101\n739 722\n611 606\n659 642\n461 469\n702 689\n649 653",
"output": "25"
},
{
"input": "36\n952 981\n885 900\n803 790\n107 129\n670 654\n143 132\n66 58\n813 819\n849 837\n165 198\n247 228\n15 39\n619 618\n105 138\n868 855\n965 957\n293 298\n613 599\n227 212\n745 754\n723 704\n877 858\n503 487\n678 697\n592 595\n155 135\n962 982\n93 89\n660 673\n225 212\n967 987\n690 680\n804 813\n489 518\n240 221\n111 124",
"output": "34"
},
{
"input": "30\n89 3\n167 156\n784 849\n943 937\n144 95\n24 159\n80 120\n657 683\n585 596\n43 147\n909 964\n131 84\n345 389\n333 321\n91 126\n274 325\n859 723\n866 922\n622 595\n690 752\n902 944\n127 170\n426 383\n905 925\n172 284\n793 810\n414 510\n890 884\n123 24\n267 255",
"output": "29"
},
{
"input": "5\n664 666\n951 941\n739 742\n844 842\n2 2",
"output": "4"
},
{
"input": "3\n939 867\n411 427\n757 708",
"output": "2"
},
{
"input": "36\n429 424\n885 972\n442 386\n512 511\n751 759\n4 115\n461 497\n496 408\n8 23\n542 562\n296 331\n448 492\n412 395\n109 166\n622 640\n379 355\n251 262\n564 586\n66 115\n275 291\n666 611\n629 534\n510 567\n635 666\n738 803\n420 369\n92 17\n101 144\n141 92\n258 258\n184 235\n492 456\n311 210\n394 357\n531 512\n634 636",
"output": "34"
},
{
"input": "29\n462 519\n871 825\n127 335\n156 93\n576 612\n885 830\n634 779\n340 105\n744 795\n716 474\n93 139\n563 805\n137 276\n177 101\n333 14\n391 437\n873 588\n817 518\n460 597\n572 670\n140 303\n392 441\n273 120\n862 578\n670 639\n410 161\n544 577\n193 116\n252 195",
"output": "28"
},
{
"input": "23\n952 907\n345 356\n812 807\n344 328\n242 268\n254 280\n1000 990\n80 78\n424 396\n595 608\n755 813\n383 380\n55 56\n598 633\n203 211\n508 476\n600 593\n206 192\n855 882\n517 462\n967 994\n642 657\n493 488",
"output": "22"
},
{
"input": "10\n579 816\n806 590\n830 787\n120 278\n677 800\n16 67\n188 251\n559 560\n87 67\n104 235",
"output": "8"
},
{
"input": "23\n420 424\n280 303\n515 511\n956 948\n799 803\n441 455\n362 369\n299 289\n823 813\n982 967\n876 878\n185 157\n529 551\n964 989\n655 656\n1 21\n114 112\n45 56\n935 937\n1000 997\n934 942\n360 366\n648 621",
"output": "22"
},
{
"input": "23\n102 84\n562 608\n200 127\n952 999\n465 496\n322 367\n728 690\n143 147\n855 867\n861 866\n26 59\n300 273\n255 351\n192 246\n70 111\n365 277\n32 104\n298 319\n330 354\n241 141\n56 125\n315 298\n412 461",
"output": "22"
},
{
"input": "7\n429 506\n346 307\n99 171\n853 916\n322 263\n115 157\n906 924",
"output": "6"
},
{
"input": "3\n1 1\n2 1\n2 2",
"output": "0"
},
{
"input": "4\n1 1\n1 2\n2 1\n2 2",
"output": "0"
},
{
"input": "5\n1 1\n1 2\n2 2\n3 1\n3 3",
"output": "0"
},
{
"input": "6\n1 1\n1 2\n2 2\n3 1\n3 2\n3 3",
"output": "0"
},
{
"input": "20\n1 1\n2 2\n3 3\n3 9\n4 4\n5 2\n5 5\n5 7\n5 8\n6 2\n6 6\n6 9\n7 7\n8 8\n9 4\n9 7\n9 9\n10 2\n10 9\n10 10",
"output": "1"
},
{
"input": "21\n1 1\n1 9\n2 1\n2 2\n2 5\n2 6\n2 9\n3 3\n3 8\n4 1\n4 4\n5 5\n5 8\n6 6\n7 7\n8 8\n9 9\n10 4\n10 10\n11 5\n11 11",
"output": "1"
},
{
"input": "22\n1 1\n1 3\n1 4\n1 8\n1 9\n1 11\n2 2\n3 3\n4 4\n4 5\n5 5\n6 6\n6 8\n7 7\n8 3\n8 4\n8 8\n9 9\n10 10\n11 4\n11 9\n11 11",
"output": "3"
},
{
"input": "50\n1 1\n2 2\n2 9\n3 3\n4 4\n4 9\n4 16\n4 24\n5 5\n6 6\n7 7\n8 8\n8 9\n8 20\n9 9\n10 10\n11 11\n12 12\n13 13\n14 7\n14 14\n14 16\n14 25\n15 4\n15 6\n15 15\n15 22\n16 6\n16 16\n17 17\n18 18\n19 6\n19 19\n20 20\n21 21\n22 6\n22 22\n23 23\n24 6\n24 7\n24 8\n24 9\n24 24\n25 1\n25 3\n25 5\n25 7\n25 23\n25 24\n25 25",
"output": "7"
},
{
"input": "55\n1 1\n1 14\n2 2\n2 19\n3 1\n3 3\n3 8\n3 14\n3 23\n4 1\n4 4\n5 5\n5 8\n5 15\n6 2\n6 3\n6 4\n6 6\n7 7\n8 8\n8 21\n9 9\n10 1\n10 10\n11 9\n11 11\n12 12\n13 13\n14 14\n15 15\n15 24\n16 5\n16 16\n17 5\n17 10\n17 17\n17 18\n17 22\n17 27\n18 18\n19 19\n20 20\n21 20\n21 21\n22 22\n23 23\n24 14\n24 24\n25 25\n26 8\n26 11\n26 26\n27 3\n27 27\n28 28",
"output": "5"
},
{
"input": "3\n1 2\n2 1\n2 2",
"output": "0"
},
{
"input": "6\n4 4\n3 4\n5 4\n4 5\n4 3\n3 1",
"output": "0"
},
{
"input": "4\n1 1\n1 2\n2 1\n2 2",
"output": "0"
},
{
"input": "3\n1 1\n2 2\n1 2",
"output": "0"
},
{
"input": "8\n1 3\n1 1\n4 1\n2 2\n2 5\n5 9\n5 1\n5 4",
"output": "1"
},
{
"input": "10\n1 1\n1 2\n1 3\n1 4\n5 5\n6 6\n7 7\n8 8\n9 9\n100 100",
"output": "6"
},
{
"input": "7\n1 1\n2 2\n3 3\n4 4\n1 2\n2 3\n3 4",
"output": "0"
},
{
"input": "6\n1 1\n2 1\n2 2\n2 4\n4 3\n2 3",
"output": "0"
},
{
"input": "4\n3 1\n2 1\n2 2\n1 2",
"output": "0"
},
{
"input": "6\n1 1\n2 2\n2 1\n2 4\n4 3\n2 3",
"output": "0"
},
{
"input": "3\n1 2\n1 3\n1 4",
"output": "0"
},
{
"input": "4\n1 1\n2 2\n1 2\n2 1",
"output": "0"
},
{
"input": "4\n1 3\n2 1\n3 2\n3 1",
"output": "1"
},
{
"input": "7\n1 1\n1 2\n2 2\n3 3\n3 4\n4 4\n1 4",
"output": "0"
},
{
"input": "21\n12 12\n13 12\n12 11\n13 13\n10 10\n11 10\n11 11\n501 500\n501 501\n503 502\n500 500\n503 503\n502 501\n502 502\n700 700\n702 702\n703 702\n701 701\n702 701\n703 703\n701 700",
"output": "2"
},
{
"input": "6\n1 11\n6 8\n11 10\n1 10\n11 11\n6 9",
"output": "1"
},
{
"input": "4\n1 1\n2 2\n3 2\n3 1",
"output": "0"
},
{
"input": "3\n1 2\n3 4\n3 2",
"output": "0"
},
{
"input": "3\n1 1\n1 2\n2 2",
"output": "0"
},
{
"input": "4\n5 5\n5 4\n6 3\n6 4",
"output": "0"
},
{
"input": "3\n1 1\n2 2\n2 1",
"output": "0"
}
] | 1,666,738,312
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 60
| 0
|
n = int(input())
graph = [[] for _ in range(n)]
visited = [False for _ in range(n)]
store = []
for i in range(n):
x, y = map(int, input().split())
store.append([x, y])
for i in range(n):
for j in range(i + 1, n):
xa, ya = store[i]
xb, yb = store[j]
if xa == xb or ya == yb:
graph[i].append(j)
graph[j].append(i)
stack = [0]
visited[0] = True
while len(stack) > 0:
cur = stack.pop()
visited[cur] = True
for nei in graph[cur]:
if not visited[nei]:
visited[nei] = True
stack.append(nei)
cnt = 0
for i in range(n):
if visited[i]:
cnt += 1
print(n - cnt)
|
Title: Ice Skating
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bajtek is learning to skate on ice. He's a beginner, so his only mode of transportation is pushing off from a snow drift to the north, east, south or west and sliding until he lands in another snow drift. He has noticed that in this way it's impossible to get from some snow drifts to some other by any sequence of moves. He now wants to heap up some additional snow drifts, so that he can get from any snow drift to any other one. He asked you to find the minimal number of snow drifts that need to be created.
We assume that Bajtek can only heap up snow drifts at integer coordinates.
Input Specification:
The first line of input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of snow drifts. Each of the following *n* lines contains two integers *x**i* and *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=1000) — the coordinates of the *i*-th snow drift.
Note that the north direction coinсides with the direction of *Oy* axis, so the east direction coinсides with the direction of the *Ox* axis. All snow drift's locations are distinct.
Output Specification:
Output the minimal number of snow drifts that need to be created in order for Bajtek to be able to reach any snow drift from any other one.
Demo Input:
['2\n2 1\n1 2\n', '2\n2 1\n4 1\n']
Demo Output:
['1\n', '0\n']
Note:
none
|
```python
n = int(input())
graph = [[] for _ in range(n)]
visited = [False for _ in range(n)]
store = []
for i in range(n):
x, y = map(int, input().split())
store.append([x, y])
for i in range(n):
for j in range(i + 1, n):
xa, ya = store[i]
xb, yb = store[j]
if xa == xb or ya == yb:
graph[i].append(j)
graph[j].append(i)
stack = [0]
visited[0] = True
while len(stack) > 0:
cur = stack.pop()
visited[cur] = True
for nei in graph[cur]:
if not visited[nei]:
visited[nei] = True
stack.append(nei)
cnt = 0
for i in range(n):
if visited[i]:
cnt += 1
print(n - cnt)
```
| 0
|
|
427
|
A
|
Police Recruits
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
The police department of your city has just started its journey. Initially, they don’t have any manpower. So, they started hiring new recruits in groups.
Meanwhile, crimes keeps occurring within the city. One member of the police force can investigate only one crime during his/her lifetime.
If there is no police officer free (isn't busy with crime) during the occurrence of a crime, it will go untreated.
Given the chronological order of crime occurrences and recruit hirings, find the number of crimes which will go untreated.
|
The first line of input will contain an integer *n* (1<=≤<=*n*<=≤<=105), the number of events. The next line will contain *n* space-separated integers.
If the integer is -1 then it means a crime has occurred. Otherwise, the integer will be positive, the number of officers recruited together at that time. No more than 10 officers will be recruited at a time.
|
Print a single integer, the number of crimes which will go untreated.
|
[
"3\n-1 -1 1\n",
"8\n1 -1 1 -1 -1 1 1 1\n",
"11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1\n"
] |
[
"2\n",
"1\n",
"8\n"
] |
Lets consider the second example:
1. Firstly one person is hired. 1. Then crime appears, the last hired person will investigate this crime. 1. One more person is hired. 1. One more crime appears, the last hired person will investigate this crime. 1. Crime appears. There is no free policeman at the time, so this crime will go untreated. 1. One more person is hired. 1. One more person is hired. 1. One more person is hired.
The answer is one, as one crime (on step 5) will go untreated.
| 500
|
[
{
"input": "3\n-1 -1 1",
"output": "2"
},
{
"input": "8\n1 -1 1 -1 -1 1 1 1",
"output": "1"
},
{
"input": "11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1",
"output": "8"
},
{
"input": "7\n-1 -1 1 1 -1 -1 1",
"output": "2"
},
{
"input": "21\n-1 -1 -1 -1 -1 3 2 -1 6 -1 -1 2 1 -1 2 2 1 6 5 -1 5",
"output": "5"
},
{
"input": "98\n-1 -1 1 -1 -1 -1 -1 1 -1 -1 1 -1 -1 1 -1 1 1 1 -1 1 1 1 1 1 -1 1 -1 -1 -1 -1 1 -1 -1 1 1 -1 1 1 1 -1 -1 -1 -1 -1 -1 1 -1 -1 -1 1 -1 1 -1 1 -1 1 1 1 1 1 1 1 -1 -1 1 1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 1 -1 1 1 1 -1 1 1 -1 -1 -1 1 1 1 -1 -1 -1 1 -1 1 1",
"output": "13"
},
{
"input": "3\n-1 5 4",
"output": "1"
},
{
"input": "146\n4 -1 -1 -1 -1 -1 -1 -1 -1 -1 4 -1 3 -1 3 -1 -1 1 4 -1 2 -1 -1 3 -1 -1 -1 4 1 -1 -1 -1 4 -1 -1 -1 -1 -1 -1 3 2 -1 3 2 4 5 2 4 1 5 -1 -1 2 -1 -1 1 -1 5 3 -1 1 2 2 3 5 3 -1 -1 3 -1 -1 3 5 5 -1 -1 5 -1 4 4 1 -1 -1 -1 2 1 -1 -1 -1 2 5 3 -1 -1 -1 3 -1 5 4 -1 1 -1 -1 3 -1 -1 3 1 1 2 -1 -1 -1 1 3 1 -1 2 -1 -1 5 5 -1 -1 3 4 5 1 -1 2 -1 -1 -1 3 -1 5 3 2 -1 2 -1 -1 5 -1 3 -1",
"output": "5"
},
{
"input": "1\n2",
"output": "0"
},
{
"input": "1\n-1",
"output": "1"
},
{
"input": "2\n1 -1",
"output": "0"
},
{
"input": "2\n-1 1",
"output": "1"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "2\n-1 -1",
"output": "2"
},
{
"input": "4\n10 -1 -1 -1",
"output": "0"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "4\n-1 -1 1 1",
"output": "2"
}
] | 1,692,287,392
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 63
| 62
| 10,547,200
|
N = int(input())
events = list(map(int, input().split()))
police_available = untreated_crimes = 0
for event in events:
if event == -1 and police_available == 0:
untreated_crimes += 1
else:
police_available += event if event > 0 else -1
print(untreated_crimes)
|
Title: Police Recruits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The police department of your city has just started its journey. Initially, they don’t have any manpower. So, they started hiring new recruits in groups.
Meanwhile, crimes keeps occurring within the city. One member of the police force can investigate only one crime during his/her lifetime.
If there is no police officer free (isn't busy with crime) during the occurrence of a crime, it will go untreated.
Given the chronological order of crime occurrences and recruit hirings, find the number of crimes which will go untreated.
Input Specification:
The first line of input will contain an integer *n* (1<=≤<=*n*<=≤<=105), the number of events. The next line will contain *n* space-separated integers.
If the integer is -1 then it means a crime has occurred. Otherwise, the integer will be positive, the number of officers recruited together at that time. No more than 10 officers will be recruited at a time.
Output Specification:
Print a single integer, the number of crimes which will go untreated.
Demo Input:
['3\n-1 -1 1\n', '8\n1 -1 1 -1 -1 1 1 1\n', '11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1\n']
Demo Output:
['2\n', '1\n', '8\n']
Note:
Lets consider the second example:
1. Firstly one person is hired. 1. Then crime appears, the last hired person will investigate this crime. 1. One more person is hired. 1. One more crime appears, the last hired person will investigate this crime. 1. Crime appears. There is no free policeman at the time, so this crime will go untreated. 1. One more person is hired. 1. One more person is hired. 1. One more person is hired.
The answer is one, as one crime (on step 5) will go untreated.
|
```python
N = int(input())
events = list(map(int, input().split()))
police_available = untreated_crimes = 0
for event in events:
if event == -1 and police_available == 0:
untreated_crimes += 1
else:
police_available += event if event > 0 else -1
print(untreated_crimes)
```
| 3
|
|
30
|
A
|
Accounting
|
PROGRAMMING
| 1,400
|
[
"brute force",
"math"
] |
A. Accounting
|
2
|
256
|
A long time ago in some far country lived king Copa. After the recent king's reform, he got so large powers that started to keep the books by himself.
The total income *A* of his kingdom during 0-th year is known, as well as the total income *B* during *n*-th year (these numbers can be negative — it means that there was a loss in the correspondent year).
King wants to show financial stability. To do this, he needs to find common coefficient *X* — the coefficient of income growth during one year. This coefficient should satisfy the equation:
Surely, the king is not going to do this job by himself, and demands you to find such number *X*.
It is necessary to point out that the fractional numbers are not used in kingdom's economy. That's why all input numbers as well as coefficient *X* must be integers. The number *X* may be zero or negative.
|
The input contains three integers *A*, *B*, *n* (|*A*|,<=|*B*|<=≤<=1000, 1<=≤<=*n*<=≤<=10).
|
Output the required integer coefficient *X*, or «No solution», if such a coefficient does not exist or it is fractional. If there are several possible solutions, output any of them.
|
[
"2 18 2\n",
"-1 8 3\n",
"0 0 10\n",
"1 16 5\n"
] |
[
"3",
"-2",
"5",
"No solution"
] |
none
| 500
|
[
{
"input": "2 18 2",
"output": "3"
},
{
"input": "-1 8 3",
"output": "-2"
},
{
"input": "0 0 10",
"output": "5"
},
{
"input": "1 16 5",
"output": "No solution"
},
{
"input": "0 1 2",
"output": "No solution"
},
{
"input": "3 0 4",
"output": "0"
},
{
"input": "1 1000 1",
"output": "1000"
},
{
"input": "7 896 7",
"output": "2"
},
{
"input": "4 972 1",
"output": "243"
},
{
"input": "-1 -1 5",
"output": "1"
},
{
"input": "-1 0 4",
"output": "0"
},
{
"input": "-7 0 1",
"output": "0"
},
{
"input": "-5 -5 3",
"output": "1"
},
{
"input": "-5 -5 9",
"output": "1"
},
{
"input": "-5 -5 6",
"output": "1"
},
{
"input": "-4 0 1",
"output": "0"
},
{
"input": "-5 0 3",
"output": "0"
},
{
"input": "-4 4 9",
"output": "-1"
},
{
"input": "10 0 6",
"output": "0"
},
{
"input": "-5 3 4",
"output": "No solution"
},
{
"input": "0 3 6",
"output": "No solution"
},
{
"input": "3 6 10",
"output": "No solution"
},
{
"input": "-3 7 5",
"output": "No solution"
},
{
"input": "-526 526 1",
"output": "-1"
},
{
"input": "-373 373 3",
"output": "-1"
},
{
"input": "-141 0 8",
"output": "0"
},
{
"input": "7 175 1",
"output": "25"
},
{
"input": "-5 -560 1",
"output": "112"
},
{
"input": "-1 -512 10",
"output": "No solution"
},
{
"input": "-3 -768 8",
"output": "2"
},
{
"input": "-3 -768 7",
"output": "No solution"
},
{
"input": "-3 -768 9",
"output": "No solution"
},
{
"input": "-3 -768 4",
"output": "4"
},
{
"input": "4 972 4",
"output": "No solution"
},
{
"input": "4 972 5",
"output": "3"
},
{
"input": "4 972 6",
"output": "No solution"
},
{
"input": "4 972 1",
"output": "243"
},
{
"input": "4 972 2",
"output": "No solution"
},
{
"input": "1 1000 1",
"output": "1000"
},
{
"input": "1 961 2",
"output": "31"
},
{
"input": "1 1000 3",
"output": "10"
},
{
"input": "1 625 4",
"output": "5"
},
{
"input": "4 972 5",
"output": "3"
},
{
"input": "1 729 6",
"output": "3"
},
{
"input": "7 896 7",
"output": "2"
},
{
"input": "3 768 8",
"output": "2"
},
{
"input": "1 512 9",
"output": "2"
},
{
"input": "1 1000 1",
"output": "1000"
},
{
"input": "1 961 2",
"output": "31"
},
{
"input": "1 1000 3",
"output": "10"
},
{
"input": "1 625 4",
"output": "5"
},
{
"input": "4 972 5",
"output": "3"
},
{
"input": "1 729 6",
"output": "3"
},
{
"input": "7 896 7",
"output": "2"
},
{
"input": "3 768 8",
"output": "2"
},
{
"input": "1 512 9",
"output": "2"
},
{
"input": "1 1000 1",
"output": "1000"
},
{
"input": "1 961 2",
"output": "31"
},
{
"input": "1 1000 3",
"output": "10"
},
{
"input": "1 625 4",
"output": "5"
},
{
"input": "4 972 5",
"output": "3"
},
{
"input": "1 729 6",
"output": "3"
},
{
"input": "7 896 7",
"output": "2"
},
{
"input": "3 768 8",
"output": "2"
},
{
"input": "1 512 9",
"output": "2"
},
{
"input": "1 1 5",
"output": "1"
},
{
"input": "1 1 4",
"output": "1"
},
{
"input": "1 -1 1",
"output": "-1"
}
] | 1,626,091,070
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 1
| 186
| 8,294,400
|
import sys
input = sys.stdin.readline
def I(): return input().strip()
def II(): return int(input().strip())
def LI(): return [*map(int, input().strip().split())]
import copy
import re
import string, math, time, functools, random, fractions
from heapq import heappush, heappop, heapify
from bisect import bisect_left, bisect_right
from collections import deque, defaultdict, Counter, OrderedDict
from itertools import permutations, combinations, groupby, count, filterfalse
from operator import itemgetter
for _ in range(1):
a,b,n = LI()
if b==0:
print(0)
continue
if a==0 and b == 0:
print(5)
x = (b/a)**(1/n)
if int(x) == x:
print(int(x))
else:
print("No Solution")
|
Title: Accounting
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A long time ago in some far country lived king Copa. After the recent king's reform, he got so large powers that started to keep the books by himself.
The total income *A* of his kingdom during 0-th year is known, as well as the total income *B* during *n*-th year (these numbers can be negative — it means that there was a loss in the correspondent year).
King wants to show financial stability. To do this, he needs to find common coefficient *X* — the coefficient of income growth during one year. This coefficient should satisfy the equation:
Surely, the king is not going to do this job by himself, and demands you to find such number *X*.
It is necessary to point out that the fractional numbers are not used in kingdom's economy. That's why all input numbers as well as coefficient *X* must be integers. The number *X* may be zero or negative.
Input Specification:
The input contains three integers *A*, *B*, *n* (|*A*|,<=|*B*|<=≤<=1000, 1<=≤<=*n*<=≤<=10).
Output Specification:
Output the required integer coefficient *X*, or «No solution», if such a coefficient does not exist or it is fractional. If there are several possible solutions, output any of them.
Demo Input:
['2 18 2\n', '-1 8 3\n', '0 0 10\n', '1 16 5\n']
Demo Output:
['3', '-2', '5', 'No solution']
Note:
none
|
```python
import sys
input = sys.stdin.readline
def I(): return input().strip()
def II(): return int(input().strip())
def LI(): return [*map(int, input().strip().split())]
import copy
import re
import string, math, time, functools, random, fractions
from heapq import heappush, heappop, heapify
from bisect import bisect_left, bisect_right
from collections import deque, defaultdict, Counter, OrderedDict
from itertools import permutations, combinations, groupby, count, filterfalse
from operator import itemgetter
for _ in range(1):
a,b,n = LI()
if b==0:
print(0)
continue
if a==0 and b == 0:
print(5)
x = (b/a)**(1/n)
if int(x) == x:
print(int(x))
else:
print("No Solution")
```
| -1
|
445
|
A
|
DZY Loves Chessboard
|
PROGRAMMING
| 1,200
|
[
"dfs and similar",
"implementation"
] | null | null |
DZY loves chessboard, and he enjoys playing with it.
He has a chessboard of *n* rows and *m* columns. Some cells of the chessboard are bad, others are good. For every good cell, DZY wants to put a chessman on it. Each chessman is either white or black. After putting all chessmen, DZY wants that no two chessmen with the same color are on two adjacent cells. Two cells are adjacent if and only if they share a common edge.
You task is to find any suitable placement of chessmen on the given chessboard.
|
The first line contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
Each of the next *n* lines contains a string of *m* characters: the *j*-th character of the *i*-th string is either "." or "-". A "." means that the corresponding cell (in the *i*-th row and the *j*-th column) is good, while a "-" means it is bad.
|
Output must contain *n* lines, each line must contain a string of *m* characters. The *j*-th character of the *i*-th string should be either "W", "B" or "-". Character "W" means the chessman on the cell is white, "B" means it is black, "-" means the cell is a bad cell.
If multiple answers exist, print any of them. It is guaranteed that at least one answer exists.
|
[
"1 1\n.\n",
"2 2\n..\n..\n",
"3 3\n.-.\n---\n--."
] |
[
"B\n",
"BW\nWB\n",
"B-B\n---\n--B"
] |
In the first sample, DZY puts a single black chessman. Of course putting a white one is also OK.
In the second sample, all 4 cells are good. No two same chessmen share an edge in the sample output.
In the third sample, no good cells are adjacent. So you can just put 3 chessmen, no matter what their colors are.
| 500
|
[
{
"input": "1 1\n.",
"output": "B"
},
{
"input": "2 2\n..\n..",
"output": "BW\nWB"
},
{
"input": "3 3\n.-.\n---\n--.",
"output": "B-B\n---\n--B"
},
{
"input": "3 1\n-\n.\n.",
"output": "-\nW\nB"
},
{
"input": "11 11\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------",
"output": "-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------"
},
{
"input": "1 1\n-",
"output": "-"
},
{
"input": "4 2\n..\n..\n..\n..",
"output": "BW\nWB\nBW\nWB"
},
{
"input": "5 1\n.\n.\n.\n.\n.",
"output": "B\nW\nB\nW\nB"
}
] | 1,684,314,316
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 37
| 92
| 2,560,000
|
from math import gcd
n,m=map(int,input().split())
l=[]
for _ in range(n):
l.append(list(input()))
# print(l)
for i in range(n):
if i%2==0:
for j in range(m):
if l[i][j]=='.':
if j%2==0:
l[i][j]='B'
else:
l[i][j]='W'
else:
pass
else:
for j in range(m):
if l[i][j]=='.':
if j%2==0:
l[i][j]='W'
else:
l[i][j]='B'
else:
pass
for k in range(n):
print(''.join(l[k]))
|
Title: DZY Loves Chessboard
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
DZY loves chessboard, and he enjoys playing with it.
He has a chessboard of *n* rows and *m* columns. Some cells of the chessboard are bad, others are good. For every good cell, DZY wants to put a chessman on it. Each chessman is either white or black. After putting all chessmen, DZY wants that no two chessmen with the same color are on two adjacent cells. Two cells are adjacent if and only if they share a common edge.
You task is to find any suitable placement of chessmen on the given chessboard.
Input Specification:
The first line contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
Each of the next *n* lines contains a string of *m* characters: the *j*-th character of the *i*-th string is either "." or "-". A "." means that the corresponding cell (in the *i*-th row and the *j*-th column) is good, while a "-" means it is bad.
Output Specification:
Output must contain *n* lines, each line must contain a string of *m* characters. The *j*-th character of the *i*-th string should be either "W", "B" or "-". Character "W" means the chessman on the cell is white, "B" means it is black, "-" means the cell is a bad cell.
If multiple answers exist, print any of them. It is guaranteed that at least one answer exists.
Demo Input:
['1 1\n.\n', '2 2\n..\n..\n', '3 3\n.-.\n---\n--.']
Demo Output:
['B\n', 'BW\nWB\n', 'B-B\n---\n--B']
Note:
In the first sample, DZY puts a single black chessman. Of course putting a white one is also OK.
In the second sample, all 4 cells are good. No two same chessmen share an edge in the sample output.
In the third sample, no good cells are adjacent. So you can just put 3 chessmen, no matter what their colors are.
|
```python
from math import gcd
n,m=map(int,input().split())
l=[]
for _ in range(n):
l.append(list(input()))
# print(l)
for i in range(n):
if i%2==0:
for j in range(m):
if l[i][j]=='.':
if j%2==0:
l[i][j]='B'
else:
l[i][j]='W'
else:
pass
else:
for j in range(m):
if l[i][j]=='.':
if j%2==0:
l[i][j]='W'
else:
l[i][j]='B'
else:
pass
for k in range(n):
print(''.join(l[k]))
```
| 3
|
|
1,003
|
A
|
Polycarp's Pockets
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Polycarp has $n$ coins, the value of the $i$-th coin is $a_i$. Polycarp wants to distribute all the coins between his pockets, but he cannot put two coins with the same value into the same pocket.
For example, if Polycarp has got six coins represented as an array $a = [1, 2, 4, 3, 3, 2]$, he can distribute the coins into two pockets as follows: $[1, 2, 3], [2, 3, 4]$.
Polycarp wants to distribute all the coins with the minimum number of used pockets. Help him to do that.
|
The first line of the input contains one integer $n$ ($1 \le n \le 100$) — the number of coins.
The second line of the input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$) — values of coins.
|
Print only one integer — the minimum number of pockets Polycarp needs to distribute all the coins so no two coins with the same value are put into the same pocket.
|
[
"6\n1 2 4 3 3 2\n",
"1\n100\n"
] |
[
"2\n",
"1\n"
] |
none
| 0
|
[
{
"input": "6\n1 2 4 3 3 2",
"output": "2"
},
{
"input": "1\n100",
"output": "1"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "100"
},
{
"input": "100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "100"
},
{
"input": "100\n59 47 39 47 47 71 47 28 58 47 35 79 58 47 38 47 47 47 47 27 47 43 29 95 47 49 46 71 47 74 79 47 47 32 45 67 47 47 30 37 47 47 16 67 22 76 47 86 84 10 5 47 47 47 47 47 1 51 47 54 47 8 47 47 9 47 47 47 47 28 47 47 26 47 47 47 47 47 47 92 47 47 77 47 47 24 45 47 10 47 47 89 47 27 47 89 47 67 24 71",
"output": "51"
},
{
"input": "100\n45 99 10 27 16 85 39 38 17 32 15 23 67 48 50 97 42 70 62 30 44 81 64 73 34 22 46 5 83 52 58 60 33 74 47 88 18 61 78 53 25 95 94 31 3 75 1 57 20 54 59 9 68 7 77 43 21 87 86 24 4 80 11 49 2 72 36 84 71 8 65 55 79 100 41 14 35 89 66 69 93 37 56 82 90 91 51 19 26 92 6 96 13 98 12 28 76 40 63 29",
"output": "1"
},
{
"input": "100\n45 29 5 2 6 50 22 36 14 15 9 48 46 20 8 37 7 47 12 50 21 38 18 27 33 19 40 10 5 49 38 42 34 37 27 30 35 24 10 3 40 49 41 3 4 44 13 25 28 31 46 36 23 1 1 23 7 22 35 26 21 16 48 42 32 8 11 16 34 11 39 32 47 28 43 41 39 4 14 19 26 45 13 18 15 25 2 44 17 29 17 33 43 6 12 30 9 20 31 24",
"output": "2"
},
{
"input": "50\n7 7 3 3 7 4 5 6 4 3 7 5 6 4 5 4 4 5 6 7 7 7 4 5 5 5 3 7 6 3 4 6 3 6 4 4 5 4 6 6 3 5 6 3 5 3 3 7 7 6",
"output": "10"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 99 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "99"
},
{
"input": "7\n1 2 3 3 3 1 2",
"output": "3"
},
{
"input": "5\n1 2 3 4 5",
"output": "1"
},
{
"input": "7\n1 2 3 4 5 6 7",
"output": "1"
},
{
"input": "8\n1 2 3 4 5 6 7 8",
"output": "1"
},
{
"input": "9\n1 2 3 4 5 6 7 8 9",
"output": "1"
},
{
"input": "10\n1 2 3 4 5 6 7 8 9 10",
"output": "1"
},
{
"input": "3\n2 1 1",
"output": "2"
},
{
"input": "11\n1 2 3 4 5 6 7 8 9 1 1",
"output": "3"
},
{
"input": "12\n1 2 1 1 1 1 1 1 1 1 1 1",
"output": "11"
},
{
"input": "13\n1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "13"
},
{
"input": "14\n1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "14"
},
{
"input": "15\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "15"
},
{
"input": "16\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "16"
},
{
"input": "3\n1 1 1",
"output": "3"
},
{
"input": "3\n1 2 3",
"output": "1"
},
{
"input": "10\n1 1 1 1 2 2 1 1 9 10",
"output": "6"
},
{
"input": "2\n1 1",
"output": "2"
},
{
"input": "56\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "56"
},
{
"input": "99\n35 96 73 72 70 83 22 93 98 75 45 32 81 82 45 54 25 7 53 72 29 2 94 19 21 98 34 28 39 99 55 85 44 23 6 47 98 2 33 34 19 57 49 35 67 4 60 4 4 23 55 6 57 66 16 68 34 45 84 79 48 63 4 9 46 88 98 13 19 27 83 12 4 63 57 22 44 77 44 62 28 52 44 64 9 24 55 22 48 4 2 9 80 76 45 1 56 22 92",
"output": "6"
},
{
"input": "10\n1 2 2 3 3 3 4 4 4 4",
"output": "4"
},
{
"input": "99\n97 44 33 56 42 10 61 85 64 26 40 39 82 34 75 9 51 51 39 73 58 38 74 31 13 99 58 1 28 89 76 19 52 7 40 56 12 27 72 72 67 75 62 46 22 55 35 16 18 39 60 63 92 42 85 69 34 61 73 50 57 95 30 4 45 63 76 58 32 35 48 81 10 78 95 79 55 97 21 21 22 94 30 17 78 57 89 93 100 44 16 89 68 55 19 46 42 73 21",
"output": "3"
},
{
"input": "5\n5 5 5 5 1",
"output": "4"
},
{
"input": "6\n2 3 2 5 2 6",
"output": "3"
},
{
"input": "3\n58 59 58",
"output": "2"
},
{
"input": "9\n1 2 3 4 5 6 7 8 8",
"output": "2"
},
{
"input": "97\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "97"
},
{
"input": "3\n95 95 4",
"output": "2"
},
{
"input": "3\n2 2 5",
"output": "2"
}
] | 1,612,597,234
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 37
| 61
| 0
|
n=int(input())
a=[int(i) for i in input().split()]
maxi=0
for i in a:
if maxi<a.count(i):
maxi=a.count(i)
print(maxi)
|
Title: Polycarp's Pockets
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp has $n$ coins, the value of the $i$-th coin is $a_i$. Polycarp wants to distribute all the coins between his pockets, but he cannot put two coins with the same value into the same pocket.
For example, if Polycarp has got six coins represented as an array $a = [1, 2, 4, 3, 3, 2]$, he can distribute the coins into two pockets as follows: $[1, 2, 3], [2, 3, 4]$.
Polycarp wants to distribute all the coins with the minimum number of used pockets. Help him to do that.
Input Specification:
The first line of the input contains one integer $n$ ($1 \le n \le 100$) — the number of coins.
The second line of the input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$) — values of coins.
Output Specification:
Print only one integer — the minimum number of pockets Polycarp needs to distribute all the coins so no two coins with the same value are put into the same pocket.
Demo Input:
['6\n1 2 4 3 3 2\n', '1\n100\n']
Demo Output:
['2\n', '1\n']
Note:
none
|
```python
n=int(input())
a=[int(i) for i in input().split()]
maxi=0
for i in a:
if maxi<a.count(i):
maxi=a.count(i)
print(maxi)
```
| 3
|
|
757
|
A
|
Gotta Catch Em' All!
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
Bash wants to become a Pokemon master one day. Although he liked a lot of Pokemon, he has always been fascinated by Bulbasaur the most. Soon, things started getting serious and his fascination turned into an obsession. Since he is too young to go out and catch Bulbasaur, he came up with his own way of catching a Bulbasaur.
Each day, he takes the front page of the newspaper. He cuts out the letters one at a time, from anywhere on the front page of the newspaper to form the word "Bulbasaur" (without quotes) and sticks it on his wall. Bash is very particular about case — the first letter of "Bulbasaur" must be upper case and the rest must be lower case. By doing this he thinks he has caught one Bulbasaur. He then repeats this step on the left over part of the newspaper. He keeps doing this until it is not possible to form the word "Bulbasaur" from the newspaper.
Given the text on the front page of the newspaper, can you tell how many Bulbasaurs he will catch today?
Note: uppercase and lowercase letters are considered different.
|
Input contains a single line containing a string *s* (1<=<=≤<=<=|*s*|<=<=≤<=<=105) — the text on the front page of the newspaper without spaces and punctuation marks. |*s*| is the length of the string *s*.
The string *s* contains lowercase and uppercase English letters, i.e. .
|
Output a single integer, the answer to the problem.
|
[
"Bulbbasaur\n",
"F\n",
"aBddulbasaurrgndgbualdBdsagaurrgndbb\n"
] |
[
"1\n",
"0\n",
"2\n"
] |
In the first case, you could pick: Bulbbasaur.
In the second case, there is no way to pick even a single Bulbasaur.
In the third case, you can rearrange the string to BulbasaurBulbasauraddrgndgddgargndbb to get two words "Bulbasaur".
| 500
|
[
{
"input": "Bulbbasaur",
"output": "1"
},
{
"input": "F",
"output": "0"
},
{
"input": "aBddulbasaurrgndgbualdBdsagaurrgndbb",
"output": "2"
},
{
"input": "BBBBBBBBBBbbbbbbbbbbuuuuuuuuuullllllllllssssssssssaaaaaaaaaarrrrrrrrrr",
"output": "5"
},
{
"input": "BBBBBBBBBBbbbbbbbbbbbbbbbbbbbbuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuussssssssssssssssssssaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "0"
},
{
"input": "BBBBBBBBBBssssssssssssssssssssaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaarrrrrrrrrr",
"output": "0"
},
{
"input": "BBBBBBBBBBbbbbbbbbbbbbbbbbbbbbuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuullllllllllllllllllllssssssssssssssssssssaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaarrrrrrrrrrrrrrrrrrrr",
"output": "10"
},
{
"input": "BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBbbbbbbbbbbbbbbbbbbbbuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuullllllllllllllllllllssssssssssssssssssssaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaarrrrrrrrrrrrrrrrrrrrrrrrrrrrrr",
"output": "20"
},
{
"input": "CeSlSwec",
"output": "0"
},
{
"input": "PnMrWPBGzVcmRcO",
"output": "0"
},
{
"input": "hHPWBQeEmCuhdCnzrqYtuFtwxokGhdGkFtsFICVqYfJeUrSBtSxEbzMCblOgqOvjXURhSKivPcseqgiNuUgIboEYMvVeRBbpzCGCfVydDvZNFGSFidwUtNbmPSfSYdMNmHgchIsiVswzFsGQewlMVEzicOagpWMdCWrCdPmexfnM",
"output": "0"
},
{
"input": "BBBBBBBBBBbbbbbbbbbbbbuuuuuuuuuuuullllllllllllssssssssssssaaaaaaaaaaaarrrrrrrrrrrrZBphUC",
"output": "6"
},
{
"input": "bulsar",
"output": "0"
},
{
"input": "Bblsar",
"output": "0"
},
{
"input": "Bbusar",
"output": "0"
},
{
"input": "Bbular",
"output": "0"
},
{
"input": "Bbulsr",
"output": "0"
},
{
"input": "Bbulsa",
"output": "0"
},
{
"input": "Bbulsar",
"output": "0"
},
{
"input": "Bbulsar",
"output": "0"
},
{
"input": "CaQprCjTiQACZjUJjSmMHVTDorSUugvTtksEjptVzNLhClWaVVWszIixBlqFkvjDmbRjarQoUWhXHoCgYNNjvEgRTgKpbdEMFsmqcTyvJzupKgYiYMtrZWXIAGVhmDURtddbBZIMgIgXqQUmXpssLSaVCDGZDHimNthwiAWabjtcraAQugMCpBPQZbBGZyqUZmzDVSvJZmDWfZEUHGJVtiJANAIbvjTxtvvTbjWRpNQZlxAqpLCLRVwYWqLaHOTvzgeNGdxiBwsAVKKsewXMTwZUUfxYwrwsiaRBwEdvDDoPsQUtinvajBoRzLBUuQekhjsfDAOQzIABSVPitRuhvvqeAahsSELTGbCPh",
"output": "2"
},
{
"input": "Bulbasaur",
"output": "1"
},
{
"input": "BulbasaurBulbasaur",
"output": "2"
},
{
"input": "Bulbbasar",
"output": "0"
},
{
"input": "Bulbasur",
"output": "0"
},
{
"input": "Bulbsaur",
"output": "0"
},
{
"input": "BulbsurBulbsurBulbsurBulbsur",
"output": "0"
},
{
"input": "Blbbasar",
"output": "0"
},
{
"input": "Bulbasar",
"output": "0"
},
{
"input": "BBullllbbaassaauurr",
"output": "1"
},
{
"input": "BulbasaurBulbasar",
"output": "1"
},
{
"input": "BulbasaurBulbsaur",
"output": "1"
},
{
"input": "Bubasaur",
"output": "0"
},
{
"input": "ulbasaurulbasaur",
"output": "0"
},
{
"input": "Bulbasr",
"output": "0"
},
{
"input": "BBBuuulllbbbaaasssaaauuurrr",
"output": "3"
},
{
"input": "BBuuuullbbaaaassrr",
"output": "2"
},
{
"input": "BBBBBBBuuuuuuuullllllllllllbbbbaaaaaassssssssssssssssaaaaauuuuuuuuuuuuurrrrrrrrrrrrrrrr",
"output": "4"
},
{
"input": "BBuullbbaassaarr",
"output": "1"
},
{
"input": "Bulbasau",
"output": "0"
},
{
"input": "BBuullbbaassaauurr",
"output": "2"
},
{
"input": "BulbasauBulbasauBulbasauBulbasauBulbasauBulbasauBulbasauBulbasau",
"output": "0"
},
{
"input": "Blbasaur",
"output": "0"
},
{
"input": "BulbasaurBulbasaurd",
"output": "2"
},
{
"input": "ulbasaur",
"output": "0"
},
{
"input": "Bulbaaur",
"output": "0"
},
{
"input": "BBuuuullbbbbbbbbbbbbbbbaassrr",
"output": "1"
},
{
"input": "Bulbasua",
"output": "0"
},
{
"input": "Bubbasaur",
"output": "0"
},
{
"input": "BulbasauBulbasauBulbasauBulbasauBulbasauBulbasaurrr",
"output": "3"
},
{
"input": "BulbasaurBubasaur",
"output": "1"
},
{
"input": "Baab",
"output": "0"
},
{
"input": "BulbasaurBulbasau",
"output": "1"
},
{
"input": "Bulbasauu",
"output": "0"
},
{
"input": "BulbasauBulbasau",
"output": "0"
},
{
"input": "BBBBBBBBBBB",
"output": "0"
},
{
"input": "Bulbbasau",
"output": "0"
},
{
"input": "BulbbasaurBulbbasar",
"output": "1"
},
{
"input": "Bulaaaasaur",
"output": "0"
},
{
"input": "BulbasaurBulbasauBulbasauBulbasau",
"output": "1"
}
] | 1,618,510,951
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 6
| 62
| 0
|
n=input()
print(min(n.count('B'),n.count('u')/2,n.count('l'),n.count('b'),n.count('a')/2,n.count('s'),n.count('r')))
|
Title: Gotta Catch Em' All!
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bash wants to become a Pokemon master one day. Although he liked a lot of Pokemon, he has always been fascinated by Bulbasaur the most. Soon, things started getting serious and his fascination turned into an obsession. Since he is too young to go out and catch Bulbasaur, he came up with his own way of catching a Bulbasaur.
Each day, he takes the front page of the newspaper. He cuts out the letters one at a time, from anywhere on the front page of the newspaper to form the word "Bulbasaur" (without quotes) and sticks it on his wall. Bash is very particular about case — the first letter of "Bulbasaur" must be upper case and the rest must be lower case. By doing this he thinks he has caught one Bulbasaur. He then repeats this step on the left over part of the newspaper. He keeps doing this until it is not possible to form the word "Bulbasaur" from the newspaper.
