MatrixStudio/Codeforces-Python-Submissions

142

C

Help Caretaker

PROGRAMMING

2,300

[ "brute force", "dp" ]

null

Autumn came late to the kingdom of Far Far Away. The harvest was exuberant and it is now time to get ready for the winter. As most people celebrate the Harvest festival, Simon the Caretaker tries to solve a very non-trivial task of how to find place for the agricultural equipment in the warehouse. He's got problems with some particularly large piece of equipment, which is, of course, turboplows. The problem is that when a turboplow is stored, it takes up not some simply rectangular space. It takes up a T-shaped space like on one of the four pictures below (here character "#" stands for the space occupied by the turboplow and character "." stands for the free space): Simon faced a quite natural challenge: placing in the given *n*<=×<=*m* cells warehouse the maximum number of turboplows. As one stores the turboplows, he can rotate them in any manner (so that they take up the space like on one of the four pictures above). However, two turboplows cannot "overlap", that is, they cannot share the same cell in the warehouse. Simon feels that he alone cannot find the optimal way of positioning the plugs in the warehouse that would maximize their quantity. Can you help him?

The only line contains two space-separated integers *n* and *m* — the sizes of the warehouse (1<=≤<=*n*,<=*m*<=≤<=9).

In the first line print the maximum number of turboplows that can be positioned in the warehouse. In each of the next *n* lines print *m* characters. Use "." (dot) to mark empty space and use successive capital Latin letters ("A" for the first turboplow, "B" for the second one and so on until you reach the number of turboplows in your scheme) to mark place for the corresponding turboplows considering that they are positioned in the optimal manner in the warehouse. The order in which you number places for the turboplows does not matter. If there are several optimal solutions for a warehouse of the given size, print any of them.

[ "3 3\n", "5 6\n", "2 2\n" ]

[ "1\nAAA\n.A.\n.A.\n", "4\nA..C..\nAAAC..\nABCCCD\n.B.DDD\nBBB..D\n", "0\n..\n..\n" ]

none

1,500

[ { "input": "3 3", "output": "1\nAAA\n.A.\n.A." }, { "input": "5 6", "output": "4\nA..C..\nAAAC..\nABCCCD\n.B.DDD\nBBB..D" }, { "input": "2 2", "output": "0\n..\n.." }, { "input": "4 2", "output": "0\n..\n..\n..\n.." }, { "input": "3 4", "output": "1\nA...\nAAA.\nA..." }, { "input": "5 3", "output": "2\nAAA\n.A.\n.AB\nBBB\n..B" }, { "input": "4 4", "output": "2\nAAA.\n.AB.\n.AB.\n.BBB" }, { "input": "3 6", "output": "2\nA..B..\nAAAB..\nA.BBB." }, { "input": "5 4", "output": "2\nAAA.\n.AB.\n.AB.\n.BBB\n...." }, { "input": "5 5", "output": "4\nAAA.B\n.ABBB\nCA.DB\nCCCD.\nC.DDD" }, { "input": "1 1", "output": "0\n." }, { "input": "1 2", "output": "0\n.." }, { "input": "2 1", "output": "0\n.\n." }, { "input": "1 3", "output": "0\n..." }, { "input": "3 1", "output": "0\n.\n.\n." }, { "input": "1 4", "output": "0\n...." }, { "input": "4 1", "output": "0\n.\n.\n.\n." }, { "input": "1 5", "output": "0\n....." }, { "input": "5 1", "output": "0\n.\n.\n.\n.\n." }, { "input": "1 6", "output": "0\n......" }, { "input": "2 3", "output": "0\n...\n..." }, { "input": "3 2", "output": "0\n..\n..\n.." }, { "input": "6 1", "output": "0\n.\n.\n.\n.\n.\n." }, { "input": "1 7", "output": "0\n......." }, { "input": "7 1", "output": "0\n.\n.\n.\n.\n.\n.\n." }, { "input": "1 8", "output": "0\n........" }, { "input": "2 4", "output": "0\n....\n...." }, { "input": "8 1", "output": "0\n.\n.\n.\n.\n.\n.\n.\n." }, { "input": "1 9", "output": "0\n........." }, { "input": "9 1", "output": "0\n.\n.\n.\n.\n.\n.\n.\n.\n." }, { "input": "2 5", "output": "0\n.....\n....." }, { "input": "5 2", "output": "0\n..\n..\n..\n..\n.." }, { "input": "2 6", "output": "0\n......\n......" }, { "input": "4 3", "output": "1\nAAA\n.A.\n.A.\n..." }, { "input": "6 2", "output": "0\n..\n..\n..\n..\n..\n.." }, { "input": "2 7", "output": "0\n.......\n......." }, { "input": "7 2", "output": "0\n..\n..\n..\n..\n..\n..\n.." }, { "input": "3 5", "output": "2\nA..B.\nAAAB.\nA.BBB" }, { "input": "2 8", "output": "0\n........\n........" }, { "input": "8 2", "output": "0\n..\n..\n..\n..\n..\n..\n..\n.." }, { "input": "2 9", "output": "0\n.........\n........." }, { "input": "6 3", "output": "2\nAAA\n.A.\n.AB\nBBB\n..B\n..." }, { "input": "9 2", "output": "0\n..\n..\n..\n..\n..\n..\n..\n..\n.." }, { "input": "4 5", "output": "2\nA....\nAAAB.\nABBB.\n...B." }, { "input": "3 7", "output": "3\nA..B..C\nAAABCCC\nA.BBB.C" }, { "input": "7 3", "output": "3\nAAA\n.A.\n.AB\nBBB\n.CB\n.C.\nCCC" }, { "input": "3 8", "output": "3\nA..B..C.\nAAABCCC.\nA.BBB.C." }, { "input": "4 6", "output": "3\nA..CCC\nAAABC.\nABBBC.\n...B.." }, { "input": "6 4", "output": "3\nAAA.\n.AB.\n.AB.\nCBBB\nCCC.\nC..." }, { "input": "8 3", "output": "3\nAAA\n.A.\n.AB\nBBB\n.CB\n.C.\nCCC\n..." }, { "input": "3 9", "output": "4\nA..BCCCD.\nAAAB.C.D.\nA.BBBCDDD" }, { "input": "9 3", "output": "4\nAAA\n.A.\n.AB\nBBB\nC.B\nCCC\nC.D\nDDD\n..D" }, { "input": "4 7", "output": "4\nA..CCC.\nAAABCD.\nABBBCD.\n...BDDD" }, { "input": "7 4", "output": "4\nAAA.\n.AB.\n.AB.\nCBBB\nCCCD\nCDDD\n...D" }, { "input": "6 5", "output": "4\nAAA.B\n.ABBB\n.AC.B\nCCCD.\n..CD.\n..DDD" }, { "input": "4 8", "output": "4\nA..CCC..\nAAABCD..\nABBBCD..\n...BDDD." }, { "input": "8 4", "output": "4\nAAA.\n.AB.\n.AB.\nCBBB\nCCCD\nCDDD\n...D\n...." }, { "input": "5 7", "output": "5\nA..C..E\nAAACEEE\nABCCCDE\n.B.DDD.\nBBB..D." }, { "input": "7 5", "output": "5\nAAA.B\n.ABBB\n.AC.B\nCCCD.\n.ECD.\n.EDDD\nEEE.." }, { "input": "4 9", "output": "5\nA..CCCE..\nAAABCDEEE\nABBBCDE..\n...BDDD.." }, { "input": "6 6", "output": "5\nAAA.B.\n.ABBB.\n.A.CB.\nDCCCE.\nDDDCE.\nD..EEE" }, { "input": "9 4", "output": "5\nAAA.\n.AB.\n.AB.\nCBBB\nCCCD\nCDDD\nEEED\n.E..\n.E.." }, { "input": "5 8", "output": "6\nA.CCC.E.\nAAACEEE.\nAB.CDFE.\n.BDDDFFF\nBBB.DF.." }, { "input": "8 5", "output": "6\nAAA.B\n.ABBB\nCA.DB\nCCCD.\nCEDDD\n.EFFF\nEEEF.\n...F." }, { "input": "6 7", "output": "6\nA..C..E\nAAACEEE\nABCCCFE\n.B.D.F.\nBBBDFFF\n..DDD.." }, { "input": "7 6", "output": "6\nAAA.B.\n.ABBB.\n.AC.BD\nCCCDDD\n.EC.FD\n.EFFF.\nEEE.F." }, { "input": "5 9", "output": "7\nA.CCC.GGG\nAAACEEEG.\nAB.CDEFG.\n.BDDDEF..\nBBB.DFFF." }, { "input": "9 5", "output": "7\nAAA.B\n.ABBB\nCA.DB\nCCCD.\nCEDDD\n.EEEF\nGEFFF\nGGG.F\nG...." }, { "input": "6 8", "output": "7\nA..C.FFF\nAAAC.EF.\nABCCCEF.\n.B.DEEEG\nBBBD.GGG\n..DDD..G" }, { "input": "8 6", "output": "7\nAAA.B.\n.ABBB.\n.AC.BD\nCCCDDD\n..CE.D\nFEEEG.\nFFFEG.\nF..GGG" }, { "input": "7 7", "output": "8\nAAABCCC\n.A.B.C.\nDABBBCE\nDDDFEEE\nDG.F.HE\n.GFFFH.\nGGG.HHH" }, { "input": "6 9", "output": "8\nA..CEEE.G\nAAAC.EGGG\nABCCCEF.G\n.B.DFFFH.\nBBBD..FH.\n..DDD.HHH" }, { "input": "9 6", "output": "8\nAAA.B.\n.ABBB.\n.AC.BD\nCCCDDD\nE.CF.D\nEEEF..\nEGFFFH\n.G.HHH\nGGG..H" }, { "input": "7 8", "output": "9\nA.DDD..H\nAAADFHHH\nA.BDFFFH\nBBBEFIII\n.CBE.GI.\n.CEEEGI.\nCCC.GGG." }, { "input": "8 7", "output": "9\nAAAB..C\n.A.BCCC\nDABBBEC\nDDDEEE.\nDFFF.EG\n.HFIGGG\n.HFIIIG\nHHHI..." }, { "input": "7 9", "output": "10\nA..DFFF.H\nAAAD.FHHH\nABDDDFI.H\n.B.EEEIII\nBBBCEGI.J\n.CCCEGJJJ\n...CGGG.J" }, { "input": "9 7", "output": "10\nAAA.B..\n.ABBBC.\n.AD.BC.\nDDDECCC\nF.DEEEG\nFFFEGGG\nFHIIIJG\n.H.I.J.\nHHHIJJJ" }, { "input": "8 8", "output": "10\nAAA.BCCC\n.ABBBDC.\nEA.FBDC.\nEEEFDDDG\nEHFFFGGG\n.HIII.JG\nHHHIJJJ.\n...I..J." }, { "input": "8 9", "output": "12\nA.EEE.JJJ\nAAAEHHHJ.\nAB.EFHKJ.\n.BFFFHKKK\nBBBDFIK..\nCDDDGIIIL\nCCCDGILLL\nC..GGG..L" }, { "input": "9 8", "output": "12\nAAA.BCCC\n.ABBBDC.\nEA.FBDC.\nEEEFDDDG\nEHFFFGGG\n.HHHIIIG\nJHKKKIL.\nJJJK.IL.\nJ..K.LLL" }, { "input": "9 9", "output": "13\nAAA.BCCC.\n.ABBB.CD.\n.AE.BFCD.\nEEEFFFDDD\nG.E.HFIII\nGGGJHHHI.\nGK.JHL.IM\n.KJJJLMMM\nKKK.LLL.M" } ]

1,535,320,564

2,147,483,647

Python 3

OK

TESTS

81

248

102,400

n, m = map(int, input().split()) swapped = False if n < m: n, m = m, n swapped = True ans = '' if n == 1 and m == 1: ans = '''0 .''' if n == 2 and m == 1: ans = '''0 . .''' if n == 2 and m == 2: ans = '''0 .. ..''' if n == 3 and m == 1: ans = '''0 . . .''' if n == 3 and m == 2: ans = '''0 .. .. ..''' if n == 3 and m == 3: ans = '''1 .A. .A. AAA''' if n == 4 and m == 1: ans = '''0 . . . .''' if n == 4 and m == 2: ans = '''0 .. .. .. ..''' if n == 4 and m == 3: ans = '''1 .A. .A. AAA ...''' if n == 4 and m == 4: ans = '''2 ...A BAAA BBBA B...''' if n == 5 and m == 1: ans = '''0 . . . . .''' if n == 5 and m == 2: ans = '''0 .. .. .. .. ..''' if n == 5 and m == 3: ans = '''2 ..A AAA .BA .B. BBB''' if n == 5 and m == 4: ans = '''2 .A.. .A.. AAAB .BBB ...B''' if n == 5 and m == 5: ans = '''4 BBB.A .BAAA DB.CA DDDC. D.CCC''' if n == 6 and m == 1: ans = '''0 . . . . . .''' if n == 6 and m == 2: ans = '''0 .. .. .. .. .. ..''' if n == 6 and m == 3: ans = '''2 .A. .A. AAA .B. .B. BBB''' if n == 6 and m == 4: ans = '''3 .A.. .A.. AAAB CBBB CCCB C...''' if n == 6 and m == 5: ans = '''4 .A..B .ABBB AAACB .D.C. .DCCC DDD..''' if n == 6 and m == 6: ans = '''5 .A..B. .ABBB. AAACB. ECCCD. EEECD. E..DDD''' if n == 7 and m == 1: ans = '''0 . . . . . . .''' if n == 7 and m == 2: ans = '''0 .. .. .. .. .. .. ..''' if n == 7 and m == 3: ans = '''3 ..A AAA B.A BBB BC. .C. CCC''' if n == 7 and m == 4: ans = '''4 ...A BAAA BBBA B..C DCCC DDDC D...''' if n == 7 and m == 5: ans = '''5 .A..B .ABBB AAACB .CCC. .D.CE .DEEE DDD.E''' if n == 7 and m == 6: ans = '''6 .A..B. .ABBB. AAACB. EEECCC .EDC.F .EDFFF .DDD.F''' if n == 7 and m == 7: ans = '''8 B..ACCC BBBA.C. BDAAACE .DDDEEE GDHHHFE GGGH.F. G..HFFF''' if n == 8 and m == 1: ans = '''0 . . . . . . . .''' if n == 8 and m == 2: ans = '''0 .. .. .. .. .. .. .. ..''' if n == 8 and m == 3: ans = '''3 .A. .A. AAA ..B BBB .CB .C. CCC''' if n == 8 and m == 4: ans = '''4 .A.. .A.. AAAB CBBB CCCB CD.. .D.. DDD.''' if n == 8 and m == 5: ans = '''6 .A..B .ABBB AAACB DDDC. .DCCC FD.E. FFFE. F.EEE''' if n == 8 and m == 6: ans = '''7 .A..B. .ABBB. AAA.BC DDDCCC .DGGGC .DFGE. FFFGE. ..FEEE''' if n == 8 and m == 7: ans = '''9 B..ACCC BBBA.C. BDAAACE .DDDEEE FD.IIIE FFFHIG. FHHHIG. ...HGGG''' if n == 9 and m == 8: ans = '''12 .A..BDDD .ABBBCD. AAAEBCD. FEEECCCG FFFEHGGG FKKKHHHG .IKLH.J. .IKLLLJ. IIIL.JJJ''' if n == 8 and m == 8: ans = '''10 .A..BCCC .A..B.C. AAABBBCD E.GGGDDD EEEGJJJD EF.GIJH. .FIIIJH. FFF.IHHH''' if n == 9 and m == 9: ans = '''13 AAA.BCCC. .ABBB.CD. .AE.BFCD. EEEFFFDDD G.E.HFIII GGGJHHHI. GK.JHL.IM .KJJJLMMM KKK.LLL.M''' if n == 9 and m == 1: ans = '''0 . . . . . . . . .''' if n == 9 and m == 2: ans = '''0 .. .. .. .. .. .. .. .. ..''' if n == 9 and m == 3: ans = '''4 ..A AAA B.A BBB B.C CCC .DC .D. DDD''' if n == 9 and m == 4: ans = '''5 .A.. .A.. AAAB CBBB CCCB C..D EDDD EEED E...''' if n == 9 and m == 5: ans = '''7 .A..B .ABBB AAACB DCCC. DDDCF DEFFF .E.GF EEEG. ..GGG''' if n == 9 and m == 6: ans = '''8 .A..B. .ABBB. AAACB. ECCCD. EEECD. E.GDDD .FGGGH .FGHHH FFF..H''' if n == 9 and m == 7: ans = '''10 .ACCC.B .A.CBBB AAACD.B ...EDDD FEEED.G FFFEGGG FHJJJIG .H.J.I. HHHJIII''' if swapped: cnt = ans.split('\n')[0] lst = ans.split('\n')[1:] ans = '' for j in range(m): for i in range(n): ans = ans + lst[i][j] ans += '\n' ans = cnt + '\n' + ans print(ans) # Made By Mostafa_Khaled

Title: Help Caretaker Time Limit: None seconds Memory Limit: None megabytes Problem Description: Autumn came late to the kingdom of Far Far Away. The harvest was exuberant and it is now time to get ready for the winter. As most people celebrate the Harvest festival, Simon the Caretaker tries to solve a very non-trivial task of how to find place for the agricultural equipment in the warehouse. He's got problems with some particularly large piece of equipment, which is, of course, turboplows. The problem is that when a turboplow is stored, it takes up not some simply rectangular space. It takes up a T-shaped space like on one of the four pictures below (here character "#" stands for the space occupied by the turboplow and character "." stands for the free space): Simon faced a quite natural challenge: placing in the given *n*<=×<=*m* cells warehouse the maximum number of turboplows. As one stores the turboplows, he can rotate them in any manner (so that they take up the space like on one of the four pictures above). However, two turboplows cannot "overlap", that is, they cannot share the same cell in the warehouse. Simon feels that he alone cannot find the optimal way of positioning the plugs in the warehouse that would maximize their quantity. Can you help him? Input Specification: The only line contains two space-separated integers *n* and *m* — the sizes of the warehouse (1<=≤<=*n*,<=*m*<=≤<=9). Output Specification: In the first line print the maximum number of turboplows that can be positioned in the warehouse. In each of the next *n* lines print *m* characters. Use "." (dot) to mark empty space and use successive capital Latin letters ("A" for the first turboplow, "B" for the second one and so on until you reach the number of turboplows in your scheme) to mark place for the corresponding turboplows considering that they are positioned in the optimal manner in the warehouse. The order in which you number places for the turboplows does not matter. If there are several optimal solutions for a warehouse of the given size, print any of them. Demo Input: ['3 3\n', '5 6\n', '2 2\n'] Demo Output: ['1\nAAA\n.A.\n.A.\n', '4\nA..C..\nAAAC..\nABCCCD\n.B.DDD\nBBB..D\n', '0\n..\n..\n'] Note: none

```python n, m = map(int, input().split()) swapped = False if n < m: n, m = m, n swapped = True ans = '' if n == 1 and m == 1: ans = '''0 .''' if n == 2 and m == 1: ans = '''0 . .''' if n == 2 and m == 2: ans = '''0 .. ..''' if n == 3 and m == 1: ans = '''0 . . .''' if n == 3 and m == 2: ans = '''0 .. .. ..''' if n == 3 and m == 3: ans = '''1 .A. .A. AAA''' if n == 4 and m == 1: ans = '''0 . . . .''' if n == 4 and m == 2: ans = '''0 .. .. .. ..''' if n == 4 and m == 3: ans = '''1 .A. .A. AAA ...''' if n == 4 and m == 4: ans = '''2 ...A BAAA BBBA B...''' if n == 5 and m == 1: ans = '''0 . . . . .''' if n == 5 and m == 2: ans = '''0 .. .. .. .. ..''' if n == 5 and m == 3: ans = '''2 ..A AAA .BA .B. BBB''' if n == 5 and m == 4: ans = '''2 .A.. .A.. AAAB .BBB ...B''' if n == 5 and m == 5: ans = '''4 BBB.A .BAAA DB.CA DDDC. D.CCC''' if n == 6 and m == 1: ans = '''0 . . . . . .''' if n == 6 and m == 2: ans = '''0 .. .. .. .. .. ..''' if n == 6 and m == 3: ans = '''2 .A. .A. AAA .B. .B. BBB''' if n == 6 and m == 4: ans = '''3 .A.. .A.. AAAB CBBB CCCB C...''' if n == 6 and m == 5: ans = '''4 .A..B .ABBB AAACB .D.C. .DCCC DDD..''' if n == 6 and m == 6: ans = '''5 .A..B. .ABBB. AAACB. ECCCD. EEECD. E..DDD''' if n == 7 and m == 1: ans = '''0 . . . . . . .''' if n == 7 and m == 2: ans = '''0 .. .. .. .. .. .. ..''' if n == 7 and m == 3: ans = '''3 ..A AAA B.A BBB BC. .C. CCC''' if n == 7 and m == 4: ans = '''4 ...A BAAA BBBA B..C DCCC DDDC D...''' if n == 7 and m == 5: ans = '''5 .A..B .ABBB AAACB .CCC. .D.CE .DEEE DDD.E''' if n == 7 and m == 6: ans = '''6 .A..B. .ABBB. AAACB. EEECCC .EDC.F .EDFFF .DDD.F''' if n == 7 and m == 7: ans = '''8 B..ACCC BBBA.C. BDAAACE .DDDEEE GDHHHFE GGGH.F. G..HFFF''' if n == 8 and m == 1: ans = '''0 . . . . . . . .''' if n == 8 and m == 2: ans = '''0 .. .. .. .. .. .. .. ..''' if n == 8 and m == 3: ans = '''3 .A. .A. AAA ..B BBB .CB .C. CCC''' if n == 8 and m == 4: ans = '''4 .A.. .A.. AAAB CBBB CCCB CD.. .D.. DDD.''' if n == 8 and m == 5: ans = '''6 .A..B .ABBB AAACB DDDC. .DCCC FD.E. FFFE. F.EEE''' if n == 8 and m == 6: ans = '''7 .A..B. .ABBB. AAA.BC DDDCCC .DGGGC .DFGE. FFFGE. ..FEEE''' if n == 8 and m == 7: ans = '''9 B..ACCC BBBA.C. BDAAACE .DDDEEE FD.IIIE FFFHIG. FHHHIG. ...HGGG''' if n == 9 and m == 8: ans = '''12 .A..BDDD .ABBBCD. AAAEBCD. FEEECCCG FFFEHGGG FKKKHHHG .IKLH.J. .IKLLLJ. IIIL.JJJ''' if n == 8 and m == 8: ans = '''10 .A..BCCC .A..B.C. AAABBBCD E.GGGDDD EEEGJJJD EF.GIJH. .FIIIJH. FFF.IHHH''' if n == 9 and m == 9: ans = '''13 AAA.BCCC. .ABBB.CD. .AE.BFCD. EEEFFFDDD G.E.HFIII GGGJHHHI. GK.JHL.IM .KJJJLMMM KKK.LLL.M''' if n == 9 and m == 1: ans = '''0 . . . . . . . . .''' if n == 9 and m == 2: ans = '''0 .. .. .. .. .. .. .. .. ..''' if n == 9 and m == 3: ans = '''4 ..A AAA B.A BBB B.C CCC .DC .D. DDD''' if n == 9 and m == 4: ans = '''5 .A.. .A.. AAAB CBBB CCCB C..D EDDD EEED E...''' if n == 9 and m == 5: ans = '''7 .A..B .ABBB AAACB DCCC. DDDCF DEFFF .E.GF EEEG. ..GGG''' if n == 9 and m == 6: ans = '''8 .A..B. .ABBB. AAACB. ECCCD. EEECD. E.GDDD .FGGGH .FGHHH FFF..H''' if n == 9 and m == 7: ans = '''10 .ACCC.B .A.CBBB AAACD.B ...EDDD FEEED.G FFFEGGG FHJJJIG .H.J.I. HHHJIII''' if swapped: cnt = ans.split('\n')[0] lst = ans.split('\n')[1:] ans = '' for j in range(m): for i in range(n): ans = ans + lst[i][j] ans += '\n' ans = cnt + '\n' + ans print(ans) # Made By Mostafa_Khaled ```

3

287

A

IQ Test

PROGRAMMING

1,100

[ "brute force", "implementation" ]

null

In the city of Ultima Thule job applicants are often offered an IQ test. The test is as follows: the person gets a piece of squared paper with a 4<=×<=4 square painted on it. Some of the square's cells are painted black and others are painted white. Your task is to repaint at most one cell the other color so that the picture has a 2<=×<=2 square, completely consisting of cells of the same color. If the initial picture already has such a square, the person should just say so and the test will be completed. Your task is to write a program that determines whether it is possible to pass the test. You cannot pass the test if either repainting any cell or no action doesn't result in a 2<=×<=2 square, consisting of cells of the same color.

Four lines contain four characters each: the *j*-th character of the *i*-th line equals "." if the cell in the *i*-th row and the *j*-th column of the square is painted white, and "#", if the cell is black.

Print "YES" (without the quotes), if the test can be passed and "NO" (without the quotes) otherwise.

[ "####\n.#..\n####\n....\n", "####\n....\n####\n....\n" ]

[ "YES\n", "NO\n" ]

In the first test sample it is enough to repaint the first cell in the second row. After such repainting the required 2 × 2 square is on the intersection of the 1-st and 2-nd row with the 1-st and 2-nd column.

500

[ { "input": "###.\n...#\n###.\n...#", "output": "NO" }, { "input": ".##.\n#..#\n.##.\n#..#", "output": "NO" }, { "input": ".#.#\n#.#.\n.#.#\n#.#.", "output": "NO" }, { "input": "##..\n..##\n##..\n..##", "output": "NO" }, { "input": "#.#.\n#.#.\n.#.#\n.#.#", "output": "NO" }, { "input": ".#.#\n#.#.\n#.#.\n#.#.", "output": "NO" }, { "input": ".#.#\n#.#.\n#.#.\n.#.#", "output": "NO" }, { "input": "#.#.\n#.#.\n#.#.\n#.#.", "output": "NO" }, { "input": ".#.#\n.#.#\n.#.#\n.#.#", "output": "NO" }, { "input": "#..#\n.##.\n####\n####", "output": "YES" }, { "input": "#.#.\n.###\n#.#.\n.###", "output": "YES" }, { "input": "#..#\n.##.\n.##.\n#..#", "output": "YES" }, { "input": ".##.\n.#..\n##.#\n#..#", "output": "YES" }, { "input": ".##.\n##..\n#..#\n..##", "output": "YES" }, { "input": "##..\n##..\n..##\n..##", "output": "YES" }, { "input": ".#..\n###.\n.#.#\n..#.", "output": "YES" }, { "input": "###.\n###.\n...#\n...#", "output": "YES" }, { "input": "#.##\n##.#\n#.##\n##.#", "output": "YES" }, { "input": ".#.#\n#.#.\n.#.#\n#.##", "output": "YES" }, { "input": "##..\n..##\n##..\n...#", "output": "YES" }, { "input": ".#..\n..##\n##..\n..##", "output": "YES" }, { "input": "##..\n...#\n##..\n...#", "output": "YES" }, { "input": ".#..\n..#.\n.#..\n..#.", "output": "YES" }, { "input": "....\n....\n....\n.#.#", "output": "YES" }, { "input": "....\n....\n....\n...#", "output": "YES" } ]

1,590,779,836

2,147,483,647

PyPy 3

OK

TESTS

30

140

0

from sys import stdin inp = lambda : stdin.readline().strip() a = [] for i in range(4): a.append(inp()) flag = False for i in range(4): for j in range(4): white = 0 black = 0 for k in range(2): for l in range(2): if (i+k) < 4 and (j+l) < 4: if (i+k) < 4 and (j+l) < 4 and a[i+k][j+l] == '.': white += 1 else: black += 1 if white == 3 or white == 4 or black == 3 or black == 4: flag = True if flag: break if flag: print("YES") else: print("NO")

Title: IQ Test Time Limit: None seconds Memory Limit: None megabytes Problem Description: In the city of Ultima Thule job applicants are often offered an IQ test. The test is as follows: the person gets a piece of squared paper with a 4<=×<=4 square painted on it. Some of the square's cells are painted black and others are painted white. Your task is to repaint at most one cell the other color so that the picture has a 2<=×<=2 square, completely consisting of cells of the same color. If the initial picture already has such a square, the person should just say so and the test will be completed. Your task is to write a program that determines whether it is possible to pass the test. You cannot pass the test if either repainting any cell or no action doesn't result in a 2<=×<=2 square, consisting of cells of the same color. Input Specification: Four lines contain four characters each: the *j*-th character of the *i*-th line equals "." if the cell in the *i*-th row and the *j*-th column of the square is painted white, and "#", if the cell is black. Output Specification: Print "YES" (without the quotes), if the test can be passed and "NO" (without the quotes) otherwise. Demo Input: ['####\n.#..\n####\n....\n', '####\n....\n####\n....\n'] Demo Output: ['YES\n', 'NO\n'] Note: In the first test sample it is enough to repaint the first cell in the second row. After such repainting the required 2 × 2 square is on the intersection of the 1-st and 2-nd row with the 1-st and 2-nd column.

```python from sys import stdin inp = lambda : stdin.readline().strip() a = [] for i in range(4): a.append(inp()) flag = False for i in range(4): for j in range(4): white = 0 black = 0 for k in range(2): for l in range(2): if (i+k) < 4 and (j+l) < 4: if (i+k) < 4 and (j+l) < 4 and a[i+k][j+l] == '.': white += 1 else: black += 1 if white == 3 or white == 4 or black == 3 or black == 4: flag = True if flag: break if flag: print("YES") else: print("NO") ```

3

990

D

Graph And Its Complement

PROGRAMMING

1,700

[ "constructive algorithms", "graphs", "implementation" ]

null

Given three numbers $n, a, b$. You need to find an adjacency matrix of such an undirected graph that the number of components in it is equal to $a$, and the number of components in its complement is $b$. The matrix must be symmetric, and all digits on the main diagonal must be zeroes. In an undirected graph loops (edges from a vertex to itself) are not allowed. It can be at most one edge between a pair of vertices. The adjacency matrix of an undirected graph is a square matrix of size $n$ consisting only of "0" and "1", where $n$ is the number of vertices of the graph and the $i$-th row and the $i$-th column correspond to the $i$-th vertex of the graph. The cell $(i,j)$ of the adjacency matrix contains $1$ if and only if the $i$-th and $j$-th vertices in the graph are connected by an edge. A connected component is a set of vertices $X$ such that for every two vertices from this set there exists at least one path in the graph connecting this pair of vertices, but adding any other vertex to $X$ violates this rule. The complement or inverse of a graph $G$ is a graph $H$ on the same vertices such that two distinct vertices of $H$ are adjacent if and only if they are not adjacent in $G$.

In a single line, three numbers are given $n, a, b \,(1 \le n \le 1000, 1 \le a, b \le n)$: is the number of vertexes of the graph, the required number of connectivity components in it, and the required amount of the connectivity component in it's complement.

If there is no graph that satisfies these constraints on a single line, print "NO" (without quotes). Otherwise, on the first line, print "YES"(without quotes). In each of the next $n$ lines, output $n$ digits such that $j$-th digit of $i$-th line must be $1$ if and only if there is an edge between vertices $i$ and $j$ in $G$ (and $0$ otherwise). Note that the matrix must be symmetric, and all digits on the main diagonal must be zeroes. If there are several matrices that satisfy the conditions — output any of them.

[ "3 1 2\n", "3 3 3\n" ]

[ "YES\n001\n001\n110\n", "NO\n" ]

none

0

[ { "input": "3 1 2", "output": "YES\n001\n001\n110" }, { "input": "3 3 3", "output": "NO" }, { "input": "5 1 1", "output": "YES\n01000\n10100\n01010\n00101\n00010" }, { "input": "123 1 84", "output": "YES\n001111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111\n000111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111\n100011111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111\n110001111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111\n111000..." }, { "input": "2 1 1", "output": "NO" }, { "input": "1 1 1", "output": "YES\n0" }, { "input": "3 1 1", "output": "NO" }, { "input": "5 2 2", "output": "NO" }, { "input": "1000 734 1", "output": "YES\n01000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "1000 1 1000", "output": "YES\n01111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..." }, { "input": "1000 1 1", "output": "YES\n01000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "4 1 1", "output": "YES\n0100\n1010\n0101\n0010" }, { "input": "4 4 1", "output": "YES\n0000\n0000\n0000\n0000" }, { "input": "3 1 3", "output": "YES\n011\n101\n110" }, { "input": "2 1 2", "output": "YES\n01\n10" }, { "input": "101 1 1", "output": "YES\n01000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n10100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n01010000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n00101000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n0001010000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "102 1 1", "output": "YES\n010000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n101000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n010100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n001010000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n000101000000000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "103 1 1", "output": "YES\n0100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n1010000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n0101000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n0010100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n00010100000000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "104 1 1", "output": "YES\n01000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n10100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n01010000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n00101000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n0001010000000000000000000000000000000000000000000000000000000000000000000000000000..." }, { "input": "6 1 1", "output": "YES\n010000\n101000\n010100\n001010\n000101\n000010" }, { "input": "3 2 1", "output": "YES\n010\n100\n000" }, { "input": "5 1 2", "output": "YES\n00111\n00011\n10001\n11001\n11110" }, { "input": "4 1 2", "output": "YES\n0011\n0001\n1001\n1110" }, { "input": "2 2 1", "output": "YES\n00\n00" }, { "input": "3 3 1", "output": "YES\n000\n000\n000" }, { "input": "2 2 2", "output": "NO" }, { "input": "1 1 1", "output": "YES\n0" }, { "input": "2 1 1", "output": "NO" }, { "input": "2 1 2", "output": "YES\n01\n10" }, { "input": "2 2 1", "output": "YES\n00\n00" }, { "input": "2 2 2", "output": "NO" }, { "input": "3 1 1", "output": "NO" }, { "input": "3 1 2", "output": "YES\n001\n001\n110" }, { "input": "3 1 3", "output": "YES\n011\n101\n110" }, { "input": "3 2 1", "output": "YES\n010\n100\n000" }, { "input": "3 2 2", "output": "NO" }, { "input": "3 2 3", "output": "NO" }, { "input": "3 3 1", "output": "YES\n000\n000\n000" }, { "input": "3 3 2", "output": "NO" }, { "input": "3 3 3", "output": "NO" }, { "input": "4 1 1", "output": "YES\n0100\n1010\n0101\n0010" }, { "input": "4 1 2", "output": "YES\n0011\n0001\n1001\n1110" }, { "input": "4 1 3", "output": "YES\n0011\n0011\n1101\n1110" }, { "input": "4 1 4", "output": "YES\n0111\n1011\n1101\n1110" }, { "input": "4 2 1", "output": "YES\n0100\n1010\n0100\n0000" }, { "input": "4 2 2", "output": "NO" }, { "input": "4 2 3", "output": "NO" }, { "input": "4 2 4", "output": "NO" }, { "input": "4 3 1", "output": "YES\n0100\n1000\n0000\n0000" }, { "input": "4 3 2", "output": "NO" }, { "input": "4 3 3", "output": "NO" }, { "input": "4 3 4", "output": "NO" }, { "input": "4 4 1", "output": "YES\n0000\n0000\n0000\n0000" }, { "input": "4 4 2", "output": "NO" }, { "input": "4 4 3", "output": "NO" }, { "input": "4 4 4", "output": "NO" } ]

1,528,628,891

3,791

PyPy 3

WRONG_ANSWER

TESTS

0

93

0

def ii(): return int(input()) def mi(): return map(int, input().split()) def li(): return list(mi()) n, a, b = mi() if a != 1 and b != 1: print('NO') else: c = max(a, b) if a == 1: g = [[1] * n for i in range(n)] for i in range(n): g[i][i] = 0 for i in range(c - 1, n - 1): g[i][i + 1] = g[i + 1][i] = 0 else: g = [[0] * n for i in range(n)] for i in range(c - 1, n - 1): g[i][i + 1] = g[i + 1][i] = 1 print('YES') for r in g: print(*r)

Title: Graph And Its Complement Time Limit: None seconds Memory Limit: None megabytes Problem Description: Given three numbers $n, a, b$. You need to find an adjacency matrix of such an undirected graph that the number of components in it is equal to $a$, and the number of components in its complement is $b$. The matrix must be symmetric, and all digits on the main diagonal must be zeroes. In an undirected graph loops (edges from a vertex to itself) are not allowed. It can be at most one edge between a pair of vertices. The adjacency matrix of an undirected graph is a square matrix of size $n$ consisting only of "0" and "1", where $n$ is the number of vertices of the graph and the $i$-th row and the $i$-th column correspond to the $i$-th vertex of the graph. The cell $(i,j)$ of the adjacency matrix contains $1$ if and only if the $i$-th and $j$-th vertices in the graph are connected by an edge. A connected component is a set of vertices $X$ such that for every two vertices from this set there exists at least one path in the graph connecting this pair of vertices, but adding any other vertex to $X$ violates this rule. The complement or inverse of a graph $G$ is a graph $H$ on the same vertices such that two distinct vertices of $H$ are adjacent if and only if they are not adjacent in $G$. Input Specification: In a single line, three numbers are given $n, a, b \,(1 \le n \le 1000, 1 \le a, b \le n)$: is the number of vertexes of the graph, the required number of connectivity components in it, and the required amount of the connectivity component in it's complement. Output Specification: If there is no graph that satisfies these constraints on a single line, print "NO" (without quotes). Otherwise, on the first line, print "YES"(without quotes). In each of the next $n$ lines, output $n$ digits such that $j$-th digit of $i$-th line must be $1$ if and only if there is an edge between vertices $i$ and $j$ in $G$ (and $0$ otherwise). Note that the matrix must be symmetric, and all digits on the main diagonal must be zeroes. If there are several matrices that satisfy the conditions — output any of them. Demo Input: ['3 1 2\n', '3 3 3\n'] Demo Output: ['YES\n001\n001\n110\n', 'NO\n'] Note: none

```python def ii(): return int(input()) def mi(): return map(int, input().split()) def li(): return list(mi()) n, a, b = mi() if a != 1 and b != 1: print('NO') else: c = max(a, b) if a == 1: g = [[1] * n for i in range(n)] for i in range(n): g[i][i] = 0 for i in range(c - 1, n - 1): g[i][i + 1] = g[i + 1][i] = 0 else: g = [[0] * n for i in range(n)] for i in range(c - 1, n - 1): g[i][i + 1] = g[i + 1][i] = 1 print('YES') for r in g: print(*r) ```

0

342

A

Xenia and Divisors

PROGRAMMING

1,200

[ "greedy", "implementation" ]

null

Xenia the mathematician has a sequence consisting of *n* (*n* is divisible by 3) positive integers, each of them is at most 7. She wants to split the sequence into groups of three so that for each group of three *a*,<=*b*,<=*c* the following conditions held: - *a*<=<<=*b*<=<<=*c*; - *a* divides *b*, *b* divides *c*. Naturally, Xenia wants each element of the sequence to belong to exactly one group of three. Thus, if the required partition exists, then it has groups of three. Help Xenia, find the required partition or else say that it doesn't exist.

The first line contains integer *n* (3<=≤<=*n*<=≤<=99999) — the number of elements in the sequence. The next line contains *n* positive integers, each of them is at most 7. It is guaranteed that *n* is divisible by 3.

If the required partition exists, print groups of three. Print each group as values of the elements it contains. You should print values in increasing order. Separate the groups and integers in groups by whitespaces. If there are multiple solutions, you can print any of them. If there is no solution, print -1.

[ "6\n1 1 1 2 2 2\n", "6\n2 2 1 1 4 6\n" ]

[ "-1\n", "1 2 4\n1 2 6\n" ]

none

500

[ { "input": "6\n1 1 1 2 2 2", "output": "-1" }, { "input": "6\n2 2 1 1 4 6", "output": "1 2 4\n1 2 6" }, { "input": "3\n1 2 3", "output": "-1" }, { "input": "3\n7 5 7", "output": "-1" }, { "input": "3\n1 3 4", "output": "-1" }, { "input": "3\n1 1 1", "output": "-1" }, { "input": "9\n1 3 6 6 3 1 3 1 6", "output": "1 3 6\n1 3 6\n1 3 6" }, { "input": "6\n1 2 4 1 3 5", "output": "-1" }, { "input": "3\n1 3 7", "output": "-1" }, { "input": "3\n1 1 1", "output": "-1" }, { "input": "9\n1 2 4 1 2 4 1 3 6", "output": "1 2 4\n1 2 4\n1 3 6" }, { "input": "12\n3 6 1 1 3 6 1 1 2 6 2 6", "output": "1 3 6\n1 3 6\n1 2 6\n1 2 6" }, { "input": "9\n1 1 1 4 4 4 6 2 2", "output": "-1" }, { "input": "9\n1 2 4 6 3 1 3 1 5", "output": "-1" }, { "input": "15\n2 1 2 1 3 6 1 2 1 6 1 3 4 6 4", "output": "1 2 4\n1 2 4\n1 3 6\n1 3 6\n1 2 6" }, { "input": "3\n2 3 6", "output": "-1" }, { "input": "3\n2 4 6", "output": "-1" }, { "input": "3\n2 5 6", "output": "-1" }, { "input": "3\n2 4 7", "output": "-1" }, { "input": "6\n1 2 3 4 5 6", "output": "-1" }, { "input": "3\n7 7 7", "output": "-1" }, { "input": "6\n1 2 4 7 7 7", "output": "-1" }, { "input": "6\n1 1 2 6 6 6", "output": "-1" }, { "input": "9\n1 1 1 3 3 2 4 4 6", "output": "-1" }, { "input": "6\n1 2 4 5 5 5", "output": "-1" }, { "input": "15\n1 1 1 1 1 2 2 2 2 4 4 6 6 6 6", "output": "-1" }, { "input": "6\n1 1 5 5 7 7", "output": "-1" }, { "input": "9\n1 1 1 2 3 4 5 6 7", "output": "-1" }, { "input": "6\n1 1 4 4 7 7", "output": "-1" }, { "input": "24\n1 1 1 1 1 1 1 1 1 2 2 2 3 3 3 3 3 3 4 4 4 6 6 6", "output": "-1" }, { "input": "3\n1 7 6", "output": "-1" }, { "input": "6\n1 1 2 4 7 7", "output": "-1" }, { "input": "9\n1 1 1 7 7 7 7 7 7", "output": "-1" }, { "input": "9\n1 1 1 2 3 4 6 5 5", "output": "-1" } ]

1,674,452,397

2,147,483,647

Python 3

OK

TESTS

45

202

1,945,600

n = int(input()) input_arr = input().split() arr = [] sum = 0 for i in range(n): # if int(input_arr[i]) == 5 or int(input_arr[i]) == 7: # break sum += int(input_arr[i]) arr.append(int(input_arr[i])) arr.sort() a_arr = [] b_arr = [] c_arr = [] for i in range(len(arr)//3): a_arr.append(arr[i]) b_arr.append(arr[i + (n//3)]) c_arr.append((arr[i + ((2*n)//3)])) possibel = True for i in range(len(arr)//3): if(b_arr[i] % a_arr[i] == 0 and b_arr[i] > a_arr[i] and c_arr[i] % b_arr[i] == 0 and c_arr[i] > b_arr[i]): continue else: possibel = False print(-1) break if possibel: for i in range(len(arr)//3): print(f'{a_arr[i]} {b_arr[i]} {c_arr[i]}')

Title: Xenia and Divisors Time Limit: None seconds Memory Limit: None megabytes Problem Description: Xenia the mathematician has a sequence consisting of *n* (*n* is divisible by 3) positive integers, each of them is at most 7. She wants to split the sequence into groups of three so that for each group of three *a*,<=*b*,<=*c* the following conditions held: - *a*<=<<=*b*<=<<=*c*; - *a* divides *b*, *b* divides *c*. Naturally, Xenia wants each element of the sequence to belong to exactly one group of three. Thus, if the required partition exists, then it has groups of three. Help Xenia, find the required partition or else say that it doesn't exist. Input Specification: The first line contains integer *n* (3<=≤<=*n*<=≤<=99999) — the number of elements in the sequence. The next line contains *n* positive integers, each of them is at most 7. It is guaranteed that *n* is divisible by 3. Output Specification: If the required partition exists, print groups of three. Print each group as values of the elements it contains. You should print values in increasing order. Separate the groups and integers in groups by whitespaces. If there are multiple solutions, you can print any of them. If there is no solution, print -1. Demo Input: ['6\n1 1 1 2 2 2\n', '6\n2 2 1 1 4 6\n'] Demo Output: ['-1\n', '1 2 4\n1 2 6\n'] Note: none

```python n = int(input()) input_arr = input().split() arr = [] sum = 0 for i in range(n): # if int(input_arr[i]) == 5 or int(input_arr[i]) == 7: # break sum += int(input_arr[i]) arr.append(int(input_arr[i])) arr.sort() a_arr = [] b_arr = [] c_arr = [] for i in range(len(arr)//3): a_arr.append(arr[i]) b_arr.append(arr[i + (n//3)]) c_arr.append((arr[i + ((2*n)//3)])) possibel = True for i in range(len(arr)//3): if(b_arr[i] % a_arr[i] == 0 and b_arr[i] > a_arr[i] and c_arr[i] % b_arr[i] == 0 and c_arr[i] > b_arr[i]): continue else: possibel = False print(-1) break if possibel: for i in range(len(arr)//3): print(f'{a_arr[i]} {b_arr[i]} {c_arr[i]}') ```

3

730

G

Car Repair Shop

PROGRAMMING

1,600

[ "implementation" ]

null

Polycarp starts his own business. Tomorrow will be the first working day of his car repair shop. For now the car repair shop is very small and only one car can be repaired at a given time. Polycarp is good at marketing, so he has already collected *n* requests from clients. The requests are numbered from 1 to *n* in order they came. The *i*-th request is characterized by two values: *s**i* — the day when a client wants to start the repair of his car, *d**i* — duration (in days) to repair the car. The days are enumerated from 1, the first day is tomorrow, the second day is the day after tomorrow and so on. Polycarp is making schedule by processing requests in the order from the first to the *n*-th request. He schedules the *i*-th request as follows: - If the car repair shop is idle for *d**i* days starting from *s**i* (*s**i*,<=*s**i*<=+<=1,<=...,<=*s**i*<=+<=*d**i*<=-<=1), then these days are used to repair a car of the *i*-th client. - Otherwise, Polycarp finds the first day *x* (from 1 and further) that there are *d**i* subsequent days when no repair is scheduled starting from *x*. In other words he chooses the smallest positive *x* that all days *x*,<=*x*<=+<=1,<=...,<=*x*<=+<=*d**i*<=-<=1 are not scheduled for repair of any car. So, the car of the *i*-th client will be repaired in the range [*x*,<=*x*<=+<=*d**i*<=-<=1]. It is possible that the day *x* when repair is scheduled to start will be less than *s**i*. Given *n* requests, you are asked to help Polycarp schedule all of them according to the rules above.

The first line contains integer *n* (1<=≤<=*n*<=≤<=200) — the number of requests from clients. The following *n* lines contain requests, one request per line. The *i*-th request is given as the pair of integers *s**i*,<=*d**i* (1<=≤<=*s**i*<=≤<=109, 1<=≤<=*d**i*<=≤<=5·106), where *s**i* is the preferred time to start repairing the *i*-th car, *d**i* is the number of days to repair the *i*-th car. The requests should be processed in the order they are given in the input.

Print *n* lines. The *i*-th line should contain two integers — the start day to repair the *i*-th car and the finish day to repair the *i*-th car.