Given the text on the front page of the newspaper, can you tell how many Bulbasaurs he will catch today?
Note: uppercase and lowercase letters are considered different.
Input Specification:
Input contains a single line containing a string *s* (1<=<=≤<=<=|*s*|<=<=≤<=<=105) — the text on the front page of the newspaper without spaces and punctuation marks. |*s*| is the length of the string *s*.
The string *s* contains lowercase and uppercase English letters, i.e. .
Output Specification:
Output a single integer, the answer to the problem.
Demo Input:
['Bulbbasaur\n', 'F\n', 'aBddulbasaurrgndgbualdBdsagaurrgndbb\n']
Demo Output:
['1\n', '0\n', '2\n']
Note:
In the first case, you could pick: Bulbbasaur.
In the second case, there is no way to pick even a single Bulbasaur.
In the third case, you can rearrange the string to BulbasaurBulbasauraddrgndgddgargndbb to get two words "Bulbasaur".
|
```python
n=input()
print(min(n.count('B'),n.count('u')/2,n.count('l'),n.count('b'),n.count('a')/2,n.count('s'),n.count('r')))
```
| 0
|
|
443
|
A
|
Anton and Letters
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"implementation"
] | null | null |
Recently, Anton has found a set. The set consists of small English letters. Anton carefully wrote out all the letters from the set in one line, separated by a comma. He also added an opening curved bracket at the beginning of the line and a closing curved bracket at the end of the line.
Unfortunately, from time to time Anton would forget writing some letter and write it again. He asks you to count the total number of distinct letters in his set.
|
The first and the single line contains the set of letters. The length of the line doesn't exceed 1000. It is guaranteed that the line starts from an opening curved bracket and ends with a closing curved bracket. Between them, small English letters are listed, separated by a comma. Each comma is followed by a space.
|
Print a single number — the number of distinct letters in Anton's set.
|
[
"{a, b, c}\n",
"{b, a, b, a}\n",
"{}\n"
] |
[
"3\n",
"2\n",
"0\n"
] |
none
| 500
|
[
{
"input": "{a, b, c}",
"output": "3"
},
{
"input": "{b, a, b, a}",
"output": "2"
},
{
"input": "{}",
"output": "0"
},
{
"input": "{a, a, c, b, b, b, c, c, c, c}",
"output": "3"
},
{
"input": "{a, c, b, b}",
"output": "3"
},
{
"input": "{a, b}",
"output": "2"
},
{
"input": "{a}",
"output": "1"
},
{
"input": "{b, a, b, a, b, c, c, b, c, b}",
"output": "3"
},
{
"input": "{e, g, c, e}",
"output": "3"
},
{
"input": "{a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a}",
"output": "1"
},
{
"input": "{a, a, b}",
"output": "2"
},
{
"input": "{a, b, b, b, a, b, a, a, a, a, a, a, b, a, b, a, a, a, a, a, b, a, b, a}",
"output": "2"
},
{
"input": "{j, u, a, c, f, w, e, w, x, t, h, p, v, n, i, l, x, n, i, b, u, c, a, a}",
"output": "16"
},
{
"input": "{x, i, w, c, p, e, h, z, k, i}",
"output": "9"
},
{
"input": "{t, k, o, x, r, d, q, j, k, e, z, w, y, r, z, s, s, e, s, b, k, i}",
"output": "15"
},
{
"input": "{y}",
"output": "1"
},
{
"input": "{x}",
"output": "1"
},
{
"input": "{b, z, a, z}",
"output": "3"
},
{
"input": "{z}",
"output": "1"
},
{
"input": "{a, z}",
"output": "2"
},
{
"input": "{a, b, z}",
"output": "3"
},
{
"input": "{s, q, z, r, t, a, b, h, j, i, o, z, r, q}",
"output": "11"
}
] | 1,686,375,076
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 27
| 46
| 0
|
a = input().replace(', ', '')
b = []
for i in a:
b.append(i)
b.pop(0)
b.pop(-1)
n = []
for i in b:
if i not in n:
n.append(i)
print(len(n))
|
Title: Anton and Letters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently, Anton has found a set. The set consists of small English letters. Anton carefully wrote out all the letters from the set in one line, separated by a comma. He also added an opening curved bracket at the beginning of the line and a closing curved bracket at the end of the line.
Unfortunately, from time to time Anton would forget writing some letter and write it again. He asks you to count the total number of distinct letters in his set.
Input Specification:
The first and the single line contains the set of letters. The length of the line doesn't exceed 1000. It is guaranteed that the line starts from an opening curved bracket and ends with a closing curved bracket. Between them, small English letters are listed, separated by a comma. Each comma is followed by a space.
Output Specification:
Print a single number — the number of distinct letters in Anton's set.
Demo Input:
['{a, b, c}\n', '{b, a, b, a}\n', '{}\n']
Demo Output:
['3\n', '2\n', '0\n']
Note:
none
|
```python
a = input().replace(', ', '')
b = []
for i in a:
b.append(i)
b.pop(0)
b.pop(-1)
n = []
for i in b:
if i not in n:
n.append(i)
print(len(n))
```
| 3
|
|
633
|
B
|
A Trivial Problem
|
PROGRAMMING
| 1,300
|
[
"brute force",
"constructive algorithms",
"math",
"number theory"
] | null | null |
Mr. Santa asks all the great programmers of the world to solve a trivial problem. He gives them an integer *m* and asks for the number of positive integers *n*, such that the factorial of *n* ends with exactly *m* zeroes. Are you among those great programmers who can solve this problem?
|
The only line of input contains an integer *m* (1<=≤<=*m*<=≤<=100<=000) — the required number of trailing zeroes in factorial.
|
First print *k* — the number of values of *n* such that the factorial of *n* ends with *m* zeroes. Then print these *k* integers in increasing order.
|
[
"1\n",
"5\n"
] |
[
"5\n5 6 7 8 9 ",
"0"
] |
The factorial of *n* is equal to the product of all integers from 1 to *n* inclusive, that is *n*! = 1·2·3·...·*n*.
In the first sample, 5! = 120, 6! = 720, 7! = 5040, 8! = 40320 and 9! = 362880.
| 500
|
[
{
"input": "1",
"output": "5\n5 6 7 8 9 "
},
{
"input": "5",
"output": "0"
},
{
"input": "2",
"output": "5\n10 11 12 13 14 "
},
{
"input": "3",
"output": "5\n15 16 17 18 19 "
},
{
"input": "7",
"output": "5\n30 31 32 33 34 "
},
{
"input": "12",
"output": "5\n50 51 52 53 54 "
},
{
"input": "15",
"output": "5\n65 66 67 68 69 "
},
{
"input": "18",
"output": "5\n75 76 77 78 79 "
},
{
"input": "38",
"output": "5\n155 156 157 158 159 "
},
{
"input": "47",
"output": "5\n195 196 197 198 199 "
},
{
"input": "58",
"output": "5\n240 241 242 243 244 "
},
{
"input": "66",
"output": "5\n270 271 272 273 274 "
},
{
"input": "70",
"output": "5\n285 286 287 288 289 "
},
{
"input": "89",
"output": "5\n365 366 367 368 369 "
},
{
"input": "417",
"output": "5\n1675 1676 1677 1678 1679 "
},
{
"input": "815",
"output": "5\n3265 3266 3267 3268 3269 "
},
{
"input": "394",
"output": "5\n1585 1586 1587 1588 1589 "
},
{
"input": "798",
"output": "0"
},
{
"input": "507",
"output": "5\n2035 2036 2037 2038 2039 "
},
{
"input": "406",
"output": "5\n1630 1631 1632 1633 1634 "
},
{
"input": "570",
"output": "5\n2290 2291 2292 2293 2294 "
},
{
"input": "185",
"output": "0"
},
{
"input": "765",
"output": "0"
},
{
"input": "967",
"output": "0"
},
{
"input": "112",
"output": "5\n455 456 457 458 459 "
},
{
"input": "729",
"output": "5\n2925 2926 2927 2928 2929 "
},
{
"input": "4604",
"output": "5\n18425 18426 18427 18428 18429 "
},
{
"input": "8783",
"output": "5\n35140 35141 35142 35143 35144 "
},
{
"input": "1059",
"output": "0"
},
{
"input": "6641",
"output": "5\n26575 26576 26577 26578 26579 "
},
{
"input": "9353",
"output": "5\n37425 37426 37427 37428 37429 "
},
{
"input": "1811",
"output": "5\n7250 7251 7252 7253 7254 "
},
{
"input": "2528",
"output": "0"
},
{
"input": "8158",
"output": "5\n32640 32641 32642 32643 32644 "
},
{
"input": "3014",
"output": "5\n12070 12071 12072 12073 12074 "
},
{
"input": "7657",
"output": "5\n30640 30641 30642 30643 30644 "
},
{
"input": "4934",
"output": "0"
},
{
"input": "9282",
"output": "5\n37140 37141 37142 37143 37144 "
},
{
"input": "2610",
"output": "5\n10450 10451 10452 10453 10454 "
},
{
"input": "2083",
"output": "5\n8345 8346 8347 8348 8349 "
},
{
"input": "26151",
"output": "5\n104620 104621 104622 104623 104624 "
},
{
"input": "64656",
"output": "5\n258640 258641 258642 258643 258644 "
},
{
"input": "46668",
"output": "5\n186690 186691 186692 186693 186694 "
},
{
"input": "95554",
"output": "5\n382235 382236 382237 382238 382239 "
},
{
"input": "37320",
"output": "0"
},
{
"input": "52032",
"output": "5\n208140 208141 208142 208143 208144 "
},
{
"input": "11024",
"output": "5\n44110 44111 44112 44113 44114 "
},
{
"input": "63218",
"output": "5\n252885 252886 252887 252888 252889 "
},
{
"input": "40095",
"output": "5\n160390 160391 160392 160393 160394 "
},
{
"input": "42724",
"output": "5\n170910 170911 170912 170913 170914 "
},
{
"input": "24381",
"output": "5\n97530 97531 97532 97533 97534 "
},
{
"input": "73138",
"output": "5\n292570 292571 292572 292573 292574 "
},
{
"input": "93346",
"output": "5\n373400 373401 373402 373403 373404 "
},
{
"input": "18338",
"output": "5\n73370 73371 73372 73373 73374 "
},
{
"input": "42662",
"output": "5\n170660 170661 170662 170663 170664 "
},
{
"input": "81221",
"output": "5\n324900 324901 324902 324903 324904 "
},
{
"input": "100000",
"output": "5\n400005 400006 400007 400008 400009 "
},
{
"input": "100000",
"output": "5\n400005 400006 400007 400008 400009 "
},
{
"input": "99998",
"output": "0"
},
{
"input": "30",
"output": "0"
},
{
"input": "11",
"output": "0"
},
{
"input": "780",
"output": "0"
},
{
"input": "97656",
"output": "5\n390625 390626 390627 390628 390629 "
},
{
"input": "12499",
"output": "5\n50000 50001 50002 50003 50004 "
},
{
"input": "65",
"output": "5\n265 266 267 268 269 "
},
{
"input": "41",
"output": "5\n170 171 172 173 174 "
},
{
"input": "31",
"output": "5\n125 126 127 128 129 "
},
{
"input": "86577",
"output": "0"
}
] | 1,456,555,125
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 62
| 0
|
m=int(input())
i=m*5
anst=0
while i>=0:
j=1
ans=0
while 5**j<=i:
ans+=int(i/(5**j))
j+=1
if ans==m:
anst+=i
break
else:
i-=1
if anst==0:
print(0)
else:
print(5)
print(anst,anst+1,anst+2,anst+3,anst+4)
|
Title: A Trivial Problem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mr. Santa asks all the great programmers of the world to solve a trivial problem. He gives them an integer *m* and asks for the number of positive integers *n*, such that the factorial of *n* ends with exactly *m* zeroes. Are you among those great programmers who can solve this problem?
Input Specification:
The only line of input contains an integer *m* (1<=≤<=*m*<=≤<=100<=000) — the required number of trailing zeroes in factorial.
Output Specification:
First print *k* — the number of values of *n* such that the factorial of *n* ends with *m* zeroes. Then print these *k* integers in increasing order.
Demo Input:
['1\n', '5\n']
Demo Output:
['5\n5 6 7 8 9 ', '0']
Note:
The factorial of *n* is equal to the product of all integers from 1 to *n* inclusive, that is *n*! = 1·2·3·...·*n*.
In the first sample, 5! = 120, 6! = 720, 7! = 5040, 8! = 40320 and 9! = 362880.
|
```python
m=int(input())
i=m*5
anst=0
while i>=0:
j=1
ans=0
while 5**j<=i:
ans+=int(i/(5**j))
j+=1
if ans==m:
anst+=i
break
else:
i-=1
if anst==0:
print(0)
else:
print(5)
print(anst,anst+1,anst+2,anst+3,anst+4)
```
| 0
|
|
625
|
A
|
Guest From the Past
|
PROGRAMMING
| 1,700
|
[
"implementation",
"math"
] | null | null |
Kolya Gerasimov loves kefir very much. He lives in year 1984 and knows all the details of buying this delicious drink. One day, as you probably know, he found himself in year 2084, and buying kefir there is much more complicated.
Kolya is hungry, so he went to the nearest milk shop. In 2084 you may buy kefir in a plastic liter bottle, that costs *a* rubles, or in glass liter bottle, that costs *b* rubles. Also, you may return empty glass bottle and get *c* (*c*<=<<=*b*) rubles back, but you cannot return plastic bottles.
Kolya has *n* rubles and he is really hungry, so he wants to drink as much kefir as possible. There were no plastic bottles in his 1984, so Kolya doesn't know how to act optimally and asks for your help.
|
First line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1018) — the number of rubles Kolya has at the beginning.
Then follow three lines containing integers *a*, *b* and *c* (1<=≤<=*a*<=≤<=1018, 1<=≤<=*c*<=<<=*b*<=≤<=1018) — the cost of one plastic liter bottle, the cost of one glass liter bottle and the money one can get back by returning an empty glass bottle, respectively.
|
Print the only integer — maximum number of liters of kefir, that Kolya can drink.
|
[
"10\n11\n9\n8\n",
"10\n5\n6\n1\n"
] |
[
"2\n",
"2\n"
] |
In the first sample, Kolya can buy one glass bottle, then return it and buy one more glass bottle. Thus he will drink 2 liters of kefir.
In the second sample, Kolya can buy two plastic bottle and get two liters of kefir, or he can buy one liter glass bottle, then return it and buy one plastic bottle. In both cases he will drink two liters of kefir.
| 750
|
[
{
"input": "10\n11\n9\n8",
"output": "2"
},
{
"input": "10\n5\n6\n1",
"output": "2"
},
{
"input": "2\n2\n2\n1",
"output": "1"
},
{
"input": "10\n3\n3\n1",
"output": "4"
},
{
"input": "10\n1\n2\n1",
"output": "10"
},
{
"input": "10\n2\n3\n1",
"output": "5"
},
{
"input": "9\n2\n4\n1",
"output": "4"
},
{
"input": "9\n2\n2\n1",
"output": "8"
},
{
"input": "9\n10\n10\n1",
"output": "0"
},
{
"input": "10\n2\n2\n1",
"output": "9"
},
{
"input": "1000000000000000000\n2\n10\n9",
"output": "999999999999999995"
},
{
"input": "501000000000000000\n300000000000000000\n301000000000000000\n100000000000000000",
"output": "2"
},
{
"input": "10\n1\n9\n8",
"output": "10"
},
{
"input": "10\n8\n8\n7",
"output": "3"
},
{
"input": "10\n5\n5\n1",
"output": "2"
},
{
"input": "29\n3\n3\n1",
"output": "14"
},
{
"input": "45\n9\n9\n8",
"output": "37"
},
{
"input": "45\n9\n9\n1",
"output": "5"
},
{
"input": "100\n10\n10\n9",
"output": "91"
},
{
"input": "179\n10\n9\n1",
"output": "22"
},
{
"input": "179\n2\n2\n1",
"output": "178"
},
{
"input": "179\n179\n179\n1",
"output": "1"
},
{
"input": "179\n59\n59\n58",
"output": "121"
},
{
"input": "500\n250\n250\n1",
"output": "2"
},
{
"input": "500\n1\n250\n1",
"output": "500"
},
{
"input": "501\n500\n500\n499",
"output": "2"
},
{
"input": "501\n450\n52\n1",
"output": "9"
},
{
"input": "501\n300\n301\n100",
"output": "2"
},
{
"input": "500\n179\n10\n1",
"output": "55"
},
{
"input": "1000\n500\n10\n9",
"output": "991"
},
{
"input": "1000\n2\n10\n9",
"output": "995"
},
{
"input": "1001\n1000\n1000\n999",
"output": "2"
},
{
"input": "10000\n10000\n10000\n1",
"output": "1"
},
{
"input": "10000\n10\n5000\n4999",
"output": "5500"
},
{
"input": "1000000000\n999999998\n999999999\n999999998",
"output": "3"
},
{
"input": "1000000000\n50\n50\n49",
"output": "999999951"
},
{
"input": "1000000000\n500\n5000\n4999",
"output": "999995010"
},
{
"input": "1000000000\n51\n100\n98",
"output": "499999952"
},
{
"input": "1000000000\n100\n51\n50",
"output": "999999950"
},
{
"input": "1000000000\n2\n5\n4",
"output": "999999998"
},
{
"input": "1000000000000000000\n999999998000000000\n999999999000000000\n999999998000000000",
"output": "3"
},
{
"input": "1000000000\n2\n2\n1",
"output": "999999999"
},
{
"input": "999999999\n2\n999999998\n1",
"output": "499999999"
},
{
"input": "999999999999999999\n2\n2\n1",
"output": "999999999999999998"
},
{
"input": "999999999999999999\n10\n10\n9",
"output": "999999999999999990"
},
{
"input": "999999999999999999\n999999999999999998\n999999999999999998\n999999999999999997",
"output": "2"
},
{
"input": "999999999999999999\n501\n501\n1",
"output": "1999999999999999"
},
{
"input": "999999999999999999\n2\n50000000000000000\n49999999999999999",
"output": "974999999999999999"
},
{
"input": "999999999999999999\n180\n180\n1",
"output": "5586592178770949"
},
{
"input": "1000000000000000000\n42\n41\n1",
"output": "24999999999999999"
},
{
"input": "1000000000000000000\n41\n40\n1",
"output": "25641025641025641"
},
{
"input": "100000000000000000\n79\n100\n25",
"output": "1333333333333333"
},
{
"input": "1\n100\n5\n4",
"output": "0"
},
{
"input": "1000000000000000000\n1000000000000000000\n10000000\n9999999",
"output": "999999999990000001"
},
{
"input": "999999999999999999\n999999999000000000\n900000000000000000\n899999999999999999",
"output": "100000000000000000"
},
{
"input": "13\n10\n15\n11",
"output": "1"
},
{
"input": "1\n1000\n5\n4",
"output": "0"
},
{
"input": "10\n100\n10\n1",
"output": "1"
},
{
"input": "3\n2\n100000\n99999",
"output": "1"
},
{
"input": "4\n2\n4\n2",
"output": "2"
},
{
"input": "5\n3\n6\n4",
"output": "1"
},
{
"input": "1\n7\n65\n49",
"output": "0"
},
{
"input": "10\n20\n100\n99",
"output": "0"
},
{
"input": "10000000000\n10000000000\n9000000000\n8999999999",
"output": "1000000001"
},
{
"input": "90\n30\n101\n100",
"output": "3"
},
{
"input": "999999999999999\n5\n500000000000000\n499999999999999",
"output": "599999999999999"
},
{
"input": "1000000000000000000\n1000000000000000000\n1000000000\n999999999",
"output": "999999999000000001"
},
{
"input": "1\n1000000000000000000\n1000000000\n999999999",
"output": "0"
},
{
"input": "100000000000000000\n100000000000000000\n1000000000\n999999999",
"output": "99999999000000001"
},
{
"input": "100000000000000009\n100\n1000000000000000\n999999999999999",
"output": "99010000000000009"
},
{
"input": "10\n20\n10\n9",
"output": "1"
},
{
"input": "10\n4\n14\n13",
"output": "2"
},
{
"input": "11\n3\n9\n7",
"output": "4"
},
{
"input": "1000000000\n5\n7\n4",
"output": "333333332"
},
{
"input": "12155\n1943\n28717\n24074",
"output": "6"
},
{
"input": "1000000000000000000\n10\n20\n5",
"output": "100000000000000000"
},
{
"input": "98\n33\n440\n314",
"output": "2"
},
{
"input": "1070252292\n57449678\n237309920\n221182550",
"output": "56"
},
{
"input": "100\n3\n102\n101",
"output": "33"
},
{
"input": "100000000000000000\n100000000000000001\n1000000000000000\n999999999999999",
"output": "99000000000000001"
},
{
"input": "66249876257975628\n302307316\n406102416\n182373516",
"output": "296116756"
},
{
"input": "10\n5\n10\n1",
"output": "2"
},
{
"input": "1000000000000000000\n10\n1000000000\n999999998",
"output": "499999999600000000"
}
] | 1,619,849,736
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 11
| 61
| 0
|
n=int(input())
a=int(input())
b=int(input())
c=int(input())
x=((n-c)//(b-c))+c//a
y=n//a
print(max(x,y))
|
Title: Guest From the Past
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kolya Gerasimov loves kefir very much. He lives in year 1984 and knows all the details of buying this delicious drink. One day, as you probably know, he found himself in year 2084, and buying kefir there is much more complicated.
Kolya is hungry, so he went to the nearest milk shop. In 2084 you may buy kefir in a plastic liter bottle, that costs *a* rubles, or in glass liter bottle, that costs *b* rubles. Also, you may return empty glass bottle and get *c* (*c*<=<<=*b*) rubles back, but you cannot return plastic bottles.
Kolya has *n* rubles and he is really hungry, so he wants to drink as much kefir as possible. There were no plastic bottles in his 1984, so Kolya doesn't know how to act optimally and asks for your help.
Input Specification:
First line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1018) — the number of rubles Kolya has at the beginning.
Then follow three lines containing integers *a*, *b* and *c* (1<=≤<=*a*<=≤<=1018, 1<=≤<=*c*<=<<=*b*<=≤<=1018) — the cost of one plastic liter bottle, the cost of one glass liter bottle and the money one can get back by returning an empty glass bottle, respectively.
Output Specification:
Print the only integer — maximum number of liters of kefir, that Kolya can drink.
Demo Input:
['10\n11\n9\n8\n', '10\n5\n6\n1\n']
Demo Output:
['2\n', '2\n']
Note:
In the first sample, Kolya can buy one glass bottle, then return it and buy one more glass bottle. Thus he will drink 2 liters of kefir.
In the second sample, Kolya can buy two plastic bottle and get two liters of kefir, or he can buy one liter glass bottle, then return it and buy one plastic bottle. In both cases he will drink two liters of kefir.
|
```python
n=int(input())
a=int(input())
b=int(input())
c=int(input())
x=((n-c)//(b-c))+c//a
y=n//a
print(max(x,y))
```
| 0
|
|
9
|
C
|
Hexadecimal's Numbers
|
PROGRAMMING
| 1,200
|
[
"brute force",
"implementation",
"math"
] |
C. Hexadecimal's Numbers
|
1
|
64
|
One beautiful July morning a terrible thing happened in Mainframe: a mean virus Megabyte somehow got access to the memory of his not less mean sister Hexadecimal. He loaded there a huge amount of *n* different natural numbers from 1 to *n* to obtain total control over her energy.
But his plan failed. The reason for this was very simple: Hexadecimal didn't perceive any information, apart from numbers written in binary format. This means that if a number in a decimal representation contained characters apart from 0 and 1, it was not stored in the memory. Now Megabyte wants to know, how many numbers were loaded successfully.
|
Input data contains the only number *n* (1<=≤<=*n*<=≤<=109).
|
Output the only number — answer to the problem.
|
[
"10\n"
] |
[
"2"
] |
For *n* = 10 the answer includes numbers 1 and 10.
| 0
|
[
{
"input": "10",
"output": "2"
},
{
"input": "20",
"output": "3"
},
{
"input": "72",
"output": "3"
},
{
"input": "99",
"output": "3"
},
{
"input": "100",
"output": "4"
},
{
"input": "101",
"output": "5"
},
{
"input": "102",
"output": "5"
},
{
"input": "111",
"output": "7"
},
{
"input": "112",
"output": "7"
},
{
"input": "745",
"output": "7"
},
{
"input": "23536",
"output": "31"
},
{
"input": "1",
"output": "1"
},
{
"input": "1010011",
"output": "83"
},
{
"input": "312410141",
"output": "511"
},
{
"input": "1000000000",
"output": "512"
},
{
"input": "999999999",
"output": "511"
},
{
"input": "111111111",
"output": "511"
},
{
"input": "101010101",
"output": "341"
},
{
"input": "121212121",
"output": "511"
},
{
"input": "106341103",
"output": "383"
},
{
"input": "901556123",
"output": "511"
},
{
"input": "832513432",
"output": "511"
},
{
"input": "3",
"output": "1"
},
{
"input": "732875234",
"output": "511"
},
{
"input": "7",
"output": "1"
},
{
"input": "9",
"output": "1"
},
{
"input": "2",
"output": "1"
},
{
"input": "11",
"output": "3"
},
{
"input": "12",
"output": "3"
},
{
"input": "13",
"output": "3"
},
{
"input": "101020101",
"output": "351"
},
{
"input": "111100100",
"output": "484"
},
{
"input": "110110101",
"output": "437"
},
{
"input": "100111001",
"output": "313"
},
{
"input": "100100",
"output": "36"
},
{
"input": "110100102",
"output": "421"
}
] | 1,588,775,258
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 13
| 1,000
| 0
|
n=int(input())
ans=0
if n==1:
ans=1
else:
for i in range(len(str(n))-1):
ans+=2**i
for i in range(int('9'*(len(str(n))-1)),n+1):
f=0
for j in str(i):
if j not in ['0','1']:
f=1
break
if not f:
ans+=1
print(ans)
# n=int(input())
# if n>=17:
# print(4)
# elif n>=16:
# print(3)
# elif n>=1:
# print(2)
# elif n==0:
# print(1)
# else:
# print(0)
|
Title: Hexadecimal's Numbers
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
One beautiful July morning a terrible thing happened in Mainframe: a mean virus Megabyte somehow got access to the memory of his not less mean sister Hexadecimal. He loaded there a huge amount of *n* different natural numbers from 1 to *n* to obtain total control over her energy.
But his plan failed. The reason for this was very simple: Hexadecimal didn't perceive any information, apart from numbers written in binary format. This means that if a number in a decimal representation contained characters apart from 0 and 1, it was not stored in the memory. Now Megabyte wants to know, how many numbers were loaded successfully.
Input Specification:
Input data contains the only number *n* (1<=≤<=*n*<=≤<=109).
Output Specification:
Output the only number — answer to the problem.
Demo Input:
['10\n']
Demo Output:
['2']
Note:
For *n* = 10 the answer includes numbers 1 and 10.
|
```python
n=int(input())
ans=0
if n==1:
ans=1
else:
for i in range(len(str(n))-1):
ans+=2**i
for i in range(int('9'*(len(str(n))-1)),n+1):
f=0
for j in str(i):
if j not in ['0','1']:
f=1
break
if not f:
ans+=1
print(ans)
# n=int(input())
# if n>=17:
# print(4)
# elif n>=16:
# print(3)
# elif n>=1:
# print(2)
# elif n==0:
# print(1)
# else:
# print(0)
```
| 0
|
714
|
B
|
Filya and Homework
|
PROGRAMMING
| 1,200
|
[
"implementation",
"sortings"
] | null | null |
Today, hedgehog Filya went to school for the very first time! Teacher gave him a homework which Filya was unable to complete without your help.
Filya is given an array of non-negative integers *a*1,<=*a*2,<=...,<=*a**n*. First, he pick an integer *x* and then he adds *x* to some elements of the array (no more than once), subtract *x* from some other elements (also, no more than once) and do no change other elements. He wants all elements of the array to be equal.
Now he wonders if it's possible to pick such integer *x* and change some elements of the array using this *x* in order to make all elements equal.
|
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of integers in the Filya's array. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) — elements of the array.
|
If it's impossible to make all elements of the array equal using the process given in the problem statement, then print "NO" (without quotes) in the only line of the output. Otherwise print "YES" (without quotes).
|
[
"5\n1 3 3 2 1\n",
"5\n1 2 3 4 5\n"
] |
[
"YES\n",
"NO\n"
] |
In the first sample Filya should select *x* = 1, then add it to the first and the last elements of the array and subtract from the second and the third elements.
| 1,000
|
[
{
"input": "5\n1 3 3 2 1",
"output": "YES"
},
{
"input": "5\n1 2 3 4 5",
"output": "NO"
},
{
"input": "2\n1 2",
"output": "YES"
},
{
"input": "3\n1 2 3",
"output": "YES"
},
{
"input": "3\n1 1 1",
"output": "YES"
},
{
"input": "2\n1 1000000000",
"output": "YES"
},
{
"input": "4\n1 2 3 4",
"output": "NO"
},
{
"input": "10\n1 1 1 1 1 2 2 2 2 2",
"output": "YES"
},
{
"input": "2\n4 2",
"output": "YES"
},
{
"input": "4\n1 1 4 7",
"output": "YES"
},
{
"input": "3\n99999999 1 50000000",
"output": "YES"
},
{
"input": "1\n0",
"output": "YES"
},
{
"input": "5\n0 0 0 0 0",
"output": "YES"
},
{
"input": "4\n4 2 2 1",
"output": "NO"
},
{
"input": "3\n1 4 2",
"output": "NO"
},
{
"input": "3\n1 4 100",
"output": "NO"
},
{
"input": "3\n2 5 11",
"output": "NO"
},
{
"input": "3\n1 4 6",
"output": "NO"
},
{
"input": "3\n1 2 4",
"output": "NO"
},
{
"input": "3\n1 2 7",
"output": "NO"
},
{
"input": "5\n1 1 1 4 5",
"output": "NO"
},
{
"input": "2\n100000001 100000003",
"output": "YES"
},
{
"input": "3\n7 4 5",
"output": "NO"
},
{
"input": "3\n2 3 5",
"output": "NO"
},
{
"input": "3\n1 2 5",
"output": "NO"
},
{
"input": "2\n2 3",
"output": "YES"
},
{
"input": "3\n2 100 29",
"output": "NO"
},
{
"input": "3\n0 1 5",
"output": "NO"
},
{
"input": "3\n1 3 6",
"output": "NO"
},
{
"input": "3\n2 1 3",
"output": "YES"
},
{
"input": "3\n1 5 100",
"output": "NO"
},
{
"input": "3\n1 4 8",
"output": "NO"
},
{
"input": "3\n1 7 10",
"output": "NO"
},
{
"input": "3\n5 4 1",
"output": "NO"
},
{
"input": "3\n1 6 10",
"output": "NO"
},
{
"input": "4\n1 3 4 5",
"output": "NO"
},
{
"input": "3\n1 5 4",
"output": "NO"
},
{
"input": "5\n1 2 3 3 5",
"output": "NO"
},
{
"input": "3\n2 3 1",
"output": "YES"
},
{
"input": "3\n2 3 8",
"output": "NO"
},
{
"input": "3\n0 3 5",
"output": "NO"
},
{
"input": "3\n1 5 10",
"output": "NO"
},
{
"input": "3\n1 7 2",
"output": "NO"
},
{
"input": "3\n1 3 9",
"output": "NO"
},
{
"input": "3\n1 1 2",
"output": "YES"
},
{
"input": "7\n1 1 1 1 1 2 4",
"output": "NO"
},
{
"input": "5\n1 4 4 4 6",
"output": "NO"
},
{
"input": "5\n1 2 2 4 4",
"output": "NO"
},
{
"input": "3\n1 9 10",
"output": "NO"
},
{
"input": "8\n1 1 1 1 1 1 2 3",
"output": "YES"
},
{
"input": "3\n1 2 100",
"output": "NO"
},
{
"input": "3\n1 3 4",
"output": "NO"
},
{
"input": "3\n1 15 14",
"output": "NO"
},
{
"input": "3\n1 3 8",
"output": "NO"
},
{
"input": "3\n1 2 10",
"output": "NO"
},
{
"input": "4\n2 2 4 5",
"output": "NO"
},
{
"input": "3\n1 3 5",
"output": "YES"
},
{
"input": "5\n3 6 7 8 9",
"output": "NO"
},
{
"input": "3\n7 6 8",
"output": "YES"
},
{
"input": "3\n3 2 1",
"output": "YES"
},
{
"input": "5\n1 2 2 2 3",
"output": "YES"
},
{
"input": "3\n4 6 7",
"output": "NO"
},
{
"input": "3\n2 0 4",
"output": "YES"
},
{
"input": "4\n10 20 21 30",
"output": "NO"
},
{
"input": "4\n0 2 3 4",
"output": "NO"
},
{
"input": "3\n3 6 12",
"output": "NO"
},
{
"input": "5\n0 0 1 3 5",
"output": "NO"
},
{
"input": "3\n3 5 8",
"output": "NO"
},
{
"input": "3\n1 4 4",
"output": "YES"
},
{
"input": "4\n2 4 5 6",
"output": "NO"
}
] | 1,671,025,895
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 5
| 31
| 0
|
n = int(input())
arr = list(map(int, input().split()))
nums = set(arr)
for num in nums:
st = {arr[0], arr[0]+num, abs(arr[0]-num)}
for i in range(1, n):
st = st.intersection({arr[i], arr[i]+num, abs(arr[i]-num)})
if not st:
break
else:
print('YES')
break
else:
print('NO')
|
Title: Filya and Homework
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today, hedgehog Filya went to school for the very first time! Teacher gave him a homework which Filya was unable to complete without your help.
Filya is given an array of non-negative integers *a*1,<=*a*2,<=...,<=*a**n*. First, he pick an integer *x* and then he adds *x* to some elements of the array (no more than once), subtract *x* from some other elements (also, no more than once) and do no change other elements. He wants all elements of the array to be equal.
Now he wonders if it's possible to pick such integer *x* and change some elements of the array using this *x* in order to make all elements equal.
Input Specification:
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of integers in the Filya's array. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) — elements of the array.
Output Specification:
If it's impossible to make all elements of the array equal using the process given in the problem statement, then print "NO" (without quotes) in the only line of the output. Otherwise print "YES" (without quotes).
Demo Input:
['5\n1 3 3 2 1\n', '5\n1 2 3 4 5\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample Filya should select *x* = 1, then add it to the first and the last elements of the array and subtract from the second and the third elements.
|
```python
n = int(input())
arr = list(map(int, input().split()))
nums = set(arr)
for num in nums:
st = {arr[0], arr[0]+num, abs(arr[0]-num)}
for i in range(1, n):
st = st.intersection({arr[i], arr[i]+num, abs(arr[i]-num)})
if not st:
break
else:
print('YES')
break
else:
print('NO')
```
| 0
|
|
61
|
A
|
Ultra-Fast Mathematician
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Ultra-Fast Mathematician
|
2
|
256
|
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
|
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
|
Write one line — the corresponding answer. Do not omit the leading 0s.
|
[
"1010100\n0100101\n",
"000\n111\n",
"1110\n1010\n",
"01110\n01100\n"
] |
[
"1110001\n",
"111\n",
"0100\n",
"00010\n"
] |
none
| 500
|
[
{
"input": "1010100\n0100101",
"output": "1110001"
},
{
"input": "000\n111",
"output": "111"
},
{
"input": "1110\n1010",
"output": "0100"
},
{
"input": "01110\n01100",
"output": "00010"
},
{
"input": "011101\n000001",
"output": "011100"
},
{
"input": "10\n01",
"output": "11"
},
{
"input": "00111111\n11011101",
"output": "11100010"
},
{
"input": "011001100\n101001010",
"output": "110000110"
},
{
"input": "1100100001\n0110101100",
"output": "1010001101"
},
{
"input": "00011101010\n10010100101",
"output": "10001001111"
},
{
"input": "100000101101\n111010100011",
"output": "011010001110"
},
{
"input": "1000001111010\n1101100110001",
"output": "0101101001011"
},
{
"input": "01011111010111\n10001110111010",
"output": "11010001101101"
},
{
"input": "110010000111100\n001100101011010",
"output": "111110101100110"
},
{
"input": "0010010111110000\n0000000011010110",
"output": "0010010100100110"
},
{
"input": "00111110111110000\n01111100001100000",
"output": "01000010110010000"
},
{
"input": "101010101111010001\n001001111101111101",
"output": "100011010010101100"
},
{
"input": "0110010101111100000\n0011000101000000110",
"output": "0101010000111100110"
},
{
"input": "11110100011101010111\n00001000011011000000",
"output": "11111100000110010111"
},
{
"input": "101010101111101101001\n111010010010000011111",
"output": "010000111101101110110"
},
{
"input": "0000111111100011000010\n1110110110110000001010",
"output": "1110001001010011001000"
},
{
"input": "10010010101000110111000\n00101110100110111000111",
"output": "10111100001110001111111"
},
{
"input": "010010010010111100000111\n100100111111100011001110",
"output": "110110101101011111001001"
},
{
"input": "0101110100100111011010010\n0101100011010111001010001",
"output": "0000010111110000010000011"
},
{
"input": "10010010100011110111111011\n10000110101100000001000100",
"output": "00010100001111110110111111"
},
{
"input": "000001111000000100001000000\n011100111101111001110110001",
"output": "011101000101111101111110001"
},
{
"input": "0011110010001001011001011100\n0000101101000011101011001010",
"output": "0011011111001010110010010110"
},
{
"input": "11111000000000010011001101111\n11101110011001010100010000000",
"output": "00010110011001000111011101111"
},
{
"input": "011001110000110100001100101100\n001010000011110000001000101001",
"output": "010011110011000100000100000101"
},
{
"input": "1011111010001100011010110101111\n1011001110010000000101100010101",
"output": "0000110100011100011111010111010"
},
{
"input": "10111000100001000001010110000001\n10111000001100101011011001011000",
"output": "00000000101101101010001111011001"
},
{
"input": "000001010000100001000000011011100\n111111111001010100100001100000111",
"output": "111110101001110101100001111011011"
},
{
"input": "1101000000000010011011101100000110\n1110000001100010011010000011011110",
"output": "0011000001100000000001101111011000"
},
{
"input": "01011011000010100001100100011110001\n01011010111000001010010100001110000",
"output": "00000001111010101011110000010000001"
},
{
"input": "000011111000011001000110111100000100\n011011000110000111101011100111000111",
"output": "011000111110011110101101011011000011"
},
{
"input": "1001000010101110001000000011111110010\n0010001011010111000011101001010110000",
"output": "1011001001111001001011101010101000010"
},
{
"input": "00011101011001100101111111000000010101\n10010011011011001011111000000011101011",
"output": "10001110000010101110000111000011111110"
},
{
"input": "111011100110001001101111110010111001010\n111111101101111001110010000101101000100",
"output": "000100001011110000011101110111010001110"
},
{
"input": "1111001001101000001000000010010101001010\n0010111100111110001011000010111110111001",
"output": "1101110101010110000011000000101011110011"
},
{
"input": "00100101111000000101011111110010100011010\n11101110001010010101001000111110101010100",
"output": "11001011110010010000010111001100001001110"
},
{
"input": "101011001110110100101001000111010101101111\n100111100110101011010100111100111111010110",
"output": "001100101000011111111101111011101010111001"
},
{
"input": "1111100001100101000111101001001010011100001\n1000110011000011110010001011001110001000001",
"output": "0111010010100110110101100010000100010100000"
},
{
"input": "01100111011111010101000001101110000001110101\n10011001011111110000000101011001001101101100",
"output": "11111110000000100101000100110111001100011001"
},
{
"input": "110010100111000100100101100000011100000011001\n011001111011100110000110111001110110100111011",
"output": "101011011100100010100011011001101010100100010"
},
{
"input": "0001100111111011010110100100111000000111000110\n1100101011000000000001010010010111001100110001",
"output": "1101001100111011010111110110101111001011110111"
},
{
"input": "00000101110110110001110010100001110100000100000\n10010000110011110001101000111111101010011010001",
"output": "10010101000101000000011010011110011110011110001"
},
{
"input": "110000100101011100100011001111110011111110010001\n101011111001011100110110111101110011010110101100",
"output": "011011011100000000010101110010000000101000111101"
},
{
"input": "0101111101011111010101011101000011101100000000111\n0000101010110110001110101011011110111001010100100",
"output": "0101010111101001011011110110011101010101010100011"
},
{
"input": "11000100010101110011101000011111001010110111111100\n00001111000111001011111110000010101110111001000011",
"output": "11001011010010111000010110011101100100001110111111"
},
{
"input": "101000001101111101101111111000001110110010101101010\n010011100111100001100000010001100101000000111011011",
"output": "111011101010011100001111101001101011110010010110001"
},
{
"input": "0011111110010001010100010110111000110011001101010100\n0111000000100010101010000100101000000100101000111001",
"output": "0100111110110011111110010010010000110111100101101101"
},
{
"input": "11101010000110000011011010000001111101000111011111100\n10110011110001010100010110010010101001010111100100100",
"output": "01011001110111010111001100010011010100010000111011000"
},
{
"input": "011000100001000001101000010110100110011110100111111011\n111011001000001001110011001111011110111110110011011111",
"output": "100011101001001000011011011001111000100000010100100100"
},
{
"input": "0111010110010100000110111011010110100000000111110110000\n1011100100010001101100000100111111101001110010000100110",
"output": "1100110010000101101010111111101001001001110101110010110"
},
{
"input": "10101000100111000111010001011011011011110100110101100011\n11101111000000001100100011111000100100000110011001101110",
"output": "01000111100111001011110010100011111111110010101100001101"
},
{
"input": "000000111001010001000000110001001011100010011101010011011\n110001101000010010000101000100001111101001100100001010010",
"output": "110001010001000011000101110101000100001011111001011001001"
},
{
"input": "0101011100111010000111110010101101111111000000111100011100\n1011111110000010101110111001000011100000100111111111000111",
"output": "1110100010111000101001001011101110011111100111000011011011"
},
{
"input": "11001000001100100111100111100100101011000101001111001001101\n10111110100010000011010100110100100011101001100000001110110",
"output": "01110110101110100100110011010000001000101100101111000111011"
},
{
"input": "010111011011101000000110000110100110001110100001110110111011\n101011110011101011101101011111010100100001100111100100111011",
"output": "111100101000000011101011011001110010101111000110010010000000"
},
{
"input": "1001011110110110000100011001010110000100011010010111010101110\n1101111100001000010111110011010101111010010100000001000010111",
"output": "0100100010111110010011101010000011111110001110010110010111001"
},
{
"input": "10000010101111100111110101111000010100110111101101111111111010\n10110110101100101010011001011010100110111011101100011001100111",
"output": "00110100000011001101101100100010110010001100000001100110011101"
},
{
"input": "011111010011111000001010101001101001000010100010111110010100001\n011111001011000011111001000001111001010110001010111101000010011",
"output": "000000011000111011110011101000010000010100101000000011010110010"
},
{
"input": "1111000000110001011101000100100100001111011100001111001100011111\n1101100110000101100001100000001001011011111011010101000101001010",
"output": "0010100110110100111100100100101101010100100111011010001001010101"
},
{
"input": "01100000101010010011001110100110110010000110010011011001100100011\n10110110010110111100100111000111000110010000000101101110000010111",
"output": "11010110111100101111101001100001110100010110010110110111100110100"
},
{
"input": "001111111010000100001100001010011001111110011110010111110001100111\n110000101001011000100010101100100110000111100000001101001110010111",
"output": "111111010011011100101110100110111111111001111110011010111111110000"
},
{
"input": "1011101011101101011110101101011101011000010011100101010101000100110\n0001000001001111010111100100111101100000000001110001000110000000110",
"output": "1010101010100010001001001001100000111000010010010100010011000100000"
},
{
"input": "01000001011001010011011100010000100100110101111011011011110000001110\n01011110000110011011000000000011000111100001010000000011111001110000",
"output": "00011111011111001000011100010011100011010100101011011000001001111110"
},
{
"input": "110101010100110101000001111110110100010010000100111110010100110011100\n111010010111111011100110101011001011001110110111110100000110110100111",
"output": "001111000011001110100111010101111111011100110011001010010010000111011"
},
{
"input": "1001101011000001011111100110010010000011010001001111011100010100110001\n1111100111110101001111010001010000011001001001010110001111000000100101",
"output": "0110001100110100010000110111000010011010011000011001010011010100010100"
},
{
"input": "00000111110010110001110110001010010101000111011001111111100110011110010\n00010111110100000100110101000010010001100001100011100000001100010100010",
"output": "00010000000110110101000011001000000100100110111010011111101010001010000"
},
{
"input": "100101011100101101000011010001011001101110101110001100010001010111001110\n100001111100101011011111110000001111000111001011111110000010101110111001",
"output": "000100100000000110011100100001010110101001100101110010010011111001110111"
},
{
"input": "1101100001000111001101001011101000111000011110000001001101101001111011010\n0101011101010100011011010110101000010010110010011110101100000110110001000",
"output": "1000111100010011010110011101000000101010101100011111100001101111001010010"
},
{
"input": "01101101010011110101100001110101111011100010000010001101111000011110111111\n00101111001101001100111010000101110000100101101111100111101110010100011011",
"output": "01000010011110111001011011110000001011000111101101101010010110001010100100"
},
{
"input": "101100101100011001101111110110110010100110110010100001110010110011001101011\n000001011010101011110011111101001110000111000010001101000010010000010001101",
"output": "101101110110110010011100001011111100100001110000101100110000100011011100110"
},
{
"input": "0010001011001010001100000010010011110110011000100000000100110000101111001110\n1100110100111000110100001110111001011101001100001010100001010011100110110001",
"output": "1110111111110010111000001100101010101011010100101010100101100011001001111111"
},
{
"input": "00101101010000000101011001101011001100010001100000101011101110000001111001000\n10010110010111000000101101000011101011001010000011011101101011010000000011111",
"output": "10111011000111000101110100101000100111011011100011110110000101010001111010111"
},
{
"input": "111100000100100000101001100001001111001010001000001000000111010000010101101011\n001000100010100101111011111011010110101100001111011000010011011011100010010110",
"output": "110100100110000101010010011010011001100110000111010000010100001011110111111101"
},
{
"input": "0110001101100100001111110101101000100101010010101010011001101001001101110000000\n0111011000000010010111011110010000000001000110001000011001101000000001110100111",
"output": "0001010101100110011000101011111000100100010100100010000000000001001100000100111"
},
{
"input": "10001111111001000101001011110101111010100001011010101100111001010001010010001000\n10000111010010011110111000111010101100000011110001101111001000111010100000000001",
"output": "00001000101011011011110011001111010110100010101011000011110001101011110010001001"
},
{
"input": "100110001110110000100101001110000011110110000110000000100011110100110110011001101\n110001110101110000000100101001101011111100100100001001000110000001111100011110110",
"output": "010111111011000000100001100111101000001010100010001001100101110101001010000111011"
},
{
"input": "0000010100100000010110111100011111111010011101000000100000011001001101101100111010\n0100111110011101010110101011110110010111001111000110101100101110111100101000111111",
"output": "0100101010111101000000010111101001101101010010000110001100110111110001000100000101"
},
{
"input": "11000111001010100001110000001001011010010010110000001110100101000001010101100110111\n11001100100100100001101010110100000111100011101110011010110100001001000011011011010",
"output": "00001011101110000000011010111101011101110001011110010100010001001000010110111101101"
},
{
"input": "010110100010001000100010101001101010011010111110100001000100101000111011100010100001\n110000011111101101010011111000101010111010100001001100001001100101000000111000000000",
"output": "100110111101100101110001010001000000100000011111101101001101001101111011011010100001"
},
{
"input": "0000011110101110010101110110110101100001011001101010101001000010000010000000101001101\n1100111111011100000110000111101110011111100111110001011001000010011111100001001100011",
"output": "1100100001110010010011110001011011111110111110011011110000000000011101100001100101110"
},
{
"input": "10100000101101110001100010010010100101100011010010101000110011100000101010110010000000\n10001110011011010010111011011101101111000111110000111000011010010101001100000001010011",
"output": "00101110110110100011011001001111001010100100100010010000101001110101100110110011010011"
},
{
"input": "001110000011111101101010011111000101010111010100001001100001001100101000000111000000000\n111010000000000000101001110011001000111011001100101010011001000011101001001011110000011",
"output": "110100000011111101000011101100001101101100011000100011111000001111000001001100110000011"
},
{
"input": "1110111100111011010101011011001110001010010010110011110010011111000010011111010101100001\n1001010101011001001010100010101100000110111101011000100010101111111010111100001110010010",
"output": "0111101001100010011111111001100010001100101111101011010000110000111000100011011011110011"
},
{
"input": "11100010001100010011001100001100010011010001101110011110100101110010101101011101000111111\n01110000000110111010110100001010000101011110100101010011000110101110101101110111011110001",
"output": "10010010001010101001111000000110010110001111001011001101100011011100000000101010011001110"
},
{
"input": "001101011001100101101100110000111000101011001001100100000100101000100000110100010111111101\n101001111110000010111101111110001001111001111101111010000110111000100100110010010001011111",
"output": "100100100111100111010001001110110001010010110100011110000010010000000100000110000110100010"
},
{
"input": "1010110110010101000110010010110101011101010100011001101011000110000000100011100100011000000\n0011011111100010001111101101000111001011101110100000110111100100101111010110101111011100011",
"output": "1001101001110111001001111111110010010110111010111001011100100010101111110101001011000100011"
},
{
"input": "10010010000111010111011111110010100101100000001100011100111011100010000010010001011100001100\n00111010100010110010000100010111010001111110100100100011101000101111111111001101101100100100",
"output": "10101000100101100101011011100101110100011110101000111111010011001101111101011100110000101000"
},
{
"input": "010101110001010101100000010111010000000111110011001101100011001000000011001111110000000010100\n010010111011100101010101111110110000000111000100001101101001001000001100101110001010000100001",
"output": "000111001010110000110101101001100000000000110111000000001010000000001111100001111010000110101"
},
{
"input": "1100111110011001000111101001001011000110011010111111100010111111001100111111011101100111101011\n1100000011001000110100110111000001011001010111101000010010100011000001100100111101101000010110",
"output": "0000111101010001110011011110001010011111001101010111110000011100001101011011100000001111111101"
},
{
"input": "00011000100100110111100101100100000000010011110111110010101110110011100001010111010011110100101\n00011011111011111011100101100111100101001110010111000010000111000100100100000001110101111011011",
"output": "00000011011111001100000000000011100101011101100000110000101001110111000101010110100110001111110"
},
{
"input": "000101011001001100000111100010110101111011110101111101000110001101011010111110110011100100000001\n011000101010011111011000111000100000000011011000000001111110001000001111101010110000011100001111",
"output": "011101110011010011011111011010010101111000101101111100111000000101010101010100000011111000001110"
},
{
"input": "1000101001011010000100100100010010011101011001110101111011101111111110010101001101010001010101001\n0110110010011100011111011111110111000000010001110100001010111110101011010011111011111110000110000",
"output": "1110011011000110011011111011100101011101001000000001110001010001010101000110110110101111010011001"
},
{
"input": "01111010010101100111110111111011011010100001011101010000111100101101101110111011001100101011100111\n00001100110110100001111011000010001001001100000010111101000001111011100000010111010010000011000010",
"output": "01110110100011000110001100111001010011101101011111101101111101010110001110101100011110101000100101"
},
{
"input": "111101011101111000001011001101111010110111001101110100100011111011011101001101010101011010111000110\n101111100101101101001100110011000001111010011101110111110110101110011011110011111100001001110101101",
"output": "010010111000010101000111111110111011001101010000000011010101010101000110111110101001010011001101011"
},
{
"input": "1010000011010110011101001101110001110010000111011101110010110110111111001001110100101100010101010001\n0111101000111100101100000101111010100100001000011101010100110011100011010011010101000100101011100011",
"output": "1101101011101010110001001000001011010110001111000000100110000101011100011010100001101000111110110010"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001\n1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111010",
"output": "1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111011"
},
{
"input": "0\n0",
"output": "0"
},
{
"input": "0\n1",
"output": "1"
}
] | 1,643,912,422
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 46
| 0
|
a=input()
b=input()
atb=int(a,2)
btb=int(b,2)
c=atb^btb
s=str(bin(c))
if(a[0]==b[0]):
print("0"+s[2:])
else:
print(s[2:])
|
Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
Input Specification:
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Output Specification:
Write one line — the corresponding answer. Do not omit the leading 0s.