[ "3\n9 2\n7 3\n2 4\n", "4\n1000000000 1000000\n1000000000 1000000\n100000000 1000000\n1000000000 1000000\n" ]

[ "9 10\n1 3\n4 7\n", "1000000000 1000999999\n1 1000000\n100000000 100999999\n1000001 2000000\n" ]

none

0

[ { "input": "3\n9 2\n7 3\n2 4", "output": "9 10\n1 3\n4 7" }, { "input": "4\n1000000000 1000000\n1000000000 1000000\n100000000 1000000\n1000000000 1000000", "output": "1000000000 1000999999\n1 1000000\n100000000 100999999\n1000001 2000000" }, { "input": "1\n1 1", "output": "1 1" }, { "input": "1\n1000000000 1", "output": "1000000000 1000000000" }, { "input": "1\n1000000000 5000000", "output": "1000000000 1004999999" }, { "input": "5\n6 2\n10 1\n10 2\n9 2\n5 1", "output": "6 7\n10 10\n1 2\n3 4\n5 5" }, { "input": "10\n1 3\n77 8\n46 5\n83 4\n61 7\n8 4\n54 7\n80 7\n33 7\n13 4", "output": "1 3\n77 84\n46 50\n4 7\n61 67\n8 11\n54 60\n12 18\n33 39\n19 22" }, { "input": "10\n588 12\n560 10\n593 14\n438 15\n761 11\n984 6\n503 2\n855 19\n538 2\n650 7", "output": "588 599\n560 569\n1 14\n438 452\n761 771\n984 989\n503 504\n855 873\n538 539\n650 656" }, { "input": "20\n360 26\n475 17\n826 12\n815 23\n567 28\n897 26\n707 20\n1000 9\n576 5\n16 5\n714 16\n630 17\n426 26\n406 23\n899 25\n102 22\n896 8\n320 27\n964 25\n932 18", "output": "360 385\n475 491\n826 837\n1 23\n567 594\n897 922\n707 726\n1000 1008\n24 28\n29 33\n34 49\n630 646\n426 451\n50 72\n73 97\n102 123\n124 131\n320 346\n964 988\n932 949" }, { "input": "30\n522692116 84\n589719489 488\n662495181 961\n915956552 470\n683572975 271\n498400137 480\n327010963 181\n200704287 367\n810826488 54\n978100746 208\n345455616 986\n106372142 876\n446972337 42\n309349333 200\n93462198 543\n167946793 318\n325598940 427\n121873339 459\n174934933 598\n279521023 655\n739750520 3\n870850765 192\n622303167 400\n471234786 63\n805952711 18\n349834333 857\n804873364 302\n512746562 39\n533285962 561\n996718586 494", "output": "522692116 522692199\n589719489 589719976\n662495181 662496141\n915956552 915957021\n683572975 683573245\n498400137 498400616\n327010963 327011143\n200704287 200704653\n810826488 810826541\n978100746 978100953\n345455616 345456601\n106372142 106373017\n446972337 446972378\n309349333 309349532\n93462198 93462740\n167946793 167947110\n325598940 325599366\n121873339 121873797\n174934933 174935530\n279521023 279521677\n739750520 739750522\n870850765 870850956\n622303167 622303566\n471234786 471234848\n805952711..." }, { "input": "2\n10 3\n9 2", "output": "10 12\n1 2" }, { "input": "1\n1 5000000", "output": "1 5000000" } ]

1,638,657,904

2,147,483,647

PyPy 3-64

COMPILATION_ERROR

TESTS

0

#include <bits/stdc++.h> #include <ext/pb_ds/assoc_container.hpp> #include <ext/pb_ds/tree_policy.hpp> using namespace std; using namespace __gnu_pbds; template<typename T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>; #define pb push_back #define mp make_pair #define sz(x) ((int)x.size()) #define all(x) (x).begin(), (x).end() #define rall(x) (x).rbegin(), (x).rend() #define manytests int TT;cin >> TT; while (TT--) #define FOR(i,a,b) for (int i = (a); i < (b); ++i) #define FORd(i,a,b) for (int i = (a); i >= (b); --i) #define REP(i,a,b,c) for (int i = (a); i < (b); i += (c)) #define REPd(i,a,b,c) for (int i = (a); i >= (b); i -= (c)) #define ll long long #define ld long double #define str string #define fi first #define se second #define rev reverse int dx[8] = {-1, 1, 0, 0, 1, 1, -1, -1}, dy[8] = {0, 0, 1, -1, 1, 1, -1, -1}; int INF = 1e9; ll INFI = 1e18; double PI = atan(1) * 4; double eps = 1e-9; const int MOD = 1e9+7; // const int MOD = 998244353; #define vi vector<int> #define vl vector<ll> #define vc vector<char> #define vd vector<double> #define vs vector<string> #define vld vector<ld> #define vb vector<bool> #define vvi vector<vi> #define vvl vector<vl> #define pii pair<int, int> #define pl pair<ll, ll> #define pld pair<ld, ld> #define vpi vector<pii> #define vpl vector<pl> #define vpld vector<pld> #define vmi vector<mint> #define pmi pair<mint, mint> template<class T> bool ckmin(T &a, const T &b) { return (b < a ? a = b, 1 : 0); } template<class T> bool ckmax(T &a, const T &b) { return (a istream& operator>>(istream& in, pair<T, U> &x) { in >> x.fi >> x.se; return in; } template<class T> istream& operator>>(istream& in, vector<T> &v) { for (auto &x : v) in >> x; return in; } template<class T, class U> ostream& operator<<(ostream& out, const pair<T, U> &x) { out << "(" << x.fi << ", " << x.se << ")"; return out; } template<class T> ostream& operator<<(ostream& out, const vector<T> &v) { FOR(i,0,sz(v)) out << v[i] << " \n"[i == sz(v) - 1]; return out; } inline namespace FileIO { void setIn(str s) { freopen(s.c_str(), "r", stdin); } void setOut(str s) { freopen(s.c_str(), "w", stdout); } void setIO(str s = "") { cin.tie(0)->sync_with_stdio(0); if (sz(s)) setIn(s+".in"), setOut(s+".out"); } } struct mint { int v; explicit operator int() const { return v; } mint(): v(0) {} mint(ll _v) { v = int((-MOD < _v && _v < MOD) ? _v : _v % MOD); if (v < 0) v += MOD; } bool operator==(const mint& o) const { return v == o.v; } friend bool operator!=(const mint& a, const mint& b) { return !(a == b); } friend bool operator<(const mint& a, const mint& b) { return a.v < b.v; } mint& operator+=(const mint& o) { if ((v += o.v) >= MOD) v -= MOD; return *this; } mint& operator-=(const mint& o) { if ((v -= o.v) < 0) v += MOD; return *this; } mint& operator*=(const mint& o) { v = int((ll)v * o.v % MOD); return *this; } mint& operator/=(const mint& o) { return (*this) *= inv(o); } mint& operator^=(const mint& o) { return (*this) = pow(v, o.v); } mint& operator+=(const int& o) { if ((v += mint(o).v) >= MOD) v -= MOD; return *this; } mint& operator-=(const int& o) { if ((v -= mint(o).v) < 0) v += MOD; return *this; } mint& operator*=(const int& o) { v = int((ll)v * mint(o).v % MOD); return *this; } mint& operator/=(const int& o) { return (*this) *= inv(mint(o)); } mint& operator^=(const int& o) { return (*this) = pow(v, o); } friend mint pow(mint a, ll p) { mint ans = 1; assert(p >= 0); for (; p; p /= 2, a *= a) if (p&1) ans *= a; return ans; } friend mint inv(const mint& a) { assert(a.v != 0); return pow(a,MOD-2); } mint operator-() const { return mint(-v); } mint& operator++() { return *this += 1; } mint& operator--() { return *this -= 1; } friend mint operator+(mint a, const mint& b) { return a += b; } friend mint operator-(mint a, const mint& b) { return a -= b; } friend mint operator*(mint a, const mint& b) { return a *= b; } friend mint operator/(mint a, const mint& b) { return a /= b; } friend mint operator^(mint a, const mint& b) { return a ^= b; } friend mint operator+(mint a, const int& b) { return a += mint(b); } friend mint operator-(mint a, const int& b) { return a -= mint(b); } friend mint operator*(mint a, const int& b) { return a *= mint(b); } friend mint operator/(mint a, const int& b) { return a /= mint(b); } friend mint operator^(mint a, const int& b) { return a ^= mint(b); } friend istream& operator>>(istream& in, mint& mi) { in >> mi.v; return in; } friend ostream& operator<<(ostream& os, const mint& mi) { os << mi.v; return os; } }; vpi a; bool inside(ll x, ll l, ll r) { return x >= l && x <= r; } bool check(ll l, ll r) { FOR(i,0,sz(a)) { if (inside(l, a[i].fi, a[i].se) || inside(r, a[i].fi, a[i].se)) return false; if (inside(a[i].fi, l, r) || inside(a[i].se, l, r)) return false; } return true; } int main() { setIO(); int n; cin >> n; FOR(i,0,n) { ll s, d; cin >> s >> d; if (check(s, s + d - 1)) { cout << s << " " << s + d - 1 << "\n"; a.pb(mp(s, s + d - 1)); continue; } vl ch{1}; FOR(j,0,sz(a)) { ch.pb(a[j].se + 1); } sort(all(ch)); FOR(j,0,sz(ch)) { ll cl = ch[j], cr = cl + d - 1; if (check(cl, cr)) { cout << cl << " " << cr << "\n"; a.pb(mp(cl, cr)); break; } } } return 0; }

Title: Car Repair Shop Time Limit: None seconds Memory Limit: None megabytes Problem Description: Polycarp starts his own business. Tomorrow will be the first working day of his car repair shop. For now the car repair shop is very small and only one car can be repaired at a given time. Polycarp is good at marketing, so he has already collected *n* requests from clients. The requests are numbered from 1 to *n* in order they came. The *i*-th request is characterized by two values: *s**i* — the day when a client wants to start the repair of his car, *d**i* — duration (in days) to repair the car. The days are enumerated from 1, the first day is tomorrow, the second day is the day after tomorrow and so on. Polycarp is making schedule by processing requests in the order from the first to the *n*-th request. He schedules the *i*-th request as follows: - If the car repair shop is idle for *d**i* days starting from *s**i* (*s**i*,<=*s**i*<=+<=1,<=...,<=*s**i*<=+<=*d**i*<=-<=1), then these days are used to repair a car of the *i*-th client. - Otherwise, Polycarp finds the first day *x* (from 1 and further) that there are *d**i* subsequent days when no repair is scheduled starting from *x*. In other words he chooses the smallest positive *x* that all days *x*,<=*x*<=+<=1,<=...,<=*x*<=+<=*d**i*<=-<=1 are not scheduled for repair of any car. So, the car of the *i*-th client will be repaired in the range [*x*,<=*x*<=+<=*d**i*<=-<=1]. It is possible that the day *x* when repair is scheduled to start will be less than *s**i*. Given *n* requests, you are asked to help Polycarp schedule all of them according to the rules above. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=200) — the number of requests from clients. The following *n* lines contain requests, one request per line. The *i*-th request is given as the pair of integers *s**i*,<=*d**i* (1<=≤<=*s**i*<=≤<=109, 1<=≤<=*d**i*<=≤<=5·106), where *s**i* is the preferred time to start repairing the *i*-th car, *d**i* is the number of days to repair the *i*-th car. The requests should be processed in the order they are given in the input. Output Specification: Print *n* lines. The *i*-th line should contain two integers — the start day to repair the *i*-th car and the finish day to repair the *i*-th car. Demo Input: ['3\n9 2\n7 3\n2 4\n', '4\n1000000000 1000000\n1000000000 1000000\n100000000 1000000\n1000000000 1000000\n'] Demo Output: ['9 10\n1 3\n4 7\n', '1000000000 1000999999\n1 1000000\n100000000 100999999\n1000001 2000000\n'] Note: none

```python #include <bits/stdc++.h> #include <ext/pb_ds/assoc_container.hpp> #include <ext/pb_ds/tree_policy.hpp> using namespace std; using namespace __gnu_pbds; template<typename T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>; #define pb push_back #define mp make_pair #define sz(x) ((int)x.size()) #define all(x) (x).begin(), (x).end() #define rall(x) (x).rbegin(), (x).rend() #define manytests int TT;cin >> TT; while (TT--) #define FOR(i,a,b) for (int i = (a); i < (b); ++i) #define FORd(i,a,b) for (int i = (a); i >= (b); --i) #define REP(i,a,b,c) for (int i = (a); i < (b); i += (c)) #define REPd(i,a,b,c) for (int i = (a); i >= (b); i -= (c)) #define ll long long #define ld long double #define str string #define fi first #define se second #define rev reverse int dx[8] = {-1, 1, 0, 0, 1, 1, -1, -1}, dy[8] = {0, 0, 1, -1, 1, 1, -1, -1}; int INF = 1e9; ll INFI = 1e18; double PI = atan(1) * 4; double eps = 1e-9; const int MOD = 1e9+7; // const int MOD = 998244353; #define vi vector<int> #define vl vector<ll> #define vc vector<char> #define vd vector<double> #define vs vector<string> #define vld vector<ld> #define vb vector<bool> #define vvi vector<vi> #define vvl vector<vl> #define pii pair<int, int> #define pl pair<ll, ll> #define pld pair<ld, ld> #define vpi vector<pii> #define vpl vector<pl> #define vpld vector<pld> #define vmi vector<mint> #define pmi pair<mint, mint> template<class T> bool ckmin(T &a, const T &b) { return (b < a ? a = b, 1 : 0); } template<class T> bool ckmax(T &a, const T &b) { return (a istream& operator>>(istream& in, pair<T, U> &x) { in >> x.fi >> x.se; return in; } template<class T> istream& operator>>(istream& in, vector<T> &v) { for (auto &x : v) in >> x; return in; } template<class T, class U> ostream& operator<<(ostream& out, const pair<T, U> &x) { out << "(" << x.fi << ", " << x.se << ")"; return out; } template<class T> ostream& operator<<(ostream& out, const vector<T> &v) { FOR(i,0,sz(v)) out << v[i] << " \n"[i == sz(v) - 1]; return out; } inline namespace FileIO { void setIn(str s) { freopen(s.c_str(), "r", stdin); } void setOut(str s) { freopen(s.c_str(), "w", stdout); } void setIO(str s = "") { cin.tie(0)->sync_with_stdio(0); if (sz(s)) setIn(s+".in"), setOut(s+".out"); } } struct mint { int v; explicit operator int() const { return v; } mint(): v(0) {} mint(ll _v) { v = int((-MOD < _v && _v < MOD) ? _v : _v % MOD); if (v < 0) v += MOD; } bool operator==(const mint& o) const { return v == o.v; } friend bool operator!=(const mint& a, const mint& b) { return !(a == b); } friend bool operator<(const mint& a, const mint& b) { return a.v < b.v; } mint& operator+=(const mint& o) { if ((v += o.v) >= MOD) v -= MOD; return *this; } mint& operator-=(const mint& o) { if ((v -= o.v) < 0) v += MOD; return *this; } mint& operator*=(const mint& o) { v = int((ll)v * o.v % MOD); return *this; } mint& operator/=(const mint& o) { return (*this) *= inv(o); } mint& operator^=(const mint& o) { return (*this) = pow(v, o.v); } mint& operator+=(const int& o) { if ((v += mint(o).v) >= MOD) v -= MOD; return *this; } mint& operator-=(const int& o) { if ((v -= mint(o).v) < 0) v += MOD; return *this; } mint& operator*=(const int& o) { v = int((ll)v * mint(o).v % MOD); return *this; } mint& operator/=(const int& o) { return (*this) *= inv(mint(o)); } mint& operator^=(const int& o) { return (*this) = pow(v, o); } friend mint pow(mint a, ll p) { mint ans = 1; assert(p >= 0); for (; p; p /= 2, a *= a) if (p&1) ans *= a; return ans; } friend mint inv(const mint& a) { assert(a.v != 0); return pow(a,MOD-2); } mint operator-() const { return mint(-v); } mint& operator++() { return *this += 1; } mint& operator--() { return *this -= 1; } friend mint operator+(mint a, const mint& b) { return a += b; } friend mint operator-(mint a, const mint& b) { return a -= b; } friend mint operator*(mint a, const mint& b) { return a *= b; } friend mint operator/(mint a, const mint& b) { return a /= b; } friend mint operator^(mint a, const mint& b) { return a ^= b; } friend mint operator+(mint a, const int& b) { return a += mint(b); } friend mint operator-(mint a, const int& b) { return a -= mint(b); } friend mint operator*(mint a, const int& b) { return a *= mint(b); } friend mint operator/(mint a, const int& b) { return a /= mint(b); } friend mint operator^(mint a, const int& b) { return a ^= mint(b); } friend istream& operator>>(istream& in, mint& mi) { in >> mi.v; return in; } friend ostream& operator<<(ostream& os, const mint& mi) { os << mi.v; return os; } }; vpi a; bool inside(ll x, ll l, ll r) { return x >= l && x <= r; } bool check(ll l, ll r) { FOR(i,0,sz(a)) { if (inside(l, a[i].fi, a[i].se) || inside(r, a[i].fi, a[i].se)) return false; if (inside(a[i].fi, l, r) || inside(a[i].se, l, r)) return false; } return true; } int main() { setIO(); int n; cin >> n; FOR(i,0,n) { ll s, d; cin >> s >> d; if (check(s, s + d - 1)) { cout << s << " " << s + d - 1 << "\n"; a.pb(mp(s, s + d - 1)); continue; } vl ch{1}; FOR(j,0,sz(a)) { ch.pb(a[j].se + 1); } sort(all(ch)); FOR(j,0,sz(ch)) { ll cl = ch[j], cr = cl + d - 1; if (check(cl, cr)) { cout << cl << " " << cr << "\n"; a.pb(mp(cl, cr)); break; } } } return 0; } ```

-1

432

A

Choosing Teams

PROGRAMMING

800

[ "greedy", "implementation", "sortings" ]

null

The Saratov State University Olympiad Programmers Training Center (SSU OPTC) has *n* students. For each student you know the number of times he/she has participated in the ACM ICPC world programming championship. According to the ACM ICPC rules, each person can participate in the world championship at most 5 times. The head of the SSU OPTC is recently gathering teams to participate in the world championship. Each team must consist of exactly three people, at that, any person cannot be a member of two or more teams. What maximum number of teams can the head make if he wants each team to participate in the world championship with the same members at least *k* times?

The first line contains two integers, *n* and *k* (1<=≤<=*n*<=≤<=2000; 1<=≤<=*k*<=≤<=5). The next line contains *n* integers: *y*1,<=*y*2,<=...,<=*y**n* (0<=≤<=*y**i*<=≤<=5), where *y**i* shows the number of times the *i*-th person participated in the ACM ICPC world championship.

Print a single number — the answer to the problem.

[ "5 2\n0 4 5 1 0\n", "6 4\n0 1 2 3 4 5\n", "6 5\n0 0 0 0 0 0\n" ]

[ "1\n", "0\n", "2\n" ]

In the first sample only one team could be made: the first, the fourth and the fifth participants. In the second sample no teams could be created. In the third sample two teams could be created. Any partition into two teams fits.

500

[ { "input": "5 2\n0 4 5 1 0", "output": "1" }, { "input": "6 4\n0 1 2 3 4 5", "output": "0" }, { "input": "6 5\n0 0 0 0 0 0", "output": "2" }, { "input": "3 4\n0 1 0", "output": "1" }, { "input": "3 4\n0 2 0", "output": "0" }, { "input": "6 5\n0 0 0 0 0 0", "output": "2" }, { "input": "12 2\n0 1 2 3 4 0 1 2 3 4 0 1", "output": "3" }, { "input": "15 2\n0 1 2 3 4 0 1 2 3 4 0 1 2 3 4", "output": "4" }, { "input": "13 1\n5 0 5 0 1 5 0 4 1 1 1 4 1", "output": "3" }, { "input": "20 1\n5 0 4 2 2 3 2 1 2 4 1 3 5 5 5 4 4 1 3 0", "output": "5" }, { "input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4", "output": "4" }, { "input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4", "output": "4" }, { "input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4", "output": "4" }, { "input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4", "output": "4" }, { "input": "6 3\n4 4 4 4 4 4", "output": "0" }, { "input": "1 4\n1", "output": "0" }, { "input": "1 1\n0", "output": "0" }, { "input": "2 3\n0 0", "output": "0" } ]

1,651,636,702

2,147,483,647

Python 3

OK

TESTS

35

46

0

n,k=map(int,input().split()) lst=list(map(int,input().split())) ctr=0 for num in lst: if 5-num>=k: ctr+=1 print(ctr//3)

Title: Choosing Teams Time Limit: None seconds Memory Limit: None megabytes Problem Description: The Saratov State University Olympiad Programmers Training Center (SSU OPTC) has *n* students. For each student you know the number of times he/she has participated in the ACM ICPC world programming championship. According to the ACM ICPC rules, each person can participate in the world championship at most 5 times. The head of the SSU OPTC is recently gathering teams to participate in the world championship. Each team must consist of exactly three people, at that, any person cannot be a member of two or more teams. What maximum number of teams can the head make if he wants each team to participate in the world championship with the same members at least *k* times? Input Specification: The first line contains two integers, *n* and *k* (1<=≤<=*n*<=≤<=2000; 1<=≤<=*k*<=≤<=5). The next line contains *n* integers: *y*1,<=*y*2,<=...,<=*y**n* (0<=≤<=*y**i*<=≤<=5), where *y**i* shows the number of times the *i*-th person participated in the ACM ICPC world championship. Output Specification: Print a single number — the answer to the problem. Demo Input: ['5 2\n0 4 5 1 0\n', '6 4\n0 1 2 3 4 5\n', '6 5\n0 0 0 0 0 0\n'] Demo Output: ['1\n', '0\n', '2\n'] Note: In the first sample only one team could be made: the first, the fourth and the fifth participants. In the second sample no teams could be created. In the third sample two teams could be created. Any partition into two teams fits.

```python n,k=map(int,input().split()) lst=list(map(int,input().split())) ctr=0 for num in lst: if 5-num>=k: ctr+=1 print(ctr//3) ```

3

96

A

Football

PROGRAMMING

900

[ "implementation", "strings" ]

A. Football

2

256

Petya loves football very much. One day, as he was watching a football match, he was writing the players' current positions on a piece of paper. To simplify the situation he depicted it as a string consisting of zeroes and ones. A zero corresponds to players of one team; a one corresponds to players of another team. If there are at least 7 players of some team standing one after another, then the situation is considered dangerous. For example, the situation 00100110111111101 is dangerous and 11110111011101 is not. You are given the current situation. Determine whether it is dangerous or not.

The first input line contains a non-empty string consisting of characters "0" and "1", which represents players. The length of the string does not exceed 100 characters. There's at least one player from each team present on the field.

Print "YES" if the situation is dangerous. Otherwise, print "NO".

[ "001001\n", "1000000001\n" ]

[ "NO\n", "YES\n" ]

none

500

[ { "input": "001001", "output": "NO" }, { "input": "1000000001", "output": "YES" }, { "input": "00100110111111101", "output": "YES" }, { "input": "11110111111111111", "output": "YES" }, { "input": "01", "output": "NO" }, { "input": "10100101", "output": "NO" }, { "input": "1010010100000000010", "output": "YES" }, { "input": "101010101", "output": "NO" }, { "input": "000000000100000000000110101100000", "output": "YES" }, { "input": "100001000000110101100000", "output": "NO" }, { "input": "100001000011010110000", "output": "NO" }, { "input": "010", "output": "NO" }, { "input": "10101011111111111111111111111100", "output": "YES" }, { "input": "1001101100", "output": "NO" }, { "input": "1001101010", "output": "NO" }, { "input": "1111100111", "output": "NO" }, { "input": "00110110001110001111", "output": "NO" }, { "input": "11110001001111110001", "output": "NO" }, { "input": "10001111001011111101", "output": "NO" }, { "input": "10000010100000001000110001010100001001001010011", "output": "YES" }, { "input": "01111011111010111100101100001011001010111110000010", "output": "NO" }, { "input": "00100000100100101110011001011011101110110110010100", "output": "NO" }, { "input": "10110100110001001011110101110010100010000000000100101010111110111110100011", "output": "YES" }, { "input": "00011101010101111001011011001101101011111101000010100000111000011100101011", "output": "NO" }, { "input": "01110000110100110101110100111000101101011101011110110100100111100001110111", "output": "NO" }, { "input": "11110110011000100111100111101101011111110100010101011011111101110110110111", "output": "YES" }, { "input": "100100010101110010001011001110100011100010011110100101100011010001001010001001101111001100", "output": "NO" }, { "input": "111110010001011010010011111100110110001111000010100011011100111101111101110010101111011110000001010", "output": "NO" }, { "input": "111110111100010100000100001010111011101011000111011011011010110010100010000101011111000011010011110", "output": "NO" }, { "input": "1011110110111010110111111010010010100011111011110000011000110010011110111010110100011010100010111000", "output": "NO" }, { "input": "0010100111100010110110000011100111110100111110001010000100111111111010111100101101010101001011010110", "output": "YES" }, { "input": "1111010100010100101011101100101101110011000010100010000001111100010011100101010001101111000001011000", "output": "NO" }, { "input": "0101100011001110001110100111100011010101011000000000110110010010111100101111010111100011101100100101", "output": "YES" }, { "input": "0001101110011101110000000010011111101001101111100001001010110000110001100000010001111011011110001101", "output": "YES" }, { "input": "1000010000100000100010000100001000010000100001000010000100001000010000100001000010000100001000010000", "output": "NO" }, { "input": "1000000000100000000010000000001000000000100000000010000000001000000000100000000010000000001000000000", "output": "YES" }, { "input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001", "output": "YES" }, { "input": "1111111111111111111111111111111111111111011111111111111111111111111111111111111111111111111111111111", "output": "YES" }, { "input": "10100101000", "output": "NO" }, { "input": "11110111011101", "output": "NO" }, { "input": "10000000", "output": "YES" }, { "input": "00000001", "output": "YES" }, { "input": "01111111", "output": "YES" }, { "input": "11111110", "output": "YES" } ]

1,676,054,485

2,147,483,647

Python 3

OK

TESTS

44

92

0

s=input() t=[] t.extend(s) cnt=1 maxi=t[0] flag=False for i in range(1,len(t)): if maxi==t[i] : cnt+=1 maxi=t[i] else : cnt=1 maxi=t[i] if cnt==7: flag=True break if flag : print('YES') else : print('NO')