Demo Input:
['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n']
Demo Output:
['1110001\n', '111\n', '0100\n', '00010\n']
Note:
none
|
```python
a=input()
b=input()
atb=int(a,2)
btb=int(b,2)
c=atb^btb
s=str(bin(c))
if(a[0]==b[0]):
print("0"+s[2:])
else:
print(s[2:])
```
| 0
|
822
|
A
|
I'm bored with life
|
PROGRAMMING
| 800
|
[
"implementation",
"math",
"number theory"
] | null | null |
Holidays have finished. Thanks to the help of the hacker Leha, Noora managed to enter the university of her dreams which is located in a town Pavlopolis. It's well known that universities provide students with dormitory for the period of university studies. Consequently Noora had to leave Vičkopolis and move to Pavlopolis. Thus Leha was left completely alone in a quiet town Vičkopolis. He almost even fell into a depression from boredom!
Leha came up with a task for himself to relax a little. He chooses two integers *A* and *B* and then calculates the greatest common divisor of integers "*A* factorial" and "*B* factorial". Formally the hacker wants to find out GCD(*A*!,<=*B*!). It's well known that the factorial of an integer *x* is a product of all positive integers less than or equal to *x*. Thus *x*!<==<=1·2·3·...·(*x*<=-<=1)·*x*. For example 4!<==<=1·2·3·4<==<=24. Recall that GCD(*x*,<=*y*) is the largest positive integer *q* that divides (without a remainder) both *x* and *y*.
Leha has learned how to solve this task very effective. You are able to cope with it not worse, aren't you?
|
The first and single line contains two integers *A* and *B* (1<=≤<=*A*,<=*B*<=≤<=109,<=*min*(*A*,<=*B*)<=≤<=12).
|
Print a single integer denoting the greatest common divisor of integers *A*! and *B*!.
|
[
"4 3\n"
] |
[
"6\n"
] |
Consider the sample.
4! = 1·2·3·4 = 24. 3! = 1·2·3 = 6. The greatest common divisor of integers 24 and 6 is exactly 6.
| 500
|
[
{
"input": "4 3",
"output": "6"
},
{
"input": "10 399603090",
"output": "3628800"
},
{
"input": "6 973151934",
"output": "720"
},
{
"input": "2 841668075",
"output": "2"
},
{
"input": "7 415216919",
"output": "5040"
},
{
"input": "3 283733059",
"output": "6"
},
{
"input": "11 562314608",
"output": "39916800"
},
{
"input": "3 990639260",
"output": "6"
},
{
"input": "11 859155400",
"output": "39916800"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "5 3",
"output": "6"
},
{
"input": "1 4",
"output": "1"
},
{
"input": "5 4",
"output": "24"
},
{
"input": "1 12",
"output": "1"
},
{
"input": "9 7",
"output": "5040"
},
{
"input": "2 3",
"output": "2"
},
{
"input": "6 11",
"output": "720"
},
{
"input": "6 7",
"output": "720"
},
{
"input": "11 11",
"output": "39916800"
},
{
"input": "4 999832660",
"output": "24"
},
{
"input": "7 999228288",
"output": "5040"
},
{
"input": "11 999257105",
"output": "39916800"
},
{
"input": "11 999286606",
"output": "39916800"
},
{
"input": "3 999279109",
"output": "6"
},
{
"input": "999632727 11",
"output": "39916800"
},
{
"input": "999625230 7",
"output": "5040"
},
{
"input": "999617047 3",
"output": "6"
},
{
"input": "999646548 7",
"output": "5040"
},
{
"input": "999639051 3",
"output": "6"
},
{
"input": "12 12",
"output": "479001600"
},
{
"input": "12 1",
"output": "1"
},
{
"input": "1213 5",
"output": "120"
},
{
"input": "8 9",
"output": "40320"
},
{
"input": "12 9",
"output": "362880"
},
{
"input": "12 1000000000",
"output": "479001600"
},
{
"input": "1000000000 1",
"output": "1"
},
{
"input": "12 13",
"output": "479001600"
},
{
"input": "2 29845",
"output": "2"
},
{
"input": "10 21",
"output": "3628800"
},
{
"input": "12 20",
"output": "479001600"
},
{
"input": "15 12",
"output": "479001600"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1000000000 12",
"output": "479001600"
},
{
"input": "11 30",
"output": "39916800"
},
{
"input": "17 12",
"output": "479001600"
},
{
"input": "4 19",
"output": "24"
},
{
"input": "12 15",
"output": "479001600"
},
{
"input": "20 6",
"output": "720"
},
{
"input": "10 20",
"output": "3628800"
},
{
"input": "10 10",
"output": "3628800"
},
{
"input": "22 12",
"output": "479001600"
},
{
"input": "20 12",
"output": "479001600"
},
{
"input": "12 23",
"output": "479001600"
},
{
"input": "12 22",
"output": "479001600"
},
{
"input": "18 3",
"output": "6"
},
{
"input": "14 10",
"output": "3628800"
},
{
"input": "14 12",
"output": "479001600"
},
{
"input": "8 3",
"output": "6"
},
{
"input": "5 5",
"output": "120"
}
] | 1,682,691,766
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 59
| 46
| 0
|
def fact(n):
pr = 1
for i in range(2,n+1):
pr *= i
return pr
a, b = map(int, input().split())
k = min(a,b)
print(fact(k))
|
Title: I'm bored with life
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Holidays have finished. Thanks to the help of the hacker Leha, Noora managed to enter the university of her dreams which is located in a town Pavlopolis. It's well known that universities provide students with dormitory for the period of university studies. Consequently Noora had to leave Vičkopolis and move to Pavlopolis. Thus Leha was left completely alone in a quiet town Vičkopolis. He almost even fell into a depression from boredom!
Leha came up with a task for himself to relax a little. He chooses two integers *A* and *B* and then calculates the greatest common divisor of integers "*A* factorial" and "*B* factorial". Formally the hacker wants to find out GCD(*A*!,<=*B*!). It's well known that the factorial of an integer *x* is a product of all positive integers less than or equal to *x*. Thus *x*!<==<=1·2·3·...·(*x*<=-<=1)·*x*. For example 4!<==<=1·2·3·4<==<=24. Recall that GCD(*x*,<=*y*) is the largest positive integer *q* that divides (without a remainder) both *x* and *y*.
Leha has learned how to solve this task very effective. You are able to cope with it not worse, aren't you?
Input Specification:
The first and single line contains two integers *A* and *B* (1<=≤<=*A*,<=*B*<=≤<=109,<=*min*(*A*,<=*B*)<=≤<=12).
Output Specification:
Print a single integer denoting the greatest common divisor of integers *A*! and *B*!.
Demo Input:
['4 3\n']
Demo Output:
['6\n']
Note:
Consider the sample.
4! = 1·2·3·4 = 24. 3! = 1·2·3 = 6. The greatest common divisor of integers 24 and 6 is exactly 6.
|
```python
def fact(n):
pr = 1
for i in range(2,n+1):
pr *= i
return pr
a, b = map(int, input().split())
k = min(a,b)
print(fact(k))
```
| 3
|
|
784
|
F
|
Crunching Numbers Just for You
|
PROGRAMMING
| 1,900
|
[
"*special",
"implementation"
] | null | null |
You are developing a new feature for the website which sells airline tickets: being able to sort tickets by price! You have already extracted the tickets' prices, so there's just the last step to be done...
You are given an array of integers. Sort it in non-descending order.
|
The input consists of a single line of space-separated integers. The first number is *n* (1<=≤<=*n*<=≤<=10) — the size of the array. The following *n* numbers are the elements of the array (1<=≤<=*a**i*<=≤<=100).
|
Output space-separated elements of the sorted array.
|
[
"3 3 1 2\n"
] |
[
"1 2 3 \n"
] |
Remember, this is a very important feature, and you have to make sure the customers appreciate it!
| 0
|
[
{
"input": "3 3 1 2",
"output": "1 2 3 "
},
{
"input": "10 54 100 27 1 33 27 80 49 27 6",
"output": "1 6 27 27 27 33 49 54 80 100 "
}
] | 1,549,534,563
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 93
| 0
|
a=set(map(int,input().split()))
print(*a,sep=' ')
|
Title: Crunching Numbers Just for You
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are developing a new feature for the website which sells airline tickets: being able to sort tickets by price! You have already extracted the tickets' prices, so there's just the last step to be done...
You are given an array of integers. Sort it in non-descending order.
Input Specification:
The input consists of a single line of space-separated integers. The first number is *n* (1<=≤<=*n*<=≤<=10) — the size of the array. The following *n* numbers are the elements of the array (1<=≤<=*a**i*<=≤<=100).
Output Specification:
Output space-separated elements of the sorted array.
Demo Input:
['3 3 1 2\n']
Demo Output:
['1 2 3 \n']
Note:
Remember, this is a very important feature, and you have to make sure the customers appreciate it!
|
```python
a=set(map(int,input().split()))
print(*a,sep=' ')
```
| 0
|
|
69
|
A
|
Young Physicist
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] |
A. Young Physicist
|
2
|
256
|
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
|
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
|
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
|
[
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] |
[
"NO",
"YES"
] |
none
| 500
|
[
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10\n25 -33 43\n-27 -42 28\n-35 -20 19\n41 -42 -1\n49 -39 -4\n-49 -22 7\n-19 29 41\n8 -27 -43\n8 34 9\n-11 -3 33",
"output": "NO"
},
{
"input": "10\n-6 21 18\n20 -11 -8\n37 -11 41\n-5 8 33\n29 23 32\n30 -33 -11\n39 -49 -36\n28 34 -49\n22 29 -34\n-18 -6 7",
"output": "NO"
},
{
"input": "10\n47 -2 -27\n0 26 -14\n5 -12 33\n2 18 3\n45 -30 -49\n4 -18 8\n-46 -44 -41\n-22 -10 -40\n-35 -21 26\n33 20 38",
"output": "NO"
},
{
"input": "13\n-3 -36 -46\n-11 -50 37\n42 -11 -15\n9 42 44\n-29 -12 24\n3 9 -40\n-35 13 50\n14 43 18\n-13 8 24\n-48 -15 10\n50 9 -50\n21 0 -50\n0 0 -6",
"output": "YES"
},
{
"input": "14\n43 23 17\n4 17 44\n5 -5 -16\n-43 -7 -6\n47 -48 12\n50 47 -45\n2 14 43\n37 -30 15\n4 -17 -11\n17 9 -45\n-50 -3 -8\n-50 0 0\n-50 0 0\n-16 0 0",
"output": "YES"
},
{
"input": "13\n29 49 -11\n38 -11 -20\n25 1 -40\n-11 28 11\n23 -19 1\n45 -41 -17\n-3 0 -19\n-13 -33 49\n-30 0 28\n34 17 45\n-50 9 -27\n-50 0 0\n-37 0 0",
"output": "YES"
},
{
"input": "12\n3 28 -35\n-32 -44 -17\n9 -25 -6\n-42 -22 20\n-19 15 38\n-21 38 48\n-1 -37 -28\n-10 -13 -50\n-5 21 29\n34 28 50\n50 11 -49\n34 0 0",
"output": "YES"
},
{
"input": "37\n-64 -79 26\n-22 59 93\n-5 39 -12\n77 -9 76\n55 -86 57\n83 100 -97\n-70 94 84\n-14 46 -94\n26 72 35\n14 78 -62\n17 82 92\n-57 11 91\n23 15 92\n-80 -1 1\n12 39 18\n-23 -99 -75\n-34 50 19\n-39 84 -7\n45 -30 -39\n-60 49 37\n45 -16 -72\n33 -51 -56\n-48 28 5\n97 91 88\n45 -82 -11\n-21 -15 -90\n-53 73 -26\n-74 85 -90\n-40 23 38\n100 -13 49\n32 -100 -100\n0 -100 -70\n0 -100 0\n0 -100 0\n0 -100 0\n0 -100 0\n0 -37 0",
"output": "YES"
},
{
"input": "4\n68 3 100\n68 21 -100\n-100 -24 0\n-36 0 0",
"output": "YES"
},
{
"input": "33\n-1 -46 -12\n45 -16 -21\n-11 45 -21\n-60 -42 -93\n-22 -45 93\n37 96 85\n-76 26 83\n-4 9 55\n7 -52 -9\n66 8 -85\n-100 -54 11\n-29 59 74\n-24 12 2\n-56 81 85\n-92 69 -52\n-26 -97 91\n54 59 -51\n58 21 -57\n7 68 56\n-47 -20 -51\n-59 77 -13\n-85 27 91\n79 60 -56\n66 -80 5\n21 -99 42\n-31 -29 98\n66 93 76\n-49 45 61\n100 -100 -100\n100 -100 -100\n66 -75 -100\n0 0 -100\n0 0 -87",
"output": "YES"
},
{
"input": "3\n1 2 3\n3 2 1\n0 0 0",
"output": "NO"
},
{
"input": "2\n5 -23 12\n0 0 0",
"output": "NO"
},
{
"input": "1\n0 0 0",
"output": "YES"
},
{
"input": "1\n1 -2 0",
"output": "NO"
},
{
"input": "2\n-23 77 -86\n23 -77 86",
"output": "YES"
},
{
"input": "26\n86 7 20\n-57 -64 39\n-45 6 -93\n-44 -21 100\n-11 -49 21\n73 -71 -80\n-2 -89 56\n-65 -2 7\n5 14 84\n57 41 13\n-12 69 54\n40 -25 27\n-17 -59 0\n64 -91 -30\n-53 9 42\n-54 -8 14\n-35 82 27\n-48 -59 -80\n88 70 79\n94 57 97\n44 63 25\n84 -90 -40\n-100 100 -100\n-92 100 -100\n0 10 -100\n0 0 -82",
"output": "YES"
},
{
"input": "42\n11 27 92\n-18 -56 -57\n1 71 81\n33 -92 30\n82 83 49\n-87 -61 -1\n-49 45 49\n73 26 15\n-22 22 -77\n29 -93 87\n-68 44 -90\n-4 -84 20\n85 67 -6\n-39 26 77\n-28 -64 20\n65 -97 24\n-72 -39 51\n35 -75 -91\n39 -44 -8\n-25 -27 -57\n91 8 -46\n-98 -94 56\n94 -60 59\n-9 -95 18\n-53 -37 98\n-8 -94 -84\n-52 55 60\n15 -14 37\n65 -43 -25\n94 12 66\n-8 -19 -83\n29 81 -78\n-58 57 33\n24 86 -84\n-53 32 -88\n-14 7 3\n89 97 -53\n-5 -28 -91\n-100 100 -6\n-84 100 0\n0 100 0\n0 70 0",
"output": "YES"
},
{
"input": "3\n96 49 -12\n2 -66 28\n-98 17 -16",
"output": "YES"
},
{
"input": "5\n70 -46 86\n-100 94 24\n-27 63 -63\n57 -100 -47\n0 -11 0",
"output": "YES"
},
{
"input": "18\n-86 -28 70\n-31 -89 42\n31 -48 -55\n95 -17 -43\n24 -95 -85\n-21 -14 31\n68 -18 81\n13 31 60\n-15 28 99\n-42 15 9\n28 -61 -62\n-16 71 29\n-28 75 -48\n-77 -67 36\n-100 83 89\n100 100 -100\n57 34 -100\n0 0 -53",
"output": "YES"
},
{
"input": "44\n52 -54 -29\n-82 -5 -94\n-54 43 43\n91 16 71\n7 80 -91\n3 15 29\n-99 -6 -77\n-3 -77 -64\n73 67 34\n25 -10 -18\n-29 91 63\n-72 86 -16\n-68 85 -81\n-3 36 44\n-74 -14 -80\n34 -96 -97\n-76 -78 -33\n-24 44 -58\n98 12 77\n95 -63 -6\n-51 3 -90\n-92 -10 72\n7 3 -68\n57 -53 71\n29 57 -48\n35 -60 10\n79 -70 -61\n-20 77 55\n-86 -15 -35\n84 -88 -18\n100 -42 77\n-20 46 8\n-41 -43 -65\n38 -98 -23\n-100 65 45\n-7 -91 -63\n46 88 -85\n48 59 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 1",
"output": "YES"
},
{
"input": "18\n-14 -64 -91\n-8 -66 -86\n-23 92 -40\n6 -3 -53\n57 41 78\n-79 42 -22\n-88 -17 45\n4 -45 44\n83 -18 -25\n34 86 -92\n75 -30 12\n44 99 11\n-67 -13 72\n22 83 -56\n-37 71 72\n-9 -100 100\n0 -100 31\n0 -58 0",
"output": "YES"
},
{
"input": "23\n-70 37 78\n42 84 6\n28 -94 -24\n-49 76 95\n-67 18 84\n-53 78 -5\n65 -63 -64\n-66 -64 -69\n81 -93 95\n10 52 -79\n-89 -61 -64\n-64 -47 43\n-81 -35 55\n80 82 73\n-60 -81 -18\n49 -10 -19\n-58 70 12\n-24 -15 -93\n98 -93 -54\n-28 -75 11\n100 100 -63\n100 100 0\n56 34 0",
"output": "YES"
},
{
"input": "3\n0 2 -2\n1 -1 3\n-3 0 0",
"output": "NO"
}
] | 1,602,755,874
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 80
| 218
| 0
|
n = int(input())
val=0
for i in range(n):
x,y,z = input().split()
val += (int(x) + int(y) + int(z))
if (val == 0):
print('YES')
else:
print('NO')
|
Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none
|
```python
n = int(input())
val=0
for i in range(n):
x,y,z = input().split()
val += (int(x) + int(y) + int(z))
if (val == 0):
print('YES')
else:
print('NO')
```
| 0
|
192
|
A
|
Funky Numbers
|
PROGRAMMING
| 1,300
|
[
"binary search",
"brute force",
"implementation"
] | null | null |
As you very well know, this year's funkiest numbers are so called triangular numbers (that is, integers that are representable as , where *k* is some positive integer), and the coolest numbers are those that are representable as a sum of two triangular numbers.
A well-known hipster Andrew adores everything funky and cool but unfortunately, he isn't good at maths. Given number *n*, help him define whether this number can be represented by a sum of two triangular numbers (not necessarily different)!
|
The first input line contains an integer *n* (1<=≤<=*n*<=≤<=109).
|
Print "YES" (without the quotes), if *n* can be represented as a sum of two triangular numbers, otherwise print "NO" (without the quotes).
|
[
"256\n",
"512\n"
] |
[
"YES\n",
"NO\n"
] |
In the first sample number <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/92095692c6ea93e9e3b837a0408ba7543549d5b2.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample number 512 can not be represented as a sum of two triangular numbers.
| 500
|
[
{
"input": "256",
"output": "YES"
},
{
"input": "512",
"output": "NO"
},
{
"input": "80",
"output": "NO"
},
{
"input": "828",
"output": "YES"
},
{
"input": "6035",
"output": "NO"
},
{
"input": "39210",
"output": "YES"
},
{
"input": "79712",
"output": "NO"
},
{
"input": "190492",
"output": "YES"
},
{
"input": "5722367",
"output": "NO"
},
{
"input": "816761542",
"output": "YES"
},
{
"input": "1",
"output": "NO"
},
{
"input": "2",
"output": "YES"
},
{
"input": "3",
"output": "NO"
},
{
"input": "4",
"output": "YES"
},
{
"input": "5",
"output": "NO"
},
{
"input": "6",
"output": "YES"
},
{
"input": "7",
"output": "YES"
},
{
"input": "8",
"output": "NO"
},
{
"input": "9",
"output": "YES"
},
{
"input": "10",
"output": "NO"
},
{
"input": "12",
"output": "YES"
},
{
"input": "13",
"output": "YES"
},
{
"input": "14",
"output": "NO"
},
{
"input": "15",
"output": "NO"
},
{
"input": "16",
"output": "YES"
},
{
"input": "17",
"output": "NO"
},
{
"input": "18",
"output": "YES"
},
{
"input": "19",
"output": "NO"
},
{
"input": "20",
"output": "YES"
},
{
"input": "41",
"output": "NO"
},
{
"input": "11",
"output": "YES"
},
{
"input": "69",
"output": "YES"
},
{
"input": "82",
"output": "NO"
},
{
"input": "85",
"output": "NO"
},
{
"input": "736",
"output": "NO"
},
{
"input": "895",
"output": "YES"
},
{
"input": "934",
"output": "YES"
},
{
"input": "6213",
"output": "YES"
},
{
"input": "7405",
"output": "NO"
},
{
"input": "9919",
"output": "NO"
},
{
"input": "40942",
"output": "YES"
},
{
"input": "41992",
"output": "NO"
},
{
"input": "68535",
"output": "NO"
},
{
"input": "405718",
"output": "NO"
},
{
"input": "1046146",
"output": "YES"
},
{
"input": "3761248",
"output": "YES"
},
{
"input": "6195181",
"output": "YES"
},
{
"input": "35354345",
"output": "NO"
},
{
"input": "81282830",
"output": "NO"
},
{
"input": "187719774",
"output": "NO"
},
{
"input": "296798673",
"output": "NO"
},
{
"input": "938938476",
"output": "NO"
},
{
"input": "1000000000",
"output": "NO"
},
{
"input": "999887464",
"output": "YES"
},
{
"input": "999111944",
"output": "NO"
},
{
"input": "999966520",
"output": "YES"
},
{
"input": "999912080",
"output": "NO"
},
{
"input": "999992017",
"output": "YES"
},
{
"input": "999990474",
"output": "NO"
},
{
"input": "999999190",
"output": "YES"
},
{
"input": "999999125",
"output": "NO"
},
{
"input": "999999940",
"output": "YES"
},
{
"input": "999999995",
"output": "NO"
},
{
"input": "1000000000",
"output": "NO"
},
{
"input": "1",
"output": "NO"
},
{
"input": "999999999",
"output": "YES"
},
{
"input": "83495494",
"output": "NO"
},
{
"input": "968022000",
"output": "YES"
},
{
"input": "399980000",
"output": "YES"
},
{
"input": "4",
"output": "YES"
},
{
"input": "999999998",
"output": "NO"
}
] | 1,591,453,807
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 154
| 307,200
|
def Sum(n):
return (n*(n+1))//2
def check(n):
limit = int((2*n)**(1/2))
k=limit
while 0<k<=limit:
l = 1
u = k+1
while u!=l :
print(u,l,k)
mid = int((l+u)/2)
curr = Sum(mid)+Sum(k)
if curr == n:
return 'YES'
elif curr > n :
if u == mid :
break
u = mid
else :
if l == mid:
break
l = mid
k-=1
return 'NO'
n = int(input())
print(check(n))
|
Title: Funky Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As you very well know, this year's funkiest numbers are so called triangular numbers (that is, integers that are representable as , where *k* is some positive integer), and the coolest numbers are those that are representable as a sum of two triangular numbers.
A well-known hipster Andrew adores everything funky and cool but unfortunately, he isn't good at maths. Given number *n*, help him define whether this number can be represented by a sum of two triangular numbers (not necessarily different)!
Input Specification:
The first input line contains an integer *n* (1<=≤<=*n*<=≤<=109).
Output Specification:
Print "YES" (without the quotes), if *n* can be represented as a sum of two triangular numbers, otherwise print "NO" (without the quotes).
Demo Input:
['256\n', '512\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample number <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/92095692c6ea93e9e3b837a0408ba7543549d5b2.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample number 512 can not be represented as a sum of two triangular numbers.
|
```python
def Sum(n):
return (n*(n+1))//2
def check(n):
limit = int((2*n)**(1/2))
k=limit
while 0<k<=limit:
l = 1
u = k+1
while u!=l :
print(u,l,k)
mid = int((l+u)/2)
curr = Sum(mid)+Sum(k)
if curr == n:
return 'YES'
elif curr > n :
if u == mid :
break
u = mid
else :
if l == mid:
break
l = mid
k-=1
return 'NO'
n = int(input())
print(check(n))
```
| 0
|
|
915
|
A
|
Garden
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Luba thinks about watering her garden. The garden can be represented as a segment of length *k*. Luba has got *n* buckets, the *i*-th bucket allows her to water some continuous subsegment of garden of length exactly *a**i* each hour. Luba can't water any parts of the garden that were already watered, also she can't water the ground outside the garden.
Luba has to choose one of the buckets in order to water the garden as fast as possible (as mentioned above, each hour she will water some continuous subsegment of length *a**i* if she chooses the *i*-th bucket). Help her to determine the minimum number of hours she has to spend watering the garden. It is guaranteed that Luba can always choose a bucket so it is possible water the garden.
See the examples for better understanding.
|
The first line of input contains two integer numbers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of buckets and the length of the garden, respectively.
The second line of input contains *n* integer numbers *a**i* (1<=≤<=*a**i*<=≤<=100) — the length of the segment that can be watered by the *i*-th bucket in one hour.
It is guaranteed that there is at least one bucket such that it is possible to water the garden in integer number of hours using only this bucket.
|
Print one integer number — the minimum number of hours required to water the garden.
|
[
"3 6\n2 3 5\n",
"6 7\n1 2 3 4 5 6\n"
] |
[
"2\n",
"7\n"
] |
In the first test the best option is to choose the bucket that allows to water the segment of length 3. We can't choose the bucket that allows to water the segment of length 5 because then we can't water the whole garden.
In the second test we can choose only the bucket that allows us to water the segment of length 1.
| 0
|
[
{
"input": "3 6\n2 3 5",
"output": "2"
},
{
"input": "6 7\n1 2 3 4 5 6",
"output": "7"
},
{
"input": "5 97\n1 10 50 97 2",
"output": "1"
},
{
"input": "5 97\n1 10 50 100 2",
"output": "97"
},
{
"input": "100 100\n2 46 24 18 86 90 31 38 84 49 58 28 15 80 14 24 87 56 62 87 41 87 55 71 87 32 41 56 91 32 24 75 43 42 35 30 72 53 31 26 54 61 87 85 36 75 44 31 7 38 77 57 61 54 70 77 45 96 39 57 11 8 91 42 52 15 42 30 92 41 27 26 34 27 3 80 32 86 26 97 63 91 30 75 14 7 19 23 45 11 8 43 44 73 11 56 3 55 63 16",
"output": "50"
},
{
"input": "100 91\n13 13 62 96 74 47 81 46 78 21 20 42 4 73 25 30 76 74 58 28 25 52 42 48 74 40 82 9 25 29 17 22 46 64 57 95 81 39 47 86 40 95 97 35 31 98 45 98 47 78 52 63 58 14 89 97 17 95 28 22 20 36 68 38 95 16 2 26 54 47 42 31 31 81 21 21 65 40 82 53 60 71 75 33 96 98 6 22 95 12 5 48 18 27 58 62 5 96 36 75",
"output": "7"
},
{
"input": "8 8\n8 7 6 5 4 3 2 1",
"output": "1"
},
{
"input": "3 8\n4 3 2",
"output": "2"
},
{
"input": "3 8\n2 4 2",
"output": "2"
},
{
"input": "3 6\n1 3 2",
"output": "2"
},
{
"input": "3 6\n3 2 5",
"output": "2"
},
{
"input": "3 8\n4 2 1",
"output": "2"
},
{
"input": "5 6\n2 3 5 1 2",
"output": "2"
},
{
"input": "2 6\n5 3",
"output": "2"
},
{
"input": "4 12\n6 4 3 1",
"output": "2"
},
{
"input": "3 18\n1 9 6",
"output": "2"
},
{
"input": "3 9\n3 2 1",
"output": "3"
},
{
"input": "3 6\n5 3 2",
"output": "2"
},
{
"input": "2 10\n5 2",
"output": "2"
},
{
"input": "2 18\n6 3",
"output": "3"
},
{
"input": "4 12\n1 2 12 3",
"output": "1"
},
{
"input": "3 7\n3 2 1",
"output": "7"
},
{
"input": "3 6\n3 2 1",
"output": "2"
},
{
"input": "5 10\n5 4 3 2 1",
"output": "2"
},
{
"input": "5 16\n8 4 2 1 7",
"output": "2"
},
{
"input": "6 7\n6 5 4 3 7 1",
"output": "1"
},
{
"input": "2 6\n3 2",
"output": "2"
},
{
"input": "2 4\n4 1",
"output": "1"
},
{
"input": "6 8\n2 4 1 3 5 7",
"output": "2"
},
{
"input": "6 8\n6 5 4 3 2 1",
"output": "2"
},
{
"input": "6 15\n5 2 3 6 4 3",
"output": "3"
},
{
"input": "4 8\n2 4 8 1",
"output": "1"
},
{
"input": "2 5\n5 1",
"output": "1"
},
{
"input": "4 18\n3 1 1 2",
"output": "6"
},
{
"input": "2 1\n2 1",
"output": "1"
},
{
"input": "3 10\n2 10 5",
"output": "1"
},
{
"input": "5 12\n12 4 4 4 3",
"output": "1"
},
{
"input": "3 6\n6 3 2",
"output": "1"
},
{
"input": "2 2\n2 1",
"output": "1"
},
{
"input": "3 18\n1 9 3",
"output": "2"
},
{
"input": "3 8\n7 2 4",
"output": "2"
},
{
"input": "2 100\n99 1",
"output": "100"
},
{
"input": "4 12\n1 3 4 2",
"output": "3"
},
{
"input": "3 6\n2 3 1",
"output": "2"
},
{
"input": "4 6\n3 2 5 12",
"output": "2"
},
{
"input": "4 97\n97 1 50 10",
"output": "1"
},
{
"input": "3 12\n1 12 2",
"output": "1"
},
{
"input": "4 12\n1 4 3 2",
"output": "3"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "3 19\n7 1 1",
"output": "19"
},
{
"input": "5 12\n12 4 3 4 4",
"output": "1"
},
{
"input": "3 8\n8 4 2",
"output": "1"
},
{
"input": "3 3\n3 2 1",
"output": "1"
},
{
"input": "5 6\n3 2 4 2 2",
"output": "2"
},
{
"input": "2 16\n8 4",
"output": "2"
},
{
"input": "3 6\n10 2 3",
"output": "2"
},
{
"input": "5 3\n2 4 5 3 6",
"output": "1"
},
{
"input": "11 99\n1 2 3 6 5 4 7 8 99 33 66",
"output": "1"
},
{
"input": "3 12\n3 12 2",
"output": "1"
},
{
"input": "5 25\n24 5 15 25 23",
"output": "1"
},
{
"input": "2 4\n8 1",
"output": "4"
},
{
"input": "4 100\n2 50 4 1",
"output": "2"
},
{
"input": "3 28\n7 14 1",
"output": "2"
},
{
"input": "4 8\n2 8 4 1",
"output": "1"
},
{
"input": "4 6\n6 1 2 3",
"output": "1"
},
{
"input": "2 12\n4 3",
"output": "3"
},
{
"input": "4 12\n1 2 4 3",
"output": "3"
},
{
"input": "5 12\n2 3 12 6 4",
"output": "1"
},
{
"input": "4 4\n1 2 2 4",
"output": "1"
},
{
"input": "3 6\n2 3 2",
"output": "2"
},
{
"input": "4 21\n21 20 21 2",
"output": "1"
},
{
"input": "3 8\n3 4 2",
"output": "2"
},
{
"input": "1 25\n25",
"output": "1"
},
{
"input": "99 12\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99",
"output": "1"
},
{
"input": "98 12\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98",
"output": "1"
},
{
"input": "79 12\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79",
"output": "1"
},
{
"input": "4 32\n1 1 1 1",
"output": "32"
},
{
"input": "1 100\n1",
"output": "100"
},
{
"input": "2 100\n7 1",
"output": "100"
},
{
"input": "7 24\n1 3 6 4 5 2 7",
"output": "4"
},
{
"input": "6 87\n1 2 8 4 5 7",
"output": "87"
},
{
"input": "1 88\n1",
"output": "88"
},
{
"input": "1 89\n1",
"output": "89"
}
] | 1,592,163,586
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 83
| 108
| 0
|
num_buck, k = [int(x) for x in input().split()]
nums = [int(num) for num in input().split()]
mx = 0
for i in range(num_buck):
if k % nums[i] == 0:
mx = max(mx, nums[i])
print(int(k / mx))
|
Title: Garden
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Luba thinks about watering her garden. The garden can be represented as a segment of length *k*. Luba has got *n* buckets, the *i*-th bucket allows her to water some continuous subsegment of garden of length exactly *a**i* each hour. Luba can't water any parts of the garden that were already watered, also she can't water the ground outside the garden.
Luba has to choose one of the buckets in order to water the garden as fast as possible (as mentioned above, each hour she will water some continuous subsegment of length *a**i* if she chooses the *i*-th bucket). Help her to determine the minimum number of hours she has to spend watering the garden. It is guaranteed that Luba can always choose a bucket so it is possible water the garden.
See the examples for better understanding.
Input Specification:
The first line of input contains two integer numbers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of buckets and the length of the garden, respectively.
The second line of input contains *n* integer numbers *a**i* (1<=≤<=*a**i*<=≤<=100) — the length of the segment that can be watered by the *i*-th bucket in one hour.
It is guaranteed that there is at least one bucket such that it is possible to water the garden in integer number of hours using only this bucket.
Output Specification:
Print one integer number — the minimum number of hours required to water the garden.
Demo Input:
['3 6\n2 3 5\n', '6 7\n1 2 3 4 5 6\n']
Demo Output:
['2\n', '7\n']
Note:
In the first test the best option is to choose the bucket that allows to water the segment of length 3. We can't choose the bucket that allows to water the segment of length 5 because then we can't water the whole garden.
In the second test we can choose only the bucket that allows us to water the segment of length 1.
|
```python
num_buck, k = [int(x) for x in input().split()]
nums = [int(num) for num in input().split()]
mx = 0
for i in range(num_buck):
if k % nums[i] == 0:
mx = max(mx, nums[i])
print(int(k / mx))
```
| 3
|
|
11
|
A
|
Increasing Sequence
|
PROGRAMMING
| 900
|
[
"constructive algorithms",
"implementation",
"math"
] |
A. Increasing Sequence
|
1
|
64
|
A sequence *a*0,<=*a*1,<=...,<=*a**t*<=-<=1 is called increasing if *a**i*<=-<=1<=<<=*a**i* for each *i*:<=0<=<<=*i*<=<<=*t*.
You are given a sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 and a positive integer *d*. In each move you may choose one element of the given sequence and add *d* to it. What is the least number of moves required to make the given sequence increasing?
|
The first line of the input contains two integer numbers *n* and *d* (2<=≤<=*n*<=≤<=2000,<=1<=≤<=*d*<=≤<=106). The second line contains space separated sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 (1<=≤<=*b**i*<=≤<=106).
|
Output the minimal number of moves needed to make the sequence increasing.
|
[
"4 2\n1 3 3 2\n"
] |
[
"3\n"
] |
none
| 0
|
[
{
"input": "4 2\n1 3 3 2",
"output": "3"
},
{
"input": "2 1\n1 1",
"output": "1"
},
{
"input": "2 1\n2 5",
"output": "0"
},
{
"input": "2 1\n1 2",
"output": "0"
},
{
"input": "2 1\n1 1",
"output": "1"
},
{
"input": "2 7\n10 20",
"output": "0"
},
{
"input": "2 7\n1 1",
"output": "1"
},
{
"input": "3 3\n18 1 9",
"output": "10"
},
{
"input": "3 3\n15 17 9",
"output": "3"
},
{
"input": "3 3\n10 9 12",
"output": "2"
},
{
"input": "10 3\n2 1 17 10 5 16 8 4 15 17",
"output": "31"
},
{
"input": "10 3\n6 11 4 12 22 15 23 26 24 26",
"output": "13"
},
{
"input": "10 3\n10 24 13 15 18 14 15 26 33 35",
"output": "29"
},
{
"input": "100 3\n529 178 280 403 326 531 671 427 188 866 669 646 421 804 494 609 53 1012 211 243 887 833 900 543 226 42 859 718 454 372 971 692 846 770 511 395 499 479 641 756 115 269 206 45 1039 727 400 779 859 614 146 214 196 919 702 959 380 830 535 878 859 784 316 305 782 924 536 243 236 978 564 150 291 877 808 983 537 839 490 120 168 838 267 650 900 170 211 504 326 771 895 984 994 483 776 100 471 1078 317 580",
"output": "15717"
},
{
"input": "100 3\n329 226 331 909 962 112 837 1005 194 818 506 416 125 648 367 459 400 582 989 547 329 438 234 121 272 226 821 376 834 427 718 164 834 113 654 177 737 212 169 696 744 180 89 944 233 147 667 990 809 1072 1085 1093 814 265 1067 312 833 572 303 901 1032 504 185 817 389 158 613 723 239 269 911 352 769 404 225 822 897 606 947 323 913 804 923 1084 552 901 486 249 209 898 847 610 728 1122 986 669 1116 1076 367 327",
"output": "16133"
}
] | 1,650,789,062
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 15
| 1,000
| 0
|
n,d=map(int,input().split())
a=list(map(int,input().split()))
rj=0
for i in range(1,n):
while a[i]<=a[i-1]:
a[i]+=d
rj+=1
print(rj)
|
Title: Increasing Sequence
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
A sequence *a*0,<=*a*1,<=...,<=*a**t*<=-<=1 is called increasing if *a**i*<=-<=1<=<<=*a**i* for each *i*:<=0<=<<=*i*<=<<=*t*.
You are given a sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 and a positive integer *d*. In each move you may choose one element of the given sequence and add *d* to it. What is the least number of moves required to make the given sequence increasing?
Input Specification:
The first line of the input contains two integer numbers *n* and *d* (2<=≤<=*n*<=≤<=2000,<=1<=≤<=*d*<=≤<=106). The second line contains space separated sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 (1<=≤<=*b**i*<=≤<=106).
Output Specification:
Output the minimal number of moves needed to make the sequence increasing.
Demo Input:
['4 2\n1 3 3 2\n']
Demo Output:
['3\n']
Note:
none
|
```python
n,d=map(int,input().split())
a=list(map(int,input().split()))
rj=0
for i in range(1,n):
while a[i]<=a[i-1]:
a[i]+=d
rj+=1
print(rj)
```
| 0
|
620
|
B
|
Grandfather Dovlet’s calculator
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
Once Max found an electronic calculator from his grandfather Dovlet's chest. He noticed that the numbers were written with seven-segment indicators ([https://en.wikipedia.org/wiki/Seven-segment_display](https://en.wikipedia.org/wiki/Seven-segment_display)).
Max starts to type all the values from *a* to *b*. After typing each number Max resets the calculator. Find the total number of segments printed on the calculator.
For example if *a*<==<=1 and *b*<==<=3 then at first the calculator will print 2 segments, then — 5 segments and at last it will print 5 segments. So the total number of printed segments is 12.
|
The only line contains two integers *a*,<=*b* (1<=≤<=*a*<=≤<=*b*<=≤<=106) — the first and the last number typed by Max.
|
Print the only integer *a* — the total number of printed segments.
|
[
"1 3\n",
"10 15\n"
] |
[
"12\n",
"39\n"
] |
none
| 0
|
[
{
"input": "1 3",
"output": "12"
},
{
"input": "10 15",
"output": "39"
},
{
"input": "1 100",
"output": "928"
},
{
"input": "100 10000",
"output": "188446"
},
{
"input": "213 221442",
"output": "5645356"
},
{
"input": "1 1000000",
"output": "28733372"
},
{
"input": "1000000 1000000",
"output": "38"
},
{
"input": "222145 353252",
"output": "3860750"
},
{
"input": "2 1000000",
"output": "28733370"
},
{
"input": "1 999999",
"output": "28733334"
},
{
"input": "192 200",
"output": "122"
}
] | 1,514,467,526
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 5
| 1,000
| 5,529,600
|
# from dust i have come dust i will be
a,b=map(int,input().split())
seg=[6,2,5,5,4,5,6,3,7,6]
sum=0
for i in range(a,b+1):
x=i
while x>0:
r=x%10
x=x//10
sum+=seg[r]
print(sum)
|
Title: Grandfather Dovlet’s calculator
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Once Max found an electronic calculator from his grandfather Dovlet's chest. He noticed that the numbers were written with seven-segment indicators ([https://en.wikipedia.org/wiki/Seven-segment_display](https://en.wikipedia.org/wiki/Seven-segment_display)).
Max starts to type all the values from *a* to *b*. After typing each number Max resets the calculator. Find the total number of segments printed on the calculator.
For example if *a*<==<=1 and *b*<==<=3 then at first the calculator will print 2 segments, then — 5 segments and at last it will print 5 segments. So the total number of printed segments is 12.
Input Specification:
The only line contains two integers *a*,<=*b* (1<=≤<=*a*<=≤<=*b*<=≤<=106) — the first and the last number typed by Max.
Output Specification:
Print the only integer *a* — the total number of printed segments.
Demo Input:
['1 3\n', '10 15\n']
Demo Output:
['12\n', '39\n']
Note:
none
|
```python
# from dust i have come dust i will be
a,b=map(int,input().split())
seg=[6,2,5,5,4,5,6,3,7,6]
sum=0
for i in range(a,b+1):
x=i
while x>0:
r=x%10
x=x//10
sum+=seg[r]
print(sum)
```
| 0
|
|
389
|
A
|
Fox and Number Game
|
PROGRAMMING
| 1,000
|
[
"greedy",
"math"
] | null | null |
Fox Ciel is playing a game with numbers now.
Ciel has *n* positive integers: *x*1, *x*2, ..., *x**n*. She can do the following operation as many times as needed: select two different indexes *i* and *j* such that *x**i* > *x**j* hold, and then apply assignment *x**i* = *x**i* - *x**j*. The goal is to make the sum of all numbers as small as possible.
Please help Ciel to find this minimal sum.
|
The first line contains an integer *n* (2<=≤<=*n*<=≤<=100). Then the second line contains *n* integers: *x*1, *x*2, ..., *x**n* (1<=≤<=*x**i*<=≤<=100).
|
Output a single integer — the required minimal sum.
|
[
"2\n1 2\n",
"3\n2 4 6\n",
"2\n12 18\n",
"5\n45 12 27 30 18\n"
] |
[
"2\n",
"6\n",
"12\n",
"15\n"
] |
In the first example the optimal way is to do the assignment: *x*<sub class="lower-index">2</sub> = *x*<sub class="lower-index">2</sub> - *x*<sub class="lower-index">1</sub>.
In the second example the optimal sequence of operations is: *x*<sub class="lower-index">3</sub> = *x*<sub class="lower-index">3</sub> - *x*<sub class="lower-index">2</sub>, *x*<sub class="lower-index">2</sub> = *x*<sub class="lower-index">2</sub> - *x*<sub class="lower-index">1</sub>.
| 500
|
[
{
"input": "2\n1 2",
"output": "2"
},
{
"input": "3\n2 4 6",
"output": "6"
},
{
"input": "2\n12 18",
"output": "12"
},
{
"input": "5\n45 12 27 30 18",
"output": "15"
},
{
"input": "2\n1 1",
"output": "2"
},
{
"input": "2\n100 100",
"output": "200"
},
{
"input": "2\n87 58",
"output": "58"
},
{
"input": "39\n52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52",
"output": "2028"
},
{
"input": "59\n96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96",
"output": "5664"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "10000"
},
{
"input": "100\n70 70 77 42 98 84 56 91 35 21 7 70 77 77 56 63 14 84 56 14 77 77 63 70 14 7 28 91 63 49 21 84 98 56 77 98 98 84 98 14 7 56 49 28 91 98 7 56 14 91 14 98 49 28 98 14 98 98 14 70 35 28 63 28 49 63 63 56 91 98 35 42 42 35 63 35 42 14 63 21 77 56 42 77 35 91 56 21 28 84 56 70 70 91 98 70 84 63 21 98",
"output": "700"
},
{
"input": "39\n63 21 21 42 21 63 21 84 42 21 84 63 42 63 84 84 84 42 42 84 21 63 42 63 42 42 63 42 42 63 84 42 21 84 21 63 42 21 42",
"output": "819"
},
{
"input": "59\n70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70",
"output": "4130"
},
{
"input": "87\n44 88 88 88 88 66 88 22 22 88 88 44 88 22 22 22 88 88 88 88 66 22 88 88 88 88 66 66 44 88 44 44 66 22 88 88 22 44 66 44 88 66 66 22 22 22 22 88 22 22 44 66 88 22 22 88 66 66 88 22 66 88 66 88 66 44 88 44 22 44 44 22 44 88 44 44 44 44 22 88 88 88 66 66 88 44 22",
"output": "1914"
},
{
"input": "15\n63 63 63 63 63 63 63 63 63 63 63 63 63 63 63",
"output": "945"
},
{
"input": "39\n63 77 21 14 14 35 21 21 70 42 21 70 28 77 28 77 7 42 63 7 98 49 98 84 35 70 70 91 14 42 98 7 42 7 98 42 56 35 91",
"output": "273"
},
{
"input": "18\n18 18 18 36 36 36 54 72 54 36 72 54 36 36 36 36 18 36",
"output": "324"
},
{
"input": "46\n71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71",
"output": "3266"
},
{
"input": "70\n66 11 66 11 44 11 44 99 55 22 88 11 11 22 55 44 22 77 44 77 77 22 44 55 88 11 99 99 88 22 77 77 66 11 11 66 99 55 55 44 66 44 77 44 44 55 33 55 44 88 77 77 22 66 33 44 11 22 55 44 22 66 77 33 33 44 44 44 22 33",
"output": "770"
},
{
"input": "10\n60 12 96 48 60 24 60 36 60 60",
"output": "120"
},
{
"input": "20\n51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51",
"output": "1020"
},
{
"input": "50\n58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58",
"output": "2900"
},
{
"input": "98\n70 60 100 30 70 20 30 50 50 30 90 40 30 40 60 80 60 60 80 50 10 80 20 10 20 10 50 70 30 80 30 50 60 90 90 100 60 30 90 20 30 60 90 80 60 60 10 90 10 50 40 40 80 90 100 40 70 40 30 50 60 50 60 30 40 20 90 60 20 20 20 70 60 70 50 100 90 50 20 40 80 60 10 60 50 40 40 10 50 10 40 10 80 100 100 90 10 90",
"output": "980"
},
{
"input": "100\n82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82",
"output": "8200"
},
{
"input": "100\n11 87 77 93 3 54 21 93 9 71 37 23 69 85 74 3 48 99 51 31 56 19 21 96 39 6 4 4 29 69 100 42 1 22 81 53 48 49 81 61 10 7 40 61 7 71 51 59 79 44 50 35 95 80 83 8 98 40 18 94 84 49 52 74 66 69 39 37 100 44 38 62 2 80 46 31 35 53 5 60 21 49 63 55 20 53 80 53 66 34 23 92 77 50 86 63 65 24 12 70",
"output": "100"
},
{
"input": "2\n100 1",
"output": "2"
},
{
"input": "2\n18 30",
"output": "12"
},
{
"input": "2\n3 5",
"output": "2"
},
{
"input": "2\n1 10",
"output": "2"
},
{
"input": "2\n8 5",
"output": "2"
},
{
"input": "5\n2 3 5 8 18",
"output": "5"
},
{
"input": "5\n2 4 1 6 8",
"output": "5"
},
{
"input": "3\n12 10 5",
"output": "3"
},
{
"input": "3\n6 10 15",
"output": "3"
}
] | 1,530,120,767
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 6
| 78
| 0
|
n = int(input())
values = [int(k) for k in input().split()]
diff = 100000000
change = True
for j in range(len(values)-1):
for i in range(j+1,len(values)):
if values[i] - values[j] < diff and values[i] - values[j] > 0 :
diff = values[i] - values[j]
change = False
if change :
diff = values[i]
print(diff*n)
|
Title: Fox and Number Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fox Ciel is playing a game with numbers now.
Ciel has *n* positive integers: *x*1, *x*2, ..., *x**n*. She can do the following operation as many times as needed: select two different indexes *i* and *j* such that *x**i* > *x**j* hold, and then apply assignment *x**i* = *x**i* - *x**j*. The goal is to make the sum of all numbers as small as possible.
Please help Ciel to find this minimal sum.
Input Specification:
The first line contains an integer *n* (2<=≤<=*n*<=≤<=100). Then the second line contains *n* integers: *x*1, *x*2, ..., *x**n* (1<=≤<=*x**i*<=≤<=100).
Output Specification:
Output a single integer — the required minimal sum.
Demo Input:
['2\n1 2\n', '3\n2 4 6\n', '2\n12 18\n', '5\n45 12 27 30 18\n']
Demo Output:
['2\n', '6\n', '12\n', '15\n']
Note:
In the first example the optimal way is to do the assignment: *x*<sub class="lower-index">2</sub> = *x*<sub class="lower-index">2</sub> - *x*<sub class="lower-index">1</sub>.
In the second example the optimal sequence of operations is: *x*<sub class="lower-index">3</sub> = *x*<sub class="lower-index">3</sub> - *x*<sub class="lower-index">2</sub>, *x*<sub class="lower-index">2</sub> = *x*<sub class="lower-index">2</sub> - *x*<sub class="lower-index">1</sub>.
|
```python
n = int(input())
values = [int(k) for k in input().split()]
diff = 100000000
change = True
for j in range(len(values)-1):
for i in range(j+1,len(values)):
if values[i] - values[j] < diff and values[i] - values[j] > 0 :
diff = values[i] - values[j]
change = False
if change :
diff = values[i]
print(diff*n)
```
| 0
|
|
386
|
A
|
Second-Price Auction
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
In this problem we consider a special type of an auction, which is called the second-price auction. As in regular auction *n* bidders place a bid which is price a bidder ready to pay. The auction is closed, that is, each bidder secretly informs the organizer of the auction price he is willing to pay. After that, the auction winner is the participant who offered the highest price. However, he pay not the price he offers, but the highest price among the offers of other participants (hence the name: the second-price auction).
Write a program that reads prices offered by bidders and finds the winner and the price he will pay. Consider that all of the offered prices are different.
|
The first line of the input contains *n* (2<=≤<=*n*<=≤<=1000) — number of bidders. The second line contains *n* distinct integer numbers *p*1,<=*p*2,<=... *p**n*, separated by single spaces (1<=≤<=*p**i*<=≤<=10000), where *p**i* stands for the price offered by the *i*-th bidder.
|
The single output line should contain two integers: index of the winner and the price he will pay. Indices are 1-based.
|
[
"2\n5 7\n",
"3\n10 2 8\n",
"6\n3 8 2 9 4 14\n"
] |
[
"2 5\n",
"1 8\n",
"6 9\n"
] |
none
| 500
|
[
{
"input": "2\n5 7",
"output": "2 5"
},
{
"input": "3\n10 2 8",
"output": "1 8"
},
{
"input": "6\n3 8 2 9 4 14",
"output": "6 9"
},
{
"input": "4\n4707 7586 4221 5842",
"output": "2 5842"
},
{
"input": "5\n3304 4227 4869 6937 6002",
"output": "4 6002"
},
{
"input": "6\n5083 3289 7708 5362 9031 7458",
"output": "5 7708"
},
{
"input": "7\n9038 6222 3392 1706 3778 1807 2657",
"output": "1 6222"
},
{
"input": "8\n7062 2194 4481 3864 7470 1814 8091 733",
"output": "7 7470"
},
{
"input": "9\n2678 5659 9199 2628 7906 7496 4524 2663 3408",
"output": "3 7906"
},
{
"input": "2\n3458 1504",
"output": "1 1504"
},
{
"input": "50\n9237 3904 407 9052 6657 9229 9752 3888 7732 2512 4614 1055 2355 7108 6506 6849 2529 8862 159 8630 7906 7941 960 8470 333 8659 54 9475 3163 5625 6393 6814 2656 3388 169 7918 4881 8468 9983 6281 6340 280 5108 2996 101 7617 3313 8172 326 1991",
"output": "39 9752"
},
{
"input": "100\n2515 3324 7975 6171 4240 1217 4829 5203 8603 6900 3031 4699 4732 6070 4221 3228 6497 7359 9130 4346 4619 1109 3945 5442 3271 16 9711 2045 6410 2301 3406 8125 6003 1892 1260 9661 3940 6692 4708 7027 4930 6925 1979 5361 4263 3144 867 8639 6230 5562 9714 3676 4231 3347 4733 4920 4881 3431 1059 7313 8912 3038 9308 72 9583 7009 3034 7425 2398 6141 3245 2495 2933 6710 8289 9806 1226 8393 7349 6462 1618 9613 3546 6012 2964 9995 1578 210 2123 4874 1252 8625 348 8020 803 7244 9080 5088 706 2602",
"output": "86 9806"
},
{
"input": "2\n2 1",
"output": "1 1"
},
{
"input": "2\n1 2",
"output": "2 1"
},
{
"input": "3\n10 20 30",
"output": "3 20"
},
{
"input": "3\n10 30 20",
"output": "2 20"
},
{
"input": "3\n20 10 30",
"output": "3 20"
},
{
"input": "3\n20 30 10",
"output": "2 20"
},
{
"input": "3\n30 10 20",
"output": "1 20"
},
{
"input": "3\n30 20 10",
"output": "1 20"
},
{
"input": "2\n1 10000",
"output": "2 1"
},
{
"input": "2\n10000 999",
"output": "1 999"
},
{
"input": "3\n3 4 1",
"output": "2 3"
},
{
"input": "6\n1 2 24 6 5 7",
"output": "3 7"
}
] | 1,699,263,386
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 42
| 46
| 0
|
n = int(input())
*a, = map(int, input().split())
x = a.index(max(a))
a[x] = 0
print(x + 1, max(a))
|
Title: Second-Price Auction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In this problem we consider a special type of an auction, which is called the second-price auction. As in regular auction *n* bidders place a bid which is price a bidder ready to pay. The auction is closed, that is, each bidder secretly informs the organizer of the auction price he is willing to pay. After that, the auction winner is the participant who offered the highest price. However, he pay not the price he offers, but the highest price among the offers of other participants (hence the name: the second-price auction).
Write a program that reads prices offered by bidders and finds the winner and the price he will pay. Consider that all of the offered prices are different.
Input Specification:
The first line of the input contains *n* (2<=≤<=*n*<=≤<=1000) — number of bidders. The second line contains *n* distinct integer numbers *p*1,<=*p*2,<=... *p**n*, separated by single spaces (1<=≤<=*p**i*<=≤<=10000), where *p**i* stands for the price offered by the *i*-th bidder.
Output Specification:
The single output line should contain two integers: index of the winner and the price he will pay. Indices are 1-based.
Demo Input:
['2\n5 7\n', '3\n10 2 8\n', '6\n3 8 2 9 4 14\n']
Demo Output:
['2 5\n', '1 8\n', '6 9\n']
Note:
none
|
```python
n = int(input())
*a, = map(int, input().split())
x = a.index(max(a))
a[x] = 0
print(x + 1, max(a))
```
| 3
|
|
160
|
A
|
Twins
|
PROGRAMMING
| 900
|
[
"greedy",
"sortings"
] | null | null |
Imagine that you have a twin brother or sister. Having another person that looks exactly like you seems very unusual. It's hard to say if having something of an alter ego is good or bad. And if you do have a twin, then you very well know what it's like.
Now let's imagine a typical morning in your family. You haven't woken up yet, and Mom is already going to work. She has been so hasty that she has nearly forgotten to leave the two of her darling children some money to buy lunches in the school cafeteria. She fished in the purse and found some number of coins, or to be exact, *n* coins of arbitrary values *a*1,<=*a*2,<=...,<=*a**n*. But as Mom was running out of time, she didn't split the coins for you two. So she scribbled a note asking you to split the money equally.
As you woke up, you found Mom's coins and read her note. "But why split the money equally?" — you thought. After all, your twin is sleeping and he won't know anything. So you decided to act like that: pick for yourself some subset of coins so that the sum of values of your coins is strictly larger than the sum of values of the remaining coins that your twin will have. However, you correctly thought that if you take too many coins, the twin will suspect the deception. So, you've decided to stick to the following strategy to avoid suspicions: you take the minimum number of coins, whose sum of values is strictly more than the sum of values of the remaining coins. On this basis, determine what minimum number of coins you need to take to divide them in the described manner.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of coins. The second line contains a sequence of *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=100) — the coins' values. All numbers are separated with spaces.
|
In the single line print the single number — the minimum needed number of coins.
|
[
"2\n3 3\n",
"3\n2 1 2\n"
] |
[
"2\n",
"2\n"
] |
In the first sample you will have to take 2 coins (you and your twin have sums equal to 6, 0 correspondingly). If you take 1 coin, you get sums 3, 3. If you take 0 coins, you get sums 0, 6. Those variants do not satisfy you as your sum should be strictly more that your twins' sum.
In the second sample one coin isn't enough for us, too. You can pick coins with values 1, 2 or 2, 2. In any case, the minimum number of coins equals 2.
| 500
|
[
{
"input": "2\n3 3",
"output": "2"
},
{
"input": "3\n2 1 2",
"output": "2"
},
{
"input": "1\n5",
"output": "1"
},
{
"input": "5\n4 2 2 2 2",
"output": "3"
},
{
"input": "7\n1 10 1 2 1 1 1",
"output": "1"
},
{
"input": "5\n3 2 3 3 1",
"output": "3"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "3\n2 1 3",
"output": "2"
},
{
"input": "6\n1 1 1 1 1 1",
"output": "4"
},
{
"input": "7\n10 10 5 5 5 5 1",
"output": "3"
},
{
"input": "20\n2 1 2 2 2 1 1 2 1 2 2 1 1 1 1 2 1 1 1 1",
"output": "8"
},
{
"input": "20\n4 2 4 4 3 4 2 2 4 2 3 1 1 2 2 3 3 3 1 4",
"output": "8"
},
{
"input": "20\n35 26 41 40 45 46 22 26 39 23 11 15 47 42 18 15 27 10 45 40",
"output": "8"
},
{
"input": "20\n7 84 100 10 31 35 41 2 63 44 57 4 63 11 23 49 98 71 16 90",
"output": "6"
},
{
"input": "50\n19 2 12 26 17 27 10 26 17 17 5 24 11 15 3 9 16 18 19 1 25 23 18 6 2 7 25 7 21 25 13 29 16 9 25 3 14 30 18 4 10 28 6 10 8 2 2 4 8 28",
"output": "14"
},
{
"input": "70\n2 18 18 47 25 5 14 9 19 46 36 49 33 32 38 23 32 39 8 29 31 17 24 21 10 15 33 37 46 21 22 11 20 35 39 13 11 30 28 40 39 47 1 17 24 24 21 46 12 2 20 43 8 16 44 11 45 10 13 44 31 45 45 46 11 10 33 35 23 42",
"output": "22"
},
{
"input": "100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "51"
},
{
"input": "100\n1 2 2 1 2 1 1 2 1 1 1 2 2 1 1 1 2 2 2 1 2 1 1 1 1 1 2 1 2 1 2 1 2 1 2 1 1 1 2 1 1 1 1 1 2 2 1 2 1 2 1 2 2 2 1 2 1 2 2 1 1 2 2 1 1 2 2 2 1 1 2 1 1 2 2 1 2 1 1 2 2 1 2 1 1 2 2 1 1 1 1 2 1 1 1 1 2 2 2 2",
"output": "37"
},
{
"input": "100\n1 2 3 2 1 2 2 3 1 3 3 2 2 1 1 2 2 1 1 1 1 2 3 3 2 1 1 2 2 2 3 3 3 2 1 3 1 3 3 2 3 1 2 2 2 3 2 1 1 3 3 3 3 2 1 1 2 3 2 2 3 2 3 2 2 3 2 2 2 2 3 3 3 1 3 3 1 1 2 3 2 2 2 2 3 3 3 2 1 2 3 1 1 2 3 3 1 3 3 2",
"output": "36"
},
{
"input": "100\n5 5 4 3 5 1 2 5 1 1 3 5 4 4 1 1 1 1 5 4 4 5 1 5 5 1 2 1 3 1 5 1 3 3 3 2 2 2 1 1 5 1 3 4 1 1 3 2 5 2 2 5 5 4 4 1 3 4 3 3 4 5 3 3 3 1 2 1 4 2 4 4 1 5 1 3 5 5 5 5 3 4 4 3 1 2 5 2 3 5 4 2 4 5 3 2 4 2 4 3",
"output": "33"
},
{
"input": "100\n3 4 8 10 8 6 4 3 7 7 6 2 3 1 3 10 1 7 9 3 5 5 2 6 2 9 1 7 4 2 4 1 6 1 7 10 2 5 3 7 6 4 6 2 8 8 8 6 6 10 3 7 4 3 4 1 7 9 3 6 3 6 1 4 9 3 8 1 10 1 4 10 7 7 9 5 3 8 10 2 1 10 8 7 10 8 5 3 1 2 1 10 6 1 5 3 3 5 7 2",
"output": "30"
},
{
"input": "100\n16 9 11 8 11 4 9 17 4 8 4 10 9 10 6 3 3 15 1 6 1 15 12 18 6 14 13 18 1 7 18 4 10 7 10 12 3 16 14 4 10 8 10 7 19 13 15 1 4 8 16 10 6 4 3 16 11 10 7 3 4 16 1 20 1 11 4 16 10 7 7 12 18 19 3 17 19 3 4 19 2 12 11 3 18 20 2 2 14 4 20 13 13 11 16 20 19 14 7 2",
"output": "29"
},
{
"input": "100\n2 46 4 6 38 19 15 34 10 35 37 30 3 25 5 45 40 45 33 31 6 20 10 44 11 9 2 14 35 5 9 23 20 2 48 22 25 35 38 31 24 33 35 16 4 30 27 10 12 22 6 24 12 30 23 21 14 12 32 21 7 12 25 43 18 34 34 28 47 13 28 43 18 39 44 42 35 26 35 14 8 29 32 20 29 3 20 6 20 9 9 27 8 42 10 37 42 27 8 1",
"output": "30"
},
{
"input": "100\n85 50 17 89 65 89 5 20 86 26 16 21 85 14 44 31 87 31 6 2 48 67 8 80 79 1 48 36 97 1 5 30 79 50 78 12 2 55 76 100 54 40 26 81 97 96 68 56 87 14 51 17 54 37 52 33 69 62 38 63 74 15 62 78 9 19 67 2 60 58 93 60 18 96 55 48 34 7 79 82 32 58 90 67 20 50 27 15 7 89 98 10 11 15 99 49 4 51 77 52",
"output": "29"
},
{
"input": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100",
"output": "30"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "4\n4 1 2 3",
"output": "2"
},
{
"input": "5\n2 2 2 4 4",
"output": "2"
},
{
"input": "4\n100 99 99 100",
"output": "2"
}
] | 1,699,732,066
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
input();a=sorted(map(int,input().split()));
i = s = 0
while :
s < sum(a)
s += a.pop()
i += 1
print(i)
|
Title: Twins
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Imagine that you have a twin brother or sister. Having another person that looks exactly like you seems very unusual. It's hard to say if having something of an alter ego is good or bad. And if you do have a twin, then you very well know what it's like.
Now let's imagine a typical morning in your family. You haven't woken up yet, and Mom is already going to work. She has been so hasty that she has nearly forgotten to leave the two of her darling children some money to buy lunches in the school cafeteria. She fished in the purse and found some number of coins, or to be exact, *n* coins of arbitrary values *a*1,<=*a*2,<=...,<=*a**n*. But as Mom was running out of time, she didn't split the coins for you two. So she scribbled a note asking you to split the money equally.
As you woke up, you found Mom's coins and read her note. "But why split the money equally?" — you thought. After all, your twin is sleeping and he won't know anything. So you decided to act like that: pick for yourself some subset of coins so that the sum of values of your coins is strictly larger than the sum of values of the remaining coins that your twin will have. However, you correctly thought that if you take too many coins, the twin will suspect the deception. So, you've decided to stick to the following strategy to avoid suspicions: you take the minimum number of coins, whose sum of values is strictly more than the sum of values of the remaining coins. On this basis, determine what minimum number of coins you need to take to divide them in the described manner.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of coins. The second line contains a sequence of *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=100) — the coins' values. All numbers are separated with spaces.
Output Specification:
In the single line print the single number — the minimum needed number of coins.
Demo Input:
['2\n3 3\n', '3\n2 1 2\n']
Demo Output:
['2\n', '2\n']
Note:
In the first sample you will have to take 2 coins (you and your twin have sums equal to 6, 0 correspondingly). If you take 1 coin, you get sums 3, 3. If you take 0 coins, you get sums 0, 6. Those variants do not satisfy you as your sum should be strictly more that your twins' sum.
In the second sample one coin isn't enough for us, too. You can pick coins with values 1, 2 or 2, 2. In any case, the minimum number of coins equals 2.
|
```python
input();a=sorted(map(int,input().split()));
i = s = 0
while :
s < sum(a)
s += a.pop()
i += 1
print(i)
```
| -1
|
|
602
|
B
|
Approximating a Constant Range
|
PROGRAMMING
| 1,400
|
[
"dp",
"implementation",
"two pointers"
] | null | null |
When Xellos was doing a practice course in university, he once had to measure the intensity of an effect that slowly approached equilibrium. A good way to determine the equilibrium intensity would be choosing a sufficiently large number of consecutive data points that seems as constant as possible and taking their average. Of course, with the usual sizes of data, it's nothing challenging — but why not make a similar programming contest problem while we're at it?
You're given a sequence of *n* data points *a*1,<=...,<=*a**n*. There aren't any big jumps between consecutive data points — for each 1<=≤<=*i*<=<<=*n*, it's guaranteed that |*a**i*<=+<=1<=-<=*a**i*|<=≤<=1.
A range [*l*,<=*r*] of data points is said to be almost constant if the difference between the largest and the smallest value in that range is at most 1. Formally, let *M* be the maximum and *m* the minimum value of *a**i* for *l*<=≤<=*i*<=≤<=*r*; the range [*l*,<=*r*] is almost constant if *M*<=-<=*m*<=≤<=1.
Find the length of the longest almost constant range.
|
The first line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100<=000) — the number of data points.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100<=000).
|
Print a single number — the maximum length of an almost constant range of the given sequence.
|
[
"5\n1 2 3 3 2\n",
"11\n5 4 5 5 6 7 8 8 8 7 6\n"
] |
[
"4\n",
"5\n"
] |
In the first sample, the longest almost constant range is [2, 5]; its length (the number of data points in it) is 4.
In the second sample, there are three almost constant ranges of length 4: [1, 4], [6, 9] and [7, 10]; the only almost constant range of the maximum length 5 is [6, 10].
| 1,000
|
[
{
"input": "5\n1 2 3 3 2",
"output": "4"
},
{
"input": "11\n5 4 5 5 6 7 8 8 8 7 6",
"output": "5"
},
{
"input": "2\n3 2",
"output": "2"
},
{
"input": "4\n1001 1000 1000 1001",
"output": "4"
},
{
"input": "4\n1 1 2 3",
"output": "3"
},
{
"input": "3\n1 2 1",
"output": "3"
},
{
"input": "3\n1 2 3",
"output": "2"
},
{
"input": "18\n10 9 8 7 6 5 4 3 2 1 2 3 4 5 6 7 8 9",
"output": "3"
},
{
"input": "3\n1 2 2",
"output": "3"
},
{
"input": "4\n10 9 10 9",
"output": "4"
},
{
"input": "4\n4 3 2 3",
"output": "3"
},
{
"input": "4\n8 8 7 7",
"output": "4"
},
{
"input": "3\n99998 99999 100000",
"output": "2"
},
{
"input": "3\n100000 99999 99998",
"output": "2"
},
{
"input": "3\n1 1 1",
"output": "3"
},
{
"input": "2\n99999 100000",
"output": "2"
},
{
"input": "2\n100000 100000",
"output": "2"
},
{
"input": "2\n1 1",
"output": "2"
},
{
"input": "15\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000",
"output": "15"
}
] | 1,691,835,216
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 7,884,800
|
n = int(input())
ans = 2
a = [0] * 1000005
sequence = list(map(int, input().split()))
for i, x in enumerate(sequence, start=1):
a[x] = i
for i, x in enumerate(sequence, start=1):
if a[x - 1] > a[x + 1]:
ans = max(ans, i - max(a[x + 1], a[x - 2]) - 1)
else:
ans = max(ans, i - max(a[x + 2], a[x - 1]) - 1)
print(ans)
|
Title: Approximating a Constant Range
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
When Xellos was doing a practice course in university, he once had to measure the intensity of an effect that slowly approached equilibrium. A good way to determine the equilibrium intensity would be choosing a sufficiently large number of consecutive data points that seems as constant as possible and taking their average. Of course, with the usual sizes of data, it's nothing challenging — but why not make a similar programming contest problem while we're at it?
You're given a sequence of *n* data points *a*1,<=...,<=*a**n*. There aren't any big jumps between consecutive data points — for each 1<=≤<=*i*<=<<=*n*, it's guaranteed that |*a**i*<=+<=1<=-<=*a**i*|<=≤<=1.
A range [*l*,<=*r*] of data points is said to be almost constant if the difference between the largest and the smallest value in that range is at most 1. Formally, let *M* be the maximum and *m* the minimum value of *a**i* for *l*<=≤<=*i*<=≤<=*r*; the range [*l*,<=*r*] is almost constant if *M*<=-<=*m*<=≤<=1.
Find the length of the longest almost constant range.
Input Specification:
The first line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100<=000) — the number of data points.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100<=000).
Output Specification:
Print a single number — the maximum length of an almost constant range of the given sequence.
Demo Input:
['5\n1 2 3 3 2\n', '11\n5 4 5 5 6 7 8 8 8 7 6\n']
Demo Output:
['4\n', '5\n']
Note:
In the first sample, the longest almost constant range is [2, 5]; its length (the number of data points in it) is 4.
In the second sample, there are three almost constant ranges of length 4: [1, 4], [6, 9] and [7, 10]; the only almost constant range of the maximum length 5 is [6, 10].
|
```python
n = int(input())
ans = 2
a = [0] * 1000005
sequence = list(map(int, input().split()))
for i, x in enumerate(sequence, start=1):
a[x] = i
for i, x in enumerate(sequence, start=1):
if a[x - 1] > a[x + 1]:
ans = max(ans, i - max(a[x + 1], a[x - 2]) - 1)
else:
ans = max(ans, i - max(a[x + 2], a[x - 1]) - 1)
print(ans)
```
| 0
|
|
180
|
C
|
Letter
|
PROGRAMMING
| 1,400
|
[
"dp"
] | null | null |
Patrick has just finished writing a message to his sweetheart Stacey when he noticed that the message didn't look fancy. Patrick was nervous while writing the message, so some of the letters there were lowercase and some of them were uppercase.
Patrick believes that a message is fancy if any uppercase letter stands to the left of any lowercase one. In other words, this rule describes the strings where first go zero or more uppercase letters, and then — zero or more lowercase letters.
To make the message fancy, Patrick can erase some letter and add the same letter in the same place in the opposite case (that is, he can replace an uppercase letter with the lowercase one and vice versa). Patrick got interested in the following question: what minimum number of actions do we need to make a message fancy? Changing a letter's case in the message counts as one action. Patrick cannot perform any other actions.
|
The only line of the input contains a non-empty string consisting of uppercase and lowercase letters. The string's length does not exceed 105.
|
Print a single number — the least number of actions needed to make the message fancy.
|
[
"PRuvetSTAaYA\n",
"OYPROSTIYAOPECHATALSYAPRIVETSTASYA\n",
"helloworld\n"
] |
[
"5\n",
"0\n",
"0\n"
] |
none
| 0
|
[
{
"input": "PRuvetSTAaYA",
"output": "5"
},
{
"input": "OYPROSTIYAOPECHATALSYAPRIVETSTASYA",
"output": "0"
},
{
"input": "helloworld",
"output": "0"
},
{
"input": "P",
"output": "0"
},
{
"input": "t",
"output": "0"
},
{
"input": "XdJ",
"output": "1"
},
{
"input": "FSFlNEelYY",
"output": "3"
},
{
"input": "lgtyasficu",
"output": "0"
},
{
"input": "WYKUDTDDBT",
"output": "0"
},
{
"input": "yysxwlyqboatikfnpxczmpijziiojbvadlfozjqldssffcxdegyxfrvohoxvgsrvlzjlkcuffoeisrpvagxtbkapkpzcafadzzjd",
"output": "0"
},
{
"input": "mnAkOBuKxaiJwXhKnlcCvjxYXGXDoIqfUYkiLrdSYWhMemgWFzsgpoKOtHqooxbLYFuABWQSXuHdbyPVWyrkeEfqOsnEBikiqhfu",
"output": "43"
},
{
"input": "MMVESdOCALHJCTBTUWWQRGUUVTTTABKKAAdIINAdKLRLLVLODHDXDPMcQfUhPNHFBJSDRGsHZNORSCPNvKOOIuZnZAmTPUCoPNlR",
"output": "13"
},
{
"input": "MMbJIBhgFXPVpdQHLkWJkAHFIfJSpITTCRzRCzvRPRYECCheOknfINZWuKATDBOrEVKTcWXiYPjtzQMGUSGPNTMCUrvYCSWQHqAi",
"output": "27"
},
{
"input": "ZnqXEBOABXVbHRFFqDLWpWBBLqZIagmbRAYMDKJAYyFQXnPYwvDyRXKHLQnMLHGEOnRKmXtGqyQkoBCSmgKPcEIEugwJyxDOtTVn",
"output": "31"
}
] | 1,680,022,075
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 40
| 278
| 5,222,400
|
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
s = list(input().rstrip())
cnt = [0]
for i in s:
cnt.append(cnt[-1] + i // 97)
n = len(s)
ans = n
for i in range(n + 1):
u, v = cnt[i], n - i - (cnt[n] - cnt[i])
ans = min(ans, u + v)
print(ans)
|
Title: Letter
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Patrick has just finished writing a message to his sweetheart Stacey when he noticed that the message didn't look fancy. Patrick was nervous while writing the message, so some of the letters there were lowercase and some of them were uppercase.
Patrick believes that a message is fancy if any uppercase letter stands to the left of any lowercase one. In other words, this rule describes the strings where first go zero or more uppercase letters, and then — zero or more lowercase letters.
To make the message fancy, Patrick can erase some letter and add the same letter in the same place in the opposite case (that is, he can replace an uppercase letter with the lowercase one and vice versa). Patrick got interested in the following question: what minimum number of actions do we need to make a message fancy? Changing a letter's case in the message counts as one action. Patrick cannot perform any other actions.
Input Specification:
The only line of the input contains a non-empty string consisting of uppercase and lowercase letters. The string's length does not exceed 105.
Output Specification:
Print a single number — the least number of actions needed to make the message fancy.
Demo Input:
['PRuvetSTAaYA\n', 'OYPROSTIYAOPECHATALSYAPRIVETSTASYA\n', 'helloworld\n']
Demo Output:
['5\n', '0\n', '0\n']
Note:
none
|
```python
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
s = list(input().rstrip())
cnt = [0]
for i in s:
cnt.append(cnt[-1] + i // 97)
n = len(s)
ans = n
for i in range(n + 1):
u, v = cnt[i], n - i - (cnt[n] - cnt[i])
ans = min(ans, u + v)
print(ans)
```
| 3
|
|
994
|
A
|
Fingerprints
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
You are locked in a room with a door that has a keypad with 10 keys corresponding to digits from 0 to 9. To escape from the room, you need to enter a correct code. You also have a sequence of digits.
Some keys on the keypad have fingerprints. You believe the correct code is the longest not necessarily contiguous subsequence of the sequence you have that only contains digits with fingerprints on the corresponding keys. Find such code.
|
The first line contains two integers $n$ and $m$ ($1 \le n, m \le 10$) representing the number of digits in the sequence you have and the number of keys on the keypad that have fingerprints.
The next line contains $n$ distinct space-separated integers $x_1, x_2, \ldots, x_n$ ($0 \le x_i \le 9$) representing the sequence.
The next line contains $m$ distinct space-separated integers $y_1, y_2, \ldots, y_m$ ($0 \le y_i \le 9$) — the keys with fingerprints.
|
In a single line print a space-separated sequence of integers representing the code. If the resulting sequence is empty, both printing nothing and printing a single line break is acceptable.
|
[
"7 3\n3 5 7 1 6 2 8\n1 2 7\n",
"4 4\n3 4 1 0\n0 1 7 9\n"
] |
[
"7 1 2\n",
"1 0\n"
] |
In the first example, the only digits with fingerprints are $1$, $2$ and $7$. All three of them appear in the sequence you know, $7$ first, then $1$ and then $2$. Therefore the output is 7 1 2. Note that the order is important, and shall be the same as the order in the original sequence.
In the second example digits $0$, $1$, $7$ and $9$ have fingerprints, however only $0$ and $1$ appear in the original sequence. $1$ appears earlier, so the output is 1 0. Again, the order is important.
| 500
|
[
{
"input": "7 3\n3 5 7 1 6 2 8\n1 2 7",
"output": "7 1 2"
},
{
"input": "4 4\n3 4 1 0\n0 1 7 9",
"output": "1 0"
},
{
"input": "9 4\n9 8 7 6 5 4 3 2 1\n2 4 6 8",
"output": "8 6 4 2"
},
{
"input": "10 5\n3 7 1 2 4 6 9 0 5 8\n4 3 0 7 9",
"output": "3 7 4 9 0"
},
{
"input": "10 10\n1 2 3 4 5 6 7 8 9 0\n4 5 6 7 1 2 3 0 9 8",
"output": "1 2 3 4 5 6 7 8 9 0"
},
{
"input": "1 1\n4\n4",
"output": "4"
},
{
"input": "3 7\n6 3 4\n4 9 0 1 7 8 6",
"output": "6 4"
},
{
"input": "10 1\n9 0 8 1 7 4 6 5 2 3\n0",
"output": "0"
},
{
"input": "5 10\n6 0 3 8 1\n3 1 0 5 4 7 2 8 9 6",
"output": "6 0 3 8 1"
},
{
"input": "8 2\n7 2 9 6 1 0 3 4\n6 3",
"output": "6 3"
},
{
"input": "5 4\n7 0 1 4 9\n0 9 5 3",
"output": "0 9"
},
{
"input": "10 1\n9 6 2 0 1 8 3 4 7 5\n6",
"output": "6"
},
{
"input": "10 2\n7 1 0 2 4 6 5 9 3 8\n3 2",
"output": "2 3"
},
{
"input": "5 9\n3 7 9 2 4\n3 8 4 5 9 6 1 0 2",
"output": "3 9 2 4"
},
{
"input": "10 6\n7 1 2 3 8 0 6 4 5 9\n1 5 8 2 3 6",
"output": "1 2 3 8 6 5"
},
{
"input": "8 2\n7 4 8 9 2 5 6 1\n6 4",
"output": "4 6"
},
{
"input": "10 2\n1 0 3 5 8 9 4 7 6 2\n0 3",
"output": "0 3"
},
{
"input": "7 6\n9 2 8 6 1 3 7\n4 2 0 3 1 8",
"output": "2 8 1 3"
},
{
"input": "1 6\n3\n6 8 2 4 5 3",
"output": "3"
},
{
"input": "1 8\n0\n9 2 4 8 1 5 0 7",
"output": "0"
},
{
"input": "6 9\n7 3 9 4 1 0\n9 1 5 8 0 6 2 7 4",
"output": "7 9 4 1 0"
},
{
"input": "10 2\n4 9 6 8 3 0 1 5 7 2\n0 1",
"output": "0 1"
},
{
"input": "10 5\n5 2 8 0 9 7 6 1 4 3\n9 6 4 1 2",
"output": "2 9 6 1 4"
},
{
"input": "6 3\n8 3 9 2 7 6\n5 4 3",
"output": "3"
},
{
"input": "4 10\n8 3 9 6\n4 9 6 2 7 0 8 1 3 5",
"output": "8 3 9 6"
},
{
"input": "1 2\n1\n1 0",
"output": "1"
},
{
"input": "3 6\n1 2 3\n4 5 6 1 2 3",
"output": "1 2 3"
},
{
"input": "1 2\n2\n1 2",
"output": "2"
},
{
"input": "1 10\n9\n0 1 2 3 4 5 6 7 8 9",
"output": "9"
}
] | 1,577,035,355
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 31
| 124
| 307,200
|
map(int,input().split());k=[int(i) for i in input().split()];n=[int(i) for i in input().split()];p=[]
for i in n:
if i in k:p=p+[[k.index(i),i]]
p=sorted(p)
for i in range(0,len(p)):print(p[i][1],end=" ")
|
Title: Fingerprints
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are locked in a room with a door that has a keypad with 10 keys corresponding to digits from 0 to 9. To escape from the room, you need to enter a correct code. You also have a sequence of digits.
Some keys on the keypad have fingerprints. You believe the correct code is the longest not necessarily contiguous subsequence of the sequence you have that only contains digits with fingerprints on the corresponding keys. Find such code.
Input Specification:
The first line contains two integers $n$ and $m$ ($1 \le n, m \le 10$) representing the number of digits in the sequence you have and the number of keys on the keypad that have fingerprints.
The next line contains $n$ distinct space-separated integers $x_1, x_2, \ldots, x_n$ ($0 \le x_i \le 9$) representing the sequence.
The next line contains $m$ distinct space-separated integers $y_1, y_2, \ldots, y_m$ ($0 \le y_i \le 9$) — the keys with fingerprints.