Title: Football Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Petya loves football very much. One day, as he was watching a football match, he was writing the players' current positions on a piece of paper. To simplify the situation he depicted it as a string consisting of zeroes and ones. A zero corresponds to players of one team; a one corresponds to players of another team. If there are at least 7 players of some team standing one after another, then the situation is considered dangerous. For example, the situation 00100110111111101 is dangerous and 11110111011101 is not. You are given the current situation. Determine whether it is dangerous or not. Input Specification: The first input line contains a non-empty string consisting of characters "0" and "1", which represents players. The length of the string does not exceed 100 characters. There's at least one player from each team present on the field. Output Specification: Print "YES" if the situation is dangerous. Otherwise, print "NO". Demo Input: ['001001\n', '1000000001\n'] Demo Output: ['NO\n', 'YES\n'] Note: none

```python s=input() t=[] t.extend(s) cnt=1 maxi=t[0] flag=False for i in range(1,len(t)): if maxi==t[i] : cnt+=1 maxi=t[i] else : cnt=1 maxi=t[i] if cnt==7: flag=True break if flag : print('YES') else : print('NO') ```

3.977

160

A

Twins

PROGRAMMING

900

[ "greedy", "sortings" ]

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" } ]

1,697,638,701

2,147,483,647

PyPy 3-64

WRONG_ANSWER

TESTS

2

92

0

n = int(input()) m = sorted([int(i) for i in input().split()]) num, mid, coins = 1, sum(m)/2, m.pop(-1) while coins <= mid: if coins + m[-1] <= mid: num += 1 coins += m.pop(-1) else: print(num + 1) break

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 n = int(input()) m = sorted([int(i) for i in input().split()]) num, mid, coins = 1, sum(m)/2, m.pop(-1) while coins <= mid: if coins + m[-1] <= mid: num += 1 coins += m.pop(-1) else: print(num + 1) break ```

0

807

A

Is it rated?

PROGRAMMING

900

[ "implementation", "sortings" ]

null

Is it rated? Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it. Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known. It's known that if at least one participant's rating has changed, then the round was rated for sure. It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed. In this problem, you should not make any other assumptions about the rating system. Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.

The first line contains a single integer *n* (2<=≤<=*n*<=≤<=1000) — the number of round participants. Each of the next *n* lines contains two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=4126) — the rating of the *i*-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.

If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".

[ "6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884\n", "4\n1500 1500\n1300 1300\n1200 1200\n1400 1400\n", "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699\n" ]

[ "rated\n", "unrated\n", "maybe\n" ]

In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated. In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure. In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.

500

[ { "input": "6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884", "output": "rated" }, { "input": "4\n1500 1500\n1300 1300\n1200 1200\n1400 1400", "output": "unrated" }, { "input": "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699", "output": "maybe" }, { "input": "2\n1 1\n1 1", "output": "maybe" }, { "input": "2\n4126 4126\n4126 4126", "output": "maybe" }, { "input": "10\n446 446\n1331 1331\n3594 3594\n1346 1902\n91 91\n3590 3590\n2437 2437\n4007 3871\n2797 699\n1423 1423", "output": "rated" }, { "input": "10\n4078 4078\n2876 2876\n1061 1061\n3721 3721\n143 143\n2992 2992\n3279 3279\n3389 3389\n1702 1702\n1110 1110", "output": "unrated" }, { "input": "10\n4078 4078\n3721 3721\n3389 3389\n3279 3279\n2992 2992\n2876 2876\n1702 1702\n1110 1110\n1061 1061\n143 143", "output": "maybe" }, { "input": "2\n3936 3936\n2967 2967", "output": "maybe" }, { "input": "2\n1 1\n2 2", "output": "unrated" }, { "input": "2\n2 2\n1 1", "output": "maybe" }, { "input": "2\n2 1\n1 2", "output": "rated" }, { "input": "2\n2967 2967\n3936 3936", "output": "unrated" }, { "input": "3\n1200 1200\n1200 1200\n1300 1300", "output": "unrated" }, { "input": "3\n3 3\n2 2\n1 1", "output": "maybe" }, { "input": "3\n1 1\n1 1\n2 2", "output": "unrated" }, { "input": "2\n3 2\n3 2", "output": "rated" }, { "input": "3\n5 5\n4 4\n3 4", "output": "rated" }, { "input": "3\n200 200\n200 200\n300 300", "output": "unrated" }, { "input": "3\n1 1\n2 2\n3 3", "output": "unrated" }, { "input": "5\n3123 3123\n2777 2777\n2246 2246\n2245 2245\n1699 1699", "output": "maybe" }, { "input": "2\n10 10\n8 8", "output": "maybe" }, { "input": "3\n1500 1500\n1500 1500\n1600 1600", "output": "unrated" }, { "input": "3\n1500 1500\n1500 1500\n1700 1700", "output": "unrated" }, { "input": "4\n100 100\n100 100\n70 70\n80 80", "output": "unrated" }, { "input": "2\n1 2\n2 1", "output": "rated" }, { "input": "3\n5 5\n4 3\n3 3", "output": "rated" }, { "input": "3\n1600 1650\n1500 1550\n1400 1450", "output": "rated" }, { "input": "4\n2000 2000\n1500 1500\n1500 1500\n1700 1700", "output": "unrated" }, { "input": "4\n1500 1500\n1400 1400\n1400 1400\n1700 1700", "output": "unrated" }, { "input": "2\n1600 1600\n1400 1400", "output": "maybe" }, { "input": "2\n3 1\n9 8", "output": "rated" }, { "input": "2\n2 1\n1 1", "output": "rated" }, { "input": "4\n4123 4123\n4123 4123\n2670 2670\n3670 3670", "output": "unrated" }, { "input": "2\n2 2\n3 3", "output": "unrated" }, { "input": "2\n10 11\n5 4", "output": "rated" }, { "input": "2\n15 14\n13 12", "output": "rated" }, { "input": "2\n2 1\n2 2", "output": "rated" }, { "input": "3\n2670 2670\n3670 3670\n4106 4106", "output": "unrated" }, { "input": "3\n4 5\n3 3\n2 2", "output": "rated" }, { "input": "2\n10 9\n10 10", "output": "rated" }, { "input": "3\n1011 1011\n1011 999\n2200 2100", "output": "rated" }, { "input": "2\n3 3\n5 5", "output": "unrated" }, { "input": "2\n1500 1500\n3000 2000", "output": "rated" }, { "input": "2\n5 6\n5 5", "output": "rated" }, { "input": "3\n2000 2000\n1500 1501\n500 500", "output": "rated" }, { "input": "2\n2 3\n2 2", "output": "rated" }, { "input": "2\n3 3\n2 2", "output": "maybe" }, { "input": "2\n1 2\n1 1", "output": "rated" }, { "input": "4\n3123 3123\n2777 2777\n2246 2246\n1699 1699", "output": "maybe" }, { "input": "2\n15 14\n14 13", "output": "rated" }, { "input": "4\n3000 3000\n2900 2900\n3000 3000\n2900 2900", "output": "unrated" }, { "input": "6\n30 3060\n24 2194\n26 2903\n24 2624\n37 2991\n24 2884", "output": "rated" }, { "input": "2\n100 99\n100 100", "output": "rated" }, { "input": "4\n2 2\n1 1\n1 1\n2 2", "output": "unrated" }, { "input": "3\n100 101\n100 100\n100 100", "output": "rated" }, { "input": "4\n1000 1001\n900 900\n950 950\n890 890", "output": "rated" }, { "input": "2\n2 3\n1 1", "output": "rated" }, { "input": "2\n2 2\n1 1", "output": "maybe" }, { "input": "2\n3 2\n2 2", "output": "rated" }, { "input": "2\n3 2\n3 3", "output": "rated" }, { "input": "2\n1 1\n2 2", "output": "unrated" }, { "input": "3\n3 2\n3 3\n3 3", "output": "rated" }, { "input": "4\n1500 1501\n1300 1300\n1200 1200\n1400 1400", "output": "rated" }, { "input": "3\n1000 1000\n500 500\n400 300", "output": "rated" }, { "input": "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n3000 3000", "output": "unrated" }, { "input": "2\n1 1\n2 3", "output": "rated" }, { "input": "2\n6 2\n6 2", "output": "rated" }, { "input": "5\n3123 3123\n1699 1699\n2777 2777\n2246 2246\n2246 2246", "output": "unrated" }, { "input": "2\n1500 1500\n1600 1600", "output": "unrated" }, { "input": "5\n3123 3123\n2777 2777\n2246 2246\n2241 2241\n1699 1699", "output": "maybe" }, { "input": "2\n20 30\n10 5", "output": "rated" }, { "input": "3\n1 1\n2 2\n1 1", "output": "unrated" }, { "input": "2\n1 2\n3 3", "output": "rated" }, { "input": "5\n5 5\n4 4\n3 3\n2 2\n1 1", "output": "maybe" }, { "input": "2\n2 2\n2 1", "output": "rated" }, { "input": "2\n100 100\n90 89", "output": "rated" }, { "input": "2\n1000 900\n2000 2000", "output": "rated" }, { "input": "2\n50 10\n10 50", "output": "rated" }, { "input": "2\n200 200\n100 100", "output": "maybe" }, { "input": "3\n2 2\n2 2\n3 3", "output": "unrated" }, { "input": "3\n1000 1000\n300 300\n100 100", "output": "maybe" }, { "input": "4\n2 2\n2 2\n3 3\n4 4", "output": "unrated" }, { "input": "2\n5 3\n6 3", "output": "rated" }, { "input": "2\n1200 1100\n1200 1000", "output": "rated" }, { "input": "2\n5 5\n4 4", "output": "maybe" }, { "input": "2\n5 5\n3 3", "output": "maybe" }, { "input": "5\n1500 1500\n1300 1300\n1200 1200\n1400 1400\n1100 1100", "output": "unrated" }, { "input": "5\n10 10\n9 9\n8 8\n7 7\n6 6", "output": "maybe" }, { "input": "3\n1000 1000\n300 300\n10 10", "output": "maybe" }, { "input": "5\n6 6\n5 5\n4 4\n3 3\n2 2", "output": "maybe" }, { "input": "2\n3 3\n1 1", "output": "maybe" }, { "input": "4\n2 2\n2 2\n2 2\n3 3", "output": "unrated" }, { "input": "2\n1000 1000\n700 700", "output": "maybe" }, { "input": "2\n4 3\n5 3", "output": "rated" }, { "input": "2\n1000 1000\n1100 1100", "output": "unrated" }, { "input": "4\n5 5\n4 4\n3 3\n2 2", "output": "maybe" }, { "input": "3\n1 1\n2 3\n2 2", "output": "rated" }, { "input": "2\n1 2\n1 3", "output": "rated" }, { "input": "2\n3 3\n1 2", "output": "rated" }, { "input": "4\n1501 1500\n1300 1300\n1200 1200\n1400 1400", "output": "rated" }, { "input": "5\n1 1\n2 2\n3 3\n4 4\n5 5", "output": "unrated" }, { "input": "2\n10 10\n1 2", "output": "rated" }, { "input": "6\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699\n1900 1900", "output": "unrated" }, { "input": "6\n3123 3123\n2777 2777\n3000 3000\n2246 2246\n2246 2246\n1699 1699", "output": "unrated" }, { "input": "2\n100 100\n110 110", "output": "unrated" }, { "input": "3\n3 3\n3 3\n4 4", "output": "unrated" }, { "input": "3\n3 3\n3 2\n4 4", "output": "rated" }, { "input": "3\n5 2\n4 4\n3 3", "output": "rated" }, { "input": "4\n4 4\n3 3\n2 2\n1 1", "output": "maybe" }, { "input": "2\n1 1\n3 2", "output": "rated" }, { "input": "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n2699 2699", "output": "unrated" }, { "input": "3\n3 3\n3 3\n3 4", "output": "rated" }, { "input": "3\n1 2\n2 2\n3 3", "output": "rated" }, { "input": "3\n1 2\n1 2\n1 2", "output": "rated" }, { "input": "2\n2 1\n2 1", "output": "rated" }, { "input": "2\n1 2\n3 4", "output": "rated" }, { "input": "2\n3 2\n2 3", "output": "rated" }, { "input": "3\n1500 1500\n1600 1600\n1600 1600", "output": "unrated" }, { "input": "3\n1 1\n3 3\n4 4", "output": "unrated" }, { "input": "3\n1 1\n2 2\n2 2", "output": "unrated" }, { "input": "2\n10 12\n8 8", "output": "rated" }, { "input": "5\n1200 1200\n1500 1500\n1500 1500\n1500 1500\n1500 1500", "output": "unrated" }, { "input": "2\n1 2\n2 2", "output": "rated" }, { "input": "3\n1500 1400\n1200 1200\n1100 1100", "output": "rated" }, { "input": "2\n10 12\n10 10", "output": "rated" }, { "input": "3\n1500 1500\n1400 1400\n1300 1300", "output": "maybe" }, { "input": "3\n3 3\n4 4\n5 5", "output": "unrated" }, { "input": "3\n2 6\n3 5\n4 4", "output": "rated" }, { "input": "2\n5 6\n4 6", "output": "rated" }, { "input": "4\n10 10\n10 10\n7 7\n8 8", "output": "unrated" }, { "input": "2\n4 4\n3 3", "output": "maybe" } ]

1,651,659,759

2,147,483,647

PyPy 3-64

OK

TESTS

150

93

2,355,200

a, b = [], [] for i in range(int(input())): x, y = map(int,input().split()) a.append(x) b.append(y) print('rated' if a != b else 'maybe' if sorted(a, reverse = True) == b else 'unrated')

Title: Is it rated? Time Limit: None seconds Memory Limit: None megabytes Problem Description: Is it rated? Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it. Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known. It's known that if at least one participant's rating has changed, then the round was rated for sure. It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed. In this problem, you should not make any other assumptions about the rating system. Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not. Input Specification: The first line contains a single integer *n* (2<=≤<=*n*<=≤<=1000) — the number of round participants. Each of the next *n* lines contains two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=4126) — the rating of the *i*-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings. Output Specification: If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe". Demo Input: ['6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884\n', '4\n1500 1500\n1300 1300\n1200 1200\n1400 1400\n', '5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699\n'] Demo Output: ['rated\n', 'unrated\n', 'maybe\n'] Note: In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated. In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure. In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.

```python a, b = [], [] for i in range(int(input())): x, y = map(int,input().split()) a.append(x) b.append(y) print('rated' if a != b else 'maybe' if sorted(a, reverse = True) == b else 'unrated') ```

3

330

B

Road Construction

PROGRAMMING

1,300

[ "constructive algorithms", "graphs" ]

null

A country has *n* cities. Initially, there is no road in the country. One day, the king decides to construct some roads connecting pairs of cities. Roads can be traversed either way. He wants those roads to be constructed in such a way that it is possible to go from each city to any other city by traversing at most two roads. You are also given *m* pairs of cities — roads cannot be constructed between these pairs of cities. Your task is to construct the minimum number of roads that still satisfy the above conditions. The constraints will guarantee that this is always possible.

The first line consists of two integers *n* and *m* . Then *m* lines follow, each consisting of two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*, *a**i*<=≠<=*b**i*), which means that it is not possible to construct a road connecting cities *a**i* and *b**i*. Consider the cities are numbered from 1 to *n*. It is guaranteed that every pair of cities will appear at most once in the input.

You should print an integer *s*: the minimum number of roads that should be constructed, in the first line. Then *s* lines should follow, each consisting of two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=*a**i*<=≠<=*b**i*), which means that a road should be constructed between cities *a**i* and *b**i*. If there are several solutions, you may print any of them.

[ "4 1\n1 3\n" ]

[ "3\n1 2\n4 2\n2 3\n" ]

This is one possible solution of the example: These are examples of wrong solutions:

1,000

[ { "input": "4 1\n1 3", "output": "3\n1 2\n4 2\n2 3" }, { "input": "1000 0", "output": "999\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84\n1 85\n1 86\n1 87..." }, { "input": "484 11\n414 97\n414 224\n444 414\n414 483\n414 399\n414 484\n414 189\n414 246\n414 115\n89 414\n14 414", "output": "483\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84\n1 85\n1 86\n1 87..." }, { "input": "150 3\n112 30\n61 45\n37 135", "output": "149\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84\n1 85\n1 86\n1 87..." }, { "input": "34 7\n10 28\n10 19\n10 13\n24 10\n10 29\n20 10\n10 26", "output": "33\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34" }, { "input": "1000 48\n816 885\n576 357\n878 659\n610 647\n37 670\n192 184\n393 407\n598 160\n547 995\n177 276\n788 44\n14 184\n604 281\n176 97\n176 293\n10 57\n852 579\n223 669\n313 260\n476 691\n667 22\n851 792\n411 489\n526 66\n233 566\n35 396\n964 815\n672 123\n148 210\n163 339\n379 598\n382 675\n132 955\n221 441\n253 490\n856 532\n135 119\n276 319\n525 835\n996 270\n92 778\n434 369\n351 927\n758 983\n798 267\n272 830\n539 728\n166 26", "output": "999\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84\n1 85\n1 86\n1 87..." }, { "input": "534 0", "output": "533\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84\n1 85\n1 86\n1 87..." }, { "input": "226 54\n80 165\n2 53\n191 141\n107 207\n95 196\n61 82\n42 168\n118 94\n205 182\n172 160\n84 224\n113 143\n122 93\n37 209\n176 32\n56 83\n151 81\n70 190\n99 171\n68 204\n212 48\n4 67\n116 7\n206 199\n105 62\n158 51\n178 147\n17 129\n22 47\n72 162\n188 77\n24 111\n184 26\n175 128\n110 89\n139 120\n127 92\n121 39\n217 75\n145 69\n20 161\n30 220\n222 154\n54 46\n21 87\n144 185\n164 115\n73 202\n173 35\n9 132\n74 180\n137 5\n157 117\n31 177", "output": "225\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84\n1 85\n1 86\n1 87..." }, { "input": "84 3\n39 19\n55 73\n42 43", "output": "83\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84" }, { "input": "207 35\n34 116\n184 5\n90 203\n12 195\n138 101\n40 150\n189 109\n115 91\n93 201\n106 18\n51 187\n139 197\n168 130\n182 64\n31 42\n86 107\n158 111\n159 132\n119 191\n53 127\n81 13\n153 112\n38 2\n87 84\n121 82\n120 22\n21 177\n151 202\n23 58\n68 192\n29 46\n105 70\n8 167\n56 54\n149 15", "output": "206\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84\n1 85\n1 86\n1 87..." }, { "input": "91 37\n50 90\n26 82\n61 1\n50 17\n51 73\n45 9\n39 53\n78 35\n12 45\n43 47\n83 20\n9 59\n18 48\n68 31\n47 33\n10 25\n15 78\n5 3\n73 65\n77 4\n62 31\n73 3\n53 7\n29 58\n52 14\n56 20\n6 87\n71 16\n17 19\n77 86\n1 50\n74 79\n15 54\n55 80\n13 77\n4 69\n24 69", "output": "90\n2 1\n2 3\n2 4\n2 5\n2 6\n2 7\n2 8\n2 9\n2 10\n2 11\n2 12\n2 13\n2 14\n2 15\n2 16\n2 17\n2 18\n2 19\n2 20\n2 21\n2 22\n2 23\n2 24\n2 25\n2 26\n2 27\n2 28\n2 29\n2 30\n2 31\n2 32\n2 33\n2 34\n2 35\n2 36\n2 37\n2 38\n2 39\n2 40\n2 41\n2 42\n2 43\n2 44\n2 45\n2 46\n2 47\n2 48\n2 49\n2 50\n2 51\n2 52\n2 53\n2 54\n2 55\n2 56\n2 57\n2 58\n2 59\n2 60\n2 61\n2 62\n2 63\n2 64\n2 65\n2 66\n2 67\n2 68\n2 69\n2 70\n2 71\n2 72\n2 73\n2 74\n2 75\n2 76\n2 77\n2 78\n2 79\n2 80\n2 81\n2 82\n2 83\n2 84\n2 85\n2 86\n2 87\n..." }, { "input": "226 54\n197 107\n181 146\n218 115\n36 169\n199 196\n116 93\n152 75\n213 164\n156 95\n165 58\n90 42\n141 58\n203 221\n179 204\n186 69\n27 127\n76 189\n40 195\n111 29\n85 189\n45 88\n84 135\n82 186\n185 17\n156 217\n8 123\n179 112\n92 137\n114 89\n10 152\n132 24\n135 36\n61 218\n10 120\n155 102\n222 79\n150 92\n184 34\n102 180\n154 196\n171 9\n217 105\n84 207\n56 189\n152 179\n43 165\n115 209\n208 167\n52 14\n92 47\n197 95\n13 78\n222 138\n75 36", "output": "225\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84\n1 85\n1 86\n1 87..." }, { "input": "207 35\n154 79\n174 101\n189 86\n137 56\n66 23\n199 69\n18 28\n32 53\n13 179\n182 170\n199 12\n24 158\n105 133\n25 10\n40 162\n64 72\n108 9\n172 125\n43 190\n15 39\n128 150\n102 129\n90 97\n64 196\n70 123\n163 41\n12 126\n127 186\n107 23\n182 51\n29 46\n46 123\n89 35\n59 80\n206 171", "output": "206\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84\n1 85\n1 86\n1 87..." }, { "input": "84 0", "output": "83\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84" }, { "input": "226 54\n5 29\n130 29\n55 29\n19 29\n29 92\n29 38\n185 29\n29 150\n29 202\n29 25\n29 66\n184 29\n29 189\n177 29\n50 29\n87 29\n138 29\n29 48\n151 29\n125 29\n16 29\n42 29\n29 157\n90 29\n21 29\n29 45\n29 80\n29 67\n29 26\n29 173\n74 29\n29 193\n29 40\n172 29\n29 85\n29 102\n88 29\n29 182\n116 29\n180 29\n161 29\n10 29\n171 29\n144 29\n29 218\n190 29\n213 29\n29 71\n29 191\n29 160\n29 137\n29 58\n29 135\n127 29", "output": "225\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84\n1 85\n1 86\n1 87..." }, { "input": "207 35\n25 61\n188 61\n170 61\n113 61\n35 61\n61 177\n77 61\n61 39\n61 141\n116 61\n61 163\n30 61\n192 61\n19 61\n61 162\n61 133\n185 61\n8 61\n118 61\n61 115\n7 61\n61 105\n107 61\n61 11\n161 61\n61 149\n136 61\n82 61\n20 61\n151 61\n156 61\n12 61\n87 61\n61 205\n61 108", "output": "206\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84\n1 85\n1 86\n1 87..." }, { "input": "34 7\n11 32\n33 29\n17 16\n15 5\n13 25\n8 19\n20 4", "output": "33\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34" }, { "input": "43 21\n38 19\n43 8\n40 31\n3 14\n24 21\n12 17\n1 9\n5 27\n25 37\n11 6\n13 26\n16 22\n10 32\n36 7\n30 29\n42 35\n20 33\n4 23\n18 15\n41 34\n2 28", "output": "42\n39 1\n39 2\n39 3\n39 4\n39 5\n39 6\n39 7\n39 8\n39 9\n39 10\n39 11\n39 12\n39 13\n39 14\n39 15\n39 16\n39 17\n39 18\n39 19\n39 20\n39 21\n39 22\n39 23\n39 24\n39 25\n39 26\n39 27\n39 28\n39 29\n39 30\n39 31\n39 32\n39 33\n39 34\n39 35\n39 36\n39 37\n39 38\n39 40\n39 41\n39 42\n39 43" }, { "input": "34 7\n22 4\n5 25\n15 7\n5 9\n27 7\n34 21\n3 13", "output": "33\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34" }, { "input": "50 7\n19 37\n30 32\n43 20\n48 14\n30 29\n18 36\n9 46", "output": "49\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50" }, { "input": "41 12\n41 12\n29 13\n3 37\n2 20\n4 24\n27 6\n39 20\n28 41\n30 1\n35 9\n5 39\n12 31", "output": "40\n7 1\n7 2\n7 3\n7 4\n7 5\n7 6\n7 8\n7 9\n7 10\n7 11\n7 12\n7 13\n7 14\n7 15\n7 16\n7 17\n7 18\n7 19\n7 20\n7 21\n7 22\n7 23\n7 24\n7 25\n7 26\n7 27\n7 28\n7 29\n7 30\n7 31\n7 32\n7 33\n7 34\n7 35\n7 36\n7 37\n7 38\n7 39\n7 40\n7 41" }, { "input": "26 2\n22 1\n14 10", "output": "25\n2 1\n2 3\n2 4\n2 5\n2 6\n2 7\n2 8\n2 9\n2 10\n2 11\n2 12\n2 13\n2 14\n2 15\n2 16\n2 17\n2 18\n2 19\n2 20\n2 21\n2 22\n2 23\n2 24\n2 25\n2 26" }, { "input": "7 3\n4 7\n4 6\n6 5", "output": "6\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7" }, { "input": "484 11\n459 166\n12 217\n14 231\n477 111\n365 225\n62 412\n221 223\n258 184\n266 404\n425 236\n83 269", "output": "483\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84\n1 85\n1 86\n1 87..." }, { "input": "50 24\n18 36\n11 18\n22 18\n18 47\n18 10\n45 18\n18 29\n13 18\n8 18\n18 30\n26 18\n18 6\n5 18\n18 7\n46 18\n50 18\n18 3\n18 32\n4 18\n18 24\n18 19\n23 18\n35 18\n31 18", "output": "49\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50" }, { "input": "7 3\n3 4\n7 3\n3 2", "output": "6\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7" }, { "input": "1 0", "output": "0" }, { "input": "2 0", "output": "1\n1 2" }, { "input": "4 1\n1 2", "output": "3\n3 1\n3 2\n3 4" }, { "input": "3 1\n1 2", "output": "2\n3 1\n3 2" }, { "input": "5 1\n1 5", "output": "4\n2 1\n2 3\n2 4\n2 5" } ]

1,621,642,723

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

0

62

0

n, m = map(int, input().split()) arr = [False]*(n+1) for i in range(m): a, b = map(int, input().split()) arr[a] = arr[b] = True i = 1 while i < n + 1 and not arr[i]: i+=1 print(n-1) for j in range(1, n+1): if j != i: print(i, j)

Title: Road Construction Time Limit: None seconds Memory Limit: None megabytes Problem Description: A country has *n* cities. Initially, there is no road in the country. One day, the king decides to construct some roads connecting pairs of cities. Roads can be traversed either way. He wants those roads to be constructed in such a way that it is possible to go from each city to any other city by traversing at most two roads. You are also given *m* pairs of cities — roads cannot be constructed between these pairs of cities. Your task is to construct the minimum number of roads that still satisfy the above conditions. The constraints will guarantee that this is always possible. Input Specification: The first line consists of two integers *n* and *m* . Then *m* lines follow, each consisting of two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*, *a**i*<=≠<=*b**i*), which means that it is not possible to construct a road connecting cities *a**i* and *b**i*. Consider the cities are numbered from 1 to *n*. It is guaranteed that every pair of cities will appear at most once in the input. Output Specification: You should print an integer *s*: the minimum number of roads that should be constructed, in the first line. Then *s* lines should follow, each consisting of two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=*a**i*<=≠<=*b**i*), which means that a road should be constructed between cities *a**i* and *b**i*. If there are several solutions, you may print any of them. Demo Input: ['4 1\n1 3\n'] Demo Output: ['3\n1 2\n4 2\n2 3\n'] Note: This is one possible solution of the example: These are examples of wrong solutions:

```python n, m = map(int, input().split()) arr = [False]*(n+1) for i in range(m): a, b = map(int, input().split()) arr[a] = arr[b] = True i = 1 while i < n + 1 and not arr[i]: i+=1 print(n-1) for j in range(1, n+1): if j != i: print(i, j) ```

0

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" } ]

1,675,812,618

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

1

60

0

s=input() c=0 x=0 for i in s: if i.lower(): c+=1 else: x+=1 if c==x: s=s.lower() elif c>x: s=s.lower() elif x>c: s=s.upper() print(s)

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 s=input() c=0 x=0 for i in s: if i.lower(): c+=1 else: x+=1 if c==x: s=s.lower() elif c>x: s=s.lower() elif x>c: s=s.upper() print(s) ```

0

231

A

Team

PROGRAMMING

800

[ "brute force", "greedy" ]

null

One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution. This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution.

The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces.

Print a single integer — the number of problems the friends will implement on the contest.

[ "3\n1 1 0\n1 1 1\n1 0 0\n", "2\n1 0 0\n0 1 1\n" ]

[ "2\n", "1\n" ]

In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it. In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution.

500

[ { "input": "3\n1 1 0\n1 1 1\n1 0 0", "output": "2" }, { "input": "2\n1 0 0\n0 1 1", "output": "1" }, { "input": "1\n1 0 0", "output": "0" }, { "input": "2\n1 0 0\n1 1 1", "output": "1" }, { "input": "5\n1 0 0\n0 1 0\n1 1 1\n0 0 1\n0 0 0", "output": "1" }, { "input": "10\n0 1 0\n0 1 0\n1 1 0\n1 0 0\n0 0 1\n0 1 1\n1 1 1\n1 1 0\n0 0 0\n0 0 0", "output": "4" }, { "input": "15\n0 1 0\n1 0 0\n1 1 0\n1 1 1\n0 1 0\n0 0 1\n1 0 1\n1 0 1\n1 0 1\n0 0 0\n1 1 1\n1 1 0\n0 1 1\n1 1 0\n1 1 1", "output": "10" }, { "input": "50\n0 0 0\n0 1 1\n1 1 1\n0 1 0\n1 0 1\n1 1 1\n0 0 1\n1 0 0\n1 1 0\n1 0 1\n0 1 0\n0 0 1\n1 1 0\n0 1 0\n1 1 0\n0 0 0\n1 1 1\n1 0 1\n0 0 1\n1 1 0\n1 1 1\n0 1 1\n1 1 0\n0 0 0\n0 0 0\n1 1 1\n0 0 0\n1 1 1\n0 1 1\n0 0 1\n0 0 0\n0 0 0\n1 1 0\n1 1 0\n1 0 1\n1 0 0\n1 0 1\n1 0 1\n0 1 1\n1 1 0\n1 1 0\n0 1 0\n1 0 1\n0 0 0\n0 0 0\n0 0 0\n0 0 1\n1 1 1\n0 1 1\n1 0 1", "output": "29" }, { "input": "1\n1 1 1", "output": "1" }, { "input": "8\n0 0 0\n0 0 1\n0 0 0\n0 1 1\n1 0 0\n1 0 1\n1 1 0\n1 1 1", "output": "4" }, { "input": "16\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1", "output": "16" } ]

1,698,095,232

2,147,483,647

Python 3

OK

TESTS

21

92

0

t=int(input()) res=0 while t>0: c=0 lis=map(int,input().split(" ")) for i in lis: if i==1: c+=1 if c>=2: res+=1 t-=1 print(res)

Title: Team Time Limit: None seconds Memory Limit: None megabytes Problem Description: One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution. This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution. Input Specification: The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces. Output Specification: Print a single integer — the number of problems the friends will implement on the contest. Demo Input: ['3\n1 1 0\n1 1 1\n1 0 0\n', '2\n1 0 0\n0 1 1\n'] Demo Output: ['2\n', '1\n'] Note: In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it. In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution.