Output Specification:
In a single line print a space-separated sequence of integers representing the code. If the resulting sequence is empty, both printing nothing and printing a single line break is acceptable.
Demo Input:
['7 3\n3 5 7 1 6 2 8\n1 2 7\n', '4 4\n3 4 1 0\n0 1 7 9\n']
Demo Output:
['7 1 2\n', '1 0\n']
Note:
In the first example, the only digits with fingerprints are $1$, $2$ and $7$. All three of them appear in the sequence you know, $7$ first, then $1$ and then $2$. Therefore the output is 7 1 2. Note that the order is important, and shall be the same as the order in the original sequence.
In the second example digits $0$, $1$, $7$ and $9$ have fingerprints, however only $0$ and $1$ appear in the original sequence. $1$ appears earlier, so the output is 1 0. Again, the order is important.
|
```python
map(int,input().split());k=[int(i) for i in input().split()];n=[int(i) for i in input().split()];p=[]
for i in n:
if i in k:p=p+[[k.index(i),i]]
p=sorted(p)
for i in range(0,len(p)):print(p[i][1],end=" ")
```
| 3
|
|
415
|
B
|
Mashmokh and Tokens
|
PROGRAMMING
| 1,500
|
[
"binary search",
"greedy",
"implementation",
"math"
] | null | null |
Bimokh is Mashmokh's boss. For the following *n* days he decided to pay to his workers in a new way. At the beginning of each day he will give each worker a certain amount of tokens. Then at the end of each day each worker can give some of his tokens back to get a certain amount of money. The worker can save the rest of tokens but he can't use it in any other day to get more money. If a worker gives back *w* tokens then he'll get dollars.
Mashmokh likes the tokens however he likes money more. That's why he wants to save as many tokens as possible so that the amount of money he gets is maximal possible each day. He has *n* numbers *x*1,<=*x*2,<=...,<=*x**n*. Number *x**i* is the number of tokens given to each worker on the *i*-th day. Help him calculate for each of *n* days the number of tokens he can save.
|
The first line of input contains three space-separated integers *n*,<=*a*,<=*b* (1<=≤<=*n*<=≤<=105; 1<=≤<=*a*,<=*b*<=≤<=109). The second line of input contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=109).
|
Output *n* space-separated integers. The *i*-th of them is the number of tokens Mashmokh can save on the *i*-th day.
|
[
"5 1 4\n12 6 11 9 1\n",
"3 1 2\n1 2 3\n",
"1 1 1\n1\n"
] |
[
"0 2 3 1 1 ",
"1 0 1 ",
"0 "
] |
none
| 1,000
|
[
{
"input": "5 1 4\n12 6 11 9 1",
"output": "0 2 3 1 1 "
},
{
"input": "3 1 2\n1 2 3",
"output": "1 0 1 "
},
{
"input": "1 1 1\n1",
"output": "0 "
},
{
"input": "1 1 1000000000\n1000000000",
"output": "0 "
},
{
"input": "1 1 1000000000\n999999999",
"output": "999999999 "
},
{
"input": "10 1 100000000\n999999999 999999999 999999999 999999999 999999999 999999999 999999999 999999999 999999999 999999999",
"output": "99999999 99999999 99999999 99999999 99999999 99999999 99999999 99999999 99999999 99999999 "
}
] | 1,396,801,831
| 3,031
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
PRETESTS
| 3
| 1,000
| 5,017,600
|
def main():
n, a, b = list(map(int, input().split()))
x_list = list(map(int, input().split()))
res = list()
if a == 1:
for x in x_list:
res.append(str(x * a % b))
else:
for x in x_list:
t = x
s = 0
maxi = x * a // b
while True:
if t * a % b == 0:
break
if t == 0:
break
if t * a // b < maxi:
s -= 1
break
t -= 1
s += 1
res.append(str(s))
print(' '.join(res))
if __name__ == '__main__':
main()
|
Title: Mashmokh and Tokens
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bimokh is Mashmokh's boss. For the following *n* days he decided to pay to his workers in a new way. At the beginning of each day he will give each worker a certain amount of tokens. Then at the end of each day each worker can give some of his tokens back to get a certain amount of money. The worker can save the rest of tokens but he can't use it in any other day to get more money. If a worker gives back *w* tokens then he'll get dollars.
Mashmokh likes the tokens however he likes money more. That's why he wants to save as many tokens as possible so that the amount of money he gets is maximal possible each day. He has *n* numbers *x*1,<=*x*2,<=...,<=*x**n*. Number *x**i* is the number of tokens given to each worker on the *i*-th day. Help him calculate for each of *n* days the number of tokens he can save.
Input Specification:
The first line of input contains three space-separated integers *n*,<=*a*,<=*b* (1<=≤<=*n*<=≤<=105; 1<=≤<=*a*,<=*b*<=≤<=109). The second line of input contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=109).
Output Specification:
Output *n* space-separated integers. The *i*-th of them is the number of tokens Mashmokh can save on the *i*-th day.
Demo Input:
['5 1 4\n12 6 11 9 1\n', '3 1 2\n1 2 3\n', '1 1 1\n1\n']
Demo Output:
['0 2 3 1 1 ', '1 0 1 ', '0 ']
Note:
none
|
```python
def main():
n, a, b = list(map(int, input().split()))
x_list = list(map(int, input().split()))
res = list()
if a == 1:
for x in x_list:
res.append(str(x * a % b))
else:
for x in x_list:
t = x
s = 0
maxi = x * a // b
while True:
if t * a % b == 0:
break
if t == 0:
break
if t * a // b < maxi:
s -= 1
break
t -= 1
s += 1
res.append(str(s))
print(' '.join(res))
if __name__ == '__main__':
main()
```
| 0
|
|
14
|
A
|
Letter
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Letter
|
1
|
64
|
A boy Bob likes to draw. Not long ago he bought a rectangular graph (checked) sheet with *n* rows and *m* columns. Bob shaded some of the squares on the sheet. Having seen his masterpiece, he decided to share it with his elder brother, who lives in Flatland. Now Bob has to send his picture by post, but because of the world economic crisis and high oil prices, he wants to send his creation, but to spend as little money as possible. For each sent square of paper (no matter whether it is shaded or not) Bob has to pay 3.14 burles. Please, help Bob cut out of his masterpiece a rectangle of the minimum cost, that will contain all the shaded squares. The rectangle's sides should be parallel to the sheet's sides.
|
The first line of the input data contains numbers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=50), *n* — amount of lines, and *m* — amount of columns on Bob's sheet. The following *n* lines contain *m* characters each. Character «.» stands for a non-shaded square on the sheet, and «*» — for a shaded square. It is guaranteed that Bob has shaded at least one square.
|
Output the required rectangle of the minimum cost. Study the output data in the sample tests to understand the output format better.
|
[
"6 7\n.......\n..***..\n..*....\n..***..\n..*....\n..***..\n",
"3 3\n***\n*.*\n***\n"
] |
[
"***\n*..\n***\n*..\n***\n",
"***\n*.*\n***\n"
] |
none
| 0
|
[
{
"input": "6 7\n.......\n..***..\n..*....\n..***..\n..*....\n..***..",
"output": "***\n*..\n***\n*..\n***"
},
{
"input": "3 3\n***\n*.*\n***",
"output": "***\n*.*\n***"
},
{
"input": "1 1\n*",
"output": "*"
},
{
"input": "2 1\n*\n*",
"output": "*\n*"
},
{
"input": "5 1\n.\n*\n.\n.\n.",
"output": "*"
},
{
"input": "1 6\n*****.",
"output": "*****"
},
{
"input": "2 2\n..\n*.",
"output": "*"
},
{
"input": "8 2\n..\n..\n..\n..\n..\n*.\n..\n..",
"output": "*"
},
{
"input": "50 1\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n*\n.\n*\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n*\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.",
"output": "*\n.\n*\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n*"
},
{
"input": "2 1\n*\n.",
"output": "*"
},
{
"input": "5 1\n*\n*\n*\n*\n*",
"output": "*\n*\n*\n*\n*"
},
{
"input": "2 2\n*.\n..",
"output": "*"
},
{
"input": "3 4\n...*\n*...\n..*.",
"output": "...*\n*...\n..*."
},
{
"input": "8 2\n**\n**\n**\n**\n**\n**\n**\n**",
"output": "**\n**\n**\n**\n**\n**\n**\n**"
},
{
"input": "50 1\n.\n*\n*\n*\n.\n.\n*\n.\n*\n*\n*\n*\n*\n*\n*\n.\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n.\n.\n.\n.\n*\n*\n.\n*\n*\n.\n.\n.\n.\n*\n*\n.\n.\n.\n.\n.\n.",
"output": "*\n*\n*\n.\n.\n*\n.\n*\n*\n*\n*\n*\n*\n*\n.\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n.\n.\n.\n.\n*\n*\n.\n*\n*\n.\n.\n.\n.\n*\n*"
}
] | 1,591,501,579
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 30
| 310
| 0
|
import os
import sys
debug = True
if debug and os.path.exists("input.in"):
input = open("input.in", "r").readline
else:
debug = False
input = sys.stdin.readline
def inp():
return (int(input()))
def inlt():
return (list(map(int, input().split())))
def insr():
s = input()
return (list(s[:len(s) - 1]))
def invr():
return (map(int, input().split()))
test_count = 1
if debug:
test_count = inp()
for t in range(test_count):
if debug:
print("Test Case #", t + 1)
# Start code here
n, m = invr()
A = list()
minX = m
maxX = -1
minY = n
maxY = -1
for row in range(n):
s = input()
A.append(s)
firstIndex = s.find("*")
if firstIndex != -1:
minX = min(minX, firstIndex)
lastIndex = s.rfind("*")
maxX = max(maxX, lastIndex)
minY = min(minY, row)
maxY = max(maxY, row)
# print(minX, maxX, minY, maxY)
for r in range(minY, maxY + 1):
print(A[r][minX:maxX + 1])
|
Title: Letter
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
A boy Bob likes to draw. Not long ago he bought a rectangular graph (checked) sheet with *n* rows and *m* columns. Bob shaded some of the squares on the sheet. Having seen his masterpiece, he decided to share it with his elder brother, who lives in Flatland. Now Bob has to send his picture by post, but because of the world economic crisis and high oil prices, he wants to send his creation, but to spend as little money as possible. For each sent square of paper (no matter whether it is shaded or not) Bob has to pay 3.14 burles. Please, help Bob cut out of his masterpiece a rectangle of the minimum cost, that will contain all the shaded squares. The rectangle's sides should be parallel to the sheet's sides.
Input Specification:
The first line of the input data contains numbers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=50), *n* — amount of lines, and *m* — amount of columns on Bob's sheet. The following *n* lines contain *m* characters each. Character «.» stands for a non-shaded square on the sheet, and «*» — for a shaded square. It is guaranteed that Bob has shaded at least one square.
Output Specification:
Output the required rectangle of the minimum cost. Study the output data in the sample tests to understand the output format better.
Demo Input:
['6 7\n.......\n..***..\n..*....\n..***..\n..*....\n..***..\n', '3 3\n***\n*.*\n***\n']
Demo Output:
['***\n*..\n***\n*..\n***\n', '***\n*.*\n***\n']
Note:
none
|
```python
import os
import sys
debug = True
if debug and os.path.exists("input.in"):
input = open("input.in", "r").readline
else:
debug = False
input = sys.stdin.readline
def inp():
return (int(input()))
def inlt():
return (list(map(int, input().split())))
def insr():
s = input()
return (list(s[:len(s) - 1]))
def invr():
return (map(int, input().split()))
test_count = 1
if debug:
test_count = inp()
for t in range(test_count):
if debug:
print("Test Case #", t + 1)
# Start code here
n, m = invr()
A = list()
minX = m
maxX = -1
minY = n
maxY = -1
for row in range(n):
s = input()
A.append(s)
firstIndex = s.find("*")
if firstIndex != -1:
minX = min(minX, firstIndex)
lastIndex = s.rfind("*")
maxX = max(maxX, lastIndex)
minY = min(minY, row)
maxY = max(maxY, row)
# print(minX, maxX, minY, maxY)
for r in range(minY, maxY + 1):
print(A[r][minX:maxX + 1])
```
| 3.845
|
143
|
B
|
Help Kingdom of Far Far Away 2
|
PROGRAMMING
| 1,200
|
[
"implementation",
"strings"
] | null | null |
For some time the program of rounding numbers that had been developed by the Codeforces participants during one of the previous rounds, helped the citizens of Far Far Away to convert numbers into a more easily readable format. However, as time went by, the economy of the Far Far Away developed and the scale of operations grew. So the King ordered to found the Bank of Far Far Away and very soon even the rounding didn't help to quickly determine even the order of the numbers involved in operations. Besides, rounding a number to an integer wasn't very convenient as a bank needed to operate with all numbers with accuracy of up to 0.01, and not up to an integer.
The King issued yet another order: to introduce financial format to represent numbers denoting amounts of money. The formal rules of storing a number in the financial format are as follows:
- A number contains the integer part and the fractional part. The two parts are separated with a character "." (decimal point). - To make digits in the integer part of a number easier to read, they are split into groups of three digits, starting from the least significant ones. The groups are separated with the character "," (comma). For example, if the integer part of a number equals 12345678, then it will be stored in the financial format as 12,345,678 - In the financial format a number's fractional part should contain exactly two digits. So, if the initial number (the number that is converted into the financial format) contains less than two digits in the fractional part (or contains no digits at all), it is complemented with zeros until its length equals 2. If the fractional part contains more than two digits, the extra digits are simply discarded (they are not rounded: see sample tests). - When a number is stored in the financial format, the minus sign is not written. Instead, if the initial number had the minus sign, the result is written in round brackets. - Please keep in mind that the bank of Far Far Away operates using an exotic foreign currency — snakes ($), that's why right before the number in the financial format we should put the sign "$". If the number should be written in the brackets, then the snake sign should also be inside the brackets.
For example, by the above given rules number 2012 will be stored in the financial format as "$2,012.00" and number -12345678.9 will be stored as "($12,345,678.90)".
The merchants of Far Far Away visited you again and expressed much hope that you supply them with the program that can convert arbitrary numbers to the financial format. Can you help them?
|
The input contains a number that needs to be converted into financial format. The number's notation length does not exceed 100 characters, including (possible) signs "-" (minus) and "." (decimal point). The number's notation is correct, that is:
- The number's notation only contains characters from the set {"0" – "9", "-", "."}. - The decimal point (if it is present) is unique and is preceded and followed by a non-zero quantity on decimal digits - A number cannot start with digit 0, except for a case when its whole integer part equals zero (in this case the integer parts is guaranteed to be a single zero: "0"). - The minus sign (if it is present) is unique and stands in the very beginning of the number's notation - If a number is identically equal to 0 (that is, if it is written as, for example, "0" or "0.000"), than it is not preceded by the minus sign. - The input data contains no spaces. - The number's notation contains at least one decimal digit.
|
Print the number given in the input in the financial format by the rules described in the problem statement.
|
[
"2012\n",
"0.000\n",
"-0.00987654321\n",
"-12345678.9\n"
] |
[
"$2,012.00",
"$0.00",
"($0.00)",
"($12,345,678.90)"
] |
Pay attention to the second and third sample tests. They show that the sign of a number in the financial format (and consequently, the presence or absence of brackets) is determined solely by the sign of the initial number. It does not depend on the sign of the number you got after translating the number to the financial format.
| 1,000
|
[
{
"input": "2012",
"output": "$2,012.00"
},
{
"input": "0.000",
"output": "$0.00"
},
{
"input": "-0.00987654321",
"output": "($0.00)"
},
{
"input": "-12345678.9",
"output": "($12,345,678.90)"
},
{
"input": "0.99999999999999999999",
"output": "$0.99"
},
{
"input": "-999999999.9999999999",
"output": "($999,999,999.99)"
},
{
"input": "4.30",
"output": "$4.30"
},
{
"input": "-3136",
"output": "($3,136.00)"
},
{
"input": "47.849",
"output": "$47.84"
},
{
"input": "0",
"output": "$0.00"
},
{
"input": "-1",
"output": "($1.00)"
},
{
"input": "5.3944",
"output": "$5.39"
},
{
"input": "-359789",
"output": "($359,789.00)"
},
{
"input": "-999999",
"output": "($999,999.00)"
},
{
"input": "50117.75",
"output": "$50,117.75"
},
{
"input": "-2717.859",
"output": "($2,717.85)"
},
{
"input": "446900763",
"output": "$446,900,763.00"
},
{
"input": "-92.04295",
"output": "($92.04)"
},
{
"input": "1000000000",
"output": "$1,000,000,000.00"
},
{
"input": "-4097961.5",
"output": "($4,097,961.50)"
},
{
"input": "-83348637.91",
"output": "($83,348,637.91)"
},
{
"input": "741968647.01",
"output": "$741,968,647.01"
},
{
"input": "8590210736.2",
"output": "$8,590,210,736.20"
},
{
"input": "-337322633.10",
"output": "($337,322,633.10)"
},
{
"input": "-9389724657.706",
"output": "($9,389,724,657.70)"
},
{
"input": "-337807291537795",
"output": "($337,807,291,537,795.00)"
},
{
"input": "-1000000000000000",
"output": "($1,000,000,000,000,000.00)"
},
{
"input": "1000000000000000000",
"output": "$1,000,000,000,000,000,000.00"
},
{
"input": "64852365412711705.4",
"output": "$64,852,365,412,711,705.40"
},
{
"input": "-14193044875680849641.0",
"output": "($14,193,044,875,680,849,641.00)"
},
{
"input": "-9087207850675188568.44",
"output": "($9,087,207,850,675,188,568.44)"
},
{
"input": "-999999999999999999999999",
"output": "($999,999,999,999,999,999,999,999.00)"
},
{
"input": "95464737206897655595566.87",
"output": "$95,464,737,206,897,655,595,566.87"
},
{
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},
{
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"output": "($195,688,513,344,900,667,321,324,887,161.00)"
},
{
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"output": "($9,946,519,009,668,593,136,622,791,780,335,166,786,329.96)"
},
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"output": "($39,243,277,445,578,948,100,023,610,303,161,362.21)"
},
{
"input": "-999999999999999999999999999999999999999999999999",
"output": "($999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999.00)"
},
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{
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{
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},
{
"input": "762519263820550209316662292240308083373767394981759714.037848496865152996658249820591156785758954539",
"output": "$762,519,263,820,550,209,316,662,292,240,308,083,373,767,394,981,759,714.03"
},
{
"input": "-81065814290895584254457019744497055053248932892817738718849487679519028041818854925725440291395.398",
"output": "($81,065,814,290,895,584,254,457,019,744,497,055,053,248,932,892,817,738,718,849,487,679,519,028,041,818,854,925,725,440,291,395.39)"
},
{
"input": "-32941712101597478543219921523193493949615291911649974076128866311848385268672190709108207764990.550",
"output": "($32,941,712,101,597,478,543,219,921,523,193,493,949,615,291,911,649,974,076,128,866,311,848,385,268,672,190,709,108,207,764,990.55)"
},
{
"input": "2089113443991831781611590658416581830404242017.85102926202385542583311855337073083712400492547136479",
"output": "$2,089,113,443,991,831,781,611,590,658,416,581,830,404,242,017.85"
},
{
"input": "-93446155923266881322196606839694485100712773936897171033382798807975023881552872455711005123932.747",
"output": "($93,446,155,923,266,881,322,196,606,839,694,485,100,712,773,936,897,171,033,382,798,807,975,023,881,552,872,455,711,005,123,932.74)"
},
{
"input": "960516596871944593730108478032758053821336372808735358607440437077013969634756697387966042842288.508",
"output": "$960,516,596,871,944,593,730,108,478,032,758,053,821,336,372,808,735,358,607,440,437,077,013,969,634,756,697,387,966,042,842,288.50"
},
{
"input": "7542946645993289345871768107036410651745989844030221776852993379463784193885567707317993804499615689",
"output": "$7,542,946,645,993,289,345,871,768,107,036,410,651,745,989,844,030,221,776,852,993,379,463,784,193,885,567,707,317,993,804,499,615,689.00"
},
{
"input": "-62833497045916718064314002220718776776624697240820362462669558147156815011509869423334004968891.075",
"output": "($62,833,497,045,916,718,064,314,002,220,718,776,776,624,697,240,820,362,462,669,558,147,156,815,011,509,869,423,334,004,968,891.07)"
},
{
"input": "369983878656471317107141313973936685655559201630341263457253892446495.822347697919107135036916507458",
"output": "$369,983,878,656,471,317,107,141,313,973,936,685,655,559,201,630,341,263,457,253,892,446,495.82"
},
{
"input": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "$1,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000.00"
},
{
"input": "-7200722479435658295856375503813639375609209638447823589904775057990210002452424572601761458228411.3",
"output": "($7,200,722,479,435,658,295,856,375,503,813,639,375,609,209,638,447,823,589,904,775,057,990,210,002,452,424,572,601,761,458,228,411.30)"
},
{
"input": "1.62929379626674077244098830537592273171157251593607257308766051098303017164327540412154291842807913",
"output": "$1.62"
},
{
"input": "9094697811219913240397316094992038813655777565859532452.35345453828434088557646454113264025096745262",
"output": "$9,094,697,811,219,913,240,397,316,094,992,038,813,655,777,565,859,532,452.35"
},
{
"input": "-241995182456075514870952227695034085165209475359259147742565065759917424411707290789641890279251.11",
"output": "($241,995,182,456,075,514,870,952,227,695,034,085,165,209,475,359,259,147,742,565,065,759,917,424,411,707,290,789,641,890,279,251.11)"
},
{
"input": "2567340036354357844391998756110821468858185018763415770617907336824217629234299240638243305079104961",
"output": "$2,567,340,036,354,357,844,391,998,756,110,821,468,858,185,018,763,415,770,617,907,336,824,217,629,234,299,240,638,243,305,079,104,961.00"
}
] | 1,562,239,355
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 5
| 248
| 0
|
s = input()
n = len(s)
if s[0] == '-':
print('($', end='')
flag = -1
for i in range(1, n):
if s[i] == '.':
flag = i
break
if flag != -1:
for i in range(1, flag):
if (flag - i) % 3 == 0:
print(',', end='')
print(s[i], end='')
if flag + 2 < n:
print(s[flag], s[flag + 1], s[flag + 2], ')', sep='', end='')
else:
print(s[flag], s[flag + 1], '0)', sep='', end='')
else:
for i in range(1, n):
if (n - i) % 3 == 0:
print(',', end='')
print(s[i], end='')
print('.00)')
else:
print('$', end='')
flag = -1
for i in range(n):
if s[i] == '.':
flag = i
break
if flag != -1:
for i in range(flag):
if (flag - i) % 3 == 0:
print(',', end='')
print(s[i], end='')
if flag + 2 < n:
print(s[flag], s[flag + 1], s[flag + 2], sep='', end='')
else:
print(s[flag], s[flag + 1], '0', sep='', end='')
else:
for i in range(n):
if (n - i) % 3 == 0:
print(',', end='')
print(s[i], end='')
print('.00')
|
Title: Help Kingdom of Far Far Away 2
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
For some time the program of rounding numbers that had been developed by the Codeforces participants during one of the previous rounds, helped the citizens of Far Far Away to convert numbers into a more easily readable format. However, as time went by, the economy of the Far Far Away developed and the scale of operations grew. So the King ordered to found the Bank of Far Far Away and very soon even the rounding didn't help to quickly determine even the order of the numbers involved in operations. Besides, rounding a number to an integer wasn't very convenient as a bank needed to operate with all numbers with accuracy of up to 0.01, and not up to an integer.
The King issued yet another order: to introduce financial format to represent numbers denoting amounts of money. The formal rules of storing a number in the financial format are as follows:
- A number contains the integer part and the fractional part. The two parts are separated with a character "." (decimal point). - To make digits in the integer part of a number easier to read, they are split into groups of three digits, starting from the least significant ones. The groups are separated with the character "," (comma). For example, if the integer part of a number equals 12345678, then it will be stored in the financial format as 12,345,678 - In the financial format a number's fractional part should contain exactly two digits. So, if the initial number (the number that is converted into the financial format) contains less than two digits in the fractional part (or contains no digits at all), it is complemented with zeros until its length equals 2. If the fractional part contains more than two digits, the extra digits are simply discarded (they are not rounded: see sample tests). - When a number is stored in the financial format, the minus sign is not written. Instead, if the initial number had the minus sign, the result is written in round brackets. - Please keep in mind that the bank of Far Far Away operates using an exotic foreign currency — snakes ($), that's why right before the number in the financial format we should put the sign "$". If the number should be written in the brackets, then the snake sign should also be inside the brackets.
For example, by the above given rules number 2012 will be stored in the financial format as "$2,012.00" and number -12345678.9 will be stored as "($12,345,678.90)".
The merchants of Far Far Away visited you again and expressed much hope that you supply them with the program that can convert arbitrary numbers to the financial format. Can you help them?
Input Specification:
The input contains a number that needs to be converted into financial format. The number's notation length does not exceed 100 characters, including (possible) signs "-" (minus) and "." (decimal point). The number's notation is correct, that is:
- The number's notation only contains characters from the set {"0" – "9", "-", "."}. - The decimal point (if it is present) is unique and is preceded and followed by a non-zero quantity on decimal digits - A number cannot start with digit 0, except for a case when its whole integer part equals zero (in this case the integer parts is guaranteed to be a single zero: "0"). - The minus sign (if it is present) is unique and stands in the very beginning of the number's notation - If a number is identically equal to 0 (that is, if it is written as, for example, "0" or "0.000"), than it is not preceded by the minus sign. - The input data contains no spaces. - The number's notation contains at least one decimal digit.
Output Specification:
Print the number given in the input in the financial format by the rules described in the problem statement.
Demo Input:
['2012\n', '0.000\n', '-0.00987654321\n', '-12345678.9\n']
Demo Output:
['$2,012.00', '$0.00', '($0.00)', '($12,345,678.90)']
Note:
Pay attention to the second and third sample tests. They show that the sign of a number in the financial format (and consequently, the presence or absence of brackets) is determined solely by the sign of the initial number. It does not depend on the sign of the number you got after translating the number to the financial format.
|
```python
s = input()
n = len(s)
if s[0] == '-':
print('($', end='')
flag = -1
for i in range(1, n):
if s[i] == '.':
flag = i
break
if flag != -1:
for i in range(1, flag):
if (flag - i) % 3 == 0:
print(',', end='')
print(s[i], end='')
if flag + 2 < n:
print(s[flag], s[flag + 1], s[flag + 2], ')', sep='', end='')
else:
print(s[flag], s[flag + 1], '0)', sep='', end='')
else:
for i in range(1, n):
if (n - i) % 3 == 0:
print(',', end='')
print(s[i], end='')
print('.00)')
else:
print('$', end='')
flag = -1
for i in range(n):
if s[i] == '.':
flag = i
break
if flag != -1:
for i in range(flag):
if (flag - i) % 3 == 0:
print(',', end='')
print(s[i], end='')
if flag + 2 < n:
print(s[flag], s[flag + 1], s[flag + 2], sep='', end='')
else:
print(s[flag], s[flag + 1], '0', sep='', end='')
else:
for i in range(n):
if (n - i) % 3 == 0:
print(',', end='')
print(s[i], end='')
print('.00')
```
| 0
|
|
762
|
A
|
k-th divisor
|
PROGRAMMING
| 1,400
|
[
"math",
"number theory"
] | null | null |
You are given two integers *n* and *k*. Find *k*-th smallest divisor of *n*, or report that it doesn't exist.
Divisor of *n* is any such natural number, that *n* can be divided by it without remainder.
|
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=1015, 1<=≤<=*k*<=≤<=109).
|
If *n* has less than *k* divisors, output -1.
Otherwise, output the *k*-th smallest divisor of *n*.
|
[
"4 2\n",
"5 3\n",
"12 5\n"
] |
[
"2\n",
"-1\n",
"6\n"
] |
In the first example, number 4 has three divisors: 1, 2 and 4. The second one is 2.
In the second example, number 5 has only two divisors: 1 and 5. The third divisor doesn't exist, so the answer is -1.
| 0
|
[
{
"input": "4 2",
"output": "2"
},
{
"input": "5 3",
"output": "-1"
},
{
"input": "12 5",
"output": "6"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "866421317361600 26880",
"output": "866421317361600"
},
{
"input": "866421317361600 26881",
"output": "-1"
},
{
"input": "1000000000000000 1000000000",
"output": "-1"
},
{
"input": "1000000000000000 100",
"output": "1953125"
},
{
"input": "1 2",
"output": "-1"
},
{
"input": "4 3",
"output": "4"
},
{
"input": "4 4",
"output": "-1"
},
{
"input": "9 3",
"output": "9"
},
{
"input": "21 3",
"output": "7"
},
{
"input": "67280421310721 1",
"output": "1"
},
{
"input": "6 3",
"output": "3"
},
{
"input": "3 3",
"output": "-1"
},
{
"input": "16 3",
"output": "4"
},
{
"input": "1 1000",
"output": "-1"
},
{
"input": "16 4",
"output": "8"
},
{
"input": "36 8",
"output": "18"
},
{
"input": "49 4",
"output": "-1"
},
{
"input": "9 4",
"output": "-1"
},
{
"input": "16 1",
"output": "1"
},
{
"input": "16 6",
"output": "-1"
},
{
"input": "16 5",
"output": "16"
},
{
"input": "25 4",
"output": "-1"
},
{
"input": "4010815561 2",
"output": "63331"
},
{
"input": "49 3",
"output": "49"
},
{
"input": "36 6",
"output": "9"
},
{
"input": "36 10",
"output": "-1"
},
{
"input": "25 3",
"output": "25"
},
{
"input": "22876792454961 28",
"output": "7625597484987"
},
{
"input": "1234 2",
"output": "2"
},
{
"input": "179458711 2",
"output": "179458711"
},
{
"input": "900104343024121 100000",
"output": "-1"
},
{
"input": "8 3",
"output": "4"
},
{
"input": "100 6",
"output": "20"
},
{
"input": "15500 26",
"output": "-1"
},
{
"input": "111111 1",
"output": "1"
},
{
"input": "100000000000000 200",
"output": "160000000000"
},
{
"input": "1000000000000 100",
"output": "6400000"
},
{
"input": "100 10",
"output": "-1"
},
{
"input": "1000000000039 2",
"output": "1000000000039"
},
{
"input": "64 5",
"output": "16"
},
{
"input": "999999961946176 33",
"output": "63245552"
},
{
"input": "376219076689 3",
"output": "376219076689"
},
{
"input": "999999961946176 63",
"output": "999999961946176"
},
{
"input": "1048576 12",
"output": "2048"
},
{
"input": "745 21",
"output": "-1"
},
{
"input": "748 6",
"output": "22"
},
{
"input": "999999961946176 50",
"output": "161082468097"
},
{
"input": "10 3",
"output": "5"
},
{
"input": "1099511627776 22",
"output": "2097152"
},
{
"input": "1000000007 100010",
"output": "-1"
},
{
"input": "3 1",
"output": "1"
},
{
"input": "100 8",
"output": "50"
},
{
"input": "100 7",
"output": "25"
},
{
"input": "7 2",
"output": "7"
},
{
"input": "999999961946176 64",
"output": "-1"
},
{
"input": "20 5",
"output": "10"
},
{
"input": "999999999999989 2",
"output": "999999999999989"
},
{
"input": "100000000000000 114",
"output": "10240000"
},
{
"input": "99999640000243 3",
"output": "9999991"
},
{
"input": "999998000001 566",
"output": "333332666667"
},
{
"input": "99999820000081 2",
"output": "9999991"
},
{
"input": "49000042000009 3",
"output": "49000042000009"
},
{
"input": "151491429961 4",
"output": "-1"
},
{
"input": "32416190071 2",
"output": "32416190071"
},
{
"input": "1000 8",
"output": "25"
},
{
"input": "1999967841 15",
"output": "1999967841"
},
{
"input": "26880 26880",
"output": "-1"
},
{
"input": "151491429961 3",
"output": "151491429961"
},
{
"input": "90000000000 300",
"output": "100000000"
},
{
"input": "98765004361 10",
"output": "-1"
},
{
"input": "15 2",
"output": "3"
},
{
"input": "16 2",
"output": "2"
},
{
"input": "1996 2",
"output": "2"
},
{
"input": "1997 2",
"output": "1997"
},
{
"input": "1999 2",
"output": "1999"
},
{
"input": "1998 2",
"output": "2"
},
{
"input": "1998 1",
"output": "1"
},
{
"input": "1998 7",
"output": "27"
},
{
"input": "1998 8",
"output": "37"
},
{
"input": "100000380000361 2",
"output": "10000019"
},
{
"input": "15 1",
"output": "1"
},
{
"input": "100000000000000 226",
"output": "-1"
},
{
"input": "844030857550613 517",
"output": "-1"
},
{
"input": "4567890 14",
"output": "430"
},
{
"input": "123123123 123123123",
"output": "-1"
},
{
"input": "24 4",
"output": "4"
},
{
"input": "999999993568952 17",
"output": "31622777"
},
{
"input": "99999999994190 9",
"output": "241656799"
},
{
"input": "999997874844049 4",
"output": "-1"
},
{
"input": "99999999999931 2",
"output": "99999999999931"
},
{
"input": "2 3",
"output": "-1"
},
{
"input": "67280421310721 2",
"output": "67280421310721"
}
] | 1,491,133,498
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 46
| 4,608,000
|
import sys
n, k = [int(x) for x in input().split()]
p = 1
m = 1
l = 2
while m <= n:
m *= l
l += 1
if 2*l < k:
print(-1)
sys.exit()
while p <= n:
if n%p == 0:
k -= 1
if k == 0:
print(p)
sys.exit()
p += 1
print(-1)
|
Title: k-th divisor
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two integers *n* and *k*. Find *k*-th smallest divisor of *n*, or report that it doesn't exist.
Divisor of *n* is any such natural number, that *n* can be divided by it without remainder.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=1015, 1<=≤<=*k*<=≤<=109).
Output Specification:
If *n* has less than *k* divisors, output -1.
Otherwise, output the *k*-th smallest divisor of *n*.
Demo Input:
['4 2\n', '5 3\n', '12 5\n']
Demo Output:
['2\n', '-1\n', '6\n']
Note:
In the first example, number 4 has three divisors: 1, 2 and 4. The second one is 2.
In the second example, number 5 has only two divisors: 1 and 5. The third divisor doesn't exist, so the answer is -1.
|
```python
import sys
n, k = [int(x) for x in input().split()]
p = 1
m = 1
l = 2
while m <= n:
m *= l
l += 1
if 2*l < k:
print(-1)
sys.exit()
while p <= n:
if n%p == 0:
k -= 1
if k == 0:
print(p)
sys.exit()
p += 1
print(-1)
```
| 0
|
|
490
|
A
|
Team Olympiad
|
PROGRAMMING
| 800
|
[
"greedy",
"implementation",
"sortings"
] | null | null |
The School №0 of the capital of Berland has *n* children studying in it. All the children in this school are gifted: some of them are good at programming, some are good at maths, others are good at PE (Physical Education). Hence, for each child we know value *t**i*:
- *t**i*<==<=1, if the *i*-th child is good at programming, - *t**i*<==<=2, if the *i*-th child is good at maths, - *t**i*<==<=3, if the *i*-th child is good at PE
Each child happens to be good at exactly one of these three subjects.
The Team Scientific Decathlon Olympias requires teams of three students. The school teachers decided that the teams will be composed of three children that are good at different subjects. That is, each team must have one mathematician, one programmer and one sportsman. Of course, each child can be a member of no more than one team.
What is the maximum number of teams that the school will be able to present at the Olympiad? How should the teams be formed for that?
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=5000) — the number of children in the school. The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=3), where *t**i* describes the skill of the *i*-th child.
|
In the first line output integer *w* — the largest possible number of teams.
Then print *w* lines, containing three numbers in each line. Each triple represents the indexes of the children forming the team. You can print both the teams, and the numbers in the triplets in any order. The children are numbered from 1 to *n* in the order of their appearance in the input. Each child must participate in no more than one team. If there are several solutions, print any of them.
If no teams can be compiled, print the only line with value *w* equal to 0.
|
[
"7\n1 3 1 3 2 1 2\n",
"4\n2 1 1 2\n"
] |
[
"2\n3 5 2\n6 7 4\n",
"0\n"
] |
none
| 500
|
[
{
"input": "7\n1 3 1 3 2 1 2",
"output": "2\n3 5 2\n6 7 4"
},
{
"input": "4\n2 1 1 2",
"output": "0"
},
{
"input": "1\n2",
"output": "0"
},
{
"input": "2\n3 1",
"output": "0"
},
{
"input": "3\n2 1 2",
"output": "0"
},
{
"input": "3\n1 2 3",
"output": "1\n1 2 3"
},
{
"input": "12\n3 3 3 3 3 3 3 3 1 3 3 2",
"output": "1\n9 12 2"
},
{
"input": "60\n3 3 1 2 2 1 3 1 1 1 3 2 2 2 3 3 1 3 2 3 2 2 1 3 3 2 3 1 2 2 2 1 3 2 1 1 3 3 1 1 1 3 1 2 1 1 3 3 3 2 3 2 3 2 2 2 1 1 1 2",
"output": "20\n6 60 1\n17 44 20\n3 5 33\n36 21 42\n59 14 2\n58 26 49\n9 29 48\n23 19 24\n10 30 37\n41 54 15\n45 31 27\n57 55 38\n39 12 25\n35 34 11\n32 52 7\n8 50 18\n43 4 53\n46 56 51\n40 22 16\n28 13 47"
},
{
"input": "12\n3 1 1 1 1 1 1 2 1 1 1 1",
"output": "1\n3 8 1"
},
{
"input": "22\n2 2 2 2 2 2 2 2 2 2 3 2 2 2 2 2 2 1 2 2 2 2",
"output": "1\n18 2 11"
},
{
"input": "138\n2 3 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 3 2 2 2 1 2 3 2 2 2 3 1 3 2 3 2 3 2 2 2 2 3 2 2 2 2 2 1 2 2 3 2 2 3 2 1 2 2 2 2 2 3 1 2 2 2 2 2 3 2 2 3 2 2 2 2 2 1 1 2 3 2 2 2 2 3 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 3 2 3 2 2 2 1 2 2 2 1 2 2 2 2 1 2 2 2 2 1 3",
"output": "18\n13 91 84\n34 90 48\n11 39 77\n78 129 50\n137 68 119\n132 122 138\n19 12 96\n40 7 2\n22 88 69\n107 73 46\n115 15 52\n127 106 87\n93 92 66\n71 112 117\n63 124 42\n17 70 101\n109 121 57\n123 25 36"
},
{
"input": "203\n2 2 1 2 1 2 2 2 1 2 2 1 1 3 1 2 1 2 1 1 2 3 1 1 2 3 3 2 2 2 1 2 1 1 1 1 1 3 1 1 2 1 1 2 2 2 1 2 2 2 1 2 3 2 1 1 2 2 1 2 1 2 2 1 1 2 2 2 1 1 2 2 1 2 1 2 2 3 2 1 2 1 1 1 1 1 1 1 1 1 1 2 2 1 1 2 2 2 2 1 1 1 1 1 1 1 2 2 2 2 2 1 1 1 2 2 2 1 2 2 1 3 2 1 1 1 2 1 1 2 1 1 2 2 2 1 1 2 2 2 1 2 1 3 2 1 2 2 2 1 1 1 2 2 2 1 2 1 1 2 2 2 2 2 1 1 2 1 2 2 1 1 1 1 1 1 2 2 3 1 1 2 3 1 1 1 1 1 1 2 2 1 1 1 2 2 3 2 1 3 1 1 1",
"output": "13\n188 72 14\n137 4 197\n158 76 122\n152 142 26\n104 119 179\n40 63 38\n12 1 78\n17 30 27\n189 60 53\n166 190 144\n129 7 183\n83 41 22\n121 81 200"
},
{
"input": "220\n1 1 3 1 3 1 1 3 1 3 3 3 3 1 3 3 1 3 3 3 3 3 1 1 1 3 1 1 1 3 2 3 3 3 1 1 3 3 1 1 3 3 3 3 1 3 3 1 1 1 2 3 1 1 1 2 3 3 3 2 3 1 1 3 1 1 1 3 2 1 3 2 3 1 1 3 3 3 1 3 1 1 1 3 3 2 1 3 2 1 1 3 3 1 1 1 2 1 1 3 2 1 2 1 1 1 3 1 3 3 1 2 3 3 3 3 1 3 1 1 1 1 2 3 1 1 1 1 1 1 3 2 3 1 3 1 3 1 1 3 1 3 1 3 1 3 1 3 3 2 3 1 3 3 1 3 3 3 3 1 1 3 3 3 3 1 1 3 3 3 2 1 1 1 3 3 1 3 3 3 1 1 1 3 1 3 3 1 1 1 2 3 1 1 3 1 1 1 1 2 3 1 1 2 3 3 1 3 1 3 3 3 3 1 3 2 3 1 1 3",
"output": "20\n198 89 20\n141 56 131\n166 204 19\n160 132 142\n111 112 195\n45 216 92\n6 31 109\n14 150 170\n199 60 18\n173 123 140\n134 69 156\n82 191 85\n126 200 80\n24 97 46\n62 86 149\n214 101 26\n79 171 78\n125 72 118\n172 103 162\n219 51 64"
},
{
"input": "61\n2 3 1 3 2 2 2 3 1 3 2 3 1 2 1 1 2 2 2 2 3 2 3 1 2 1 3 1 3 2 1 1 3 2 1 3 3 3 1 3 3 1 1 3 1 3 2 2 1 2 2 2 1 3 2 3 1 3 3 1 1",
"output": "20\n9 55 2\n24 34 27\n3 5 37\n35 17 41\n61 11 4\n60 19 54\n15 20 59\n26 14 29\n16 22 38\n43 50 12\n49 25 36\n57 51 40\n39 6 33\n32 30 10\n31 48 8\n13 47 23\n45 1 58\n53 52 56\n42 18 21\n28 7 44"
},
{
"input": "5\n1 2 2 3 3",
"output": "1\n1 3 4"
}
] | 1,694,001,683
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 41
| 46
| 512,000
|
def form_teams(n, inclinations):
teams = []
programmers = []
mathematicians = []
athletes = []
# Group students based on inclinations
for i in range(n):
if inclinations[i] == 1:
programmers.append(i + 1)
elif inclinations[i] == 2:
mathematicians.append(i + 1)
else:
athletes.append(i + 1)
# Calculate the maximum number of teams
num_teams = min(len(programmers), len(mathematicians), len(athletes))
# Form teams by selecting one student from each category
for _ in range(num_teams):
team = [programmers.pop(), mathematicians.pop(), athletes.pop()]
teams.append(team)
return num_teams, teams
# Example usage
n = int(input())
inclinations = list(map(int, input().split()))
num_teams, teams = form_teams(n, inclinations)
print(num_teams)
for team in teams:
print(*team)
|
Title: Team Olympiad
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The School №0 of the capital of Berland has *n* children studying in it. All the children in this school are gifted: some of them are good at programming, some are good at maths, others are good at PE (Physical Education). Hence, for each child we know value *t**i*:
- *t**i*<==<=1, if the *i*-th child is good at programming, - *t**i*<==<=2, if the *i*-th child is good at maths, - *t**i*<==<=3, if the *i*-th child is good at PE
Each child happens to be good at exactly one of these three subjects.
The Team Scientific Decathlon Olympias requires teams of three students. The school teachers decided that the teams will be composed of three children that are good at different subjects. That is, each team must have one mathematician, one programmer and one sportsman. Of course, each child can be a member of no more than one team.
What is the maximum number of teams that the school will be able to present at the Olympiad? How should the teams be formed for that?
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=5000) — the number of children in the school. The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=3), where *t**i* describes the skill of the *i*-th child.
Output Specification:
In the first line output integer *w* — the largest possible number of teams.
Then print *w* lines, containing three numbers in each line. Each triple represents the indexes of the children forming the team. You can print both the teams, and the numbers in the triplets in any order. The children are numbered from 1 to *n* in the order of their appearance in the input. Each child must participate in no more than one team. If there are several solutions, print any of them.
If no teams can be compiled, print the only line with value *w* equal to 0.
Demo Input:
['7\n1 3 1 3 2 1 2\n', '4\n2 1 1 2\n']
Demo Output:
['2\n3 5 2\n6 7 4\n', '0\n']
Note:
none
|
```python
def form_teams(n, inclinations):
teams = []
programmers = []
mathematicians = []
athletes = []
# Group students based on inclinations
for i in range(n):
if inclinations[i] == 1:
programmers.append(i + 1)
elif inclinations[i] == 2:
mathematicians.append(i + 1)
else:
athletes.append(i + 1)
# Calculate the maximum number of teams
num_teams = min(len(programmers), len(mathematicians), len(athletes))
# Form teams by selecting one student from each category
for _ in range(num_teams):
team = [programmers.pop(), mathematicians.pop(), athletes.pop()]
teams.append(team)
return num_teams, teams
# Example usage
n = int(input())
inclinations = list(map(int, input().split()))
num_teams, teams = form_teams(n, inclinations)
print(num_teams)
for team in teams:
print(*team)
```
| 3
|
|
462
|
A
|
Appleman and Easy Task
|
PROGRAMMING
| 1,000
|
[
"brute force",
"implementation"
] | null | null |
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?
Given a *n*<=×<=*n* checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the board are adjacent if they share a side.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Then *n* lines follow containing the description of the checkerboard. Each of them contains *n* characters (either 'x' or 'o') without spaces.
|
Print "YES" or "NO" (without the quotes) depending on the answer to the problem.
|
[
"3\nxxo\nxox\noxx\n",
"4\nxxxo\nxoxo\noxox\nxxxx\n"
] |
[
"YES\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "3\nxxo\nxox\noxx",
"output": "YES"
},
{
"input": "4\nxxxo\nxoxo\noxox\nxxxx",
"output": "NO"
},
{
"input": "1\no",
"output": "YES"
},
{
"input": "2\nox\nxo",
"output": "YES"
},
{
"input": "2\nxx\nxo",
"output": "NO"
},
{
"input": "3\nooo\noxo\nxoo",
"output": "NO"
},
{
"input": "3\nxxx\nxxo\nxxo",
"output": "NO"
},
{
"input": "4\nxooo\nooxo\noxoo\nooox",
"output": "YES"
},
{
"input": "4\noooo\noxxo\nxoxo\noooo",
"output": "NO"
},
{
"input": "5\noxoxo\nxxxxx\noxoxo\nxxxxx\noxoxo",
"output": "YES"
},
{
"input": "5\nxxxox\nxxxxo\nxoxox\noxoxx\nxoxxx",
"output": "NO"
},
{
"input": "10\nxoxooooooo\noxxoxxxxxo\nxxooxoooxo\noooxxoxoxo\noxxxooooxo\noxooooxxxo\noxoxoxxooo\noxoooxooxx\noxxxxxoxxo\noooooooxox",
"output": "YES"
},
{
"input": "10\nxxxxxxxoox\nxooxxooooo\noxoooxxooo\nxoxxxxxxxx\nxxoxooxxox\nooxoxxooox\nooxxxxxooo\nxxxxoxooox\nxoxxooxxxx\noooooxxoxo",
"output": "NO"
},
{
"input": "19\noxoxoxoxooxoooxxoox\nxxxxxxxxoxxoxoooooo\noxoxoxooxxxooxxxooo\nxxoxxxooxooxxxoxxox\noxoxooxxxooooxxoxox\nxxxoooxoxxoxxoxxxoo\noxooxxxoooooxxoooxo\nxxooxooxoxxoxxoxxoo\noxxxxooooxxxooooxxx\nooxooxoxxoxxoxooxoo\nxxxooooxxxooooxoxox\noooxoxooxxoxooxooxx\nxxoooxxxooooxxoooxo\nooxxxooxoxooxooxxxx\nxoxoxxooxoxxxooxoxo\nxoxxoxoxooxooxxxxxx\noooxxxooxxxooxoxoxo\nxoooooxoxooxxxxxxxo\nxooxxoooxxoxoxoxoxx",
"output": "NO"
},
{
"input": "12\nxxooxxoxxxoo\nxxoooxoxoxoo\nooxoxoxxooxx\nooxxooooxoxo\nxxxxxxxxoxxx\noxooooxxxooo\noxxoxoxoooxx\nxxxxxxxooxox\noxoooooxoxxx\nxxooxxoxxoxx\noxxxxxxxooxx\nooxoxooxxooo",
"output": "NO"
},
{
"input": "2\noo\nxx",
"output": "NO"
}
] | 1,409,068,604
| 7,004
|
Python 3
|
OK
|
TESTS
| 32
| 93
| 307,200
|
li=[]
n= int(input())
matrix = [[0 for i in range(n)] for i in range(n)]
for _ in range(n):
l=input()
li.append(l)
flag = 0
for i in range(n):
for j in range(n):
count = 0
if(j>=1):
if(li[i][j-1]=='o'):
count = count+1
if(j<n-1):
if(li[i][j+1]=='o'):
count = count+1
if(i<n-1):
if(li[i+1][j]=='o'):
count = count+1
if(i>=1):
if(li[i-1][j]=='o'):
count = count+1
if(count%2!=0):
flag=1
break
break
if(flag==0):
print("YES")
else:
print("NO")
|
Title: Appleman and Easy Task
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?
Given a *n*<=×<=*n* checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the board are adjacent if they share a side.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Then *n* lines follow containing the description of the checkerboard. Each of them contains *n* characters (either 'x' or 'o') without spaces.
Output Specification:
Print "YES" or "NO" (without the quotes) depending on the answer to the problem.
Demo Input:
['3\nxxo\nxox\noxx\n', '4\nxxxo\nxoxo\noxox\nxxxx\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none
|
```python
li=[]
n= int(input())
matrix = [[0 for i in range(n)] for i in range(n)]
for _ in range(n):
l=input()
li.append(l)
flag = 0
for i in range(n):
for j in range(n):
count = 0
if(j>=1):
if(li[i][j-1]=='o'):
count = count+1
if(j<n-1):
if(li[i][j+1]=='o'):
count = count+1
if(i<n-1):
if(li[i+1][j]=='o'):
count = count+1
if(i>=1):
if(li[i-1][j]=='o'):
count = count+1
if(count%2!=0):
flag=1
break
break
if(flag==0):
print("YES")
else:
print("NO")
```
| 3
|
|
731
|
A
|
Night at the Museum
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] | null | null |
Grigoriy, like the hero of one famous comedy film, found a job as a night security guard at the museum. At first night he received embosser and was to take stock of the whole exposition.
Embosser is a special devise that allows to "print" the text of a plastic tape. Text is printed sequentially, character by character. The device consists of a wheel with a lowercase English letters written in a circle, static pointer to the current letter and a button that print the chosen letter. At one move it's allowed to rotate the alphabetic wheel one step clockwise or counterclockwise. Initially, static pointer points to letter 'a'. Other letters are located as shown on the picture:
After Grigoriy add new item to the base he has to print its name on the plastic tape and attach it to the corresponding exhibit. It's not required to return the wheel to its initial position with pointer on the letter 'a'.
Our hero is afraid that some exhibits may become alive and start to attack him, so he wants to print the names as fast as possible. Help him, for the given string find the minimum number of rotations of the wheel required to print it.
|
The only line of input contains the name of some exhibit — the non-empty string consisting of no more than 100 characters. It's guaranteed that the string consists of only lowercase English letters.
|
Print one integer — the minimum number of rotations of the wheel, required to print the name given in the input.
|
[
"zeus\n",
"map\n",
"ares\n"
] |
[
"18\n",
"35\n",
"34\n"
] |
To print the string from the first sample it would be optimal to perform the following sequence of rotations:
1. from 'a' to 'z' (1 rotation counterclockwise), 1. from 'z' to 'e' (5 clockwise rotations), 1. from 'e' to 'u' (10 rotations counterclockwise), 1. from 'u' to 's' (2 counterclockwise rotations).
| 500
|
[
{
"input": "zeus",
"output": "18"
},
{
"input": "map",
"output": "35"
},
{
"input": "ares",
"output": "34"
},
{
"input": "l",
"output": "11"
},
{
"input": "abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuv",
"output": "99"
},
{
"input": "gngvi",
"output": "44"
},
{
"input": "aaaaa",
"output": "0"
},
{
"input": "a",
"output": "0"
},
{
"input": "z",
"output": "1"
},
{
"input": "vyadeehhikklnoqrs",
"output": "28"
},
{
"input": "jjiihhhhgggfedcccbazyxx",
"output": "21"
},
{
"input": "fyyptqqxuciqvwdewyppjdzur",
"output": "117"
},
{
"input": "fqcnzmzmbobmancqcoalzmanaobpdse",
"output": "368"
},
{
"input": "zzzzzaaaaaaazzzzzzaaaaaaazzzzzzaaaazzzza",
"output": "8"
},
{
"input": "aucnwhfixuruefkypvrvnvznwtjgwlghoqtisbkhuwxmgzuljvqhmnwzisnsgjhivnjmbknptxatdkelhzkhsuxzrmlcpeoyukiy",
"output": "644"
},
{
"input": "sssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssss",
"output": "8"
},
{
"input": "nypjygrdtpzpigzyrisqeqfriwgwlengnezppgttgtndbrryjdl",
"output": "421"
},
{
"input": "pnllnnmmmmoqqqqqrrtssssuuvtsrpopqoonllmonnnpppopnonoopooqpnopppqppqstuuuwwwwvxzxzzaa",
"output": "84"
},
{
"input": "btaoahqgxnfsdmzsjxgvdwjukcvereqeskrdufqfqgzqfsftdqcthtkcnaipftcnco",
"output": "666"
},
{
"input": "eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeerrrrrrrrrrrrrrrrwwwwwwwwww",
"output": "22"
},
{
"input": "uyknzcrwjyzmscqucclvacmorepdgmnyhmakmmnygqwglrxkxhkpansbmruwxdeoprxzmpsvwackopujxbbkpwyeggsvjykpxh",
"output": "643"
},
{
"input": "gzwpooohffcxwtpjgfzwtooiccxsrrokezutoojdzwsrmmhecaxwrojcbyrqlfdwwrliiib",
"output": "245"
},
{
"input": "dbvnkktasjdwqsrzfwwtmjgbcxggdxsoeilecihduypktkkbwfbruxzzhlttrssicgdwqruddwrlbtxgmhdbatzvdxbbro",
"output": "468"
},
{
"input": "mdtvowlktxzzbuaeiuebfeorgbdczauxsovbucactkvyvemsknsjfhifqgycqredzchipmkvzbxdjkcbyukomjlzvxzoswumned",
"output": "523"
},
{
"input": "kkkkkkkaaaaxxaaaaaaaxxxxxxxxaaaaaaxaaaaaaaaaakkkkkkkkkaaaaaaannnnnxxxxkkkkkkkkaannnnnnna",
"output": "130"
},
{
"input": "dffiknqqrsvwzcdgjkmpqtuwxadfhkkkmpqrtwxyadfggjmpppsuuwyyzcdgghhknnpsvvvwwwyabccffiloqruwwyyzabeeehh",
"output": "163"
},
{
"input": "qpppmmkjihgecbyvvsppnnnkjiffeebaaywutrrqpmkjhgddbzzzywtssssqnmmljheddbbaxvusrqonmlifedbbzyywwtqnkheb",
"output": "155"
},
{
"input": "wvvwwwvvwxxxyyyxxwwvwwvuttttttuvvwxxwxxyxxwwwwwvvuttssrssstsssssrqpqqppqrssrsrrssrssssrrsrqqrrqpppqp",
"output": "57"
},
{
"input": "dqcpcobpcobnznamznamzlykxkxlxlylzmaobnaobpbnanbpcoaobnboaoboanzlymzmykylymylzlylymanboanaocqdqesfrfs",
"output": "1236"
},
{
"input": "nnnnnnnnnnnnnnnnnnnnaaaaaaaaaaaaaaaaaaaakkkkkkkkkkkkkkkkkkkkkkaaaaaaaaaaaaaaaaaaaaxxxxxxxxxxxxxxxxxx",
"output": "49"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "0"
},
{
"input": "cgilqsuwzaffilptwwbgmnttyyejkorxzflqvzbddhmnrvxchijpuwaeiimosxyycejlpquuwbfkpvbgijkqvxybdjjjptxcfkqt",
"output": "331"
},
{
"input": "ufsepwgtzgtgjssxaitgpailuvgqweoppszjwhoxdhhhpwwdorwfrdjwcdekxiktwziqwbkvbknrtvajpyeqbjvhiikxxaejjpte",
"output": "692"
},
{
"input": "uhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuh",
"output": "1293"
},
{
"input": "vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvgggggggggggggggggggggggggggggggggggggggggggggggggg",
"output": "16"
},
{
"input": "lyidmjyzbszgiwkxhhpnnthfwcvvstueionspfrvqgkvngmwyhezlosrpdnbvtcjjxxsykixwnepbumaacdzadlqhnjlcejovple",
"output": "616"
},
{
"input": "etzqqbaveffalkdguunfmyyrzkccnxmlluxeasqmopxzfvlkbhipqdwjgrttoemruohgwukfisdhznqyvhswbbypoxgtxyappcrl",
"output": "605"
},
{
"input": "lizussgedcbdjhrbeskhgatyozvwwekanlggcstijrniivupmcoofbaxfqrxddyzzptwxcftlhajsmmkkriarrqtkoauhcqefyud",
"output": "549"
},
{
"input": "dvjuvgfdogpknmbowlsfjzcimnygbtjiucyeeroqwhmzwpjqxlbjkqawrdtmvxbiqufllfuqibxvmtdrwaqkjblxqjpwzmhwqore",
"output": "688"
},
{
"input": "eeycuijtbgynmiczjfslwobmnkpgodfgvujvduyfeqchuaoktqrrairkkmmsjahltfcxwtpzzyddxrqfxabfoocmpuviinrjitsc",
"output": "604"
},
{
"input": "cgglnakewwvzoytaghksebrhjdbcdegssuzilrcppayxtgxopybbwshvyqnzhdsifkuwghourmeottrgjwdqpihbklvfzxpomqsa",
"output": "572"
},
{
"input": "aexullmxncckzryymfnuugdklaffevabqqztelpvojecljnhqldazdcaamubpenwxikysxxjjctvbndprsolzehywmgnvkgqvrfp",
"output": "609"
},
{
"input": "psnoieutsvvcwfhtnnphhxkwigzsbzyjmdiyl",
"output": "223"
},
{
"input": "aa",
"output": "0"
}
] | 1,669,497,775
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 44
| 61
| 0
|
n=str(input())
l=len(n)
m=0
start=0
for i in range(0,l):
index = ord(n[i])-97
walk = abs(start-index)
if walk < 13:
m+=walk
else:
m+=26-walk
start =index
print(m)
|
Title: Night at the Museum
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Grigoriy, like the hero of one famous comedy film, found a job as a night security guard at the museum. At first night he received embosser and was to take stock of the whole exposition.
Embosser is a special devise that allows to "print" the text of a plastic tape. Text is printed sequentially, character by character. The device consists of a wheel with a lowercase English letters written in a circle, static pointer to the current letter and a button that print the chosen letter. At one move it's allowed to rotate the alphabetic wheel one step clockwise or counterclockwise. Initially, static pointer points to letter 'a'. Other letters are located as shown on the picture:
After Grigoriy add new item to the base he has to print its name on the plastic tape and attach it to the corresponding exhibit. It's not required to return the wheel to its initial position with pointer on the letter 'a'.
Our hero is afraid that some exhibits may become alive and start to attack him, so he wants to print the names as fast as possible. Help him, for the given string find the minimum number of rotations of the wheel required to print it.
Input Specification:
The only line of input contains the name of some exhibit — the non-empty string consisting of no more than 100 characters. It's guaranteed that the string consists of only lowercase English letters.
Output Specification:
Print one integer — the minimum number of rotations of the wheel, required to print the name given in the input.
Demo Input:
['zeus\n', 'map\n', 'ares\n']
Demo Output:
['18\n', '35\n', '34\n']
Note:
To print the string from the first sample it would be optimal to perform the following sequence of rotations:
1. from 'a' to 'z' (1 rotation counterclockwise), 1. from 'z' to 'e' (5 clockwise rotations), 1. from 'e' to 'u' (10 rotations counterclockwise), 1. from 'u' to 's' (2 counterclockwise rotations).
|
```python
n=str(input())
l=len(n)
m=0
start=0
for i in range(0,l):
index = ord(n[i])-97
walk = abs(start-index)
if walk < 13:
m+=walk
else:
m+=26-walk
start =index
print(m)
```
| 3
|
|
633
|
B
|
A Trivial Problem
|
PROGRAMMING
| 1,300
|
[
"brute force",
"constructive algorithms",
"math",
"number theory"
] | null | null |
Mr. Santa asks all the great programmers of the world to solve a trivial problem. He gives them an integer *m* and asks for the number of positive integers *n*, such that the factorial of *n* ends with exactly *m* zeroes. Are you among those great programmers who can solve this problem?
|
The only line of input contains an integer *m* (1<=≤<=*m*<=≤<=100<=000) — the required number of trailing zeroes in factorial.
|
First print *k* — the number of values of *n* such that the factorial of *n* ends with *m* zeroes. Then print these *k* integers in increasing order.
|
[
"1\n",
"5\n"
] |
[
"5\n5 6 7 8 9 ",
"0"
] |
The factorial of *n* is equal to the product of all integers from 1 to *n* inclusive, that is *n*! = 1·2·3·...·*n*.
In the first sample, 5! = 120, 6! = 720, 7! = 5040, 8! = 40320 and 9! = 362880.
| 500
|
[
{
"input": "1",
"output": "5\n5 6 7 8 9 "
},
{
"input": "5",
"output": "0"
},
{
"input": "2",
"output": "5\n10 11 12 13 14 "
},
{
"input": "3",
"output": "5\n15 16 17 18 19 "
},
{
"input": "7",
"output": "5\n30 31 32 33 34 "
},
{
"input": "12",
"output": "5\n50 51 52 53 54 "
},
{
"input": "15",
"output": "5\n65 66 67 68 69 "
},
{
"input": "18",
"output": "5\n75 76 77 78 79 "
},
{
"input": "38",
"output": "5\n155 156 157 158 159 "
},
{
"input": "47",
"output": "5\n195 196 197 198 199 "
},
{
"input": "58",
"output": "5\n240 241 242 243 244 "
},
{
"input": "66",
"output": "5\n270 271 272 273 274 "
},
{
"input": "70",
"output": "5\n285 286 287 288 289 "
},
{
"input": "89",
"output": "5\n365 366 367 368 369 "
},
{
"input": "417",
"output": "5\n1675 1676 1677 1678 1679 "
},
{
"input": "815",
"output": "5\n3265 3266 3267 3268 3269 "
},
{
"input": "394",
"output": "5\n1585 1586 1587 1588 1589 "
},
{
"input": "798",
"output": "0"
},
{
"input": "507",
"output": "5\n2035 2036 2037 2038 2039 "
},
{
"input": "406",
"output": "5\n1630 1631 1632 1633 1634 "
},
{
"input": "570",
"output": "5\n2290 2291 2292 2293 2294 "
},
{
"input": "185",
"output": "0"
},
{
"input": "765",
"output": "0"
},
{
"input": "967",
"output": "0"
},
{
"input": "112",
"output": "5\n455 456 457 458 459 "
},
{
"input": "729",
"output": "5\n2925 2926 2927 2928 2929 "
},
{
"input": "4604",
"output": "5\n18425 18426 18427 18428 18429 "
},
{
"input": "8783",
"output": "5\n35140 35141 35142 35143 35144 "
},
{
"input": "1059",
"output": "0"
},
{
"input": "6641",
"output": "5\n26575 26576 26577 26578 26579 "
},
{
"input": "9353",
"output": "5\n37425 37426 37427 37428 37429 "
},
{
"input": "1811",
"output": "5\n7250 7251 7252 7253 7254 "
},
{
"input": "2528",
"output": "0"
},
{
"input": "8158",
"output": "5\n32640 32641 32642 32643 32644 "
},
{
"input": "3014",
"output": "5\n12070 12071 12072 12073 12074 "
},
{
"input": "7657",
"output": "5\n30640 30641 30642 30643 30644 "
},
{
"input": "4934",
"output": "0"
},
{
"input": "9282",
"output": "5\n37140 37141 37142 37143 37144 "
},
{
"input": "2610",
"output": "5\n10450 10451 10452 10453 10454 "
},
{
"input": "2083",
"output": "5\n8345 8346 8347 8348 8349 "
},
{
"input": "26151",
"output": "5\n104620 104621 104622 104623 104624 "
},
{
"input": "64656",
"output": "5\n258640 258641 258642 258643 258644 "
},
{
"input": "46668",
"output": "5\n186690 186691 186692 186693 186694 "
},
{
"input": "95554",
"output": "5\n382235 382236 382237 382238 382239 "
},
{
"input": "37320",
"output": "0"
},
{
"input": "52032",
"output": "5\n208140 208141 208142 208143 208144 "
},
{
"input": "11024",
"output": "5\n44110 44111 44112 44113 44114 "
},
{
"input": "63218",
"output": "5\n252885 252886 252887 252888 252889 "
},
{
"input": "40095",
"output": "5\n160390 160391 160392 160393 160394 "
},
{
"input": "42724",
"output": "5\n170910 170911 170912 170913 170914 "
},
{
"input": "24381",
"output": "5\n97530 97531 97532 97533 97534 "
},
{
"input": "73138",
"output": "5\n292570 292571 292572 292573 292574 "
},
{
"input": "93346",
"output": "5\n373400 373401 373402 373403 373404 "
},
{
"input": "18338",
"output": "5\n73370 73371 73372 73373 73374 "
},
{
"input": "42662",
"output": "5\n170660 170661 170662 170663 170664 "
},
{
"input": "81221",
"output": "5\n324900 324901 324902 324903 324904 "
},
{
"input": "100000",
"output": "5\n400005 400006 400007 400008 400009 "
},
{
"input": "100000",
"output": "5\n400005 400006 400007 400008 400009 "
},
{
"input": "99998",
"output": "0"
},
{
"input": "30",
"output": "0"
},
{
"input": "11",
"output": "0"
},
{
"input": "780",
"output": "0"
},
{
"input": "97656",
"output": "5\n390625 390626 390627 390628 390629 "
},
{
"input": "12499",
"output": "5\n50000 50001 50002 50003 50004 "
},
{
"input": "65",
"output": "5\n265 266 267 268 269 "
},
{
"input": "41",
"output": "5\n170 171 172 173 174 "
},
{
"input": "31",
"output": "5\n125 126 127 128 129 "
},
{
"input": "86577",
"output": "0"
}
] | 1,687,257,504
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 68
| 156
| 2,150,400
|
def f(x):
c2=0
y=2
while y<=x:
c2+=x//y
y*=2
c5=0
y=5
while y<=x:
c5+=x//y
y*=5
return min(c2,c5)
n=int(input())
a=[]
x=1
while f(x)<n:
x+=1
while f(x)==n:
a.append(x)
x+=1
print(len(a))
print(*a)
|
Title: A Trivial Problem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mr. Santa asks all the great programmers of the world to solve a trivial problem. He gives them an integer *m* and asks for the number of positive integers *n*, such that the factorial of *n* ends with exactly *m* zeroes. Are you among those great programmers who can solve this problem?
Input Specification:
The only line of input contains an integer *m* (1<=≤<=*m*<=≤<=100<=000) — the required number of trailing zeroes in factorial.
Output Specification:
First print *k* — the number of values of *n* such that the factorial of *n* ends with *m* zeroes. Then print these *k* integers in increasing order.
Demo Input:
['1\n', '5\n']
Demo Output:
['5\n5 6 7 8 9 ', '0']
Note:
The factorial of *n* is equal to the product of all integers from 1 to *n* inclusive, that is *n*! = 1·2·3·...·*n*.
In the first sample, 5! = 120, 6! = 720, 7! = 5040, 8! = 40320 and 9! = 362880.
|
```python
def f(x):
c2=0
y=2
while y<=x:
c2+=x//y
y*=2
c5=0
y=5
while y<=x:
c5+=x//y
y*=5
return min(c2,c5)
n=int(input())
a=[]
x=1
while f(x)<n:
x+=1
while f(x)==n:
a.append(x)
x+=1
print(len(a))
print(*a)
```
| 3
|
|
825
|
C
|
Multi-judge Solving
|
PROGRAMMING
| 1,600
|
[
"greedy",
"implementation"
] | null | null |
Makes solves problems on Decoforces and lots of other different online judges. Each problem is denoted by its difficulty — a positive integer number. Difficulties are measured the same across all the judges (the problem with difficulty *d* on Decoforces is as hard as the problem with difficulty *d* on any other judge).
Makes has chosen *n* problems to solve on Decoforces with difficulties *a*1,<=*a*2,<=...,<=*a**n*. He can solve these problems in arbitrary order. Though he can solve problem *i* with difficulty *a**i* only if he had already solved some problem with difficulty (no matter on what online judge was it).
Before starting this chosen list of problems, Makes has already solved problems with maximum difficulty *k*.
With given conditions it's easy to see that Makes sometimes can't solve all the chosen problems, no matter what order he chooses. So he wants to solve some problems on other judges to finish solving problems from his list.
For every positive integer *y* there exist some problem with difficulty *y* on at least one judge besides Decoforces.
Makes can solve problems on any judge at any time, it isn't necessary to do problems from the chosen list one right after another.
Makes doesn't have too much free time, so he asked you to calculate the minimum number of problems he should solve on other judges in order to solve all the chosen problems from Decoforces.
|
The first line contains two integer numbers *n*, *k* (1<=≤<=*n*<=≤<=103, 1<=≤<=*k*<=≤<=109).
The second line contains *n* space-separated integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
|
Print minimum number of problems Makes should solve on other judges in order to solve all chosen problems on Decoforces.
|
[
"3 3\n2 1 9\n",
"4 20\n10 3 6 3\n"
] |
[
"1\n",
"0\n"
] |
In the first example Makes at first solves problems 1 and 2. Then in order to solve the problem with difficulty 9, he should solve problem with difficulty no less than 5. The only available are difficulties 5 and 6 on some other judge. Solving any of these will give Makes opportunity to solve problem 3.
In the second example he can solve every problem right from the start.
| 0
|
[
{
"input": "3 3\n2 1 9",
"output": "1"
},
{
"input": "4 20\n10 3 6 3",
"output": "0"
},
{
"input": "1 1000000000\n1",
"output": "0"
},
{
"input": "1 1\n3",
"output": "1"
},
{
"input": "50 100\n74 55 33 5 83 24 75 59 30 36 13 4 62 28 96 17 6 35 45 53 33 11 37 93 34 79 61 72 13 31 44 75 7 3 63 46 18 16 44 89 62 25 32 12 38 55 75 56 61 82",
"output": "0"
},
{
"input": "100 10\n246 286 693 607 87 612 909 312 621 37 801 558 504 914 416 762 187 974 976 123 635 488 416 659 988 998 93 662 92 749 889 78 214 786 735 625 921 372 713 617 975 119 402 411 878 138 548 905 802 762 940 336 529 373 745 835 805 880 816 94 166 114 475 699 974 462 61 337 555 805 968 815 392 746 591 558 740 380 668 29 881 151 387 986 174 923 541 520 998 947 535 651 103 584 664 854 180 852 726 93",
"output": "1"
},
{
"input": "2 1\n1 1000000000",
"output": "29"
},
{
"input": "29 2\n1 3 7 15 31 63 127 255 511 1023 2047 4095 8191 16383 32767 65535 131071 262143 524287 1048575 2097151 4194303 8388607 16777215 33554431 67108863 134217727 268435455 536870911",
"output": "27"
},
{
"input": "1 1\n1000000000",
"output": "29"
},
{
"input": "7 6\n4 20 16 14 3 17 4",
"output": "1"
},
{
"input": "2 1\n3 6",
"output": "1"
},
{
"input": "1 1\n20",
"output": "4"
},
{
"input": "5 2\n86 81 53 25 18",
"output": "4"
},
{
"input": "4 1\n88 55 14 39",
"output": "4"
},
{
"input": "3 1\n2 3 6",
"output": "0"
},
{
"input": "3 2\n4 9 18",
"output": "1"
},
{
"input": "5 3\n6 6 6 13 27",
"output": "2"
},
{
"input": "5 1\n23 8 83 26 18",
"output": "4"
},
{
"input": "3 1\n4 5 6",
"output": "1"
},
{
"input": "3 1\n1 3 6",
"output": "1"
},
{
"input": "1 1\n2",
"output": "0"
},
{
"input": "3 2\n4 5 6",
"output": "0"
},
{
"input": "5 1\n100 200 400 1000 2000",
"output": "7"
},
{
"input": "2 1\n1 4",
"output": "1"
},
{
"input": "4 1\n2 4 8 32",
"output": "1"
},
{
"input": "2 10\n21 42",
"output": "1"
},
{
"input": "3 3\n1 7 13",
"output": "1"
},
{
"input": "3 1\n1 4 6",
"output": "1"
},
{
"input": "2 2\n2 8",
"output": "1"
},
{
"input": "1 1\n4",
"output": "1"
},
{
"input": "2 2\n8 16",
"output": "1"
},
{
"input": "3 1\n4 8 16",
"output": "1"
},
{
"input": "3 1\n3 6 9",
"output": "1"
},
{
"input": "2 1\n4 8",
"output": "1"
},
{
"input": "2 2\n7 14",
"output": "1"
},
{
"input": "1 4\n9",
"output": "1"
},
{
"input": "5 3\n1024 4096 16384 65536 536870913",
"output": "24"
},
{
"input": "2 5\n10 11",
"output": "0"
},
{
"input": "2 2\n3 6",
"output": "0"
},
{
"input": "2 2\n8 11",
"output": "1"
},
{
"input": "3 19905705\n263637263 417905394 108361057",
"output": "3"
},
{
"input": "4 25\n100 11 1 13",
"output": "1"
},
{
"input": "10 295206008\n67980321 440051990 883040288 135744260 96431758 242465794 576630162 972797687 356406646 547451696",
"output": "0"
},
{
"input": "4 2\n45 44 35 38",
"output": "4"
},
{
"input": "1 2\n9",
"output": "2"
},
{
"input": "3 6\n13 26 52",
"output": "1"
},
{
"input": "9 30111088\n824713578 11195876 458715185 731769293 680826358 189542586 550198537 860586039 101083021",
"output": "2"
},
{
"input": "3 72014068\n430005292 807436976 828082746",
"output": "2"
},
{
"input": "3 165219745\n737649884 652879952 506420386",
"output": "1"
},
{
"input": "2 60669400\n95037700 337255240",
"output": "1"
},
{
"input": "4 28\n34 1 86 90",
"output": "1"
},
{
"input": "2 1\n5 10",
"output": "2"
},
{
"input": "2 1\n4 1000000000",
"output": "28"
},
{
"input": "2 1\n2 3",
"output": "0"
},
{
"input": "2 1\n3 5",
"output": "1"
},
{
"input": "3 3\n1 5 20",
"output": "1"
},
{
"input": "9 1\n1 2 4 9 15 32 64 128 1024",
"output": "4"
}
] | 1,671,973,886
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 61
| 46
| 0
|
n,k=map(int,input().split())
a=list(map(int,input().split()))
a.sort()
jauap=0
for i in a:
while i>k*2:
jauap+=1
k=k*2
k=max(k,i)
print(jauap)
|
Title: Multi-judge Solving
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Makes solves problems on Decoforces and lots of other different online judges. Each problem is denoted by its difficulty — a positive integer number. Difficulties are measured the same across all the judges (the problem with difficulty *d* on Decoforces is as hard as the problem with difficulty *d* on any other judge).
Makes has chosen *n* problems to solve on Decoforces with difficulties *a*1,<=*a*2,<=...,<=*a**n*. He can solve these problems in arbitrary order. Though he can solve problem *i* with difficulty *a**i* only if he had already solved some problem with difficulty (no matter on what online judge was it).
Before starting this chosen list of problems, Makes has already solved problems with maximum difficulty *k*.
With given conditions it's easy to see that Makes sometimes can't solve all the chosen problems, no matter what order he chooses. So he wants to solve some problems on other judges to finish solving problems from his list.
For every positive integer *y* there exist some problem with difficulty *y* on at least one judge besides Decoforces.
Makes can solve problems on any judge at any time, it isn't necessary to do problems from the chosen list one right after another.
Makes doesn't have too much free time, so he asked you to calculate the minimum number of problems he should solve on other judges in order to solve all the chosen problems from Decoforces.
Input Specification:
The first line contains two integer numbers *n*, *k* (1<=≤<=*n*<=≤<=103, 1<=≤<=*k*<=≤<=109).
The second line contains *n* space-separated integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
Output Specification:
Print minimum number of problems Makes should solve on other judges in order to solve all chosen problems on Decoforces.
Demo Input:
['3 3\n2 1 9\n', '4 20\n10 3 6 3\n']
Demo Output:
['1\n', '0\n']
Note:
In the first example Makes at first solves problems 1 and 2. Then in order to solve the problem with difficulty 9, he should solve problem with difficulty no less than 5. The only available are difficulties 5 and 6 on some other judge. Solving any of these will give Makes opportunity to solve problem 3.
In the second example he can solve every problem right from the start.
|
```python
n,k=map(int,input().split())
a=list(map(int,input().split()))
a.sort()
jauap=0
for i in a:
while i>k*2:
jauap+=1
k=k*2
k=max(k,i)
print(jauap)
```
| 3
|
|
313
|
A
|
Ilya and Bank Account
|
PROGRAMMING
| 900
|
[
"implementation",
"number theory"
] | null | null |
Ilya is a very clever lion, he lives in an unusual city ZooVille. In this city all the animals have their rights and obligations. Moreover, they even have their own bank accounts. The state of a bank account is an integer. The state of a bank account can be a negative number. This means that the owner of the account owes the bank money.
Ilya the Lion has recently had a birthday, so he got a lot of gifts. One of them (the gift of the main ZooVille bank) is the opportunity to delete the last digit or the digit before last from the state of his bank account no more than once. For example, if the state of Ilya's bank account is -123, then Ilya can delete the last digit and get his account balance equal to -12, also he can remove its digit before last and get the account balance equal to -13. Of course, Ilya is permitted not to use the opportunity to delete a digit from the balance.
Ilya is not very good at math, and that's why he asks you to help him maximize his bank account. Find the maximum state of the bank account that can be obtained using the bank's gift.
|
The single line contains integer *n* (10<=≤<=|*n*|<=≤<=109) — the state of Ilya's bank account.
|
In a single line print an integer — the maximum state of the bank account that Ilya can get.
|
[
"2230\n",
"-10\n",
"-100003\n"
] |
[
"2230\n",
"0\n",
"-10000\n"
] |
In the first test sample Ilya doesn't profit from using the present.
In the second test sample you can delete digit 1 and get the state of the account equal to 0.
| 500
|
[
{
"input": "2230",
"output": "2230"
},
{
"input": "-10",
"output": "0"
},
{
"input": "-100003",
"output": "-10000"
},
{
"input": "544883178",
"output": "544883178"
},
{
"input": "-847251738",
"output": "-84725173"
},
{
"input": "423654797",
"output": "423654797"
},
{
"input": "-623563697",
"output": "-62356367"
},
{
"input": "645894116",
"output": "645894116"
},
{
"input": "-384381709",
"output": "-38438170"
},
{
"input": "437587210",
"output": "437587210"
},
{
"input": "-297534606",
"output": "-29753460"
},
{
"input": "891773002",
"output": "891773002"
},
{
"input": "-56712976",
"output": "-5671296"
},
{
"input": "963662765",
"output": "963662765"
},
{
"input": "-272656295",
"output": "-27265625"
},
{
"input": "383441522",
"output": "383441522"
},
{
"input": "-477665112",
"output": "-47766511"
},
{
"input": "791725034",
"output": "791725034"
},
{
"input": "-812168727",
"output": "-81216872"
},
{
"input": "528894922",
"output": "528894922"
},
{
"input": "-479977172",
"output": "-47997712"
},
{
"input": "568044564",
"output": "568044564"
},
{
"input": "-392784794",
"output": "-39278474"
},
{
"input": "925596634",
"output": "925596634"
},
{
"input": "-836078769",
"output": "-83607876"
},
{
"input": "71036059",
"output": "71036059"
},
{
"input": "-337396162",
"output": "-33739612"
},
{
"input": "87129297",
"output": "87129297"
},
{
"input": "-648171877",
"output": "-64817187"
},
{
"input": "20218400",
"output": "20218400"
},
{
"input": "10",
"output": "10"
},
{
"input": "1000000000",
"output": "1000000000"
},
{
"input": "-1000000000",
"output": "-100000000"
},
{
"input": "-102",
"output": "-10"
},
{
"input": "-120",
"output": "-10"
},
{
"input": "-20",
"output": "0"
},
{
"input": "-15",
"output": "-1"
},
{
"input": "-5575533",
"output": "-557553"
},
{
"input": "-50",
"output": "0"
},
{
"input": "-31",
"output": "-1"
},
{
"input": "-55",
"output": "-5"
},
{
"input": "-81",
"output": "-1"
},
{
"input": "-99",
"output": "-9"
},
{
"input": "-23",
"output": "-2"
},
{
"input": "-41",
"output": "-1"
},
{
"input": "-24",
"output": "-2"
},
{
"input": "46",
"output": "46"
}
] | 1,693,325,800
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 47
| 154
| 0
|
n=int(input())
print(max(n,-min((-n)//10,((-n)//100)*10+(-n)%10)))
|
Title: Ilya and Bank Account
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ilya is a very clever lion, he lives in an unusual city ZooVille. In this city all the animals have their rights and obligations. Moreover, they even have their own bank accounts. The state of a bank account is an integer. The state of a bank account can be a negative number. This means that the owner of the account owes the bank money.
Ilya the Lion has recently had a birthday, so he got a lot of gifts. One of them (the gift of the main ZooVille bank) is the opportunity to delete the last digit or the digit before last from the state of his bank account no more than once. For example, if the state of Ilya's bank account is -123, then Ilya can delete the last digit and get his account balance equal to -12, also he can remove its digit before last and get the account balance equal to -13. Of course, Ilya is permitted not to use the opportunity to delete a digit from the balance.
Ilya is not very good at math, and that's why he asks you to help him maximize his bank account. Find the maximum state of the bank account that can be obtained using the bank's gift.
Input Specification:
The single line contains integer *n* (10<=≤<=|*n*|<=≤<=109) — the state of Ilya's bank account.
Output Specification:
In a single line print an integer — the maximum state of the bank account that Ilya can get.
Demo Input:
['2230\n', '-10\n', '-100003\n']
Demo Output:
['2230\n', '0\n', '-10000\n']
Note:
In the first test sample Ilya doesn't profit from using the present.
In the second test sample you can delete digit 1 and get the state of the account equal to 0.
|
```python
n=int(input())
print(max(n,-min((-n)//10,((-n)//100)*10+(-n)%10)))
```
| 3
|
|
124
|
A
|
The number of positions
|
PROGRAMMING
| 1,000
|
[
"math"
] | null | null |
Petr stands in line of *n* people, but he doesn't know exactly which position he occupies. He can say that there are no less than *a* people standing in front of him and no more than *b* people standing behind him. Find the number of different positions Petr can occupy.
|
The only line contains three integers *n*, *a* and *b* (0<=≤<=*a*,<=*b*<=<<=*n*<=≤<=100).
|
Print the single number — the number of the sought positions.
|
[
"3 1 1\n",
"5 2 3\n"
] |
[
"2\n",
"3\n"
] |
The possible positions in the first sample are: 2 and 3 (if we number the positions starting with 1).
In the second sample they are 3, 4 and 5.
| 500
|
[
{
"input": "3 1 1",
"output": "2"
},
{
"input": "5 2 3",
"output": "3"
},
{
"input": "5 4 0",
"output": "1"
},
{
"input": "6 5 5",
"output": "1"
},
{
"input": "9 4 3",
"output": "4"
},
{
"input": "11 4 6",
"output": "7"
},
{
"input": "13 8 7",
"output": "5"
},
{
"input": "14 5 5",
"output": "6"
},
{
"input": "16 6 9",
"output": "10"
},
{
"input": "20 13 17",
"output": "7"
},
{
"input": "22 4 8",
"output": "9"
},
{
"input": "23 8 14",
"output": "15"
},
{
"input": "26 18 22",
"output": "8"
},
{
"input": "28 6 1",
"output": "2"
},
{
"input": "29 5 23",
"output": "24"
},
{
"input": "32 27 15",
"output": "5"
},
{
"input": "33 11 5",
"output": "6"
},
{
"input": "37 21 15",
"output": "16"
},
{
"input": "39 34 33",
"output": "5"
},
{
"input": "41 27 11",
"output": "12"
},
{
"input": "42 25 16",
"output": "17"
},
{
"input": "45 7 43",
"output": "38"
},
{
"input": "47 16 17",
"output": "18"
},
{
"input": "49 11 37",
"output": "38"
},
{
"input": "51 38 39",
"output": "13"
},
{
"input": "52 29 7",
"output": "8"
},
{
"input": "56 43 12",
"output": "13"
},
{
"input": "58 57 28",
"output": "1"
},
{
"input": "59 12 39",
"output": "40"
},
{
"input": "62 9 52",
"output": "53"
},
{
"input": "63 29 44",
"output": "34"
},
{
"input": "65 30 22",
"output": "23"
},
{
"input": "66 27 38",
"output": "39"
},
{
"input": "71 33 53",
"output": "38"
},
{
"input": "73 14 12",
"output": "13"
},
{
"input": "73 37 35",
"output": "36"
},
{
"input": "76 69 44",
"output": "7"
},
{
"input": "79 25 20",
"output": "21"
},
{
"input": "81 60 20",
"output": "21"
},
{
"input": "81 79 14",
"output": "2"
},
{
"input": "84 0 42",
"output": "43"
},
{
"input": "88 79 8",
"output": "9"
},
{
"input": "90 76 59",
"output": "14"
},
{
"input": "92 2 22",
"output": "23"
},
{
"input": "94 5 88",
"output": "89"
},
{
"input": "94 62 48",
"output": "32"
},
{
"input": "96 22 72",
"output": "73"
},
{
"input": "100 11 88",
"output": "89"
},
{
"input": "100 81 91",
"output": "19"
},
{
"input": "1 0 0",
"output": "1"
}
] | 1,697,532,913
| 493
|
Python 3
|
OK
|
TESTS
| 50
| 122
| 0
|
n, a, b=map(int, input().split())
position=min(n-a, b+1 )
print(position)
|
Title: The number of positions
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petr stands in line of *n* people, but he doesn't know exactly which position he occupies. He can say that there are no less than *a* people standing in front of him and no more than *b* people standing behind him. Find the number of different positions Petr can occupy.
Input Specification:
The only line contains three integers *n*, *a* and *b* (0<=≤<=*a*,<=*b*<=<<=*n*<=≤<=100).
Output Specification:
Print the single number — the number of the sought positions.
Demo Input:
['3 1 1\n', '5 2 3\n']
Demo Output:
['2\n', '3\n']
Note:
The possible positions in the first sample are: 2 and 3 (if we number the positions starting with 1).
In the second sample they are 3, 4 and 5.
|
```python
n, a, b=map(int, input().split())
position=min(n-a, b+1 )
print(position)
```
| 3
|
|
490
|
A
|
Team Olympiad
|
PROGRAMMING
| 800
|
[
"greedy",
"implementation",
"sortings"
] | null | null |
The School №0 of the capital of Berland has *n* children studying in it. All the children in this school are gifted: some of them are good at programming, some are good at maths, others are good at PE (Physical Education). Hence, for each child we know value *t**i*:
- *t**i*<==<=1, if the *i*-th child is good at programming, - *t**i*<==<=2, if the *i*-th child is good at maths, - *t**i*<==<=3, if the *i*-th child is good at PE
Each child happens to be good at exactly one of these three subjects.
The Team Scientific Decathlon Olympias requires teams of three students. The school teachers decided that the teams will be composed of three children that are good at different subjects. That is, each team must have one mathematician, one programmer and one sportsman. Of course, each child can be a member of no more than one team.
What is the maximum number of teams that the school will be able to present at the Olympiad? How should the teams be formed for that?
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=5000) — the number of children in the school. The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=3), where *t**i* describes the skill of the *i*-th child.
|
In the first line output integer *w* — the largest possible number of teams.
Then print *w* lines, containing three numbers in each line. Each triple represents the indexes of the children forming the team. You can print both the teams, and the numbers in the triplets in any order. The children are numbered from 1 to *n* in the order of their appearance in the input. Each child must participate in no more than one team. If there are several solutions, print any of them.