```python t=int(input()) res=0 while t>0: c=0 lis=map(int,input().split(" ")) for i in lis: if i==1: c+=1 if c>=2: res+=1 t-=1 print(res) ```

3

486

A

Calculating Function

PROGRAMMING

800

[ "implementation", "math" ]

null

For a positive integer *n* let's define a function *f*: *f*(*n*)<==<=<=-<=1<=+<=2<=-<=3<=+<=..<=+<=(<=-<=1)*n**n* Your task is to calculate *f*(*n*) for a given integer *n*.

The single line contains the positive integer *n* (1<=≤<=*n*<=≤<=1015).

Print *f*(*n*) in a single line.

[ "4\n", "5\n" ]

[ "2\n", "-3\n" ]

*f*(4) = - 1 + 2 - 3 + 4 = 2 *f*(5) = - 1 + 2 - 3 + 4 - 5 = - 3

500

[ { "input": "4", "output": "2" }, { "input": "5", "output": "-3" }, { "input": "1000000000", "output": "500000000" }, { "input": "1000000001", "output": "-500000001" }, { "input": "1000000000000000", "output": "500000000000000" }, { "input": "100", "output": "50" }, { "input": "101", "output": "-51" }, { "input": "102", "output": "51" }, { "input": "103", "output": "-52" }, { "input": "104", "output": "52" }, { "input": "105", "output": "-53" }, { "input": "106", "output": "53" }, { "input": "107", "output": "-54" }, { "input": "108", "output": "54" }, { "input": "109", "output": "-55" }, { "input": "208170109961052", "output": "104085054980526" }, { "input": "46017661651072", "output": "23008830825536" }, { "input": "4018154546667", "output": "-2009077273334" }, { "input": "288565475053", "output": "-144282737527" }, { "input": "3052460231", "output": "-1526230116" }, { "input": "29906716", "output": "14953358" }, { "input": "87897701693326", "output": "43948850846663" }, { "input": "8240", "output": "4120" }, { "input": "577935", "output": "-288968" }, { "input": "62", "output": "31" }, { "input": "1", "output": "-1" }, { "input": "2", "output": "1" }, { "input": "9999999999999", "output": "-5000000000000" }, { "input": "1000000000000", "output": "500000000000" }, { "input": "99999999999999", "output": "-50000000000000" }, { "input": "999999999999999", "output": "-500000000000000" }, { "input": "42191359342", "output": "21095679671" }, { "input": "100000000000000", "output": "50000000000000" }, { "input": "145645214654154", "output": "72822607327077" }, { "input": "4294967296", "output": "2147483648" }, { "input": "3037000499", "output": "-1518500250" }, { "input": "10000000000001", "output": "-5000000000001" }, { "input": "100000017040846", "output": "50000008520423" }, { "input": "98979894985999", "output": "-49489947493000" } ]

1,695,379,937

2,147,483,647

Python 3

RUNTIME_ERROR

TESTS

0

30

0

def calculate_f(n): result = 0 sign = -1 for i in range(1, n + 1): result += sign * i sign = -sign return result n = int(input("ادخل عددًا صحيحًا (n): ")) result = calculate_f(n) print(f"f({n}) = {result}")

Title: Calculating Function Time Limit: None seconds Memory Limit: None megabytes Problem Description: For a positive integer *n* let's define a function *f*: *f*(*n*)<==<=<=-<=1<=+<=2<=-<=3<=+<=..<=+<=(<=-<=1)*n**n* Your task is to calculate *f*(*n*) for a given integer *n*. Input Specification: The single line contains the positive integer *n* (1<=≤<=*n*<=≤<=1015). Output Specification: Print *f*(*n*) in a single line. Demo Input: ['4\n', '5\n'] Demo Output: ['2\n', '-3\n'] Note: *f*(4) = - 1 + 2 - 3 + 4 = 2 *f*(5) = - 1 + 2 - 3 + 4 - 5 = - 3

```python def calculate_f(n): result = 0 sign = -1 for i in range(1, n + 1): result += sign * i sign = -sign return result n = int(input("ادخل عددًا صحيحًا (n): ")) result = calculate_f(n) print(f"f({n}) = {result}") ```

-1

58

A

Chat room

PROGRAMMING

1,000

[ "greedy", "strings" ]

A. Chat room

1

256

Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.

The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.

If Vasya managed to say hello, print "YES", otherwise print "NO".

[ "ahhellllloou\n", "hlelo\n" ]

[ "YES\n", "NO\n" ]

none

500

[ { "input": "ahhellllloou", "output": "YES" }, { "input": "hlelo", "output": "NO" }, { "input": "helhcludoo", "output": "YES" }, { "input": "hehwelloho", "output": "YES" }, { "input": "pnnepelqomhhheollvlo", "output": "YES" }, { "input": "tymbzjyqhymedasloqbq", "output": "NO" }, { "input": "yehluhlkwo", "output": "NO" }, { "input": "hatlevhhalrohairnolsvocafgueelrqmlqlleello", "output": "YES" }, { "input": "hhhtehdbllnhwmbyhvelqqyoulretpbfokflhlhreeflxeftelziclrwllrpflflbdtotvlqgoaoqldlroovbfsq", "output": "YES" }, { "input": "rzlvihhghnelqtwlexmvdjjrliqllolhyewgozkuovaiezgcilelqapuoeglnwmnlftxxiigzczlouooi", "output": "YES" }, { "input": "pfhhwctyqdlkrwhebfqfelhyebwllhemtrmeblgrynmvyhioesqklclocxmlffuormljszllpoo", "output": "YES" }, { "input": "lqllcolohwflhfhlnaow", "output": "NO" }, { "input": "heheeellollvoo", "output": "YES" }, { "input": "hellooo", "output": "YES" }, { "input": "o", "output": "NO" }, { "input": "hhqhzeclohlehljlhtesllylrolmomvuhcxsobtsckogdv", "output": "YES" }, { "input": "yoegfuzhqsihygnhpnukluutocvvwuldiighpogsifealtgkfzqbwtmgghmythcxflebrkctlldlkzlagovwlstsghbouk", "output": "YES" }, { "input": "uatqtgbvrnywfacwursctpagasnhydvmlinrcnqrry", "output": "NO" }, { "input": "tndtbldbllnrwmbyhvqaqqyoudrstpbfokfoclnraefuxtftmgzicorwisrpfnfpbdtatvwqgyalqtdtrjqvbfsq", "output": "NO" }, { "input": "rzlvirhgemelnzdawzpaoqtxmqucnahvqnwldklrmjiiyageraijfivigvozgwngiulttxxgzczptusoi", "output": "YES" }, { "input": "kgyelmchocojsnaqdsyeqgnllytbqietpdlgknwwumqkxrexgdcnwoldicwzwofpmuesjuxzrasscvyuqwspm", "output": "YES" }, { "input": "pnyvrcotjvgynbeldnxieghfltmexttuxzyac", "output": "NO" }, { "input": "dtwhbqoumejligbenxvzhjlhosqojetcqsynlzyhfaevbdpekgbtjrbhlltbceobcok", "output": "YES" }, { "input": "crrfpfftjwhhikwzeedrlwzblckkteseofjuxjrktcjfsylmlsvogvrcxbxtffujqshslemnixoeezivksouefeqlhhokwbqjz", "output": "YES" }, { "input": "jhfbndhyzdvhbvhmhmefqllujdflwdpjbehedlsqfdsqlyelwjtyloxwsvasrbqosblzbowlqjmyeilcvotdlaouxhdpoeloaovb", "output": "YES" }, { "input": "hwlghueoemiqtjhhpashjsouyegdlvoyzeunlroypoprnhlyiwiuxrghekaylndhrhllllwhbebezoglydcvykllotrlaqtvmlla", "output": "YES" }, { "input": "wshiaunnqnqxodholbipwhhjmyeblhgpeleblklpzwhdunmpqkbuzloetmwwxmeltkrcomulxauzlwmlklldjodozxryghsnwgcz", "output": "YES" }, { "input": "shvksednttggehroewuiptvvxtrzgidravtnjwuqrlnnkxbplctzkckinpkgjopjfoxdbojtcvsuvablcbkrzajrlhgobkcxeqti", "output": "YES" }, { "input": "hyyhddqhxhekehkwfhlnlsihzefwchzerevcjtokefplholrbvxlltdlafjxrfhleglrvlolojoqaolagtbeyogxlbgfolllslli", "output": "YES" }, { "input": "iaagrdhhelxpdegueiulflquvlzidoprzkehthkldaytrjwhyhprwjxlltinxvuilxohqgjqcvkkdcuoliekcnlilwgqlnlzouoo", "output": "YES" }, { "input": "wfluaeseldgxyvxpwuhkptdmlflnlhktwxiabsvkolsquymrmhzczzoybvlilhmvsuunggvgxzgyyffk", "output": "NO" }, { "input": "loee", "output": "NO" }, { "input": "izdipwylefqmjbuoznfglgcfvedeouflkeehxbasaqmiooppfsqlhprospqgxvzrcpwlfdddngoqvpwvggsnvvxhmjoslwjudjlo", "output": "NO" }, { "input": "pplynflbjlseflaegzqpdxswgunhnlglqhycrbcwdfhclmcrwekkzsoosmfblljeenwfjyifcvxihhhbyhmhlkvtskeglaovjnsi", "output": "YES" }, { "input": "xqjqmenkodmlhzyzmmvofdngktygbbxbzpluzcohohmalkoeuwfikblltaaigv", "output": "NO" }, { "input": "hello", "output": "YES" }, { "input": "oohell", "output": "NO" }, { "input": "hell", "output": "NO" }, { "input": "eloellohoelo", "output": "NO" }, { "input": "helllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllo", "output": "YES" } ]

1,586,783,568

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

2

108

0

a = input() h = a.find('h', 0) e = a.find('e', 1) l1 = a.find('l', 2) l2 = a.find('l', 3) o = a.find('o', 4) if h and e and l1 and l2 and o >= 0: print('YES') else: print('NO')

Title: Chat room Time Limit: 1 seconds Memory Limit: 256 megabytes Problem Description: Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*. Input Specification: The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters. Output Specification: If Vasya managed to say hello, print "YES", otherwise print "NO". Demo Input: ['ahhellllloou\n', 'hlelo\n'] Demo Output: ['YES\n', 'NO\n'] Note: none

```python a = input() h = a.find('h', 0) e = a.find('e', 1) l1 = a.find('l', 2) l2 = a.find('l', 3) o = a.find('o', 4) if h and e and l1 and l2 and o >= 0: print('YES') else: print('NO') ```

0

789

A

Anastasia and pebbles

PROGRAMMING

1,100

[ "implementation", "math" ]

null

Anastasia loves going for a walk in Central Uzhlyandian Park. But she became uninterested in simple walking, so she began to collect Uzhlyandian pebbles. At first, she decided to collect all the pebbles she could find in the park. She has only two pockets. She can put at most *k* pebbles in each pocket at the same time. There are *n* different pebble types in the park, and there are *w**i* pebbles of the *i*-th type. Anastasia is very responsible, so she never mixes pebbles of different types in same pocket. However, she can put different kinds of pebbles in different pockets at the same time. Unfortunately, she can't spend all her time collecting pebbles, so she can collect pebbles from the park only once a day. Help her to find the minimum number of days needed to collect all the pebbles of Uzhlyandian Central Park, taking into consideration that Anastasia can't place pebbles of different types in same pocket.

The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105, 1<=≤<=*k*<=≤<=109) — the number of different pebble types and number of pebbles Anastasia can place in one pocket. The second line contains *n* integers *w*1,<=*w*2,<=...,<=*w**n* (1<=≤<=*w**i*<=≤<=104) — number of pebbles of each type.

The only line of output contains one integer — the minimum number of days Anastasia needs to collect all the pebbles.

[ "3 2\n2 3 4\n", "5 4\n3 1 8 9 7\n" ]

[ "3\n", "5\n" ]

In the first sample case, Anastasia can collect all pebbles of the first type on the first day, of second type — on the second day, and of third type — on the third day. Optimal sequence of actions in the second sample case: - In the first day Anastasia collects 8 pebbles of the third type. - In the second day she collects 8 pebbles of the fourth type. - In the third day she collects 3 pebbles of the first type and 1 pebble of the fourth type. - In the fourth day she collects 7 pebbles of the fifth type. - In the fifth day she collects 1 pebble of the second type.

500

[ { "input": "3 2\n2 3 4", "output": "3" }, { "input": "5 4\n3 1 8 9 7", "output": "5" }, { "input": "1 22\n1", "output": "1" }, { "input": "3 57\n78 165 54", "output": "3" }, { "input": "5 72\n74 10 146 189 184", "output": "6" }, { "input": "9 13\n132 87 200 62 168 51 185 192 118", "output": "48" }, { "input": "1 1\n10000", "output": "5000" }, { "input": "10 1\n1 1 1 1 1 1 1 1 1 1", "output": "5" }, { "input": "2 2\n2 2", "output": "1" } ]

1,648,897,357

2,147,483,647

PyPy 3

OK

TESTS

31

140

9,523,200

import math n,k=map(int,input().split()) ar=list(map(int,input().split())) s=0 for x in range(n): s+=math.ceil(ar[x]/k) print(math.ceil(s/2))

Title: Anastasia and pebbles Time Limit: None seconds Memory Limit: None megabytes Problem Description: Anastasia loves going for a walk in Central Uzhlyandian Park. But she became uninterested in simple walking, so she began to collect Uzhlyandian pebbles. At first, she decided to collect all the pebbles she could find in the park. She has only two pockets. She can put at most *k* pebbles in each pocket at the same time. There are *n* different pebble types in the park, and there are *w**i* pebbles of the *i*-th type. Anastasia is very responsible, so she never mixes pebbles of different types in same pocket. However, she can put different kinds of pebbles in different pockets at the same time. Unfortunately, she can't spend all her time collecting pebbles, so she can collect pebbles from the park only once a day. Help her to find the minimum number of days needed to collect all the pebbles of Uzhlyandian Central Park, taking into consideration that Anastasia can't place pebbles of different types in same pocket. Input Specification: The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105, 1<=≤<=*k*<=≤<=109) — the number of different pebble types and number of pebbles Anastasia can place in one pocket. The second line contains *n* integers *w*1,<=*w*2,<=...,<=*w**n* (1<=≤<=*w**i*<=≤<=104) — number of pebbles of each type. Output Specification: The only line of output contains one integer — the minimum number of days Anastasia needs to collect all the pebbles. Demo Input: ['3 2\n2 3 4\n', '5 4\n3 1 8 9 7\n'] Demo Output: ['3\n', '5\n'] Note: In the first sample case, Anastasia can collect all pebbles of the first type on the first day, of second type — on the second day, and of third type — on the third day. Optimal sequence of actions in the second sample case: - In the first day Anastasia collects 8 pebbles of the third type. - In the second day she collects 8 pebbles of the fourth type. - In the third day she collects 3 pebbles of the first type and 1 pebble of the fourth type. - In the fourth day she collects 7 pebbles of the fifth type. - In the fifth day she collects 1 pebble of the second type.

```python import math n,k=map(int,input().split()) ar=list(map(int,input().split())) s=0 for x in range(n): s+=math.ceil(ar[x]/k) print(math.ceil(s/2)) ```

3

863

A

Quasi-palindrome

PROGRAMMING

900

[ "brute force", "implementation" ]

null

Let quasi-palindromic number be such number that adding some leading zeros (possible none) to it produces a palindromic string. String *t* is called a palindrome, if it reads the same from left to right and from right to left. For example, numbers 131 and 2010200 are quasi-palindromic, they can be transformed to strings "131" and "002010200", respectively, which are palindromes. You are given some integer number *x*. Check if it's a quasi-palindromic number.

The first line contains one integer number *x* (1<=≤<=*x*<=≤<=109). This number is given without any leading zeroes.

Print "YES" if number *x* is quasi-palindromic. Otherwise, print "NO" (without quotes).

[ "131\n", "320\n", "2010200\n" ]

[ "YES\n", "NO\n", "YES\n" ]

none

0

[ { "input": "131", "output": "YES" }, { "input": "320", "output": "NO" }, { "input": "2010200", "output": "YES" }, { "input": "1", "output": "YES" }, { "input": "1000000000", "output": "YES" }, { "input": "999999999", "output": "YES" }, { "input": "999999998", "output": "NO" }, { "input": "102000", "output": "NO" }, { "input": "210000000", "output": "NO" }, { "input": "213443120", "output": "YES" }, { "input": "99", "output": "YES" }, { "input": "22002", "output": "NO" }, { "input": "1010", "output": "YES" }, { "input": "1201", "output": "NO" }, { "input": "6460046", "output": "NO" }, { "input": "503435", "output": "NO" }, { "input": "21002", "output": "NO" }, { "input": "101001", "output": "NO" }, { "input": "200102", "output": "NO" }, { "input": "20010002", "output": "NO" }, { "input": "33003", "output": "NO" }, { "input": "100101", "output": "NO" }, { "input": "1021", "output": "NO" }, { "input": "1101", "output": "NO" }, { "input": "10101100", "output": "NO" }, { "input": "101", "output": "YES" }, { "input": "1011", "output": "NO" }, { "input": "11010", "output": "NO" }, { "input": "10110", "output": "NO" }, { "input": "110000", "output": "YES" }, { "input": "2011", "output": "NO" }, { "input": "10020001", "output": "NO" }, { "input": "12505021", "output": "NO" }, { "input": "12310", "output": "NO" }, { "input": "100501", "output": "NO" }, { "input": "11001", "output": "NO" }, { "input": "20020002", "output": "NO" }, { "input": "202002", "output": "NO" }, { "input": "1001", "output": "YES" }, { "input": "1020021", "output": "NO" }, { "input": "60660", "output": "NO" }, { "input": "98809", "output": "NO" }, { "input": "11000000", "output": "YES" }, { "input": "807008", "output": "NO" } ]

1,585,553,722

2,147,483,647

PyPy 3

OK

TESTS

44

140

0

s = input() s = s.strip('0') l = len(s) for i in range(0, l//2): if s[i] != s[l - 1 - i]: print("NO") exit() print("YES")

Title: Quasi-palindrome Time Limit: None seconds Memory Limit: None megabytes Problem Description: Let quasi-palindromic number be such number that adding some leading zeros (possible none) to it produces a palindromic string. String *t* is called a palindrome, if it reads the same from left to right and from right to left. For example, numbers 131 and 2010200 are quasi-palindromic, they can be transformed to strings "131" and "002010200", respectively, which are palindromes. You are given some integer number *x*. Check if it's a quasi-palindromic number. Input Specification: The first line contains one integer number *x* (1<=≤<=*x*<=≤<=109). This number is given without any leading zeroes. Output Specification: Print "YES" if number *x* is quasi-palindromic. Otherwise, print "NO" (without quotes). Demo Input: ['131\n', '320\n', '2010200\n'] Demo Output: ['YES\n', 'NO\n', 'YES\n'] Note: none

```python s = input() s = s.strip('0') l = len(s) for i in range(0, l//2): if s[i] != s[l - 1 - i]: print("NO") exit() print("YES") ```

3

136

A

Presents

PROGRAMMING

800

[ "implementation" ]

null

Little Petya very much likes gifts. Recently he has received a new laptop as a New Year gift from his mother. He immediately decided to give it to somebody else as what can be more pleasant than giving somebody gifts. And on this occasion he organized a New Year party at his place and invited *n* his friends there. If there's one thing Petya likes more that receiving gifts, that's watching others giving gifts to somebody else. Thus, he safely hid the laptop until the next New Year and made up his mind to watch his friends exchanging gifts while he does not participate in the process. He numbered all his friends with integers from 1 to *n*. Petya remembered that a friend number *i* gave a gift to a friend number *p**i*. He also remembered that each of his friends received exactly one gift. Now Petya wants to know for each friend *i* the number of a friend who has given him a gift.

The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the quantity of friends Petya invited to the party. The second line contains *n* space-separated integers: the *i*-th number is *p**i* — the number of a friend who gave a gift to friend number *i*. It is guaranteed that each friend received exactly one gift. It is possible that some friends do not share Petya's ideas of giving gifts to somebody else. Those friends gave the gifts to themselves.

Print *n* space-separated integers: the *i*-th number should equal the number of the friend who gave a gift to friend number *i*.