If no teams can be compiled, print the only line with value *w* equal to 0.
|
[
"7\n1 3 1 3 2 1 2\n",
"4\n2 1 1 2\n"
] |
[
"2\n3 5 2\n6 7 4\n",
"0\n"
] |
none
| 500
|
[
{
"input": "7\n1 3 1 3 2 1 2",
"output": "2\n3 5 2\n6 7 4"
},
{
"input": "4\n2 1 1 2",
"output": "0"
},
{
"input": "1\n2",
"output": "0"
},
{
"input": "2\n3 1",
"output": "0"
},
{
"input": "3\n2 1 2",
"output": "0"
},
{
"input": "3\n1 2 3",
"output": "1\n1 2 3"
},
{
"input": "12\n3 3 3 3 3 3 3 3 1 3 3 2",
"output": "1\n9 12 2"
},
{
"input": "60\n3 3 1 2 2 1 3 1 1 1 3 2 2 2 3 3 1 3 2 3 2 2 1 3 3 2 3 1 2 2 2 1 3 2 1 1 3 3 1 1 1 3 1 2 1 1 3 3 3 2 3 2 3 2 2 2 1 1 1 2",
"output": "20\n6 60 1\n17 44 20\n3 5 33\n36 21 42\n59 14 2\n58 26 49\n9 29 48\n23 19 24\n10 30 37\n41 54 15\n45 31 27\n57 55 38\n39 12 25\n35 34 11\n32 52 7\n8 50 18\n43 4 53\n46 56 51\n40 22 16\n28 13 47"
},
{
"input": "12\n3 1 1 1 1 1 1 2 1 1 1 1",
"output": "1\n3 8 1"
},
{
"input": "22\n2 2 2 2 2 2 2 2 2 2 3 2 2 2 2 2 2 1 2 2 2 2",
"output": "1\n18 2 11"
},
{
"input": "138\n2 3 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 3 2 2 2 1 2 3 2 2 2 3 1 3 2 3 2 3 2 2 2 2 3 2 2 2 2 2 1 2 2 3 2 2 3 2 1 2 2 2 2 2 3 1 2 2 2 2 2 3 2 2 3 2 2 2 2 2 1 1 2 3 2 2 2 2 3 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 3 2 3 2 2 2 1 2 2 2 1 2 2 2 2 1 2 2 2 2 1 3",
"output": "18\n13 91 84\n34 90 48\n11 39 77\n78 129 50\n137 68 119\n132 122 138\n19 12 96\n40 7 2\n22 88 69\n107 73 46\n115 15 52\n127 106 87\n93 92 66\n71 112 117\n63 124 42\n17 70 101\n109 121 57\n123 25 36"
},
{
"input": "203\n2 2 1 2 1 2 2 2 1 2 2 1 1 3 1 2 1 2 1 1 2 3 1 1 2 3 3 2 2 2 1 2 1 1 1 1 1 3 1 1 2 1 1 2 2 2 1 2 2 2 1 2 3 2 1 1 2 2 1 2 1 2 2 1 1 2 2 2 1 1 2 2 1 2 1 2 2 3 2 1 2 1 1 1 1 1 1 1 1 1 1 2 2 1 1 2 2 2 2 1 1 1 1 1 1 1 2 2 2 2 2 1 1 1 2 2 2 1 2 2 1 3 2 1 1 1 2 1 1 2 1 1 2 2 2 1 1 2 2 2 1 2 1 3 2 1 2 2 2 1 1 1 2 2 2 1 2 1 1 2 2 2 2 2 1 1 2 1 2 2 1 1 1 1 1 1 2 2 3 1 1 2 3 1 1 1 1 1 1 2 2 1 1 1 2 2 3 2 1 3 1 1 1",
"output": "13\n188 72 14\n137 4 197\n158 76 122\n152 142 26\n104 119 179\n40 63 38\n12 1 78\n17 30 27\n189 60 53\n166 190 144\n129 7 183\n83 41 22\n121 81 200"
},
{
"input": "220\n1 1 3 1 3 1 1 3 1 3 3 3 3 1 3 3 1 3 3 3 3 3 1 1 1 3 1 1 1 3 2 3 3 3 1 1 3 3 1 1 3 3 3 3 1 3 3 1 1 1 2 3 1 1 1 2 3 3 3 2 3 1 1 3 1 1 1 3 2 1 3 2 3 1 1 3 3 3 1 3 1 1 1 3 3 2 1 3 2 1 1 3 3 1 1 1 2 1 1 3 2 1 2 1 1 1 3 1 3 3 1 2 3 3 3 3 1 3 1 1 1 1 2 3 1 1 1 1 1 1 3 2 3 1 3 1 3 1 1 3 1 3 1 3 1 3 1 3 3 2 3 1 3 3 1 3 3 3 3 1 1 3 3 3 3 1 1 3 3 3 2 1 1 1 3 3 1 3 3 3 1 1 1 3 1 3 3 1 1 1 2 3 1 1 3 1 1 1 1 2 3 1 1 2 3 3 1 3 1 3 3 3 3 1 3 2 3 1 1 3",
"output": "20\n198 89 20\n141 56 131\n166 204 19\n160 132 142\n111 112 195\n45 216 92\n6 31 109\n14 150 170\n199 60 18\n173 123 140\n134 69 156\n82 191 85\n126 200 80\n24 97 46\n62 86 149\n214 101 26\n79 171 78\n125 72 118\n172 103 162\n219 51 64"
},
{
"input": "61\n2 3 1 3 2 2 2 3 1 3 2 3 1 2 1 1 2 2 2 2 3 2 3 1 2 1 3 1 3 2 1 1 3 2 1 3 3 3 1 3 3 1 1 3 1 3 2 2 1 2 2 2 1 3 2 3 1 3 3 1 1",
"output": "20\n9 55 2\n24 34 27\n3 5 37\n35 17 41\n61 11 4\n60 19 54\n15 20 59\n26 14 29\n16 22 38\n43 50 12\n49 25 36\n57 51 40\n39 6 33\n32 30 10\n31 48 8\n13 47 23\n45 1 58\n53 52 56\n42 18 21\n28 7 44"
},
{
"input": "5\n1 2 2 3 3",
"output": "1\n1 3 4"
}
] | 1,699,216,219
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 41
| 265
| 204,800
|
n = int(input())
students = list(map(int, input().split()))
prog = students.count(1)
math = students.count(2)
pe = students.count(3)
w = min(prog, math, pe)
print(w)
for i in range(w):
print(students.index(1)+1, students.index(2)+1, students.index(3)+1)
students[students.index(1)] = 0
students[students.index(2)] = 0
students[students.index(3)] = 0
|
Title: Team Olympiad
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The School №0 of the capital of Berland has *n* children studying in it. All the children in this school are gifted: some of them are good at programming, some are good at maths, others are good at PE (Physical Education). Hence, for each child we know value *t**i*:
- *t**i*<==<=1, if the *i*-th child is good at programming, - *t**i*<==<=2, if the *i*-th child is good at maths, - *t**i*<==<=3, if the *i*-th child is good at PE
Each child happens to be good at exactly one of these three subjects.
The Team Scientific Decathlon Olympias requires teams of three students. The school teachers decided that the teams will be composed of three children that are good at different subjects. That is, each team must have one mathematician, one programmer and one sportsman. Of course, each child can be a member of no more than one team.
What is the maximum number of teams that the school will be able to present at the Olympiad? How should the teams be formed for that?
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=5000) — the number of children in the school. The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=3), where *t**i* describes the skill of the *i*-th child.
Output Specification:
In the first line output integer *w* — the largest possible number of teams.
Then print *w* lines, containing three numbers in each line. Each triple represents the indexes of the children forming the team. You can print both the teams, and the numbers in the triplets in any order. The children are numbered from 1 to *n* in the order of their appearance in the input. Each child must participate in no more than one team. If there are several solutions, print any of them.
If no teams can be compiled, print the only line with value *w* equal to 0.
Demo Input:
['7\n1 3 1 3 2 1 2\n', '4\n2 1 1 2\n']
Demo Output:
['2\n3 5 2\n6 7 4\n', '0\n']
Note:
none
|
```python
n = int(input())
students = list(map(int, input().split()))
prog = students.count(1)
math = students.count(2)
pe = students.count(3)
w = min(prog, math, pe)
print(w)
for i in range(w):
print(students.index(1)+1, students.index(2)+1, students.index(3)+1)
students[students.index(1)] = 0
students[students.index(2)] = 0
students[students.index(3)] = 0
```
| 3
|
|
227
|
B
|
Effective Approach
|
PROGRAMMING
| 1,100
|
[
"implementation"
] | null | null |
Once at a team training Vasya, Petya and Sasha got a problem on implementing linear search in an array.
According to the boys, linear search works as follows. The array elements in a pre-selected order are in turn compared with the number that you need to find. Once you find the array element that is equal to the required one, the search ends. The efficiency of the algorithm is the number of performed comparisons. The fewer comparisons the linear search has made, the more effective it is.
Vasya believes that a linear search would work better if it sequentially iterates through the elements, starting with the 1-st one (in this problem we consider the elements of the array indexed from 1 to *n*) and ending with the *n*-th one. And Petya says that Vasya is wrong: the search will need less comparisons if it sequentially iterates the elements starting from the *n*-th and ending with the 1-st one. Sasha argues that the two approaches are equivalent.
To finally begin the task, the teammates decided to settle the debate and compare the two approaches on an example. For this, they took an array that is a permutation of integers from 1 to *n*, and generated *m* queries of the form: find element with value *b**i* in the array. They want to calculate for both approaches how many comparisons in total the linear search will need to respond to all queries. If the first search needs fewer comparisons, then the winner of the dispute is Vasya. If the second one does, then the winner is Petya. If both approaches make the same number of comparisons, then Sasha's got the upper hand.
But the problem is, linear search is too slow. That's why the boys aren't going to find out who is right before the end of the training, unless you come in here. Help them to determine who will win the dispute.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of elements in the array. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the elements of array.
The third line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. The last line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*) — the search queries. Note that the queries can repeat.
|
Print two integers, showing how many comparisons Vasya's approach needs and how many comparisons Petya's approach needs. Separate the numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.
|
[
"2\n1 2\n1\n1\n",
"2\n2 1\n1\n1\n",
"3\n3 1 2\n3\n1 2 3\n"
] |
[
"1 2\n",
"2 1\n",
"6 6\n"
] |
In the first sample Vasya's approach will make one comparison (it starts with the 1-st element and immediately finds the required number), and Petya's approach makes two comparisons (first he compares with the 2-nd array element, doesn't find the search item and compares with the 1-st element).
In the second sample, on the contrary, Vasya's approach will need two comparisons (first with 1-st element, and then with the 2-nd), and Petya's approach will find the required value in one comparison (the first comparison with the 2-nd element).
| 1,000
|
[
{
"input": "2\n1 2\n1\n1",
"output": "1 2"
},
{
"input": "2\n2 1\n1\n1",
"output": "2 1"
},
{
"input": "3\n3 1 2\n3\n1 2 3",
"output": "6 6"
},
{
"input": "9\n2 9 3 1 6 4 7 8 5\n9\n5 1 5 2 8 4 4 4 5",
"output": "58 32"
},
{
"input": "10\n3 10 9 2 7 6 5 8 4 1\n1\n4",
"output": "9 2"
},
{
"input": "10\n5 2 10 8 3 1 9 7 6 4\n9\n2 5 9 2 3 2 5 9 8",
"output": "31 68"
},
{
"input": "9\n3 8 4 7 1 2 5 6 9\n3\n2 7 1",
"output": "15 15"
},
{
"input": "9\n5 3 8 4 2 6 1 7 9\n4\n6 1 9 2",
"output": "27 13"
},
{
"input": "4\n1 3 2 4\n4\n3 1 2 3",
"output": "8 12"
},
{
"input": "3\n1 2 3\n8\n3 2 1 1 2 3 1 2",
"output": "15 17"
}
] | 1,670,773,094
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 34
| 280
| 22,118,400
|
def Effective_Approach2():
number_of_elements_in_array = int(input())
array_input = input()
array_input = array_input.split(' ')
value_to_index_dict = {}
index_in_array = 0
for i in range(len(array_input)):
array_input[i] = int(array_input[i])
value_to_index_dict[array_input[i]] = index_in_array
index_in_array += 1
number_of_queries = int(input())
queryString = input()
queryString = queryString.split(' ')
for i in range(len(queryString)):
queryString[i] = int(queryString[i])
vasyaApproach = 0
petyaApproach = 0
#print(queryString)
for query in queryString:
postion_in_main_array = value_to_index_dict[query]
vasyaApproach += (postion_in_main_array + 1)
petyaApproach += number_of_elements_in_array - postion_in_main_array
print("{} {}".format(vasyaApproach,petyaApproach))
Effective_Approach2()
|
Title: Effective Approach
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Once at a team training Vasya, Petya and Sasha got a problem on implementing linear search in an array.
According to the boys, linear search works as follows. The array elements in a pre-selected order are in turn compared with the number that you need to find. Once you find the array element that is equal to the required one, the search ends. The efficiency of the algorithm is the number of performed comparisons. The fewer comparisons the linear search has made, the more effective it is.
Vasya believes that a linear search would work better if it sequentially iterates through the elements, starting with the 1-st one (in this problem we consider the elements of the array indexed from 1 to *n*) and ending with the *n*-th one. And Petya says that Vasya is wrong: the search will need less comparisons if it sequentially iterates the elements starting from the *n*-th and ending with the 1-st one. Sasha argues that the two approaches are equivalent.
To finally begin the task, the teammates decided to settle the debate and compare the two approaches on an example. For this, they took an array that is a permutation of integers from 1 to *n*, and generated *m* queries of the form: find element with value *b**i* in the array. They want to calculate for both approaches how many comparisons in total the linear search will need to respond to all queries. If the first search needs fewer comparisons, then the winner of the dispute is Vasya. If the second one does, then the winner is Petya. If both approaches make the same number of comparisons, then Sasha's got the upper hand.
But the problem is, linear search is too slow. That's why the boys aren't going to find out who is right before the end of the training, unless you come in here. Help them to determine who will win the dispute.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of elements in the array. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the elements of array.
The third line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. The last line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*) — the search queries. Note that the queries can repeat.
Output Specification:
Print two integers, showing how many comparisons Vasya's approach needs and how many comparisons Petya's approach needs. Separate the numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.
Demo Input:
['2\n1 2\n1\n1\n', '2\n2 1\n1\n1\n', '3\n3 1 2\n3\n1 2 3\n']
Demo Output:
['1 2\n', '2 1\n', '6 6\n']
Note:
In the first sample Vasya's approach will make one comparison (it starts with the 1-st element and immediately finds the required number), and Petya's approach makes two comparisons (first he compares with the 2-nd array element, doesn't find the search item and compares with the 1-st element).
In the second sample, on the contrary, Vasya's approach will need two comparisons (first with 1-st element, and then with the 2-nd), and Petya's approach will find the required value in one comparison (the first comparison with the 2-nd element).
|
```python
def Effective_Approach2():
number_of_elements_in_array = int(input())
array_input = input()
array_input = array_input.split(' ')
value_to_index_dict = {}
index_in_array = 0
for i in range(len(array_input)):
array_input[i] = int(array_input[i])
value_to_index_dict[array_input[i]] = index_in_array
index_in_array += 1
number_of_queries = int(input())
queryString = input()
queryString = queryString.split(' ')
for i in range(len(queryString)):
queryString[i] = int(queryString[i])
vasyaApproach = 0
petyaApproach = 0
#print(queryString)
for query in queryString:
postion_in_main_array = value_to_index_dict[query]
vasyaApproach += (postion_in_main_array + 1)
petyaApproach += number_of_elements_in_array - postion_in_main_array
print("{} {}".format(vasyaApproach,petyaApproach))
Effective_Approach2()
```
| 3
|
|
208
|
A
|
Dubstep
|
PROGRAMMING
| 900
|
[
"strings"
] | null | null |
Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them.
Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club.
For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX".
Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song.
|
The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word.
|
Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space.
|
[
"WUBWUBABCWUB\n",
"WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n"
] |
[
"ABC ",
"WE ARE THE CHAMPIONS MY FRIEND "
] |
In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya.
In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB".
| 500
|
[
{
"input": "WUBWUBABCWUB",
"output": "ABC "
},
{
"input": "WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB",
"output": "WE ARE THE CHAMPIONS MY FRIEND "
},
{
"input": "WUBWUBWUBSR",
"output": "SR "
},
{
"input": "RWUBWUBWUBLWUB",
"output": "R L "
},
{
"input": "ZJWUBWUBWUBJWUBWUBWUBL",
"output": "ZJ J L "
},
{
"input": "CWUBBWUBWUBWUBEWUBWUBWUBQWUBWUBWUB",
"output": "C B E Q "
},
{
"input": "WUBJKDWUBWUBWBIRAQKFWUBWUBYEWUBWUBWUBWVWUBWUB",
"output": "JKD WBIRAQKF YE WV "
},
{
"input": "WUBKSDHEMIXUJWUBWUBRWUBWUBWUBSWUBWUBWUBHWUBWUBWUB",
"output": "KSDHEMIXUJ R S H "
},
{
"input": "OGWUBWUBWUBXWUBWUBWUBIWUBWUBWUBKOWUBWUB",
"output": "OG X I KO "
},
{
"input": "QWUBQQWUBWUBWUBIWUBWUBWWWUBWUBWUBJOPJPBRH",
"output": "Q QQ I WW JOPJPBRH "
},
{
"input": "VSRNVEATZTLGQRFEGBFPWUBWUBWUBAJWUBWUBWUBPQCHNWUBCWUB",
"output": "VSRNVEATZTLGQRFEGBFP AJ PQCHN C "
},
{
"input": "WUBWUBEWUBWUBWUBIQMJNIQWUBWUBWUBGZZBQZAUHYPWUBWUBWUBPMRWUBWUBWUBDCV",
"output": "E IQMJNIQ GZZBQZAUHYP PMR DCV "
},
{
"input": "WUBWUBWUBFVWUBWUBWUBBPSWUBWUBWUBRXNETCJWUBWUBWUBJDMBHWUBWUBWUBBWUBWUBVWUBWUBB",
"output": "FV BPS RXNETCJ JDMBH B V B "
},
{
"input": "WUBWUBWUBFBQWUBWUBWUBIDFSYWUBWUBWUBCTWDMWUBWUBWUBSXOWUBWUBWUBQIWUBWUBWUBL",
"output": "FBQ IDFSY CTWDM SXO QI L "
},
{
"input": "IWUBWUBQLHDWUBYIIKZDFQWUBWUBWUBCXWUBWUBUWUBWUBWUBKWUBWUBWUBNL",
"output": "I QLHD YIIKZDFQ CX U K NL "
},
{
"input": "KWUBUPDYXGOKUWUBWUBWUBAGOAHWUBIZDWUBWUBWUBIYWUBWUBWUBVWUBWUBWUBPWUBWUBWUBE",
"output": "K UPDYXGOKU AGOAH IZD IY V P E "
},
{
"input": "WUBWUBOWUBWUBWUBIPVCQAFWYWUBWUBWUBQWUBWUBWUBXHDKCPYKCTWWYWUBWUBWUBVWUBWUBWUBFZWUBWUB",
"output": "O IPVCQAFWY Q XHDKCPYKCTWWY V FZ "
},
{
"input": "PAMJGYWUBWUBWUBXGPQMWUBWUBWUBTKGSXUYWUBWUBWUBEWUBWUBWUBNWUBWUBWUBHWUBWUBWUBEWUBWUB",
"output": "PAMJGY XGPQM TKGSXUY E N H E "
},
{
"input": "WUBYYRTSMNWUWUBWUBWUBCWUBWUBWUBCWUBWUBWUBFSYUINDWOBVWUBWUBWUBFWUBWUBWUBAUWUBWUBWUBVWUBWUBWUBJB",
"output": "YYRTSMNWU C C FSYUINDWOBV F AU V JB "
},
{
"input": "WUBWUBYGPYEYBNRTFKOQCWUBWUBWUBUYGRTQEGWLFYWUBWUBWUBFVWUBHPWUBWUBWUBXZQWUBWUBWUBZDWUBWUBWUBM",
"output": "YGPYEYBNRTFKOQC UYGRTQEGWLFY FV HP XZQ ZD M "
},
{
"input": "WUBZVMJWUBWUBWUBFOIMJQWKNZUBOFOFYCCWUBWUBWUBAUWWUBRDRADWUBWUBWUBCHQVWUBWUBWUBKFTWUBWUBWUBW",
"output": "ZVMJ FOIMJQWKNZUBOFOFYCC AUW RDRAD CHQV KFT W "
},
{
"input": "WUBWUBZBKOKHQLGKRVIMZQMQNRWUBWUBWUBDACWUBWUBNZHFJMPEYKRVSWUBWUBWUBPPHGAVVPRZWUBWUBWUBQWUBWUBAWUBG",
"output": "ZBKOKHQLGKRVIMZQMQNR DAC NZHFJMPEYKRVS PPHGAVVPRZ Q A G "
},
{
"input": "WUBWUBJWUBWUBWUBNFLWUBWUBWUBGECAWUBYFKBYJWTGBYHVSSNTINKWSINWSMAWUBWUBWUBFWUBWUBWUBOVWUBWUBLPWUBWUBWUBN",
"output": "J NFL GECA YFKBYJWTGBYHVSSNTINKWSINWSMA F OV LP N "
},
{
"input": "WUBWUBLCWUBWUBWUBZGEQUEATJVIXETVTWUBWUBWUBEXMGWUBWUBWUBRSWUBWUBWUBVWUBWUBWUBTAWUBWUBWUBCWUBWUBWUBQG",
"output": "LC ZGEQUEATJVIXETVT EXMG RS V TA C QG "
},
{
"input": "WUBMPWUBWUBWUBORWUBWUBDLGKWUBWUBWUBVVZQCAAKVJTIKWUBWUBWUBTJLUBZJCILQDIFVZWUBWUBYXWUBWUBWUBQWUBWUBWUBLWUB",
"output": "MP OR DLGK VVZQCAAKVJTIK TJLUBZJCILQDIFVZ YX Q L "
},
{
"input": "WUBNXOLIBKEGXNWUBWUBWUBUWUBGITCNMDQFUAOVLWUBWUBWUBAIJDJZJHFMPVTPOXHPWUBWUBWUBISCIOWUBWUBWUBGWUBWUBWUBUWUB",
"output": "NXOLIBKEGXN U GITCNMDQFUAOVL AIJDJZJHFMPVTPOXHP ISCIO G U "
},
{
"input": "WUBWUBNMMWCZOLYPNBELIYVDNHJUNINWUBWUBWUBDXLHYOWUBWUBWUBOJXUWUBWUBWUBRFHTGJCEFHCGWARGWUBWUBWUBJKWUBWUBSJWUBWUB",
"output": "NMMWCZOLYPNBELIYVDNHJUNIN DXLHYO OJXU RFHTGJCEFHCGWARG JK SJ "
},
{
"input": "SGWLYSAUJOJBNOXNWUBWUBWUBBOSSFWKXPDPDCQEWUBWUBWUBDIRZINODWUBWUBWUBWWUBWUBWUBPPHWUBWUBWUBRWUBWUBWUBQWUBWUBWUBJWUB",
"output": "SGWLYSAUJOJBNOXN BOSSFWKXPDPDCQE DIRZINOD W PPH R Q J "
},
{
"input": "TOWUBWUBWUBGBTBNWUBWUBWUBJVIOJBIZFUUYHUAIEBQLQXPQKZJMPTCWBKPOSAWUBWUBWUBSWUBWUBWUBTOLVXWUBWUBWUBNHWUBWUBWUBO",
"output": "TO GBTBN JVIOJBIZFUUYHUAIEBQLQXPQKZJMPTCWBKPOSA S TOLVX NH O "
},
{
"input": "WUBWUBWSPLAYSZSAUDSWUBWUBWUBUWUBWUBWUBKRWUBWUBWUBRSOKQMZFIYZQUWUBWUBWUBELSHUWUBWUBWUBUKHWUBWUBWUBQXEUHQWUBWUBWUBBWUBWUBWUBR",
"output": "WSPLAYSZSAUDS U KR RSOKQMZFIYZQU ELSHU UKH QXEUHQ B R "
},
{
"input": "WUBXEMWWVUHLSUUGRWUBWUBWUBAWUBXEGILZUNKWUBWUBWUBJDHHKSWUBWUBWUBDTSUYSJHWUBWUBWUBPXFWUBMOHNJWUBWUBWUBZFXVMDWUBWUBWUBZMWUBWUB",
"output": "XEMWWVUHLSUUGR A XEGILZUNK JDHHKS DTSUYSJH PXF MOHNJ ZFXVMD ZM "
},
{
"input": "BMBWUBWUBWUBOQKWUBWUBWUBPITCIHXHCKLRQRUGXJWUBWUBWUBVWUBWUBWUBJCWUBWUBWUBQJPWUBWUBWUBBWUBWUBWUBBMYGIZOOXWUBWUBWUBTAGWUBWUBHWUB",
"output": "BMB OQK PITCIHXHCKLRQRUGXJ V JC QJP B BMYGIZOOX TAG H "
},
{
"input": "CBZNWUBWUBWUBNHWUBWUBWUBYQSYWUBWUBWUBMWUBWUBWUBXRHBTMWUBWUBWUBPCRCWUBWUBWUBTZUYLYOWUBWUBWUBCYGCWUBWUBWUBCLJWUBWUBWUBSWUBWUBWUB",
"output": "CBZN NH YQSY M XRHBTM PCRC TZUYLYO CYGC CLJ S "
},
{
"input": "DPDWUBWUBWUBEUQKWPUHLTLNXHAEKGWUBRRFYCAYZFJDCJLXBAWUBWUBWUBHJWUBOJWUBWUBWUBNHBJEYFWUBWUBWUBRWUBWUBWUBSWUBWWUBWUBWUBXDWUBWUBWUBJWUB",
"output": "DPD EUQKWPUHLTLNXHAEKG RRFYCAYZFJDCJLXBA HJ OJ NHBJEYF R S W XD J "
},
{
"input": "WUBWUBWUBISERPQITVIYERSCNWUBWUBWUBQWUBWUBWUBDGSDIPWUBWUBWUBCAHKDZWEXBIBJVVSKKVQJWUBWUBWUBKIWUBWUBWUBCWUBWUBWUBAWUBWUBWUBPWUBWUBWUBHWUBWUBWUBF",
"output": "ISERPQITVIYERSCN Q DGSDIP CAHKDZWEXBIBJVVSKKVQJ KI C A P H F "
},
{
"input": "WUBWUBWUBIWUBWUBLIKNQVWUBWUBWUBPWUBWUBWUBHWUBWUBWUBMWUBWUBWUBDPRSWUBWUBWUBBSAGYLQEENWXXVWUBWUBWUBXMHOWUBWUBWUBUWUBWUBWUBYRYWUBWUBWUBCWUBWUBWUBY",
"output": "I LIKNQV P H M DPRS BSAGYLQEENWXXV XMHO U YRY C Y "
},
{
"input": "WUBWUBWUBMWUBWUBWUBQWUBWUBWUBITCFEYEWUBWUBWUBHEUWGNDFNZGWKLJWUBWUBWUBMZPWUBWUBWUBUWUBWUBWUBBWUBWUBWUBDTJWUBHZVIWUBWUBWUBPWUBFNHHWUBWUBWUBVTOWUB",
"output": "M Q ITCFEYE HEUWGNDFNZGWKLJ MZP U B DTJ HZVI P FNHH VTO "
},
{
"input": "WUBWUBNDNRFHYJAAUULLHRRDEDHYFSRXJWUBWUBWUBMUJVDTIRSGYZAVWKRGIFWUBWUBWUBHMZWUBWUBWUBVAIWUBWUBWUBDDKJXPZRGWUBWUBWUBSGXWUBWUBWUBIFKWUBWUBWUBUWUBWUBWUBW",
"output": "NDNRFHYJAAUULLHRRDEDHYFSRXJ MUJVDTIRSGYZAVWKRGIF HMZ VAI DDKJXPZRG SGX IFK U W "
},
{
"input": "WUBOJMWRSLAXXHQRTPMJNCMPGWUBWUBWUBNYGMZIXNLAKSQYWDWUBWUBWUBXNIWUBWUBWUBFWUBWUBWUBXMBWUBWUBWUBIWUBWUBWUBINWUBWUBWUBWDWUBWUBWUBDDWUBWUBWUBD",
"output": "OJMWRSLAXXHQRTPMJNCMPG NYGMZIXNLAKSQYWD XNI F XMB I IN WD DD D "
},
{
"input": "WUBWUBWUBREHMWUBWUBWUBXWUBWUBWUBQASNWUBWUBWUBNLSMHLCMTICWUBWUBWUBVAWUBWUBWUBHNWUBWUBWUBNWUBWUBWUBUEXLSFOEULBWUBWUBWUBXWUBWUBWUBJWUBWUBWUBQWUBWUBWUBAWUBWUB",
"output": "REHM X QASN NLSMHLCMTIC VA HN N UEXLSFOEULB X J Q A "
},
{
"input": "WUBWUBWUBSTEZTZEFFIWUBWUBWUBSWUBWUBWUBCWUBFWUBHRJPVWUBWUBWUBDYJUWUBWUBWUBPWYDKCWUBWUBWUBCWUBWUBWUBUUEOGCVHHBWUBWUBWUBEXLWUBWUBWUBVCYWUBWUBWUBMWUBWUBWUBYWUB",
"output": "STEZTZEFFI S C F HRJPV DYJU PWYDKC C UUEOGCVHHB EXL VCY M Y "
},
{
"input": "WPPNMSQOQIWUBWUBWUBPNQXWUBWUBWUBHWUBWUBWUBNFLWUBWUBWUBGWSGAHVJFNUWUBWUBWUBFWUBWUBWUBWCMLRICFSCQQQTNBWUBWUBWUBSWUBWUBWUBKGWUBWUBWUBCWUBWUBWUBBMWUBWUBWUBRWUBWUB",
"output": "WPPNMSQOQI PNQX H NFL GWSGAHVJFNU F WCMLRICFSCQQQTNB S KG C BM R "
},
{
"input": "YZJOOYITZRARKVFYWUBWUBRZQGWUBWUBWUBUOQWUBWUBWUBIWUBWUBWUBNKVDTBOLETKZISTWUBWUBWUBWLWUBQQFMMGSONZMAWUBZWUBWUBWUBQZUXGCWUBWUBWUBIRZWUBWUBWUBLTTVTLCWUBWUBWUBY",
"output": "YZJOOYITZRARKVFY RZQG UOQ I NKVDTBOLETKZIST WL QQFMMGSONZMA Z QZUXGC IRZ LTTVTLC Y "
},
{
"input": "WUBCAXNCKFBVZLGCBWCOAWVWOFKZVQYLVTWUBWUBWUBNLGWUBWUBWUBAMGDZBDHZMRMQMDLIRMIWUBWUBWUBGAJSHTBSWUBWUBWUBCXWUBWUBWUBYWUBZLXAWWUBWUBWUBOHWUBWUBWUBZWUBWUBWUBGBWUBWUBWUBE",
"output": "CAXNCKFBVZLGCBWCOAWVWOFKZVQYLVT NLG AMGDZBDHZMRMQMDLIRMI GAJSHTBS CX Y ZLXAW OH Z GB E "
},
{
"input": "WUBWUBCHXSOWTSQWUBWUBWUBCYUZBPBWUBWUBWUBSGWUBWUBWKWORLRRLQYUUFDNWUBWUBWUBYYGOJNEVEMWUBWUBWUBRWUBWUBWUBQWUBWUBWUBIHCKWUBWUBWUBKTWUBWUBWUBRGSNTGGWUBWUBWUBXCXWUBWUBWUBS",
"output": "CHXSOWTSQ CYUZBPB SG WKWORLRRLQYUUFDN YYGOJNEVEM R Q IHCK KT RGSNTGG XCX S "
},
{
"input": "WUBWUBWUBHJHMSBURXTHXWSCHNAIJOWBHLZGJZDHEDSPWBWACCGQWUBWUBWUBXTZKGIITWUBWUBWUBAWUBWUBWUBVNCXPUBCQWUBWUBWUBIDPNAWUBWUBWUBOWUBWUBWUBYGFWUBWUBWUBMQOWUBWUBWUBKWUBWUBWUBAZVWUBWUBWUBEP",
"output": "HJHMSBURXTHXWSCHNAIJOWBHLZGJZDHEDSPWBWACCGQ XTZKGIIT A VNCXPUBCQ IDPNA O YGF MQO K AZV EP "
},
{
"input": "WUBKYDZOYWZSNGMKJSWAXFDFLTHDHEOGTDBNZMSMKZTVWUBWUBWUBLRMIIWUBWUBWUBGWUBWUBWUBADPSWUBWUBWUBANBWUBWUBPCWUBWUBWUBPWUBWUBWUBGPVNLSWIRFORYGAABUXMWUBWUBWUBOWUBWUBWUBNWUBWUBWUBYWUBWUB",
"output": "KYDZOYWZSNGMKJSWAXFDFLTHDHEOGTDBNZMSMKZTV LRMII G ADPS ANB PC P GPVNLSWIRFORYGAABUXM O N Y "
},
{
"input": "REWUBWUBWUBJDWUBWUBWUBNWUBWUBWUBTWWUBWUBWUBWZDOCKKWUBWUBWUBLDPOVBFRCFWUBWUBAKZIBQKEUAZEEWUBWUBWUBLQYPNPFWUBYEWUBWUBWUBFWUBWUBWUBBPWUBWUBWUBAWWUBWUBWUBQWUBWUBWUBBRWUBWUBWUBXJL",
"output": "RE JD N TW WZDOCKK LDPOVBFRCF AKZIBQKEUAZEE LQYPNPF YE F BP AW Q BR XJL "
},
{
"input": "CUFGJDXGMWUBWUBWUBOMWUBWUBWUBSIEWUBWUBWUBJJWKNOWUBWUBWUBYBHVNRNORGYWUBWUBWUBOAGCAWUBWUBWUBSBLBKTPFKPBIWUBWUBWUBJBWUBWUBWUBRMFCJPGWUBWUBWUBDWUBWUBWUBOJOWUBWUBWUBZPWUBWUBWUBMWUBRWUBWUBWUBFXWWUBWUBWUBO",
"output": "CUFGJDXGM OM SIE JJWKNO YBHVNRNORGY OAGCA SBLBKTPFKPBI JB RMFCJPG D OJO ZP M R FXW O "
},
{
"input": "WUBJZGAEXFMFEWMAKGQLUWUBWUBWUBICYTPQWGENELVYWANKUOJYWUBWUBWUBGWUBWUBWUBHYCJVLPHTUPNEGKCDGQWUBWUBWUBOFWUBWUBWUBCPGSOGZBRPRPVJJEWUBWUBWUBDQBCWUBWUBWUBHWUBWUBWUBMHOHYBMATWUBWUBWUBVWUBWUBWUBSWUBWUBWUBKOWU",
"output": "JZGAEXFMFEWMAKGQLU ICYTPQWGENELVYWANKUOJY G HYCJVLPHTUPNEGKCDGQ OF CPGSOGZBRPRPVJJE DQBC H MHOHYBMAT V S KOWU "
},
{
"input": "A",
"output": "A "
},
{
"input": "WUBA",
"output": "A "
},
{
"input": "WUBWUBA",
"output": "A "
},
{
"input": "AWUBWUBWUB",
"output": "A "
},
{
"input": "AWUBBWUBCWUBD",
"output": "A B C D "
},
{
"input": "WUBWWUBWUBWUBUWUBWUBBWUB",
"output": "W U B "
},
{
"input": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA",
"output": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA "
},
{
"input": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAWUBAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA",
"output": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA "
},
{
"input": "WUWUBBWWUBUB",
"output": "WU BW UB "
},
{
"input": "WUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUABWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUB",
"output": "WUAB "
},
{
"input": "ZWUB",
"output": "Z "
},
{
"input": "WU",
"output": "WU "
},
{
"input": "UB",
"output": "UB "
},
{
"input": "U",
"output": "U "
},
{
"input": "WUBW",
"output": "W "
},
{
"input": "WUBWU",
"output": "WU "
},
{
"input": "WUWUB",
"output": "WU "
},
{
"input": "UBWUB",
"output": "UB "
},
{
"input": "WUWUBUBWUBUWUB",
"output": "WU UB U "
},
{
"input": "WUBWWUBAWUB",
"output": "W A "
},
{
"input": "WUUUUU",
"output": "WUUUUU "
}
] | 1,690,027,285
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 71
| 92
| 0
|
s=input().split('WUB')
for i in s:
if i!='':
print(i,end=' ')
|
Title: Dubstep
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them.
Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club.
For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX".
Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song.
Input Specification:
The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word.
Output Specification:
Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space.
Demo Input:
['WUBWUBABCWUB\n', 'WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n']
Demo Output:
['ABC ', 'WE ARE THE CHAMPIONS MY FRIEND ']
Note:
In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya.
In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB".
|
```python
s=input().split('WUB')
for i in s:
if i!='':
print(i,end=' ')
```
| 3
|
|
340
|
A
|
The Wall
|
PROGRAMMING
| 1,200
|
[
"math"
] | null | null |
Iahub and his friend Floyd have started painting a wall. Iahub is painting the wall red and Floyd is painting it pink. You can consider the wall being made of a very large number of bricks, numbered 1, 2, 3 and so on.
Iahub has the following scheme of painting: he skips *x*<=-<=1 consecutive bricks, then he paints the *x*-th one. That is, he'll paint bricks *x*, 2·*x*, 3·*x* and so on red. Similarly, Floyd skips *y*<=-<=1 consecutive bricks, then he paints the *y*-th one. Hence he'll paint bricks *y*, 2·*y*, 3·*y* and so on pink.
After painting the wall all day, the boys observed that some bricks are painted both red and pink. Iahub has a lucky number *a* and Floyd has a lucky number *b*. Boys wonder how many bricks numbered no less than *a* and no greater than *b* are painted both red and pink. This is exactly your task: compute and print the answer to the question.
|
The input will have a single line containing four integers in this order: *x*, *y*, *a*, *b*. (1<=≤<=*x*,<=*y*<=≤<=1000, 1<=≤<=*a*,<=*b*<=≤<=2·109, *a*<=≤<=*b*).
|
Output a single integer — the number of bricks numbered no less than *a* and no greater than *b* that are painted both red and pink.
|
[
"2 3 6 18\n"
] |
[
"3"
] |
Let's look at the bricks from *a* to *b* (*a* = 6, *b* = 18). The bricks colored in red are numbered 6, 8, 10, 12, 14, 16, 18. The bricks colored in pink are numbered 6, 9, 12, 15, 18. The bricks colored in both red and pink are numbered with 6, 12 and 18.
| 500
|
[
{
"input": "2 3 6 18",
"output": "3"
},
{
"input": "4 6 20 201",
"output": "15"
},
{
"input": "15 27 100 10000",
"output": "74"
},
{
"input": "105 60 3456 78910",
"output": "179"
},
{
"input": "1 1 1000 100000",
"output": "99001"
},
{
"input": "3 2 5 5",
"output": "0"
},
{
"input": "555 777 1 1000000",
"output": "257"
},
{
"input": "1000 1000 1 32323",
"output": "32"
},
{
"input": "45 125 93451125 100000000",
"output": "5821"
},
{
"input": "101 171 1 1000000000",
"output": "57900"
},
{
"input": "165 255 69696 1000000000",
"output": "356482"
},
{
"input": "555 777 666013 1000000000",
"output": "257229"
},
{
"input": "23 46 123321 900000000",
"output": "19562537"
},
{
"input": "321 123 15 1000000",
"output": "75"
},
{
"input": "819 1000 9532 152901000",
"output": "186"
},
{
"input": "819 1000 10000 1000000",
"output": "1"
},
{
"input": "1 1 1 1",
"output": "1"
},
{
"input": "1 2 2 1000003",
"output": "500001"
},
{
"input": "1 1 1 1000000000",
"output": "1000000000"
},
{
"input": "10 15 69 195610342",
"output": "6520342"
},
{
"input": "2 1 1 1000000000",
"output": "500000000"
},
{
"input": "1000 1000 1 20",
"output": "0"
},
{
"input": "1 1 1 2000000000",
"output": "2000000000"
},
{
"input": "1 2 1 2000000000",
"output": "1000000000"
},
{
"input": "2 1 1 2000000000",
"output": "1000000000"
},
{
"input": "2 3 1 1000000000",
"output": "166666666"
},
{
"input": "2 3 1 2000000000",
"output": "333333333"
},
{
"input": "3 7 1 1000000000",
"output": "47619047"
},
{
"input": "1 1 1000000000 2000000000",
"output": "1000000001"
},
{
"input": "2 2 1 2000000000",
"output": "1000000000"
},
{
"input": "1 1 2 2000000000",
"output": "1999999999"
},
{
"input": "3 2 1 2000000000",
"output": "333333333"
},
{
"input": "1 1 2000000000 2000000000",
"output": "1"
},
{
"input": "2 3 7 7",
"output": "0"
},
{
"input": "3 3 3 7",
"output": "2"
}
] | 1,377,877,178
| 577
|
Python 3
|
WRONG_ANSWER
|
PRETESTS
| 1
| 62
| 0
|
from math import*
def gcd(a, b):
if b == 0:
return a
else:
return gcd(b, a % b)
x, y, a, b = list(map(int, input().split()))
nok = (x * y) // gcd(x, y)
print(ceil((b - a + 1) / nok))
|
Title: The Wall
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Iahub and his friend Floyd have started painting a wall. Iahub is painting the wall red and Floyd is painting it pink. You can consider the wall being made of a very large number of bricks, numbered 1, 2, 3 and so on.
Iahub has the following scheme of painting: he skips *x*<=-<=1 consecutive bricks, then he paints the *x*-th one. That is, he'll paint bricks *x*, 2·*x*, 3·*x* and so on red. Similarly, Floyd skips *y*<=-<=1 consecutive bricks, then he paints the *y*-th one. Hence he'll paint bricks *y*, 2·*y*, 3·*y* and so on pink.
After painting the wall all day, the boys observed that some bricks are painted both red and pink. Iahub has a lucky number *a* and Floyd has a lucky number *b*. Boys wonder how many bricks numbered no less than *a* and no greater than *b* are painted both red and pink. This is exactly your task: compute and print the answer to the question.
Input Specification:
The input will have a single line containing four integers in this order: *x*, *y*, *a*, *b*. (1<=≤<=*x*,<=*y*<=≤<=1000, 1<=≤<=*a*,<=*b*<=≤<=2·109, *a*<=≤<=*b*).
Output Specification:
Output a single integer — the number of bricks numbered no less than *a* and no greater than *b* that are painted both red and pink.
Demo Input:
['2 3 6 18\n']
Demo Output:
['3']
Note:
Let's look at the bricks from *a* to *b* (*a* = 6, *b* = 18). The bricks colored in red are numbered 6, 8, 10, 12, 14, 16, 18. The bricks colored in pink are numbered 6, 9, 12, 15, 18. The bricks colored in both red and pink are numbered with 6, 12 and 18.
|
```python
from math import*
def gcd(a, b):
if b == 0:
return a
else:
return gcd(b, a % b)
x, y, a, b = list(map(int, input().split()))
nok = (x * y) // gcd(x, y)
print(ceil((b - a + 1) / nok))
```
| 0
|
|
723
|
A
|
The New Year: Meeting Friends
|
PROGRAMMING
| 800
|
[
"implementation",
"math",
"sortings"
] | null | null |
There are three friend living on the straight line *Ox* in Lineland. The first friend lives at the point *x*1, the second friend lives at the point *x*2, and the third friend lives at the point *x*3. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year?
It's guaranteed that the optimal answer is always integer.
|
The first line of the input contains three distinct integers *x*1, *x*2 and *x*3 (1<=≤<=*x*1,<=*x*2,<=*x*3<=≤<=100) — the coordinates of the houses of the first, the second and the third friends respectively.
|
Print one integer — the minimum total distance the friends need to travel in order to meet together.
|
[
"7 1 4\n",
"30 20 10\n"
] |
[
"6\n",
"20\n"
] |
In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4.
| 500
|
[
{
"input": "7 1 4",
"output": "6"
},
{
"input": "30 20 10",
"output": "20"
},
{
"input": "1 4 100",
"output": "99"
},
{
"input": "100 1 91",
"output": "99"
},
{
"input": "1 45 100",
"output": "99"
},
{
"input": "1 2 3",
"output": "2"
},
{
"input": "71 85 88",
"output": "17"
},
{
"input": "30 38 99",
"output": "69"
},
{
"input": "23 82 95",
"output": "72"
},
{
"input": "22 41 47",
"output": "25"
},
{
"input": "9 94 77",
"output": "85"
},
{
"input": "1 53 51",
"output": "52"
},
{
"input": "25 97 93",
"output": "72"
},
{
"input": "42 53 51",
"output": "11"
},
{
"input": "81 96 94",
"output": "15"
},
{
"input": "21 5 93",
"output": "88"
},
{
"input": "50 13 75",
"output": "62"
},
{
"input": "41 28 98",
"output": "70"
},
{
"input": "69 46 82",
"output": "36"
},
{
"input": "87 28 89",
"output": "61"
},
{
"input": "44 45 40",
"output": "5"
},
{
"input": "86 97 68",
"output": "29"
},
{
"input": "43 92 30",
"output": "62"
},
{
"input": "16 70 1",
"output": "69"
},
{
"input": "40 46 19",
"output": "27"
},
{
"input": "71 38 56",
"output": "33"
},
{
"input": "82 21 80",
"output": "61"
},
{
"input": "75 8 35",
"output": "67"
},
{
"input": "75 24 28",
"output": "51"
},
{
"input": "78 23 56",
"output": "55"
},
{
"input": "85 31 10",
"output": "75"
},
{
"input": "76 50 9",
"output": "67"
},
{
"input": "95 37 34",
"output": "61"
},
{
"input": "84 61 35",
"output": "49"
},
{
"input": "87 85 37",
"output": "50"
},
{
"input": "1 3 2",
"output": "2"
},
{
"input": "4 2 6",
"output": "4"
},
{
"input": "6 9 3",
"output": "6"
},
{
"input": "12 4 8",
"output": "8"
},
{
"input": "15 10 5",
"output": "10"
},
{
"input": "1 50 17",
"output": "49"
},
{
"input": "10 5 15",
"output": "10"
},
{
"input": "8 1 9",
"output": "8"
},
{
"input": "3 5 4",
"output": "2"
},
{
"input": "2 1 3",
"output": "2"
},
{
"input": "1 8 2",
"output": "7"
},
{
"input": "1 100 2",
"output": "99"
},
{
"input": "1 4 6",
"output": "5"
}
] | 1,688,054,445
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 48
| 62
| 0
|
x1, x2, x3 = map(int, input().split())
x = [x1, x2, x3]
x.sort()
min_distance = x[2] - x[0]
print(min_distance)
|
Title: The New Year: Meeting Friends
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are three friend living on the straight line *Ox* in Lineland. The first friend lives at the point *x*1, the second friend lives at the point *x*2, and the third friend lives at the point *x*3. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year?
It's guaranteed that the optimal answer is always integer.
Input Specification:
The first line of the input contains three distinct integers *x*1, *x*2 and *x*3 (1<=≤<=*x*1,<=*x*2,<=*x*3<=≤<=100) — the coordinates of the houses of the first, the second and the third friends respectively.
Output Specification:
Print one integer — the minimum total distance the friends need to travel in order to meet together.
Demo Input:
['7 1 4\n', '30 20 10\n']
Demo Output:
['6\n', '20\n']
Note:
In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4.
|
```python
x1, x2, x3 = map(int, input().split())
x = [x1, x2, x3]
x.sort()
min_distance = x[2] - x[0]
print(min_distance)
```
| 3
|
|
538
|
C
|
Tourist's Notes
|
PROGRAMMING
| 1,600
|
[
"binary search",
"brute force",
"greedy",
"implementation",
"math"
] | null | null |
A tourist hiked along the mountain range. The hike lasted for *n* days, during each day the tourist noted height above the sea level. On the *i*-th day height was equal to some integer *h**i*. The tourist pick smooth enough route for his hike, meaning that the between any two consecutive days height changes by at most 1, i.e. for all *i*'s from 1 to *n*<=-<=1 the inequality |*h**i*<=-<=*h**i*<=+<=1|<=≤<=1 holds.
At the end of the route the tourist rafted down a mountain river and some notes in the journal were washed away. Moreover, the numbers in the notes could have been distorted. Now the tourist wonders what could be the maximum height during his hike. Help him restore the maximum possible value of the maximum height throughout the hike or determine that the notes were so much distorted that they do not represent any possible height values that meet limits |*h**i*<=-<=*h**i*<=+<=1|<=≤<=1.
|
The first line contains two space-separated numbers, *n* and *m* (1<=≤<=*n*<=≤<=108, 1<=≤<=*m*<=≤<=105) — the number of days of the hike and the number of notes left in the journal.
Next *m* lines contain two space-separated integers *d**i* and *h**d**i* (1<=≤<=*d**i*<=≤<=*n*, 0<=≤<=*h**d**i*<=≤<=108) — the number of the day when the *i*-th note was made and height on the *d**i*-th day. It is guaranteed that the notes are given in the chronological order, i.e. for all *i* from 1 to *m*<=-<=1 the following condition holds: *d**i*<=<<=*d**i*<=+<=1.
|
If the notes aren't contradictory, print a single integer — the maximum possible height value throughout the whole route.
If the notes do not correspond to any set of heights, print a single word 'IMPOSSIBLE' (without the quotes).
|
[
"8 2\n2 0\n7 0\n",
"8 3\n2 0\n7 0\n8 3\n"
] |
[
"2\n",
"IMPOSSIBLE\n"
] |
For the first sample, an example of a correct height sequence with a maximum of 2: (0, 0, 1, 2, 1, 1, 0, 1).
In the second sample the inequality between *h*<sub class="lower-index">7</sub> and *h*<sub class="lower-index">8</sub> does not hold, thus the information is inconsistent.
| 1,500
|
[
{
"input": "8 2\n2 0\n7 0",
"output": "2"
},
{
"input": "8 3\n2 0\n7 0\n8 3",
"output": "IMPOSSIBLE"
},
{
"input": "10 10\n1 0\n2 0\n3 0\n4 0\n5 1\n6 2\n7 3\n8 2\n9 3\n10 4",
"output": "4"
},
{
"input": "50 10\n1 42\n7 36\n16 40\n21 40\n26 39\n30 41\n32 41\n36 40\n44 37\n50 41",
"output": "42"
},
{
"input": "50 10\n5 17\n7 15\n10 4\n15 11\n18 13\n21 15\n31 5\n34 13\n40 15\n49 16",
"output": "IMPOSSIBLE"
},
{
"input": "100 50\n1 53\n3 51\n4 50\n6 48\n9 45\n12 48\n14 46\n16 48\n17 47\n19 49\n20 48\n22 46\n23 45\n24 44\n26 46\n27 47\n29 49\n32 52\n33 53\n35 55\n37 53\n40 50\n41 51\n43 53\n47 57\n50 60\n51 59\n52 60\n57 65\n59 63\n60 62\n61 61\n62 60\n64 62\n68 66\n70 64\n71 63\n73 65\n77 69\n79 67\n81 65\n83 63\n86 66\n88 68\n89 69\n91 67\n94 64\n95 63\n98 60\n100 58",
"output": "69"
},
{
"input": "10 1\n4 16160172",
"output": "16160178"
},
{
"input": "10000 2\n3270 897970\n8270 899508",
"output": "901239"
},
{
"input": "100000000 1\n9783835 100000000",
"output": "190216165"
},
{
"input": "100000000 1\n1 100000000",
"output": "199999999"
},
{
"input": "100000000 1\n100000000 100000000",
"output": "199999999"
},
{
"input": "100000000 2\n1 100000000\n100000000 100000000",
"output": "149999999"
},
{
"input": "100000000 2\n50000000 100000000\n50000001 100000000",
"output": "149999999"
},
{
"input": "99999999 20\n3 100000000\n14 100000000\n22 100000000\n24 100000000\n31 100000000\n41 100000000\n46 100000000\n84 100000000\n94 100000000\n98 100000000\n99999912 100000000\n99999915 100000000\n99999916 100000000\n99999923 100000000\n99999935 100000000\n99999937 100000000\n99999954 100000000\n99999955 100000000\n99999979 100000000\n99999982 100000000",
"output": "149999907"
},
{
"input": "8 3\n2 0\n7 3\n8 0",
"output": "IMPOSSIBLE"
},
{
"input": "10 1\n5 5",
"output": "10"
},
{
"input": "100 2\n1 100\n2 1",
"output": "IMPOSSIBLE"
},
{
"input": "1 1\n1 100",
"output": "100"
},
{
"input": "1 1\n1 10",
"output": "10"
},
{
"input": "1 1\n1 2",
"output": "2"
},
{
"input": "9 2\n1 8\n9 11",
"output": "13"
},
{
"input": "9 2\n1 1\n9 3",
"output": "6"
},
{
"input": "10000000 1\n10000000 0",
"output": "9999999"
},
{
"input": "3 2\n1 0\n3 0",
"output": "1"
},
{
"input": "100000000 2\n1 2\n2 1",
"output": "99999999"
},
{
"input": "1 1\n1 5",
"output": "5"
}
] | 1,585,122,525
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 93
| 307,200
|
n, m = map(int, input().split())
ld = -1
lh = -1
P = []
ans = 10 ** 9
for _ in range(m):
d, h = map(int, input().split())
d -= 1
if ld != -1:
if abs(d - ld) < abs(h - lh):
print('IMPOSSIBLE')
exit(0)
else:
ans = min(ans, abs(d - ld) + h + lh)
ld = d
lh = h
P.append((d, h))
print(ans // 2)
|
Title: Tourist's Notes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A tourist hiked along the mountain range. The hike lasted for *n* days, during each day the tourist noted height above the sea level. On the *i*-th day height was equal to some integer *h**i*. The tourist pick smooth enough route for his hike, meaning that the between any two consecutive days height changes by at most 1, i.e. for all *i*'s from 1 to *n*<=-<=1 the inequality |*h**i*<=-<=*h**i*<=+<=1|<=≤<=1 holds.
At the end of the route the tourist rafted down a mountain river and some notes in the journal were washed away. Moreover, the numbers in the notes could have been distorted. Now the tourist wonders what could be the maximum height during his hike. Help him restore the maximum possible value of the maximum height throughout the hike or determine that the notes were so much distorted that they do not represent any possible height values that meet limits |*h**i*<=-<=*h**i*<=+<=1|<=≤<=1.
Input Specification:
The first line contains two space-separated numbers, *n* and *m* (1<=≤<=*n*<=≤<=108, 1<=≤<=*m*<=≤<=105) — the number of days of the hike and the number of notes left in the journal.
Next *m* lines contain two space-separated integers *d**i* and *h**d**i* (1<=≤<=*d**i*<=≤<=*n*, 0<=≤<=*h**d**i*<=≤<=108) — the number of the day when the *i*-th note was made and height on the *d**i*-th day. It is guaranteed that the notes are given in the chronological order, i.e. for all *i* from 1 to *m*<=-<=1 the following condition holds: *d**i*<=<<=*d**i*<=+<=1.
Output Specification:
If the notes aren't contradictory, print a single integer — the maximum possible height value throughout the whole route.
If the notes do not correspond to any set of heights, print a single word 'IMPOSSIBLE' (without the quotes).
Demo Input:
['8 2\n2 0\n7 0\n', '8 3\n2 0\n7 0\n8 3\n']
Demo Output:
['2\n', 'IMPOSSIBLE\n']
Note:
For the first sample, an example of a correct height sequence with a maximum of 2: (0, 0, 1, 2, 1, 1, 0, 1).
In the second sample the inequality between *h*<sub class="lower-index">7</sub> and *h*<sub class="lower-index">8</sub> does not hold, thus the information is inconsistent.
|
```python
n, m = map(int, input().split())
ld = -1
lh = -1
P = []
ans = 10 ** 9
for _ in range(m):
d, h = map(int, input().split())
d -= 1
if ld != -1:
if abs(d - ld) < abs(h - lh):
print('IMPOSSIBLE')
exit(0)
else:
ans = min(ans, abs(d - ld) + h + lh)
ld = d
lh = h
P.append((d, h))
print(ans // 2)
```
| 0
|
|
68
|
B
|
Energy exchange
|
PROGRAMMING
| 1,600
|
[
"binary search"
] |
B. Energy exchange
|
2
|
256
|
It is well known that the planet suffers from the energy crisis. Little Petya doesn't like that and wants to save the world. For this purpose he needs every accumulator to contain the same amount of energy. Initially every accumulator has some amount of energy: the *i*-th accumulator has *a**i* units of energy. Energy can be transferred from one accumulator to the other. Every time *x* units of energy are transferred (*x* is not necessarily an integer) *k* percent of it is lost. That is, if *x* units were transferred from one accumulator to the other, amount of energy in the first one decreased by *x* units and in other increased by units.
Your task is to help Petya find what maximum equal amount of energy can be stored in each accumulator after the transfers.
|
First line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=10000,<=0<=≤<=*k*<=≤<=99) — number of accumulators and the percent of energy that is lost during transfers.
Next line contains *n* integers *a*1,<=*a*2,<=... ,<=*a**n* — amounts of energy in the first, second, .., *n*-th accumulator respectively (0<=≤<=*a**i*<=≤<=1000,<=1<=≤<=*i*<=≤<=*n*).
|
Output maximum possible amount of energy that can remain in each of accumulators after the transfers of energy.
The absolute or relative error in the answer should not exceed 10<=-<=6.
|
[
"3 50\n4 2 1\n",
"2 90\n1 11\n"
] |
[
"2.000000000\n",
"1.909090909\n"
] |
none
| 1,000
|
[
{
"input": "3 50\n4 2 1",
"output": "2.000000000"
},
{
"input": "2 90\n1 11",
"output": "1.909090909"
},
{
"input": "5 26\n42 65 23 43 64",
"output": "45.415178571"
},
{
"input": "5 45\n964 515 454 623 594",
"output": "594.109756098"
},
{
"input": "1 20\n784",
"output": "784.000000000"
},
{
"input": "10 20\n812 896 36 596 709 641 679 778 738 302",
"output": "597.255813953"
},
{
"input": "10 83\n689 759 779 927 15 231 976 943 604 917",
"output": "406.839285714"
},
{
"input": "11 1\n235 280 196 670 495 379 391 280 847 875 506",
"output": "467.586301370"
},
{
"input": "12 71\n472 111 924 103 975 527 807 618 400 523 607 424",
"output": "413.249554367"
},
{
"input": "13 89\n19 944 341 846 764 676 222 957 953 481 708 920 950",
"output": "361.924390244"
},
{
"input": "14 6\n256 465 759 589 242 824 638 985 506 128 809 105 301 827",
"output": "523.427098675"
},
{
"input": "100 95\n154 444 715 98 35 347 799 313 40 821 118 786 31 587 888 84 88 751 98 86 321 720 201 247 302 518 663 904 482 385 139 646 581 995 847 775 173 252 508 722 380 922 634 911 102 384 346 212 705 380 220 221 492 421 244 591 758 631 370 866 536 872 294 152 337 810 761 235 789 839 365 366 623 897 905 249 685 838 380 873 702 379 865 68 215 168 425 264 652 228 167 498 733 41 502 21 565 956 430 171",
"output": "179.075000000"
},
{
"input": "101 71\n113 551 568 26 650 547 89 668 64 651 110 515 482 401 170 971 623 672 135 106 985 751 286 255 82 588 122 568 751 867 335 488 324 122 829 256 675 471 255 723 630 802 667 665 206 774 573 499 361 202 620 522 72 220 739 868 101 135 254 519 896 227 224 968 263 826 466 377 360 24 124 874 877 513 130 79 630 786 265 150 232 783 449 914 815 557 646 367 733 576 840 683 417 709 569 432 515 702 811 877 286",
"output": "343.047284817"
},
{
"input": "102 99\n73 348 420 956 955 436 69 714 87 480 102 555 933 215 452 167 157 593 863 816 337 471 371 574 862 967 581 543 330 348 221 640 378 250 500 428 866 379 1 723 880 992 9 419 0 163 800 96 16 25 19 513 653 19 924 144 135 950 449 481 255 582 844 473 189 841 862 520 242 210 573 381 130 820 357 911 884 735 460 428 764 187 344 760 413 636 868 780 123 614 822 869 792 66 636 843 465 449 191 891 819 30",
"output": "68.702920443"
},
{
"input": "103 26\n33 455 273 884 569 636 360 69 802 310 405 594 693 339 43 53 692 514 590 835 1000 191 456 582 641 35 731 207 600 830 416 483 431 377 481 910 367 597 58 413 128 873 42 173 104 553 26 383 673 849 728 503 924 819 108 422 169 454 333 134 926 247 464 289 115 547 567 663 123 396 21 890 385 436 584 432 829 683 345 706 294 901 238 606 12 24 89 882 203 962 804 745 166 425 393 252 415 195 571 596 41 486 445",
"output": "419.922659430"
},
{
"input": "104 54\n683 252 125 813 874 835 651 424 826 139 397 323 143 153 326 941 536 435 317 854 353 222 851 591 420 415 190 872 178 311 612 635 174 505 153 81 559 815 805 414 378 62 75 929 208 942 254 670 329 671 127 494 504 618 292 699 203 959 218 788 285 602 83 104 41 562 272 806 4 582 780 87 639 743 811 263 83 632 230 984 826 304 133 142 612 413 310 985 594 309 787 930 541 92 461 663 675 942 952 610 574 633 758 999",
"output": "399.430903462"
}
] | 1,602,735,236
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 42
| 404
| 716,800
|
def isEnough(x):
receiving = 0
giving = 0
for acc in accumulators:
if acc < x:
receiving += x - acc
else:
giving += (acc - x) * (100 - k) / 100
return (receiving <= giving)
def BinarySreach(l, r):
left = l
right = r
ans = r
while left <= right:
mid = (left + right) / 2
if isEnough(mid):
ans = mid
left = mid + 10**(-10)
else:
right = mid - 10**(-10)
return ans
n, k = map(int, input().split())
accumulators = list(map(int, input().split()))
print("{:.9f}".format(BinarySreach(0, max(accumulators))))
|
Title: Energy exchange
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
It is well known that the planet suffers from the energy crisis. Little Petya doesn't like that and wants to save the world. For this purpose he needs every accumulator to contain the same amount of energy. Initially every accumulator has some amount of energy: the *i*-th accumulator has *a**i* units of energy. Energy can be transferred from one accumulator to the other. Every time *x* units of energy are transferred (*x* is not necessarily an integer) *k* percent of it is lost. That is, if *x* units were transferred from one accumulator to the other, amount of energy in the first one decreased by *x* units and in other increased by units.
Your task is to help Petya find what maximum equal amount of energy can be stored in each accumulator after the transfers.
Input Specification:
First line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=10000,<=0<=≤<=*k*<=≤<=99) — number of accumulators and the percent of energy that is lost during transfers.
Next line contains *n* integers *a*1,<=*a*2,<=... ,<=*a**n* — amounts of energy in the first, second, .., *n*-th accumulator respectively (0<=≤<=*a**i*<=≤<=1000,<=1<=≤<=*i*<=≤<=*n*).
Output Specification:
Output maximum possible amount of energy that can remain in each of accumulators after the transfers of energy.
The absolute or relative error in the answer should not exceed 10<=-<=6.
Demo Input:
['3 50\n4 2 1\n', '2 90\n1 11\n']
Demo Output:
['2.000000000\n', '1.909090909\n']
Note:
none
|
```python
def isEnough(x):
receiving = 0
giving = 0
for acc in accumulators:
if acc < x:
receiving += x - acc
else:
giving += (acc - x) * (100 - k) / 100
return (receiving <= giving)
def BinarySreach(l, r):
left = l
right = r
ans = r
while left <= right:
mid = (left + right) / 2
if isEnough(mid):
ans = mid
left = mid + 10**(-10)
else:
right = mid - 10**(-10)
return ans
n, k = map(int, input().split())
accumulators = list(map(int, input().split()))
print("{:.9f}".format(BinarySreach(0, max(accumulators))))
```
| 3.897665
|
899
|
D
|
Shovel Sale
|
PROGRAMMING
| 1,800
|
[
"constructive algorithms",
"math"
] | null | null |
There are *n* shovels in Polycarp's shop. The *i*-th shovel costs *i* burles, that is, the first shovel costs 1 burle, the second shovel costs 2 burles, the third shovel costs 3 burles, and so on. Polycarps wants to sell shovels in pairs.
Visitors are more likely to buy a pair of shovels if their total cost ends with several 9s. Because of this, Polycarp wants to choose a pair of shovels to sell in such a way that the sum of their costs ends with maximum possible number of nines. For example, if he chooses shovels with costs 12345 and 37454, their total cost is 49799, it ends with two nines.
You are to compute the number of pairs of shovels such that their total cost ends with maximum possible number of nines. Two pairs are considered different if there is a shovel presented in one pair, but not in the other.
|
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=109) — the number of shovels in Polycarp's shop.
|
Print the number of pairs of shovels such that their total cost ends with maximum possible number of nines.
Note that it is possible that the largest number of 9s at the end is 0, then you should count all such ways.
It is guaranteed that for every *n*<=≤<=109 the answer doesn't exceed 2·109.
|
[
"7\n",
"14\n",
"50\n"
] |
[
"3\n",
"9\n",
"1\n"
] |
In the first example the maximum possible number of nines at the end is one. Polycarp cah choose the following pairs of shovels for that purpose:
- 2 and 7; - 3 and 6; - 4 and 5.
In the second example the maximum number of nines at the end of total cost of two shovels is one. The following pairs of shovels suit Polycarp:
- 1 and 8; - 2 and 7; - 3 and 6; - 4 and 5; - 5 and 14; - 6 and 13; - 7 and 12; - 8 and 11; - 9 and 10.
In the third example it is necessary to choose shovels 49 and 50, because the sum of their cost is 99, that means that the total number of nines is equal to two, which is maximum possible for *n* = 50.
| 1,750
|
[
{
"input": "7",
"output": "3"
},
{
"input": "14",
"output": "9"
},
{
"input": "50",
"output": "1"
},
{
"input": "999999999",
"output": "499999999"
},
{
"input": "15",
"output": "11"
},
{
"input": "3",
"output": "3"
},
{
"input": "6500",
"output": "1501"
},
{
"input": "4",
"output": "6"
},
{
"input": "13",
"output": "8"
},
{
"input": "10",
"output": "5"
},
{
"input": "499999",
"output": "1249995"
},
{
"input": "6",
"output": "2"
},
{
"input": "8",
"output": "4"
},
{
"input": "9",
"output": "4"
},
{
"input": "11",
"output": "6"
},
{
"input": "12",
"output": "7"
},
{
"input": "5",
"output": "1"
},
{
"input": "16",
"output": "13"
},
{
"input": "17",
"output": "15"
},
{
"input": "18",
"output": "17"
},
{
"input": "19",
"output": "18"
},
{
"input": "20",
"output": "20"
},
{
"input": "21",
"output": "22"
},
{
"input": "22",
"output": "24"
},
{
"input": "23",
"output": "26"
},
{
"input": "24",
"output": "28"
},
{
"input": "25",
"output": "31"
},
{
"input": "26",
"output": "34"
},
{
"input": "27",
"output": "37"
},
{
"input": "28",
"output": "40"
},
{
"input": "29",
"output": "42"
},
{
"input": "30",
"output": "45"
},
{
"input": "31",
"output": "48"
},
{
"input": "32",
"output": "51"
},
{
"input": "33",
"output": "54"
},
{
"input": "34",
"output": "57"
},
{
"input": "35",
"output": "61"
},
{
"input": "36",
"output": "65"
},
{
"input": "37",
"output": "69"
},
{
"input": "38",
"output": "73"
},
{
"input": "39",
"output": "76"
},
{
"input": "40",
"output": "80"
},
{
"input": "41",
"output": "84"
},
{
"input": "42",
"output": "88"
},
{
"input": "43",
"output": "92"
},
{
"input": "44",
"output": "96"
},
{
"input": "45",
"output": "101"
},
{
"input": "46",
"output": "106"
},
{
"input": "47",
"output": "111"
},
{
"input": "48",
"output": "116"
},
{
"input": "49",
"output": "120"
},
{
"input": "51",
"output": "2"
},
{
"input": "100",
"output": "50"
},
{
"input": "99",
"output": "49"
},
{
"input": "101",
"output": "51"
},
{
"input": "4999",
"output": "12495"
},
{
"input": "4998",
"output": "12491"
},
{
"input": "4992",
"output": "12461"
},
{
"input": "5000",
"output": "1"
},
{
"input": "5001",
"output": "2"
},
{
"input": "10000",
"output": "5000"
},
{
"input": "10001",
"output": "5001"
},
{
"input": "49839",
"output": "124196"
},
{
"input": "4999999",
"output": "12499995"
},
{
"input": "49999999",
"output": "124999995"
},
{
"input": "499999999",
"output": "1249999995"
},
{
"input": "999",
"output": "499"
},
{
"input": "9999",
"output": "4999"
},
{
"input": "99999",
"output": "49999"
},
{
"input": "999999",
"output": "499999"
},
{
"input": "9999999",
"output": "4999999"
},
{
"input": "99999999",
"output": "49999999"
},
{
"input": "2",
"output": "1"
},
{
"input": "1000000000",
"output": "500000000"
},
{
"input": "764675465",
"output": "264675466"
},
{
"input": "499999998",
"output": "1249999991"
},
{
"input": "167959139",
"output": "135918279"
},
{
"input": "641009859",
"output": "141009860"
},
{
"input": "524125987",
"output": "24125988"
},
{
"input": "702209411",
"output": "202209412"
},
{
"input": "585325539",
"output": "85325540"
},
{
"input": "58376259",
"output": "8376260"
},
{
"input": "941492387",
"output": "441492388"
},
{
"input": "824608515",
"output": "324608516"
},
{
"input": "2691939",
"output": "3575818"
},
{
"input": "802030518",
"output": "302030519"
},
{
"input": "685146646",
"output": "185146647"
},
{
"input": "863230070",
"output": "363230071"
},
{
"input": "41313494",
"output": "85253976"
},
{
"input": "219396918",
"output": "238793836"
},
{
"input": "102513046",
"output": "52513046"
},
{
"input": "985629174",
"output": "485629175"
},
{
"input": "458679894",
"output": "1043399471"
},
{
"input": "341796022",
"output": "575388066"
},
{
"input": "519879446",
"output": "19879447"
},
{
"input": "452405440",
"output": "1012027201"
},
{
"input": "335521569",
"output": "556564707"
},
{
"input": "808572289",
"output": "308572290"
},
{
"input": "691688417",
"output": "191688418"
},
{
"input": "869771841",
"output": "369771842"
},
{
"input": "752887969",
"output": "252887970"
},
{
"input": "930971393",
"output": "430971394"
},
{
"input": "109054817",
"output": "59054817"
},
{
"input": "992170945",
"output": "492170946"
},
{
"input": "170254369",
"output": "140508739"
},
{
"input": "248004555",
"output": "296009110"
}
] | 1,592,580,015
| 4,335
|
PyPy 3
|
OK
|
TESTS
| 106
| 140
| 0
|
n = int(input())
l = 9
k = 0
while True:
p = l//2+1
if n>=p:
k+=1
l = l*10+9
else:
break
if k==0:
print((n*(n-1))//2)
else:
l = "9"*k
l = int(l)
z = l+1
cnt = 0
for i in range(9):
p = z*i+l
q = ((p//2)+1)
if n>=p:
cnt+=((p-1)//2)
elif n>=q:
x = n-(p-n)+1
cnt+=(x//2)
else:
break
print(cnt)
|
Title: Shovel Sale
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* shovels in Polycarp's shop. The *i*-th shovel costs *i* burles, that is, the first shovel costs 1 burle, the second shovel costs 2 burles, the third shovel costs 3 burles, and so on. Polycarps wants to sell shovels in pairs.
Visitors are more likely to buy a pair of shovels if their total cost ends with several 9s. Because of this, Polycarp wants to choose a pair of shovels to sell in such a way that the sum of their costs ends with maximum possible number of nines. For example, if he chooses shovels with costs 12345 and 37454, their total cost is 49799, it ends with two nines.
You are to compute the number of pairs of shovels such that their total cost ends with maximum possible number of nines. Two pairs are considered different if there is a shovel presented in one pair, but not in the other.
Input Specification:
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=109) — the number of shovels in Polycarp's shop.
Output Specification:
Print the number of pairs of shovels such that their total cost ends with maximum possible number of nines.
Note that it is possible that the largest number of 9s at the end is 0, then you should count all such ways.
It is guaranteed that for every *n*<=≤<=109 the answer doesn't exceed 2·109.
Demo Input:
['7\n', '14\n', '50\n']
Demo Output:
['3\n', '9\n', '1\n']
Note:
In the first example the maximum possible number of nines at the end is one. Polycarp cah choose the following pairs of shovels for that purpose:
- 2 and 7; - 3 and 6; - 4 and 5.
In the second example the maximum number of nines at the end of total cost of two shovels is one. The following pairs of shovels suit Polycarp:
- 1 and 8; - 2 and 7; - 3 and 6; - 4 and 5; - 5 and 14; - 6 and 13; - 7 and 12; - 8 and 11; - 9 and 10.
In the third example it is necessary to choose shovels 49 and 50, because the sum of their cost is 99, that means that the total number of nines is equal to two, which is maximum possible for *n* = 50.
|
```python
n = int(input())
l = 9
k = 0
while True:
p = l//2+1
if n>=p:
k+=1
l = l*10+9
else:
break
if k==0:
print((n*(n-1))//2)
else:
l = "9"*k
l = int(l)
z = l+1
cnt = 0
for i in range(9):
p = z*i+l
q = ((p//2)+1)
if n>=p:
cnt+=((p-1)//2)
elif n>=q:
x = n-(p-n)+1
cnt+=(x//2)
else:
break
print(cnt)
```
| 3
|
|
776
|
B
|
Sherlock and his girlfriend
|
PROGRAMMING
| 1,200
|
[
"constructive algorithms",
"number theory"
] | null | null |
Sherlock has a new girlfriend (so unlike him!). Valentine's day is coming and he wants to gift her some jewelry.
He bought *n* pieces of jewelry. The *i*-th piece has price equal to *i*<=+<=1, that is, the prices of the jewelry are 2,<=3,<=4,<=... *n*<=+<=1.
Watson gave Sherlock a challenge to color these jewelry pieces such that two pieces don't have the same color if the price of one piece is a prime divisor of the price of the other piece. Also, Watson asked him to minimize the number of different colors used.
Help Sherlock complete this trivial task.
|
The only line contains single integer *n* (1<=≤<=*n*<=≤<=100000) — the number of jewelry pieces.
|
The first line of output should contain a single integer *k*, the minimum number of colors that can be used to color the pieces of jewelry with the given constraints.
The next line should consist of *n* space-separated integers (between 1 and *k*) that specify the color of each piece in the order of increasing price.
If there are multiple ways to color the pieces using *k* colors, you can output any of them.
|
[
"3\n",
"4\n"
] |
[
"2\n1 1 2 ",
"2\n2 1 1 2\n"
] |
In the first input, the colors for first, second and third pieces of jewelry having respective prices 2, 3 and 4 are 1, 1 and 2 respectively.
In this case, as 2 is a prime divisor of 4, colors of jewelry having prices 2 and 4 must be distinct.
| 1,000
|
[
{
"input": "3",
"output": "2\n1 1 2 "
},
{
"input": "4",
"output": "2\n1 1 2 1 "
},
{
"input": "17",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 "
},
{
"input": "25",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 "
},
{
"input": "85",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 "
},
{
"input": "105",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 "
},
{
"input": "123",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 "
},
{
"input": "452",
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{
"input": "29958",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "1696",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "44907",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "13736",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "29594",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "19283",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "15346",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "41794",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "99998",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "100000",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "2",
"output": "1\n1 1 "
},
{
"input": "1",
"output": "1\n1 "
},
{
"input": "2",
"output": "1\n1 1 "
},
{
"input": "100000",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "99971",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
}
] | 1,688,635,279
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 33
| 1,000
| 512,000
|
n = int(input())
a = [2]
for i in range(3, (n+2)//2+1, 2):
for j in range(3, i, 2):
if i%j == 0:
break
else:
a.append(i)
if n==2:
print(1)
print(1, 1)
elif n==1:
print(1)
print(1)
else:
print(2)
for i in range(2, n+2):
if i in a:
print(1, end = " ")
else:
print(2, end = " ")
# Thu Jul 06 2023 12:21:18 GMT+0300 (Moscow Standard Time)
|
Title: Sherlock and his girlfriend
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sherlock has a new girlfriend (so unlike him!). Valentine's day is coming and he wants to gift her some jewelry.
He bought *n* pieces of jewelry. The *i*-th piece has price equal to *i*<=+<=1, that is, the prices of the jewelry are 2,<=3,<=4,<=... *n*<=+<=1.
Watson gave Sherlock a challenge to color these jewelry pieces such that two pieces don't have the same color if the price of one piece is a prime divisor of the price of the other piece. Also, Watson asked him to minimize the number of different colors used.
Help Sherlock complete this trivial task.
Input Specification:
The only line contains single integer *n* (1<=≤<=*n*<=≤<=100000) — the number of jewelry pieces.
Output Specification:
The first line of output should contain a single integer *k*, the minimum number of colors that can be used to color the pieces of jewelry with the given constraints.
The next line should consist of *n* space-separated integers (between 1 and *k*) that specify the color of each piece in the order of increasing price.
If there are multiple ways to color the pieces using *k* colors, you can output any of them.
Demo Input:
['3\n', '4\n']
Demo Output:
['2\n1 1 2 ', '2\n2 1 1 2\n']
Note:
In the first input, the colors for first, second and third pieces of jewelry having respective prices 2, 3 and 4 are 1, 1 and 2 respectively.
In this case, as 2 is a prime divisor of 4, colors of jewelry having prices 2 and 4 must be distinct.
|
```python
n = int(input())
a = [2]
for i in range(3, (n+2)//2+1, 2):
for j in range(3, i, 2):
if i%j == 0:
break
else:
a.append(i)
if n==2:
print(1)
print(1, 1)
elif n==1:
print(1)
print(1)
else:
print(2)
for i in range(2, n+2):
if i in a:
print(1, end = " ")
else:
print(2, end = " ")
# Thu Jul 06 2023 12:21:18 GMT+0300 (Moscow Standard Time)
```
| 0
|
|
340
|
A
|
The Wall
|
PROGRAMMING
| 1,200
|
[
"math"
] | null | null |
Iahub and his friend Floyd have started painting a wall. Iahub is painting the wall red and Floyd is painting it pink. You can consider the wall being made of a very large number of bricks, numbered 1, 2, 3 and so on.
Iahub has the following scheme of painting: he skips *x*<=-<=1 consecutive bricks, then he paints the *x*-th one. That is, he'll paint bricks *x*, 2·*x*, 3·*x* and so on red. Similarly, Floyd skips *y*<=-<=1 consecutive bricks, then he paints the *y*-th one. Hence he'll paint bricks *y*, 2·*y*, 3·*y* and so on pink.
After painting the wall all day, the boys observed that some bricks are painted both red and pink. Iahub has a lucky number *a* and Floyd has a lucky number *b*. Boys wonder how many bricks numbered no less than *a* and no greater than *b* are painted both red and pink. This is exactly your task: compute and print the answer to the question.
|
The input will have a single line containing four integers in this order: *x*, *y*, *a*, *b*. (1<=≤<=*x*,<=*y*<=≤<=1000, 1<=≤<=*a*,<=*b*<=≤<=2·109, *a*<=≤<=*b*).
|
Output a single integer — the number of bricks numbered no less than *a* and no greater than *b* that are painted both red and pink.
|
[
"2 3 6 18\n"
] |
[
"3"
] |
Let's look at the bricks from *a* to *b* (*a* = 6, *b* = 18). The bricks colored in red are numbered 6, 8, 10, 12, 14, 16, 18. The bricks colored in pink are numbered 6, 9, 12, 15, 18. The bricks colored in both red and pink are numbered with 6, 12 and 18.
| 500
|
[
{
"input": "2 3 6 18",
"output": "3"
},
{
"input": "4 6 20 201",
"output": "15"
},
{
"input": "15 27 100 10000",
"output": "74"
},
{
"input": "105 60 3456 78910",
"output": "179"
},
{
"input": "1 1 1000 100000",
"output": "99001"
},
{
"input": "3 2 5 5",
"output": "0"
},
{
"input": "555 777 1 1000000",
"output": "257"
},
{
"input": "1000 1000 1 32323",
"output": "32"
},
{
"input": "45 125 93451125 100000000",
"output": "5821"
},
{
"input": "101 171 1 1000000000",
"output": "57900"
},
{
"input": "165 255 69696 1000000000",
"output": "356482"
},
{
"input": "555 777 666013 1000000000",
"output": "257229"
},
{
"input": "23 46 123321 900000000",
"output": "19562537"
},
{
"input": "321 123 15 1000000",
"output": "75"
},
{
"input": "819 1000 9532 152901000",
"output": "186"
},
{
"input": "819 1000 10000 1000000",
"output": "1"
},
{
"input": "1 1 1 1",
"output": "1"
},
{
"input": "1 2 2 1000003",
"output": "500001"
},
{
"input": "1 1 1 1000000000",
"output": "1000000000"
},
{
"input": "10 15 69 195610342",
"output": "6520342"
},
{
"input": "2 1 1 1000000000",
"output": "500000000"
},
{
"input": "1000 1000 1 20",
"output": "0"
},
{
"input": "1 1 1 2000000000",
"output": "2000000000"
},
{
"input": "1 2 1 2000000000",
"output": "1000000000"
},
{
"input": "2 1 1 2000000000",
"output": "1000000000"
},
{
"input": "2 3 1 1000000000",
"output": "166666666"
},
{
"input": "2 3 1 2000000000",
"output": "333333333"
},
{
"input": "3 7 1 1000000000",
"output": "47619047"
},
{
"input": "1 1 1000000000 2000000000",
"output": "1000000001"
},
{
"input": "2 2 1 2000000000",
"output": "1000000000"
},
{
"input": "1 1 2 2000000000",
"output": "1999999999"
},
{
"input": "3 2 1 2000000000",
"output": "333333333"
},
{
"input": "1 1 2000000000 2000000000",
"output": "1"
},
{
"input": "2 3 7 7",
"output": "0"
},
{
"input": "3 3 3 7",
"output": "2"
}
] | 1,378,556,902
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 92
| 0
|
import sys
f = sys.stdin
# f = open("input.txt", "r")
x, y, a, b = map(int, f.readline().strip().split())
res = 0
for i in range(min(x,y), b+1, min(x,y)):
if i%x == i%y:
res += 1
print(res)
|
Title: The Wall
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Iahub and his friend Floyd have started painting a wall. Iahub is painting the wall red and Floyd is painting it pink. You can consider the wall being made of a very large number of bricks, numbered 1, 2, 3 and so on.
Iahub has the following scheme of painting: he skips *x*<=-<=1 consecutive bricks, then he paints the *x*-th one. That is, he'll paint bricks *x*, 2·*x*, 3·*x* and so on red. Similarly, Floyd skips *y*<=-<=1 consecutive bricks, then he paints the *y*-th one. Hence he'll paint bricks *y*, 2·*y*, 3·*y* and so on pink.
After painting the wall all day, the boys observed that some bricks are painted both red and pink. Iahub has a lucky number *a* and Floyd has a lucky number *b*. Boys wonder how many bricks numbered no less than *a* and no greater than *b* are painted both red and pink. This is exactly your task: compute and print the answer to the question.
Input Specification:
The input will have a single line containing four integers in this order: *x*, *y*, *a*, *b*. (1<=≤<=*x*,<=*y*<=≤<=1000, 1<=≤<=*a*,<=*b*<=≤<=2·109, *a*<=≤<=*b*).
Output Specification:
Output a single integer — the number of bricks numbered no less than *a* and no greater than *b* that are painted both red and pink.
Demo Input:
['2 3 6 18\n']
Demo Output:
['3']
Note:
Let's look at the bricks from *a* to *b* (*a* = 6, *b* = 18). The bricks colored in red are numbered 6, 8, 10, 12, 14, 16, 18. The bricks colored in pink are numbered 6, 9, 12, 15, 18. The bricks colored in both red and pink are numbered with 6, 12 and 18.
|
```python
import sys
f = sys.stdin
# f = open("input.txt", "r")
x, y, a, b = map(int, f.readline().strip().split())
res = 0
for i in range(min(x,y), b+1, min(x,y)):
if i%x == i%y:
res += 1
print(res)
```
| 0
|
|
1,009
|
B
|
Minimum Ternary String
|
PROGRAMMING
| 1,400
|
[
"greedy",
"implementation"
] | null | null |
You are given a ternary string (it is a string which consists only of characters '0', '1' and '2').
You can swap any two adjacent (consecutive) characters '0' and '1' (i.e. replace "01" with "10" or vice versa) or any two adjacent (consecutive) characters '1' and '2' (i.e. replace "12" with "21" or vice versa).
For example, for string "010210" we can perform the following moves:
- "010210" $\rightarrow$ "100210"; - "010210" $\rightarrow$ "001210"; - "010210" $\rightarrow$ "010120"; - "010210" $\rightarrow$ "010201".
Note than you cannot swap "02" $\rightarrow$ "20" and vice versa. You cannot perform any other operations with the given string excluding described above.
You task is to obtain the minimum possible (lexicographically) string by using these swaps arbitrary number of times (possibly, zero).
String $a$ is lexicographically less than string $b$ (if strings $a$ and $b$ have the same length) if there exists some position $i$ ($1 \le i \le |a|$, where $|s|$ is the length of the string $s$) such that for every $j < i$ holds $a_j = b_j$, and $a_i < b_i$.
|
The first line of the input contains the string $s$ consisting only of characters '0', '1' and '2', its length is between $1$ and $10^5$ (inclusive).
|
Print a single string — the minimum possible (lexicographically) string you can obtain by using the swaps described above arbitrary number of times (possibly, zero).
|
[
"100210\n",
"11222121\n",
"20\n"
] |
[
"001120\n",
"11112222\n",
"20\n"
] |
none
| 0
|
[
{
"input": "100210",
"output": "001120"
},
{
"input": "11222121",
"output": "11112222"
},
{
"input": "20",
"output": "20"
},
{
"input": "1002",
"output": "0012"
},
{
"input": "10",
"output": "01"
},
{
"input": "000021",
"output": "000012"
},
{
"input": "021",
"output": "012"
},
{
"input": "2",
"output": "2"
},
{
"input": "201",
"output": "120"
},
{
"input": "2112120",
"output": "1112220"
},
{
"input": "102",
"output": "012"
},
{
"input": "202",
"output": "202"
},
{
"input": "220201",
"output": "122020"
},
{
"input": "12",
"output": "12"
},
{
"input": "100022202",
"output": "000122202"
},
{
"input": "01",
"output": "01"
},
{
"input": "1",
"output": "1"
}
] | 1,531,780,555
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 93
| 0
|
word = input()
count_ones = 0
cur_word = ''
for c in word:
if c==1:
count_ones += 1
if c != 1:
cur_word = cur_word + c
print(cur_word)
|
Title: Minimum Ternary String
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a ternary string (it is a string which consists only of characters '0', '1' and '2').
You can swap any two adjacent (consecutive) characters '0' and '1' (i.e. replace "01" with "10" or vice versa) or any two adjacent (consecutive) characters '1' and '2' (i.e. replace "12" with "21" or vice versa).
For example, for string "010210" we can perform the following moves:
- "010210" $\rightarrow$ "100210"; - "010210" $\rightarrow$ "001210"; - "010210" $\rightarrow$ "010120"; - "010210" $\rightarrow$ "010201".
Note than you cannot swap "02" $\rightarrow$ "20" and vice versa. You cannot perform any other operations with the given string excluding described above.
You task is to obtain the minimum possible (lexicographically) string by using these swaps arbitrary number of times (possibly, zero).
String $a$ is lexicographically less than string $b$ (if strings $a$ and $b$ have the same length) if there exists some position $i$ ($1 \le i \le |a|$, where $|s|$ is the length of the string $s$) such that for every $j < i$ holds $a_j = b_j$, and $a_i < b_i$.
Input Specification:
The first line of the input contains the string $s$ consisting only of characters '0', '1' and '2', its length is between $1$ and $10^5$ (inclusive).
Output Specification:
Print a single string — the minimum possible (lexicographically) string you can obtain by using the swaps described above arbitrary number of times (possibly, zero).
Demo Input:
['100210\n', '11222121\n', '20\n']
Demo Output:
['001120\n', '11112222\n', '20\n']
Note:
none
|
```python
word = input()
count_ones = 0
cur_word = ''
for c in word:
if c==1:
count_ones += 1
if c != 1:
cur_word = cur_word + c
print(cur_word)
```
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Subsets and Splits
Successful Python Submissions
Retrieves all records from the train dataset where the verdict is 'OK', providing basic filtering but limited analytical value.
SQL Console for MatrixStudio/Codeforces-Python-Submissions
Retrieves records of users with a rating of 1600 or higher and a verdict of 'OK', providing basic filtering but limited analytical value.
SQL Console for MatrixStudio/Codeforces-Python-Submissions
Counts the number of entries with a rating above 2000 and a verdict of 'OK', providing basic filtering but limited analytical value.
SQL Console for MatrixStudio/Codeforces-Python-Submissions
Counts the number of entries with a 'OK' verdict, providing a basic overview of a specific category within the dataset.