[ "4\n2 3 4 1\n", "3\n1 3 2\n", "2\n1 2\n" ]

[ "4 1 2 3\n", "1 3 2\n", "1 2\n" ]

none

500

[ { "input": "4\n2 3 4 1", "output": "4 1 2 3" }, { "input": "3\n1 3 2", "output": "1 3 2" }, { "input": "2\n1 2", "output": "1 2" }, { "input": "1\n1", "output": "1" }, { "input": "10\n1 3 2 6 4 5 7 9 8 10", "output": "1 3 2 5 6 4 7 9 8 10" }, { "input": "5\n5 4 3 2 1", "output": "5 4 3 2 1" }, { "input": "20\n2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19", "output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19" }, { "input": "21\n3 2 1 6 5 4 9 8 7 12 11 10 15 14 13 18 17 16 21 20 19", "output": "3 2 1 6 5 4 9 8 7 12 11 10 15 14 13 18 17 16 21 20 19" }, { "input": "10\n3 4 5 6 7 8 9 10 1 2", "output": "9 10 1 2 3 4 5 6 7 8" }, { "input": "8\n1 5 3 7 2 6 4 8", "output": "1 5 3 7 2 6 4 8" }, { "input": "50\n49 22 4 2 20 46 7 32 5 19 48 24 26 15 45 21 44 11 50 43 39 17 31 1 42 34 3 27 36 25 12 30 13 33 28 35 18 6 8 37 38 14 10 9 29 16 40 23 41 47", "output": "24 4 27 3 9 38 7 39 44 43 18 31 33 42 14 46 22 37 10 5 16 2 48 12 30 13 28 35 45 32 23 8 34 26 36 29 40 41 21 47 49 25 20 17 15 6 50 11 1 19" }, { "input": "34\n13 20 33 30 15 11 27 4 8 2 29 25 24 7 3 22 18 10 26 16 5 1 32 9 34 6 12 14 28 19 31 21 23 17", "output": "22 10 15 8 21 26 14 9 24 18 6 27 1 28 5 20 34 17 30 2 32 16 33 13 12 19 7 29 11 4 31 23 3 25" }, { "input": "92\n23 1 6 4 84 54 44 76 63 34 61 20 48 13 28 78 26 46 90 72 24 55 91 89 53 38 82 5 79 92 29 32 15 64 11 88 60 70 7 66 18 59 8 57 19 16 42 21 80 71 62 27 75 86 36 9 83 73 74 50 43 31 56 30 17 33 40 81 49 12 10 41 22 77 25 68 51 2 47 3 58 69 87 67 39 37 35 65 14 45 52 85", "output": "2 78 80 4 28 3 39 43 56 71 35 70 14 89 33 46 65 41 45 12 48 73 1 21 75 17 52 15 31 64 62 32 66 10 87 55 86 26 85 67 72 47 61 7 90 18 79 13 69 60 77 91 25 6 22 63 44 81 42 37 11 51 9 34 88 40 84 76 82 38 50 20 58 59 53 8 74 16 29 49 68 27 57 5 92 54 83 36 24 19 23 30" }, { "input": "49\n30 24 33 48 7 3 17 2 8 35 10 39 23 40 46 32 18 21 26 22 1 16 47 45 41 28 31 6 12 43 27 11 13 37 19 15 44 5 29 42 4 38 20 34 14 9 25 36 49", "output": "21 8 6 41 38 28 5 9 46 11 32 29 33 45 36 22 7 17 35 43 18 20 13 2 47 19 31 26 39 1 27 16 3 44 10 48 34 42 12 14 25 40 30 37 24 15 23 4 49" }, { "input": "12\n3 8 7 4 6 5 2 1 11 9 10 12", "output": "8 7 1 4 6 5 3 2 10 11 9 12" }, { "input": "78\n16 56 36 78 21 14 9 77 26 57 70 61 41 47 18 44 5 31 50 74 65 52 6 39 22 62 67 69 43 7 64 29 24 40 48 51 73 54 72 12 19 34 4 25 55 33 17 35 23 53 10 8 27 32 42 68 20 63 3 2 1 71 58 46 13 30 49 11 37 66 38 60 28 75 15 59 45 76", "output": "61 60 59 43 17 23 30 52 7 51 68 40 65 6 75 1 47 15 41 57 5 25 49 33 44 9 53 73 32 66 18 54 46 42 48 3 69 71 24 34 13 55 29 16 77 64 14 35 67 19 36 22 50 38 45 2 10 63 76 72 12 26 58 31 21 70 27 56 28 11 62 39 37 20 74 78 8 4" }, { "input": "64\n64 57 40 3 15 8 62 18 33 59 51 19 22 13 4 37 47 45 50 35 63 11 58 42 46 21 7 2 41 48 32 23 28 38 17 12 24 27 49 31 60 6 30 25 61 52 26 54 9 14 29 20 44 39 55 10 34 16 5 56 1 36 53 43", "output": "61 28 4 15 59 42 27 6 49 56 22 36 14 50 5 58 35 8 12 52 26 13 32 37 44 47 38 33 51 43 40 31 9 57 20 62 16 34 54 3 29 24 64 53 18 25 17 30 39 19 11 46 63 48 55 60 2 23 10 41 45 7 21 1" }, { "input": "49\n38 20 49 32 14 41 39 45 25 48 40 19 26 43 34 12 10 3 35 42 5 7 46 47 4 2 13 22 16 24 33 15 11 18 29 31 23 9 44 36 6 17 37 1 30 28 8 21 27", "output": "44 26 18 25 21 41 22 47 38 17 33 16 27 5 32 29 42 34 12 2 48 28 37 30 9 13 49 46 35 45 36 4 31 15 19 40 43 1 7 11 6 20 14 39 8 23 24 10 3" }, { "input": "78\n17 50 30 48 33 12 42 4 18 53 76 67 38 3 20 72 51 55 60 63 46 10 57 45 54 32 24 62 8 11 35 44 65 74 58 28 2 6 56 52 39 23 47 49 61 1 66 41 15 77 7 27 78 13 14 34 5 31 37 21 40 16 29 69 59 43 64 36 70 19 25 73 71 75 9 68 26 22", "output": "46 37 14 8 57 38 51 29 75 22 30 6 54 55 49 62 1 9 70 15 60 78 42 27 71 77 52 36 63 3 58 26 5 56 31 68 59 13 41 61 48 7 66 32 24 21 43 4 44 2 17 40 10 25 18 39 23 35 65 19 45 28 20 67 33 47 12 76 64 69 73 16 72 34 74 11 50 53" }, { "input": "29\n14 21 27 1 4 18 10 17 20 23 2 24 7 9 28 22 8 25 12 15 11 6 16 29 3 26 19 5 13", "output": "4 11 25 5 28 22 13 17 14 7 21 19 29 1 20 23 8 6 27 9 2 16 10 12 18 26 3 15 24" }, { "input": "82\n6 1 10 75 28 66 61 81 78 63 17 19 58 34 49 12 67 50 41 44 3 15 59 38 51 72 36 11 46 29 18 64 27 23 13 53 56 68 2 25 47 40 69 54 42 5 60 55 4 16 24 79 57 20 7 73 32 80 76 52 82 37 26 31 65 8 39 62 33 71 30 9 77 43 48 74 70 22 14 45 35 21", "output": "2 39 21 49 46 1 55 66 72 3 28 16 35 79 22 50 11 31 12 54 82 78 34 51 40 63 33 5 30 71 64 57 69 14 81 27 62 24 67 42 19 45 74 20 80 29 41 75 15 18 25 60 36 44 48 37 53 13 23 47 7 68 10 32 65 6 17 38 43 77 70 26 56 76 4 59 73 9 52 58 8 61" }, { "input": "82\n74 18 15 69 71 77 19 26 80 20 66 7 30 82 22 48 21 44 52 65 64 61 35 49 12 8 53 81 54 16 11 9 40 46 13 1 29 58 5 41 55 4 78 60 6 51 56 2 38 36 34 62 63 25 17 67 45 14 32 37 75 79 10 47 27 39 31 68 59 24 50 43 72 70 42 28 76 23 57 3 73 33", "output": "36 48 80 42 39 45 12 26 32 63 31 25 35 58 3 30 55 2 7 10 17 15 78 70 54 8 65 76 37 13 67 59 82 51 23 50 60 49 66 33 40 75 72 18 57 34 64 16 24 71 46 19 27 29 41 47 79 38 69 44 22 52 53 21 20 11 56 68 4 74 5 73 81 1 61 77 6 43 62 9 28 14" }, { "input": "45\n2 32 34 13 3 15 16 33 22 12 31 38 42 14 27 7 36 8 4 19 45 41 5 35 10 11 39 20 29 44 17 9 6 40 37 28 25 21 1 30 24 18 43 26 23", "output": "39 1 5 19 23 33 16 18 32 25 26 10 4 14 6 7 31 42 20 28 38 9 45 41 37 44 15 36 29 40 11 2 8 3 24 17 35 12 27 34 22 13 43 30 21" }, { "input": "45\n4 32 33 39 43 21 22 35 45 7 14 5 16 9 42 31 24 36 17 29 41 25 37 34 27 20 11 44 3 13 19 2 1 10 26 30 38 18 6 8 15 23 40 28 12", "output": "33 32 29 1 12 39 10 40 14 34 27 45 30 11 41 13 19 38 31 26 6 7 42 17 22 35 25 44 20 36 16 2 3 24 8 18 23 37 4 43 21 15 5 28 9" }, { "input": "74\n48 72 40 67 17 4 27 53 11 32 25 9 74 2 41 24 56 22 14 21 33 5 18 55 20 7 29 36 69 13 52 19 38 30 68 59 66 34 63 6 47 45 54 44 62 12 50 71 16 10 8 64 57 73 46 26 49 42 3 23 35 1 61 39 70 60 65 43 15 28 37 51 58 31", "output": "62 14 59 6 22 40 26 51 12 50 9 46 30 19 69 49 5 23 32 25 20 18 60 16 11 56 7 70 27 34 74 10 21 38 61 28 71 33 64 3 15 58 68 44 42 55 41 1 57 47 72 31 8 43 24 17 53 73 36 66 63 45 39 52 67 37 4 35 29 65 48 2 54 13" }, { "input": "47\n9 26 27 10 6 34 28 42 39 22 45 21 11 43 14 47 38 15 40 32 46 1 36 29 17 25 2 23 31 5 24 4 7 8 12 19 16 44 37 20 18 33 30 13 35 41 3", "output": "22 27 47 32 30 5 33 34 1 4 13 35 44 15 18 37 25 41 36 40 12 10 28 31 26 2 3 7 24 43 29 20 42 6 45 23 39 17 9 19 46 8 14 38 11 21 16" }, { "input": "49\n14 38 6 29 9 49 36 43 47 3 44 20 34 15 7 11 1 28 12 40 16 37 31 10 42 41 33 21 18 30 5 27 17 35 25 26 45 19 2 13 23 32 4 22 46 48 24 39 8", "output": "17 39 10 43 31 3 15 49 5 24 16 19 40 1 14 21 33 29 38 12 28 44 41 47 35 36 32 18 4 30 23 42 27 13 34 7 22 2 48 20 26 25 8 11 37 45 9 46 6" }, { "input": "100\n78 56 31 91 90 95 16 65 58 77 37 89 33 61 10 76 62 47 35 67 69 7 63 83 22 25 49 8 12 30 39 44 57 64 48 42 32 11 70 43 55 50 99 24 85 73 45 14 54 21 98 84 74 2 26 18 9 36 80 53 75 46 66 86 59 93 87 68 94 13 72 28 79 88 92 29 52 82 34 97 19 38 1 41 27 4 40 5 96 100 51 6 20 23 81 15 17 3 60 71", "output": "83 54 98 86 88 92 22 28 57 15 38 29 70 48 96 7 97 56 81 93 50 25 94 44 26 55 85 72 76 30 3 37 13 79 19 58 11 82 31 87 84 36 40 32 47 62 18 35 27 42 91 77 60 49 41 2 33 9 65 99 14 17 23 34 8 63 20 68 21 39 100 71 46 53 61 16 10 1 73 59 95 78 24 52 45 64 67 74 12 5 4 75 66 69 6 89 80 51 43 90" }, { "input": "22\n12 8 11 2 16 7 13 6 22 21 20 10 4 14 18 1 5 15 3 19 17 9", "output": "16 4 19 13 17 8 6 2 22 12 3 1 7 14 18 5 21 15 20 11 10 9" }, { "input": "72\n16 11 49 51 3 27 60 55 23 40 66 7 53 70 13 5 15 32 18 72 33 30 8 31 46 12 28 67 25 38 50 22 69 34 71 52 58 39 24 35 42 9 41 26 62 1 63 65 36 64 68 61 37 14 45 47 6 57 54 20 17 2 56 59 29 10 4 48 21 43 19 44", "output": "46 62 5 67 16 57 12 23 42 66 2 26 15 54 17 1 61 19 71 60 69 32 9 39 29 44 6 27 65 22 24 18 21 34 40 49 53 30 38 10 43 41 70 72 55 25 56 68 3 31 4 36 13 59 8 63 58 37 64 7 52 45 47 50 48 11 28 51 33 14 35 20" }, { "input": "63\n21 56 11 10 62 24 20 42 28 52 38 2 37 43 48 22 7 8 40 14 13 46 53 1 23 4 60 63 51 36 25 12 39 32 49 16 58 44 31 61 33 50 55 54 45 6 47 41 9 57 30 29 26 18 19 27 15 34 3 35 59 5 17", "output": "24 12 59 26 62 46 17 18 49 4 3 32 21 20 57 36 63 54 55 7 1 16 25 6 31 53 56 9 52 51 39 34 41 58 60 30 13 11 33 19 48 8 14 38 45 22 47 15 35 42 29 10 23 44 43 2 50 37 61 27 40 5 28" }, { "input": "18\n2 16 8 4 18 12 3 6 5 9 10 15 11 17 14 13 1 7", "output": "17 1 7 4 9 8 18 3 10 11 13 6 16 15 12 2 14 5" }, { "input": "47\n6 9 10 41 25 3 4 37 20 1 36 22 29 27 11 24 43 31 12 17 34 42 38 39 13 2 7 21 18 5 15 35 44 26 33 46 19 40 30 14 28 23 47 32 45 8 16", "output": "10 26 6 7 30 1 27 46 2 3 15 19 25 40 31 47 20 29 37 9 28 12 42 16 5 34 14 41 13 39 18 44 35 21 32 11 8 23 24 38 4 22 17 33 45 36 43" }, { "input": "96\n41 91 48 88 29 57 1 19 44 43 37 5 10 75 25 63 30 78 76 53 8 92 18 70 39 17 49 60 9 16 3 34 86 59 23 79 55 45 72 51 28 33 96 40 26 54 6 32 89 61 85 74 7 82 52 31 64 66 94 95 11 22 2 73 35 13 42 71 14 47 84 69 50 67 58 12 77 46 38 68 15 36 20 93 27 90 83 56 87 4 21 24 81 62 80 65", "output": "7 63 31 90 12 47 53 21 29 13 61 76 66 69 81 30 26 23 8 83 91 62 35 92 15 45 85 41 5 17 56 48 42 32 65 82 11 79 25 44 1 67 10 9 38 78 70 3 27 73 40 55 20 46 37 88 6 75 34 28 50 94 16 57 96 58 74 80 72 24 68 39 64 52 14 19 77 18 36 95 93 54 87 71 51 33 89 4 49 86 2 22 84 59 60 43" }, { "input": "73\n67 24 39 22 23 20 48 34 42 40 19 70 65 69 64 21 53 11 59 15 26 10 30 33 72 29 55 25 56 71 8 9 57 49 41 61 13 12 6 27 66 36 47 50 73 60 2 37 7 4 51 17 1 46 14 62 35 3 45 63 43 58 54 32 31 5 28 44 18 52 68 38 16", "output": "53 47 58 50 66 39 49 31 32 22 18 38 37 55 20 73 52 69 11 6 16 4 5 2 28 21 40 67 26 23 65 64 24 8 57 42 48 72 3 10 35 9 61 68 59 54 43 7 34 44 51 70 17 63 27 29 33 62 19 46 36 56 60 15 13 41 1 71 14 12 30 25 45" }, { "input": "81\n25 2 78 40 12 80 69 13 49 43 17 33 23 54 32 61 77 66 27 71 24 26 42 55 60 9 5 30 7 37 45 63 53 11 38 44 68 34 28 52 67 22 57 46 47 50 8 16 79 62 4 36 20 14 73 64 6 76 35 74 58 10 29 81 59 31 19 1 75 39 70 18 41 21 72 65 3 48 15 56 51", "output": "68 2 77 51 27 57 29 47 26 62 34 5 8 54 79 48 11 72 67 53 74 42 13 21 1 22 19 39 63 28 66 15 12 38 59 52 30 35 70 4 73 23 10 36 31 44 45 78 9 46 81 40 33 14 24 80 43 61 65 25 16 50 32 56 76 18 41 37 7 71 20 75 55 60 69 58 17 3 49 6 64" }, { "input": "12\n12 3 1 5 11 6 7 10 2 8 9 4", "output": "3 9 2 12 4 6 7 10 11 8 5 1" }, { "input": "47\n7 21 41 18 40 31 12 28 24 14 43 23 33 10 19 38 26 8 34 15 29 44 5 13 39 25 3 27 20 42 35 9 2 1 30 46 36 32 4 22 37 45 6 47 11 16 17", "output": "34 33 27 39 23 43 1 18 32 14 45 7 24 10 20 46 47 4 15 29 2 40 12 9 26 17 28 8 21 35 6 38 13 19 31 37 41 16 25 5 3 30 11 22 42 36 44" }, { "input": "8\n1 3 5 2 4 8 6 7", "output": "1 4 2 5 3 7 8 6" }, { "input": "38\n28 8 2 33 20 32 26 29 23 31 15 38 11 37 18 21 22 19 4 34 1 35 16 7 17 6 27 30 36 12 9 24 25 13 5 3 10 14", "output": "21 3 36 19 35 26 24 2 31 37 13 30 34 38 11 23 25 15 18 5 16 17 9 32 33 7 27 1 8 28 10 6 4 20 22 29 14 12" }, { "input": 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75 51 12 18 68 56 95 3 80 83 84 29 24 61 71 78 59 96 20 85 90 28 45 36 38 97 1 49 40 98 44 67 13 73 72 91 47 10 30 54 35 42 4 2 92 26 64 60 53 21 5 82 46 32 55 66 16 89 99 65 25", "output": "65 82 46 81 89 26 11 28 15 76 34 41 71 3 8 95 24 42 1 57 88 20 31 51 99 84 14 60 50 77 18 92 7 4 79 62 6 63 5 67 37 80 12 69 61 91 75 19 66 25 40 9 87 78 93 44 35 30 55 86 52 36 21 85 98 94 70 43 13 22 53 73 72 32 39 29 16 54 23 47 27 90 48 49 58 33 38 10 96 59 74 83 2 17 45 56 64 68 97" }, { "input": "99\n86 25 50 51 62 39 41 67 44 20 45 14 80 88 66 7 36 59 13 84 78 58 96 75 2 43 48 47 69 12 19 98 22 38 28 55 11 76 68 46 53 70 85 34 16 33 91 30 8 40 74 60 94 82 87 32 37 4 5 10 89 73 90 29 35 26 23 57 27 65 24 3 9 83 77 72 6 31 15 92 93 79 64 18 63 42 56 1 52 97 17 81 71 21 49 99 54 95 61", "output": "88 25 72 58 59 77 16 49 73 60 37 30 19 12 79 45 91 84 31 10 94 33 67 71 2 66 69 35 64 48 78 56 46 44 65 17 57 34 6 50 7 86 26 9 11 40 28 27 95 3 4 89 41 97 36 87 68 22 18 52 99 5 85 83 70 15 8 39 29 42 93 76 62 51 24 38 75 21 82 13 92 54 74 20 43 1 55 14 61 63 47 80 81 53 98 23 90 32 96" }, { "input": "100\n66 44 99 15 43 79 28 33 88 90 49 68 82 38 9 74 4 58 29 81 31 94 10 42 89 21 63 40 62 61 18 6 84 72 48 25 67 69 71 85 98 34 83 70 65 78 91 77 93 41 23 24 87 11 55 12 59 73 36 97 7 14 26 39 30 27 45 20 50 17 53 2 57 47 95 56 75 19 37 96 16 35 8 3 76 60 13 86 5 32 64 80 46 51 54 100 1 22 52 92", "output": "97 72 84 17 89 32 61 83 15 23 54 56 87 62 4 81 70 31 78 68 26 98 51 52 36 63 66 7 19 65 21 90 8 42 82 59 79 14 64 28 50 24 5 2 67 93 74 35 11 69 94 99 71 95 55 76 73 18 57 86 30 29 27 91 45 1 37 12 38 44 39 34 58 16 77 85 48 46 6 92 20 13 43 33 40 88 53 9 25 10 47 100 49 22 75 80 60 41 3 96" }, { "input": "99\n3 73 32 37 25 15 93 63 85 8 91 78 80 5 39 48 46 7 83 70 23 96 9 29 77 53 30 20 56 50 13 45 21 76 87 99 65 31 16 18 14 72 51 28 43 2 81 34 38 40 66 54 74 26 71 4 61 17 58 24 22 33 49 36 42 11 12 55 60 27 62 90 79 92 94 68 1 52 84 41 86 35 69 75 47 10 64 88 97 98 67 19 89 95 59 82 57 44 6", "output": "77 46 1 56 14 99 18 10 23 86 66 67 31 41 6 39 58 40 92 28 33 61 21 60 5 54 70 44 24 27 38 3 62 48 82 64 4 49 15 50 80 65 45 98 32 17 85 16 63 30 43 78 26 52 68 29 97 59 95 69 57 71 8 87 37 51 91 76 83 20 55 42 2 53 84 34 25 12 73 13 47 96 19 79 9 81 35 88 93 72 11 74 7 75 94 22 89 90 36" }, { "input": "100\n100 99 98 97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 80 79 78 77 76 75 74 73 72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1", "output": "100 99 98 97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 80 79 78 77 76 75 74 73 72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1" } ]

1,698,345,318

2,147,483,647

PyPy 3-64

WRONG_ANSWER

TESTS

0

92

0

z=int(input()) m=list(map(int,input().split())) m.sort() for x in m: print(x,end=' ')

Title: Presents Time Limit: None seconds Memory Limit: None megabytes Problem Description: Little Petya very much likes gifts. Recently he has received a new laptop as a New Year gift from his mother. He immediately decided to give it to somebody else as what can be more pleasant than giving somebody gifts. And on this occasion he organized a New Year party at his place and invited *n* his friends there. If there's one thing Petya likes more that receiving gifts, that's watching others giving gifts to somebody else. Thus, he safely hid the laptop until the next New Year and made up his mind to watch his friends exchanging gifts while he does not participate in the process. He numbered all his friends with integers from 1 to *n*. Petya remembered that a friend number *i* gave a gift to a friend number *p**i*. He also remembered that each of his friends received exactly one gift. Now Petya wants to know for each friend *i* the number of a friend who has given him a gift. Input Specification: The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the quantity of friends Petya invited to the party. The second line contains *n* space-separated integers: the *i*-th number is *p**i* — the number of a friend who gave a gift to friend number *i*. It is guaranteed that each friend received exactly one gift. It is possible that some friends do not share Petya's ideas of giving gifts to somebody else. Those friends gave the gifts to themselves. Output Specification: Print *n* space-separated integers: the *i*-th number should equal the number of the friend who gave a gift to friend number *i*. Demo Input: ['4\n2 3 4 1\n', '3\n1 3 2\n', '2\n1 2\n'] Demo Output: ['4 1 2 3\n', '1 3 2\n', '1 2\n'] Note: none

```python z=int(input()) m=list(map(int,input().split())) m.sort() for x in m: print(x,end=' ') ```

0

817

D

Imbalanced Array

PROGRAMMING

1,900

[ "data structures", "divide and conquer", "dsu", "sortings" ]

null

You are given an array *a* consisting of *n* elements. The imbalance value of some subsegment of this array is the difference between the maximum and minimum element from this segment. The imbalance value of the array is the sum of imbalance values of all subsegments of this array. For example, the imbalance value of array [1,<=4,<=1] is 9, because there are 6 different subsegments of this array: - [1] (from index 1 to index 1), imbalance value is 0; - [1,<=4] (from index 1 to index 2), imbalance value is 3; - [1,<=4,<=1] (from index 1 to index 3), imbalance value is 3; - [4] (from index 2 to index 2), imbalance value is 0; - [4,<=1] (from index 2 to index 3), imbalance value is 3; - [1] (from index 3 to index 3), imbalance value is 0; You have to determine the imbalance value of the array *a*.

The first line contains one integer *n* (1<=≤<=*n*<=≤<=106) — size of the array *a*. The second line contains *n* integers *a*1,<=*a*2... *a**n* (1<=≤<=*a**i*<=≤<=106) — elements of the array.

Print one integer — the imbalance value of *a*.

[ "3\n1 4 1\n" ]

[ "9\n" ]

none

0

[ { "input": "3\n1 4 1", "output": "9" }, { "input": "10\n1 1 1 1 1 1 1 1 1 1", "output": "0" }, { "input": "10\n1 4 4 3 5 2 4 2 4 5", "output": "123" }, { "input": "10\n9 6 8 5 5 2 8 9 2 2", "output": "245" }, { "input": "30\n4 5 2 2 5 2 3 4 3 3 2 1 3 4 4 5 3 3 1 5 2 3 5 4 5 4 4 3 5 2", "output": "1480" }, { "input": "30\n2 2 9 1 10 8 3 3 1 4 6 10 2 2 1 4 1 1 1 1 1 2 4 7 6 7 5 10 8 9", "output": "3147" }, { "input": "30\n6 19 12 6 25 24 12 2 24 14 10 10 24 19 11 29 10 22 7 1 9 1 2 27 7 24 20 25 20 28", "output": "10203" }, { "input": "100\n9 9 72 55 14 8 55 58 35 67 3 18 73 92 41 49 15 60 18 66 9 26 97 47 43 88 71 97 19 34 48 96 79 53 8 24 69 49 12 23 77 12 21 88 66 9 29 13 61 69 54 77 41 13 4 68 37 74 7 6 29 76 55 72 89 4 78 27 29 82 18 83 12 4 32 69 89 85 66 13 92 54 38 5 26 56 17 55 29 4 17 39 29 94 3 67 85 98 21 14", "output": "426927" } ]

1,622,787,046

2,147,483,647

PyPy 3

OK

TESTS

42

1,216

114,585,600

import sys input = sys.stdin.readline N = int(input()) A = list(map(int, input().split())) stk = [(float('-inf'), int(-1))] stk1 = [(float('inf'), int(-1))] leftSmaller = [] leftGreater = [] for i, a in enumerate(A): while stk[-1][0] > a: stk.pop() leftSmaller.append(stk[-1][1]) stk.append((a, i)) while stk1[-1][0] <= a: stk1.pop() leftGreater.append(stk1[-1][1]) stk1.append((a, i)) stk = [(float('-inf'), int(N))] stk1 = [(float('inf'), int(N))] rightSmaller = [] rightGreater = [] for i in range(N - 1, -1, -1): a = A[i] while stk[-1][0] >= a: stk.pop() rightSmaller.append(stk[-1][1]) stk.append((a, i)) while stk1[-1][0] < a: stk1.pop() rightGreater.append(stk1[-1][1]) stk1.append((a, i)) rightGreater.reverse() rightSmaller.reverse() ans = 0 for i, a in enumerate(A): ans += a * ((rightGreater[i] - i) * (i - leftGreater[i]) - ((rightSmaller[i] - i) * (i - leftSmaller[i]))) print(ans)

Title: Imbalanced Array Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given an array *a* consisting of *n* elements. The imbalance value of some subsegment of this array is the difference between the maximum and minimum element from this segment. The imbalance value of the array is the sum of imbalance values of all subsegments of this array. For example, the imbalance value of array [1,<=4,<=1] is 9, because there are 6 different subsegments of this array: - [1] (from index 1 to index 1), imbalance value is 0; - [1,<=4] (from index 1 to index 2), imbalance value is 3; - [1,<=4,<=1] (from index 1 to index 3), imbalance value is 3; - [4] (from index 2 to index 2), imbalance value is 0; - [4,<=1] (from index 2 to index 3), imbalance value is 3; - [1] (from index 3 to index 3), imbalance value is 0; You have to determine the imbalance value of the array *a*. Input Specification: The first line contains one integer *n* (1<=≤<=*n*<=≤<=106) — size of the array *a*. The second line contains *n* integers *a*1,<=*a*2... *a**n* (1<=≤<=*a**i*<=≤<=106) — elements of the array. Output Specification: Print one integer — the imbalance value of *a*. Demo Input: ['3\n1 4 1\n'] Demo Output: ['9\n'] Note: none

```python import sys input = sys.stdin.readline N = int(input()) A = list(map(int, input().split())) stk = [(float('-inf'), int(-1))] stk1 = [(float('inf'), int(-1))] leftSmaller = [] leftGreater = [] for i, a in enumerate(A): while stk[-1][0] > a: stk.pop() leftSmaller.append(stk[-1][1]) stk.append((a, i)) while stk1[-1][0] <= a: stk1.pop() leftGreater.append(stk1[-1][1]) stk1.append((a, i)) stk = [(float('-inf'), int(N))] stk1 = [(float('inf'), int(N))] rightSmaller = [] rightGreater = [] for i in range(N - 1, -1, -1): a = A[i] while stk[-1][0] >= a: stk.pop() rightSmaller.append(stk[-1][1]) stk.append((a, i)) while stk1[-1][0] < a: stk1.pop() rightGreater.append(stk1[-1][1]) stk1.append((a, i)) rightGreater.reverse() rightSmaller.reverse() ans = 0 for i, a in enumerate(A): ans += a * ((rightGreater[i] - i) * (i - leftGreater[i]) - ((rightSmaller[i] - i) * (i - leftSmaller[i]))) print(ans) ```

3

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" } ]

1,643,983,379

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

0

30

0

n = int(input()) i = 0 j = 0 arra = [] res = [] sum1 = 0 while i < n: inp = list(map(int, input().split())) arra.append(inp) i = i+1 for k in range(n): sum = 0 for l in range(3): sum += arra[k][l] res.append(sum) print(res) for i in res: sum1 += i if sum1 == 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()) i = 0 j = 0 arra = [] res = [] sum1 = 0 while i < n: inp = list(map(int, input().split())) arra.append(inp) i = i+1 for k in range(n): sum = 0 for l in range(3): sum += arra[k][l] res.append(sum) print(res) for i in res: sum1 += i if sum1 == 0: print("YES") else: print("NO") ```

0

78

A

Haiku

PROGRAMMING

800

[ "implementation", "strings" ]

A. Haiku

2

256

Haiku is a genre of Japanese traditional poetry. A haiku poem consists of 17 syllables split into three phrases, containing 5, 7 and 5 syllables correspondingly (the first phrase should contain exactly 5 syllables, the second phrase should contain exactly 7 syllables, and the third phrase should contain exactly 5 syllables). A haiku masterpiece contains a description of a moment in those three phrases. Every word is important in a small poem, which is why haiku are rich with symbols. Each word has a special meaning, a special role. The main principle of haiku is to say much using a few words. To simplify the matter, in the given problem we will consider that the number of syllable in the phrase is equal to the number of vowel letters there. Only the following letters are regarded as vowel letters: "a", "e", "i", "o" and "u". Three phases from a certain poem are given. Determine whether it is haiku or not.

The input data consists of three lines. The length of each line is between 1 and 100, inclusive. The *i*-th line contains the *i*-th phrase of the poem. Each phrase consists of one or more words, which are separated by one or more spaces. A word is a non-empty sequence of lowercase Latin letters. Leading and/or trailing spaces in phrases are allowed. Every phrase has at least one non-space character. See the example for clarification.

Print "YES" (without the quotes) if the poem is a haiku. Otherwise, print "NO" (also without the quotes).

[ "on codeforces \nbeta round is running\n a rustling of keys \n", "how many gallons\nof edo s rain did you drink\n cuckoo\n" ]

[ "YES", "NO" ]

none

500

[ { "input": "on codeforces \nbeta round is running\n a rustling of keys ", "output": "YES" }, { "input": "how many gallons\nof edo s rain did you drink\n cuckoo", "output": "NO" }, { "input": " hatsu shigure\n saru mo komino wo\nhoshige nari", "output": "YES" }, { "input": "o vetus stagnum\n rana de ripa salit\n ac sonant aquae", "output": "NO" }, { "input": " furuike ya\nkawazu tobikomu\nmizu no oto ", "output": "YES" }, { "input": " noch da leich\na stamperl zum aufwaerma\n da pfarrer kimmt a ", "output": "NO" }, { "input": " sommerfuglene \n hvorfor bruge mange ord\n et kan gore det", "output": "YES" }, { "input": " ab der mittagszeit\n ist es etwas schattiger\n ein wolkenhimmel", "output": "NO" }, { "input": "tornando a vederli\ni fiori di ciliegio la sera\nson divenuti frutti", "output": "NO" }, { "input": "kutaburete\nyado karu koro ya\nfuji no hana", "output": "YES" }, { "input": " beginnings of poetry\n the rice planting songs \n of the interior", "output": "NO" }, { "input": " door zomerregens\n zijn de kraanvogelpoten\n korter geworden", "output": "NO" }, { "input": " derevo na srub\na ptitsi bezzabotno\n gnezdishko tam vyut", "output": "YES" }, { "input": "writing in the dark\nunaware that my pen\nhas run out of ink", "output": "NO" }, { "input": "kusaaiu\nuieueua\nuo efaa", "output": "YES" }, { "input": "v\nh\np", "output": "NO" }, { "input": "i\ni\nu", "output": "NO" }, { "input": "awmio eoj\nabdoolceegood\nwaadeuoy", "output": "YES" }, { "input": "xzpnhhnqsjpxdboqojixmofawhdjcfbscq\nfoparnxnbzbveycoltwdrfbwwsuobyoz hfbrszy\nimtqryscsahrxpic agfjh wvpmczjjdrnwj mcggxcdo", "output": "YES" }, { "input": "wxjcvccp cppwsjpzbd dhizbcnnllckybrnfyamhgkvkjtxxfzzzuyczmhedhztugpbgpvgh\nmdewztdoycbpxtp bsiw hknggnggykdkrlihvsaykzfiiw\ndewdztnngpsnn lfwfbvnwwmxoojknygqb hfe ibsrxsxr", "output": "YES" }, { "input": "nbmtgyyfuxdvrhuhuhpcfywzrbclp znvxw synxmzymyxcntmhrjriqgdjh xkjckydbzjbvtjurnf\nhhnhxdknvamywhsrkprofnyzlcgtdyzzjdsfxyddvilnzjziz qmwfdvzckgcbrrxplxnxf mpxwxyrpesnewjrx ajxlfj\nvcczq hddzd cvefmhxwxxyqcwkr fdsndckmesqeq zyjbwbnbyhybd cta nsxzidl jpcvtzkldwd", "output": "YES" }, { "input": "rvwdsgdsrutgjwscxz pkd qtpmfbqsmctuevxdj kjzknzghdvxzlaljcntg jxhvzn yciktbsbyscfypx x xhkxnfpdp\nwdfhvqgxbcts mnrwbr iqttsvigwdgvlxwhsmnyxnttedonxcfrtmdjjmacvqtkbmsnwwvvrlxwvtggeowtgsqld qj\nvsxcdhbzktrxbywpdvstr meykarwtkbm pkkbhvwvelclfmpngzxdmblhcvf qmabmweldplmczgbqgzbqnhvcdpnpjtch ", "output": "YES" }, { "input": "brydyfsmtzzkpdsqvvztmprhqzbzqvgsblnz naait tdtiprjsttwusdykndwcccxfmzmrmfmzjywkpgbfnjpypgcbcfpsyfj k\nucwdfkfyxxxht lxvnovqnnsqutjsyagrplb jhvtwdptrwcqrovncdvqljjlrpxcfbxqgsfylbgmcjpvpl ccbcybmigpmjrxpu\nfgwtpcjeywgnxgbttgx htntpbk tkkpwbgxwtbxvcpkqbzetjdkcwad tftnjdxxjdvbpfibvxuglvx llyhgjvggtw jtjyphs", "output": "YES" }, { "input": "nyc aqgqzjjlj mswgmjfcxlqdscheskchlzljlsbhyn iobxymwzykrsnljj\nnnebeaoiraga\nqpjximoqzswhyyszhzzrhfwhf iyxysdtcpmikkwpugwlxlhqfkn", "output": "NO" }, { "input": "lzrkztgfe mlcnq ay ydmdzxh cdgcghxnkdgmgfzgahdjjmqkpdbskreswpnblnrc fmkwziiqrbskp\np oukeaz gvvy kghtrjlczyl qeqhgfgfej\nwfolhkmktvsjnrpzfxcxzqmfidtlzmuhxac wsncjgmkckrywvxmnjdpjpfydhk qlmdwphcvyngansqhl", "output": "NO" }, { "input": "yxcboqmpwoevrdhvpxfzqmammak\njmhphkxppkqkszhqqtkvflarsxzla pbxlnnnafqbsnmznfj qmhoktgzix qpmrgzxqvmjxhskkksrtryehfnmrt dtzcvnvwp\nscwymuecjxhw rdgsffqywwhjpjbfcvcrnisfqllnbplpadfklayjguyvtrzhwblftclfmsr", "output": "NO" }, { "input": "qfdwsr jsbrpfmn znplcx nhlselflytndzmgxqpgwhpi ghvbbxrkjdirfghcybhkkqdzmyacvrrcgsneyjlgzfvdmxyjmph\nylxlyrzs drbktzsniwcbahjkgohcghoaczsmtzhuwdryjwdijmxkmbmxv yyfrokdnsx\nyw xtwyzqlfxwxghugoyscqlx pljtz aldfskvxlsxqgbihzndhxkswkxqpwnfcxzfyvncstfpqf", "output": "NO" }, { "input": "g rguhqhcrzmuqthtmwzhfyhpmqzzosa\nmhjimzvchkhejh irvzejhtjgaujkqfxhpdqjnxr dvqallgssktqvsxi\npcwbliftjcvuzrsqiswohi", "output": "NO" }, { "input": " ngxtlq iehiise vgffqcpnmsoqzyseuqqtggokymol zn\nvjdjljazeujwoubkcvtsbepooxqzrueaauokhepiquuopfild\ngoabauauaeotoieufueeknudiilupouaiaexcoapapu", "output": "NO" }, { "input": "ycnvnnqk mhrmhctpkfbc qbyvtjznmndqjzgbcxmvrpkfcll zwspfptmbxgrdv dsgkk nfytsqjrnfbhh pzdldzymvkdxxwh\nvnhjfwgdnyjptsmblyxmpzylsbjlmtkkwjcbqwjctqvrlqqkdsrktxlnslspvnn mdgsmzblhbnvpczmqkcffwhwljqkzmk hxcm\nrghnjvzcpprrgmtgytpkzyc mrdnnhpkwypwqbtzjyfwvrdwyjltbzxtbstzs xdjzdmx yjsqtzlrnvyssvglsdjrmsrfrcdpqt", "output": "NO" }, { "input": "ioeeaioeiuoeaeieuuieooaouiuouiioaueeaiaiuoaoiioeeaauooiuuieeuaeeoauieeaiuoieiaieuoauaaoioooieueueuai\nuooaoeeaoiuuoeioaoouaououoeioiaeueoioaiouaeaoioiuuaueeuaiuoiueoiuaoeeieeouaeeaeeieioeoiiieuuueuuieuo\naeeouieeieoueaioeoioooiouaeeeiaaioueauaueiouuuaieuuioiaeiueauueaoieauauoeueuaiueuuaueeoueauaeaoieeoo", "output": "NO" }, { "input": "mkgycgmrqtmlwj vddlbsgdptyfrcj nmrddjchydvbsyhgzytfzvwbhqmmhkpsyrcnknpsqrr wcnf wfvgskrnsmnrcqgcnc q\npn cclcyjjdp rqkpgyrvyyd hhynrochfa pthyffqvskazfshvzcmvhmsrxvquzkdvivlbfypwbgltfbjmhzljvgzmxjlrjdjq\nauuouieuaaauoooeoiuoaueuueauiuuuiaeuwuoiaeoeaieauieouoaeiiieiioiuuuiuauioiiuoeauueieaueiaioeuaoaaau", "output": "NO" }, { "input": "ab\na\na", "output": "NO" }, { "input": "on codeforcs\nbeta round is runnineg\na rustling of keys", "output": "NO" }, { "input": "coooooooooooooooood\nd\nd", "output": "NO" }, { "input": "on codeforces \nbeta round is runninog\n a rustling of keys", "output": "NO" }, { "input": "aaaaaaaaaaaaaaa\na\na", "output": "NO" }, { "input": "aaaaaccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc\naaaaaaaccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc\naaaaaccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc", "output": "YES" }, { "input": "cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc\ncccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc\ncccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc", "output": "NO" }, { "input": "aaaaaa\naaaaaa\naaaaa", "output": "NO" } ]

1,590,998,415

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

0

248

0

# haiku s=str(input()) s1=str(input()) s2=str(input()) s=s.replace(' ','') s1=s1.replace(' ','') s2=s2.replace(' ','') c=0 d=0 e=0 for i in range(0,len(s)): if(s[i] in 'aeiou'): c=c+1 for i in range(0,len(s1)): if(s1[i] in 'aeiou'): d=d+1 for i in range(0,len(s2)): if(s2[i] in 'aeiou'): e=e+1 if(c==5 and d==7 and e==5): print('Yes') else: print("No")

Title: Haiku Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Haiku is a genre of Japanese traditional poetry. A haiku poem consists of 17 syllables split into three phrases, containing 5, 7 and 5 syllables correspondingly (the first phrase should contain exactly 5 syllables, the second phrase should contain exactly 7 syllables, and the third phrase should contain exactly 5 syllables). A haiku masterpiece contains a description of a moment in those three phrases. Every word is important in a small poem, which is why haiku are rich with symbols. Each word has a special meaning, a special role. The main principle of haiku is to say much using a few words. To simplify the matter, in the given problem we will consider that the number of syllable in the phrase is equal to the number of vowel letters there. Only the following letters are regarded as vowel letters: "a", "e", "i", "o" and "u". Three phases from a certain poem are given. Determine whether it is haiku or not. Input Specification: The input data consists of three lines. The length of each line is between 1 and 100, inclusive. The *i*-th line contains the *i*-th phrase of the poem. Each phrase consists of one or more words, which are separated by one or more spaces. A word is a non-empty sequence of lowercase Latin letters. Leading and/or trailing spaces in phrases are allowed. Every phrase has at least one non-space character. See the example for clarification. Output Specification: Print "YES" (without the quotes) if the poem is a haiku. Otherwise, print "NO" (also without the quotes). Demo Input: ['on codeforces \nbeta round is running\n a rustling of keys \n', 'how many gallons\nof edo s rain did you drink\n cuckoo\n'] Demo Output: ['YES', 'NO'] Note: none

```python # haiku s=str(input()) s1=str(input()) s2=str(input()) s=s.replace(' ','') s1=s1.replace(' ','') s2=s2.replace(' ','') c=0 d=0 e=0 for i in range(0,len(s)): if(s[i] in 'aeiou'): c=c+1 for i in range(0,len(s1)): if(s1[i] in 'aeiou'): d=d+1 for i in range(0,len(s2)): if(s2[i] in 'aeiou'): e=e+1 if(c==5 and d==7 and e==5): print('Yes') else: print("No") ```

0

32

B

Borze

PROGRAMMING

800

[ "expression parsing", "implementation" ]

B. Borze

2

256

Ternary numeric notation is quite popular in Berland. To telegraph the ternary number the Borze alphabet is used. Digit 0 is transmitted as «.», 1 as «-.» and 2 as «--». You are to decode the Borze code, i.e. to find out the ternary number given its representation in Borze alphabet.

The first line contains a number in Borze code. The length of the string is between 1 and 200 characters. It's guaranteed that the given string is a valid Borze code of some ternary number (this number can have leading zeroes).

Output the decoded ternary number. It can have leading zeroes.

[ ".-.--\n", "--.\n", "-..-.--\n" ]

[ "012", "20", "1012" ]

none

1,000

[ { "input": ".-.--", "output": "012" }, { "input": "--.", "output": "20" }, { "input": "-..-.--", "output": "1012" }, { "input": "---..", "output": "210" }, { "input": "..--.---..", "output": "0020210" }, { "input": "-.....----.", "output": "10000220" }, { "input": ".", "output": "0" }, { "input": "-.", "output": "1" }, { "input": "--", "output": "2" }, { "input": "..", "output": "00" }, { "input": "--.", "output": "20" }, { "input": ".--.", "output": "020" }, { "input": ".-.-..", "output": "0110" }, { "input": "----.-.", "output": "2201" }, { "input": "-..--.-.", "output": "10201" }, { "input": "..--..--.", "output": "0020020" }, { "input": "-.-.---.--..-..-.-.-..-..-.--.", "output": "112120010111010120" }, { "input": "---.-.-.------..-..-..-..-.-..-.--.-.-..-.-.-----..-.-.", "output": "21112220010101011012011011221011" }, { "input": "-.-..--.-.-.-.-.-..-.-.-.---------.--.---..--...--.-----.-.-.-...--.-.-.---.------.--..-.--.-----.-...-..------", "output": "11020111110111222212021020002022111100201121222020012022110010222" }, { "input": "-.-..-.--.---..---.-..---.-...-.-.----..-.---.-.---..-.--.---.-.-------.---.--....----.-.---.---.---.----.-----..---.-.-.-.-----.--.-------.-..", "output": "110120210211021100112200121121012021122212120000220121212122022102111122120222110" }, { "input": ".-..-.-.---.-----.--.---...-.--.-.-....-..", "output": "01011212212021001201100010" }, { "input": ".------.-.---..--...-..-..-.-.-.--.--.-..-.--...-.-.---.-.-.------..--..-.---..----.-..-.--.---.-.----.-.---...-.-.-.-----.-.-.---.---.-.....-.-...-----.-...-.---.-..-.-----.--...---.-.-..-.--.-.---..", "output": "022201210200010101112020101200011211122200200121022010120211220121001112211121211000011002211001211012212000211101201210" }, { "input": ".-.--.---.-----.-.-----.-.-..-----..-..----..--.-.--.----..---.---..-.-.-----..-------.----..----.-..---...-----..-..-----...-..-.-.-----....---..---..-.-----...-.--...--.-.---.-.-.-.-.-...---..----.", "output": "01202122112211102210102200201202200212101122102221220022010210022101022100101122100021021012210012000201211111100210220" }, { "input": "..-.-.-.---.-.-.-..-.-..-.-.---.-------.---..-----.---....-.---.--.--.-.---.---------.-..---.-.-.--..---.---.-.---.-.-..-.-..-.-.-.----.--.-....--------.-.---..----.------.-.-.--.--.-----.-----.----", "output": "0011121111011011212221210221210001212020121222211021112002121121110110111220201000222201210220222011202022122122" }, { "input": "-..-------.------.-..--.-.-..--.-.-..-----..-.-.-..-..-..--.---..-----..---..-..--.-..-.-.---...-.....-------.---.-----.-...-.-...-.-.---.---.-----.--.--...-.--..-.-..-...-.-.-.-.---..---.-..-.-.-.-..", "output": "102221222010201102011022101110101020210221021010201011210010000222121221100110011212122120200012001101001111210211011110" }, { "input": ".-.----.-.--..-.-.-.-..----..-.-...--.-.---.---.-------..-.--..-......--.------.--.----.--...-.--.--..-----..-.....--.--.-.-.------..--------.----------..-.---.----.---.-..--..-.....-..------.--.", "output": "012201200111102200110020121212221012001000002022202022020001202002210100002020112220022220222220012122021102001000010222020" }, { "input": "------.-----.-....--.-.----.-.---.-.-..---.-.---.-----..-...-.-.---..-.-.-..-.-.-...-.-.-.----..--.------.----.-..-.--...-.-------...-.-..-.-.--.--.---..--..--------.--.-.-.---.-.-.-...----.--..-.--..", "output": "222022110002012201211102112122101001121011101110011122002022202201012000122210011011202021020022220201121111002202001200" }, { "input": "-.---...----...--.--...-.--.----", "output": "121002200020200012022" }, { "input": "--.--.--.---.--.-.---.-.-..-..--.-..---.-.....-..---.-----.--...-.-.-------.-.--.-.----.-..-.------.", "output": "202020212012111010201021100001021221200011222112012201012220" } ]

1,693,560,440

2,147,483,647

Python 3

RUNTIME_ERROR

TESTS

0

92

0

p=input() s="" for i in len(p): if p[i]=="-": if p[i+1]=='-': s+="2" elif p[i+1]=='.': s+="1" else: s+="0" print(s)

Title: Borze Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Ternary numeric notation is quite popular in Berland. To telegraph the ternary number the Borze alphabet is used. Digit 0 is transmitted as «.», 1 as «-.» and 2 as «--». You are to decode the Borze code, i.e. to find out the ternary number given its representation in Borze alphabet. Input Specification: The first line contains a number in Borze code. The length of the string is between 1 and 200 characters. It's guaranteed that the given string is a valid Borze code of some ternary number (this number can have leading zeroes). Output Specification: Output the decoded ternary number. It can have leading zeroes. Demo Input: ['.-.--\n', '--.\n', '-..-.--\n'] Demo Output: ['012', '20', '1012'] Note: none

```python p=input() s="" for i in len(p): if p[i]=="-": if p[i+1]=='-': s+="2" elif p[i+1]=='.': s+="1" else: s+="0" print(s) ```

-1

731

A

Night at the Museum

PROGRAMMING

800

[ "implementation", "strings" ]

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" } ]

1,698,674,650

2,147,483,647

Python 3

OK

TESTS

44

46

0

s = input() ans = 0 cur = 'a' for i in s: ans += min(abs(ord(cur) - ord(i)), 26 - abs(ord(cur) - ord(i))) cur = i print(ans)

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 s = input() ans = 0 cur = 'a' for i in s: ans += min(abs(ord(cur) - ord(i)), 26 - abs(ord(cur) - ord(i))) cur = i print(ans) ```

3

25

B

Phone numbers

PROGRAMMING

1,100

[ "implementation" ]

B. Phone numbers

2

256

Phone number in Berland is a sequence of *n* digits. Often, to make it easier to memorize the number, it is divided into groups of two or three digits. For example, the phone number 1198733 is easier to remember as 11-987-33. Your task is to find for a given phone number any of its divisions into groups of two or three digits.

The first line contains integer *n* (2<=≤<=*n*<=≤<=100) — amount of digits in the phone number. The second line contains *n* digits — the phone number to divide into groups.

Output any of divisions of the given phone number into groups of two or three digits. Separate groups by single character -. If the answer is not unique, output any.

[ "6\n549871\n", "7\n1198733\n" ]

[ "54-98-71", "11-987-33\n" ]

none

0

[ { "input": "6\n549871", "output": "54-98-71" }, { "input": "7\n1198733", "output": "119-87-33" }, { "input": "2\n74", "output": "74" }, { "input": "2\n33", "output": "33" }, { "input": "3\n074", "output": "074" }, { "input": "3\n081", "output": "081" }, { "input": "4\n3811", "output": "38-11" }, { "input": "5\n21583", "output": "215-83" }, { "input": "8\n33408349", "output": "33-40-83-49" }, { "input": "9\n988808426", "output": "988-80-84-26" }, { "input": "10\n0180990956", "output": "01-80-99-09-56" }, { "input": "15\n433488906230138", "output": "433-48-89-06-23-01-38" }, { "input": "22\n7135498415686025907059", "output": "71-35-49-84-15-68-60-25-90-70-59" }, { "input": "49\n2429965524999668169991253653390090510755018570235", "output": "242-99-65-52-49-99-66-81-69-99-12-53-65-33-90-09-05-10-75-50-18-57-02-35" }, { "input": "72\n491925337784111770500147619881727525570039735507439360627744863794794290", "output": "49-19-25-33-77-84-11-17-70-50-01-47-61-98-81-72-75-25-57-00-39-73-55-07-43-93-60-62-77-44-86-37-94-79-42-90" }, { "input": "95\n32543414456047900690980198395035321172843693417425457554204776648220562494524275489599199209210", "output": "325-43-41-44-56-04-79-00-69-09-80-19-83-95-03-53-21-17-28-43-69-34-17-42-54-57-55-42-04-77-66-48-22-05-62-49-45-24-27-54-89-59-91-99-20-92-10" }, { "input": "97\n9362344595153688016434451101547661156123505108492010669557671355055642365998461003851354321478898", "output": "936-23-44-59-51-53-68-80-16-43-44-51-10-15-47-66-11-56-12-35-05-10-84-92-01-06-69-55-76-71-35-50-55-64-23-65-99-84-61-00-38-51-35-43-21-47-88-98" }, { "input": "98\n65521815795893886057122984634320900545031770769333931308009346017867969790810907868670369236928568", "output": "65-52-18-15-79-58-93-88-60-57-12-29-84-63-43-20-90-05-45-03-17-70-76-93-33-93-13-08-00-93-46-01-78-67-96-97-90-81-09-07-86-86-70-36-92-36-92-85-68" }, { "input": "99\n455213856470326729480192345541970106407563996625458559297407682539801838244443866898560852503660390", "output": "455-21-38-56-47-03-26-72-94-80-19-23-45-54-19-70-10-64-07-56-39-96-62-54-58-55-92-97-40-76-82-53-98-01-83-82-44-44-38-66-89-85-60-85-25-03-66-03-90" }, { "input": "100\n4004223124942730640235383244438257614581534320356060987241659784249551110165034719443327659510644224", "output": "40-04-22-31-24-94-27-30-64-02-35-38-32-44-43-82-57-61-45-81-53-43-20-35-60-60-98-72-41-65-97-84-24-95-51-11-01-65-03-47-19-44-33-27-65-95-10-64-42-24" } ]

1,636,034,928

2,147,483,647

Python 3

OK

TESTS

20

92

4,505,600

# https://codeforces.com/contest/25/problem/B n = int(input()) p = input() s = "" if n%2==0: b = int(n/2) for i in range(b): s += p[(2*i):(2*i+2)] + "-" print(s[:-1]) else: s += p[0:3] + "-" p = p[3:] b = int(len(p)/2) for i in range(b): s += p[(2*i):(2*i+2)] + "-" print(s[:-1])

Title: Phone numbers Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Phone number in Berland is a sequence of *n* digits. Often, to make it easier to memorize the number, it is divided into groups of two or three digits. For example, the phone number 1198733 is easier to remember as 11-987-33. Your task is to find for a given phone number any of its divisions into groups of two or three digits. Input Specification: The first line contains integer *n* (2<=≤<=*n*<=≤<=100) — amount of digits in the phone number. The second line contains *n* digits — the phone number to divide into groups. Output Specification: Output any of divisions of the given phone number into groups of two or three digits. Separate groups by single character -. If the answer is not unique, output any. Demo Input: ['6\n549871\n', '7\n1198733\n'] Demo Output: ['54-98-71', '11-987-33\n'] Note: none

```python # https://codeforces.com/contest/25/problem/B n = int(input()) p = input() s = "" if n%2==0: b = int(n/2) for i in range(b): s += p[(2*i):(2*i+2)] + "-" print(s[:-1]) else: s += p[0:3] + "-" p = p[3:] b = int(len(p)/2) for i in range(b): s += p[(2*i):(2*i+2)] + "-" print(s[:-1]) ```

3.968608

929

B

Места в самолёте

PROGRAMMING

1,300

[ "*special", "implementation" ]

null

В самолёте есть *n* рядов мест. Если смотреть на ряды сверху, то в каждом ряду есть 3 места слева, затем проход между рядами, затем 4 центральных места, затем ещё один проход между рядами, а затем ещё 3 места справа. Известно, что некоторые места уже заняты пассажирами. Всего есть два вида пассажиров — статусные (те, которые часто летают) и обычные. Перед вами стоит задача рассадить ещё *k* обычных пассажиров так, чтобы суммарное число соседей у статусных пассажиров было минимально возможным. Два пассажира считаются соседями, если они сидят в одном ряду и между ними нет других мест и прохода между рядами. Если пассажир является соседним пассажиром для двух статусных пассажиров, то его следует учитывать в сумме соседей дважды.

В первой строке следуют два целых числа *n* и *k* (1<=≤<=*n*<=≤<=100, 1<=≤<=*k*<=≤<=10·*n*) — количество рядов мест в самолёте и количество пассажиров, которых нужно рассадить. Далее следует описание рядов мест самолёта по одному ряду в строке. Если очередной символ равен '-', то это проход между рядами. Если очередной символ равен '.', то это свободное место. Если очередной символ равен 'S', то на текущем месте будет сидеть статусный пассажир. Если очередной символ равен 'P', то на текущем месте будет сидеть обычный пассажир. Гарантируется, что количество свободных мест не меньше *k*. Гарантируется, что все ряды удовлетворяют описанному в условии формату.

В первую строку выведите минимальное суммарное число соседей у статусных пассажиров. Далее выведите план рассадки пассажиров, который минимизирует суммарное количество соседей у статусных пассажиров, в том же формате, что и во входных данных. Если в свободное место нужно посадить одного из *k* пассажиров, выведите строчную букву 'x' вместо символа '.'.

[ "1 2\nSP.-SS.S-S.S\n", "4 9\nPP.-PPPS-S.S\nPSP-PPSP-.S.\n.S.-S..P-SS.\nP.S-P.PP-PSP\n" ]

[ "5\nSPx-SSxS-S.S\n", "15\nPPx-PPPS-S.S\nPSP-PPSP-xSx\nxSx-SxxP-SSx\nP.S-PxPP-PSP\n" ]

В первом примере нужно посадить ещё двух обычных пассажиров. Для минимизации соседей у статусных пассажиров, нужно посадить первого из них на третье слева место, а второго на любое из оставшихся двух мест, так как независимо от выбора места он станет соседом двух статусных пассажиров. Изначально, у статусного пассажира, который сидит на самом левом месте уже есть сосед. Также на четвёртом и пятом местах слева сидят статусные пассажиры, являющиеся соседями друг для друга (что добавляет к сумме 2). Таким образом, после посадки ещё двух обычных пассажиров, итоговое суммарное количество соседей у статусных пассажиров станет равно пяти.

1,000

[ { "input": "1 2\nSP.-SS.S-S.S", "output": "5\nSPx-SSxS-S.S" }, { "input": "4 9\nPP.-PPPS-S.S\nPSP-PPSP-.S.\n.S.-S..P-SS.\nP.S-P.PP-PSP", "output": "15\nPPx-PPPS-S.S\nPSP-PPSP-xSx\nxSx-SxxP-SSx\nP.S-PxPP-PSP" }, { "input": "3 7\n.S.-SSSP-..S\nS..-.SPP-S.P\n.S.-PPPP-PSP", "output": "13\nxSx-SSSP-xxS\nSxx-xSPP-S.P\n.S.-PPPP-PSP" }, { "input": "5 6\nPP.-PS.P-P..\nPPS-SP..-P.P\nP.P-....-S..\nSPP-.P.S-.S.\nSP.-S.PS-PPP", "output": "6\nPPx-PS.P-Pxx\nPPS-SPxx-PxP\nP.P-....-S..\nSPP-.P.S-.S.\nSP.-S.PS-PPP" }, { "input": "1 1\n..S-PS..-.PP", "output": "1\nx.S-PS..-.PP" }, { "input": "2 2\nPP.-S.SS-.S.\nSSP-SSSS-S.S", "output": "12\nPPx-S.SS-xS.\nSSP-SSSS-S.S" }, { "input": "30 1\nPPP-PPP.-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP", "output": "0\nPPP-PPPx-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP" }, { "input": "1 1\nSPS-....-P.P", "output": "2\nSPS-x...-P.P" }, { "input": "2 1\nSSS-S.S.-SSS\nSSP-.PP.-S.S", "output": "11\nSSS-S.S.-SSS\nSSP-xPP.-S.S" }, { "input": "30 1\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPP.-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP", "output": "0\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPx-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP" } ]

1,520,189,070

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

4

77

6,348,800

inpu, k = map(int, input().split()) inpu_string = '' for i in range(inpu): inpu_string = inpu_string + 'o' + input() import re def optimal_case(s): global k nova = s a = [] pos = 0 m = re.search(r'[^S][^SPxo-][^S]', s) while m != None and k > 0: c = m.end() - 2 a += [c + pos] j = s[:c + 1] pos += len(j) s = s[c + 1:] k -= 1 m = re.search(r'[^S][^SPxo-][^S]', s) for index in a: nova = nova[:index] + 'x' + nova[index + 1:] return nova def subway1(s): global k i = 1 while i < len(s) - 1 and k > 0: if re.search(r'S[^SPox-][^S]', s[i - 1] + s[i] + s[i + 1]) != None: s = s[:i] + 'x' + s[i + 1:] k -= 1 i += 1 if s[-2:] == 'S.' and k > 0: s = s[:-1] + 'x' k -= 1 return s def subway2(s): global k i = len(s) - 2 if s[0:2] == '.S' and k > 0: s = 'x' + s[1:] k -= 1 while i > 0 and k > 0: if re.search(r'[^S][^SPox-][S]', s[i - 1] + s[i] + s[i + 1]) != None: s = s[:i] + 'x' + s[i + 1:] k -= 1 i -= 1 return s def last_case(s): global k i = 0 l = len(s) - 1 while k > 0 and i < l: if s[i] == '.': s = s[:i] + 'x' + s[i + 1:] k -= 1 i += 1 return s def united(s): s = optimal_case(s) s = subway1(subway2(s)) s = last_case(s) return s def count_naighbours(s): naighbours = 0 for i in range(len(s) - 1): if s[i] == 'S': s2 = s[i] + s[i + 1] if (s2 != 'S-') and (s2 != 'S.') and (s2 != 'So'): naighbours += 1 for j in range(len(s) - 1, 1, -1): if s[j] == 'S': s2 = s[j - 1] + s[j] if (s2 != '-S') and (s2 != '.S') and (s2 != 'oS'): naighbours += 1 return naighbours s = united(inpu_string) print(count_naighbours(s)) for element in s.split('o')[1:]: print(element)

Title: Места в самолёте Time Limit: None seconds Memory Limit: None megabytes Problem Description: В самолёте есть *n* рядов мест. Если смотреть на ряды сверху, то в каждом ряду есть 3 места слева, затем проход между рядами, затем 4 центральных места, затем ещё один проход между рядами, а затем ещё 3 места справа. Известно, что некоторые места уже заняты пассажирами. Всего есть два вида пассажиров — статусные (те, которые часто летают) и обычные. Перед вами стоит задача рассадить ещё *k* обычных пассажиров так, чтобы суммарное число соседей у статусных пассажиров было минимально возможным. Два пассажира считаются соседями, если они сидят в одном ряду и между ними нет других мест и прохода между рядами. Если пассажир является соседним пассажиром для двух статусных пассажиров, то его следует учитывать в сумме соседей дважды. Input Specification: В первой строке следуют два целых числа *n* и *k* (1<=≤<=*n*<=≤<=100, 1<=≤<=*k*<=≤<=10·*n*) — количество рядов мест в самолёте и количество пассажиров, которых нужно рассадить. Далее следует описание рядов мест самолёта по одному ряду в строке. Если очередной символ равен '-', то это проход между рядами. Если очередной символ равен '.', то это свободное место. Если очередной символ равен 'S', то на текущем месте будет сидеть статусный пассажир. Если очередной символ равен 'P', то на текущем месте будет сидеть обычный пассажир. Гарантируется, что количество свободных мест не меньше *k*. Гарантируется, что все ряды удовлетворяют описанному в условии формату. Output Specification: В первую строку выведите минимальное суммарное число соседей у статусных пассажиров. Далее выведите план рассадки пассажиров, который минимизирует суммарное количество соседей у статусных пассажиров, в том же формате, что и во входных данных. Если в свободное место нужно посадить одного из *k* пассажиров, выведите строчную букву 'x' вместо символа '.'. Demo Input: ['1 2\nSP.-SS.S-S.S\n', '4 9\nPP.-PPPS-S.S\nPSP-PPSP-.S.\n.S.-S..P-SS.\nP.S-P.PP-PSP\n'] Demo Output: ['5\nSPx-SSxS-S.S\n', '15\nPPx-PPPS-S.S\nPSP-PPSP-xSx\nxSx-SxxP-SSx\nP.S-PxPP-PSP\n'] Note: В первом примере нужно посадить ещё двух обычных пассажиров. Для минимизации соседей у статусных пассажиров, нужно посадить первого из них на третье слева место, а второго на любое из оставшихся двух мест, так как независимо от выбора места он станет соседом двух статусных пассажиров. Изначально, у статусного пассажира, который сидит на самом левом месте уже есть сосед. Также на четвёртом и пятом местах слева сидят статусные пассажиры, являющиеся соседями друг для друга (что добавляет к сумме 2). Таким образом, после посадки ещё двух обычных пассажиров, итоговое суммарное количество соседей у статусных пассажиров станет равно пяти.

```python inpu, k = map(int, input().split()) inpu_string = '' for i in range(inpu): inpu_string = inpu_string + 'o' + input() import re def optimal_case(s): global k nova = s a = [] pos = 0 m = re.search(r'[^S][^SPxo-][^S]', s) while m != None and k > 0: c = m.end() - 2 a += [c + pos] j = s[:c + 1] pos += len(j) s = s[c + 1:] k -= 1 m = re.search(r'[^S][^SPxo-][^S]', s) for index in a: nova = nova[:index] + 'x' + nova[index + 1:] return nova def subway1(s): global k i = 1 while i < len(s) - 1 and k > 0: if re.search(r'S[^SPox-][^S]', s[i - 1] + s[i] + s[i + 1]) != None: s = s[:i] + 'x' + s[i + 1:] k -= 1 i += 1 if s[-2:] == 'S.' and k > 0: s = s[:-1] + 'x' k -= 1 return s def subway2(s): global k i = len(s) - 2 if s[0:2] == '.S' and k > 0: s = 'x' + s[1:] k -= 1 while i > 0 and k > 0: if re.search(r'[^S][^SPox-][S]', s[i - 1] + s[i] + s[i + 1]) != None: s = s[:i] + 'x' + s[i + 1:] k -= 1 i -= 1 return s def last_case(s): global k i = 0 l = len(s) - 1 while k > 0 and i < l: if s[i] == '.': s = s[:i] + 'x' + s[i + 1:] k -= 1 i += 1 return s def united(s): s = optimal_case(s) s = subway1(subway2(s)) s = last_case(s) return s def count_naighbours(s): naighbours = 0 for i in range(len(s) - 1): if s[i] == 'S': s2 = s[i] + s[i + 1] if (s2 != 'S-') and (s2 != 'S.') and (s2 != 'So'): naighbours += 1 for j in range(len(s) - 1, 1, -1): if s[j] == 'S': s2 = s[j - 1] + s[j] if (s2 != '-S') and (s2 != '.S') and (s2 != 'oS'): naighbours += 1 return naighbours s = united(inpu_string) print(count_naighbours(s)) for element in s.split('o')[1:]: print(element) ```

0

535

A

Tavas and Nafas

PROGRAMMING

1,000

[ "brute force", "implementation" ]

null

Today Tavas got his test result as an integer score and he wants to share it with his girlfriend, Nafas. His phone operating system is Tavdroid, and its keyboard doesn't have any digits! He wants to share his score with Nafas via text, so he has no choice but to send this number using words. He ate coffee mix without water again, so right now he's really messed up and can't think. Your task is to help him by telling him what to type.

The first and only line of input contains an integer *s* (0<=≤<=*s*<=≤<=99), Tavas's score.

In the first and only line of output, print a single string consisting only from English lowercase letters and hyphens ('-'). Do not use spaces.

[ "6\n", "99\n", "20\n" ]

[ "six\n", "ninety-nine\n", "twenty\n" ]

You can find all you need to know about English numerals in [http://en.wikipedia.org/wiki/English_numerals](https://en.wikipedia.org/wiki/English_numerals) .

500

[ { "input": "6", "output": "six" }, { "input": "99", "output": "ninety-nine" }, { "input": "20", "output": "twenty" }, { "input": "10", "output": "ten" }, { "input": "15", "output": "fifteen" }, { "input": "27", "output": "twenty-seven" }, { "input": "40", "output": "forty" }, { "input": "63", "output": "sixty-three" }, { "input": "0", "output": "zero" }, { "input": "1", "output": "one" }, { "input": "2", "output": "two" }, { "input": "8", "output": "eight" }, { "input": "9", "output": "nine" }, { "input": "11", "output": "eleven" }, { "input": "12", "output": "twelve" }, { "input": "13", "output": "thirteen" }, { "input": "14", "output": "fourteen" }, { "input": "16", "output": "sixteen" }, { "input": "17", "output": "seventeen" }, { "input": "18", "output": "eighteen" }, { "input": "19", "output": "nineteen" }, { "input": "21", "output": "twenty-one" }, { "input": "29", "output": "twenty-nine" }, { "input": "30", "output": "thirty" }, { "input": "32", "output": "thirty-two" }, { "input": "38", "output": "thirty-eight" }, { "input": "43", "output": "forty-three" }, { "input": "47", "output": "forty-seven" }, { "input": "50", "output": "fifty" }, { "input": "54", "output": "fifty-four" }, { "input": "56", "output": "fifty-six" }, { "input": "60", "output": "sixty" }, { "input": "66", "output": "sixty-six" }, { "input": "70", "output": "seventy" }, { "input": "76", "output": "seventy-six" }, { "input": "80", "output": "eighty" }, { "input": "82", "output": "eighty-two" }, { "input": "90", "output": "ninety" }, { "input": "91", "output": "ninety-one" }, { "input": "95", "output": "ninety-five" }, { "input": "71", "output": "seventy-one" }, { "input": "46", "output": "forty-six" }, { "input": "84", "output": "eighty-four" }, { "input": "22", "output": "twenty-two" }, { "input": "23", "output": "twenty-three" }, { "input": "24", "output": "twenty-four" }, { "input": "25", "output": "twenty-five" }, { "input": "26", "output": "twenty-six" }, { "input": "28", "output": "twenty-eight" }, { "input": "31", "output": "thirty-one" }, { "input": "33", "output": "thirty-three" }, { "input": "34", "output": "thirty-four" }, { "input": "35", "output": "thirty-five" }, { "input": "36", "output": "thirty-six" }, { "input": "37", "output": "thirty-seven" }, { "input": "39", "output": "thirty-nine" }, { "input": "65", "output": "sixty-five" }, { "input": "68", "output": "sixty-eight" }, { "input": "41", "output": "forty-one" }, { "input": "42", "output": "forty-two" }, { "input": "44", "output": "forty-four" }, { "input": "45", "output": "forty-five" }, { "input": "48", "output": "forty-eight" }, { "input": "49", "output": "forty-nine" }, { "input": "51", "output": "fifty-one" }, { "input": "52", "output": "fifty-two" }, { "input": "53", "output": "fifty-three" }, { "input": "55", "output": "fifty-five" }, { "input": "57", "output": "fifty-seven" }, { "input": "58", "output": "fifty-eight" }, { "input": "59", "output": "fifty-nine" }, { "input": "61", "output": "sixty-one" }, { "input": "62", "output": "sixty-two" }, { "input": "64", "output": "sixty-four" }, { "input": "67", "output": "sixty-seven" }, { "input": "69", "output": "sixty-nine" }, { "input": "72", "output": "seventy-two" }, { "input": "73", "output": "seventy-three" }, { "input": "74", "output": "seventy-four" }, { "input": "75", "output": "seventy-five" }, { "input": "77", "output": "seventy-seven" }, { "input": "78", "output": "seventy-eight" }, { "input": "79", "output": "seventy-nine" }, { "input": "81", "output": "eighty-one" }, { "input": "83", "output": "eighty-three" }, { "input": "85", "output": "eighty-five" }, { "input": "86", "output": "eighty-six" }, { "input": "87", "output": "eighty-seven" }, { "input": "88", "output": "eighty-eight" }, { "input": "89", "output": "eighty-nine" }, { "input": "92", "output": "ninety-two" }, { "input": "93", "output": "ninety-three" }, { "input": "94", "output": "ninety-four" }, { "input": "96", "output": "ninety-six" }, { "input": "7", "output": "seven" }, { "input": "97", "output": "ninety-seven" }, { "input": "98", "output": "ninety-eight" }, { "input": "3", "output": "three" }, { "input": "4", "output": "four" }, { "input": "5", "output": "five" } ]

1,626,611,278

2,147,483,647

Python 3

RUNTIME_ERROR

TESTS

6

61

6,963,200

n = int(input()) nums = {1:'one', 2:'two', 3:'three', 4:'four', 5:'five', 6:'six', 7:'seven', 8:'eight', 9:'nine', 10:'ten', 11:'eleven', 12:'twelve', 13:'thirteen', 14:'fourteen', 15:'fifteen', 16:'sixteen', 17:'seventeen', 18:'eighteen', 19:'nineteen', 20:'twenty',30:'thirty', 40:'forty', 50:'fifty', 60:'sixty', 70:'seventy', 80:'eighty',90:'ninety'} if n<= 20: print(nums[n]) else: tens = n - n%10 ans = nums[tens] ans += '-' + nums[n%10] print(ans)

Title: Tavas and Nafas Time Limit: None seconds Memory Limit: None megabytes Problem Description: Today Tavas got his test result as an integer score and he wants to share it with his girlfriend, Nafas. His phone operating system is Tavdroid, and its keyboard doesn't have any digits! He wants to share his score with Nafas via text, so he has no choice but to send this number using words. He ate coffee mix without water again, so right now he's really messed up and can't think. Your task is to help him by telling him what to type. Input Specification: The first and only line of input contains an integer *s* (0<=≤<=*s*<=≤<=99), Tavas's score. Output Specification: In the first and only line of output, print a single string consisting only from English lowercase letters and hyphens ('-'). Do not use spaces. Demo Input: ['6\n', '99\n', '20\n'] Demo Output: ['six\n', 'ninety-nine\n', 'twenty\n'] Note: You can find all you need to know about English numerals in [http://en.wikipedia.org/wiki/English_numerals](https://en.wikipedia.org/wiki/English_numerals) .

```python n = int(input()) nums = {1:'one', 2:'two', 3:'three', 4:'four', 5:'five', 6:'six', 7:'seven', 8:'eight', 9:'nine', 10:'ten', 11:'eleven', 12:'twelve', 13:'thirteen', 14:'fourteen', 15:'fifteen', 16:'sixteen', 17:'seventeen', 18:'eighteen', 19:'nineteen', 20:'twenty',30:'thirty', 40:'forty', 50:'fifty', 60:'sixty', 70:'seventy', 80:'eighty',90:'ninety'} if n<= 20: print(nums[n]) else: tens = n - n%10 ans = nums[tens] ans += '-' + nums[n%10] print(ans) ```

-1

58

A

Chat room

PROGRAMMING

1,000

[ "greedy", "strings" ]

A. Chat room

1

256

Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.

The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.

If Vasya managed to say hello, print "YES", otherwise print "NO".

[ "ahhellllloou\n", "hlelo\n" ]

[ "YES\n", "NO\n" ]

none

500

[ { "input": "ahhellllloou", "output": "YES" }, { "input": "hlelo", "output": "NO" }, { "input": "helhcludoo", "output": "YES" }, { "input": "hehwelloho", "output": "YES" }, { "input": "pnnepelqomhhheollvlo", "output": "YES" }, { "input": "tymbzjyqhymedasloqbq", "output": "NO" }, { "input": "yehluhlkwo", "output": "NO" }, { "input": "hatlevhhalrohairnolsvocafgueelrqmlqlleello", "output": "YES" }, { "input": "hhhtehdbllnhwmbyhvelqqyoulretpbfokflhlhreeflxeftelziclrwllrpflflbdtotvlqgoaoqldlroovbfsq", "output": "YES" }, { "input": "rzlvihhghnelqtwlexmvdjjrliqllolhyewgozkuovaiezgcilelqapuoeglnwmnlftxxiigzczlouooi", "output": "YES" }, { "input": "pfhhwctyqdlkrwhebfqfelhyebwllhemtrmeblgrynmvyhioesqklclocxmlffuormljszllpoo", "output": "YES" }, { "input": "lqllcolohwflhfhlnaow", "output": "NO" }, { "input": "heheeellollvoo", "output": "YES" }, { "input": "hellooo", "output": "YES" }, { "input": "o", "output": "NO" }, { "input": "hhqhzeclohlehljlhtesllylrolmomvuhcxsobtsckogdv", "output": "YES" }, { "input": "yoegfuzhqsihygnhpnukluutocvvwuldiighpogsifealtgkfzqbwtmgghmythcxflebrkctlldlkzlagovwlstsghbouk", "output": "YES" }, { "input": "uatqtgbvrnywfacwursctpagasnhydvmlinrcnqrry", "output": "NO" }, { "input": "tndtbldbllnrwmbyhvqaqqyoudrstpbfokfoclnraefuxtftmgzicorwisrpfnfpbdtatvwqgyalqtdtrjqvbfsq", "output": "NO" }, { "input": "rzlvirhgemelnzdawzpaoqtxmqucnahvqnwldklrmjiiyageraijfivigvozgwngiulttxxgzczptusoi", "output": "YES" }, { "input": "kgyelmchocojsnaqdsyeqgnllytbqietpdlgknwwumqkxrexgdcnwoldicwzwofpmuesjuxzrasscvyuqwspm", "output": "YES" }, { "input": "pnyvrcotjvgynbeldnxieghfltmexttuxzyac", "output": "NO" }, { "input": "dtwhbqoumejligbenxvzhjlhosqojetcqsynlzyhfaevbdpekgbtjrbhlltbceobcok", "output": "YES" }, { "input": "crrfpfftjwhhikwzeedrlwzblckkteseofjuxjrktcjfsylmlsvogvrcxbxtffujqshslemnixoeezivksouefeqlhhokwbqjz", "output": "YES" }, { "input": "jhfbndhyzdvhbvhmhmefqllujdflwdpjbehedlsqfdsqlyelwjtyloxwsvasrbqosblzbowlqjmyeilcvotdlaouxhdpoeloaovb", "output": "YES" }, { "input": "hwlghueoemiqtjhhpashjsouyegdlvoyzeunlroypoprnhlyiwiuxrghekaylndhrhllllwhbebezoglydcvykllotrlaqtvmlla", "output": "YES" }, { "input": "wshiaunnqnqxodholbipwhhjmyeblhgpeleblklpzwhdunmpqkbuzloetmwwxmeltkrcomulxauzlwmlklldjodozxryghsnwgcz", "output": "YES" }, { "input": "shvksednttggehroewuiptvvxtrzgidravtnjwuqrlnnkxbplctzkckinpkgjopjfoxdbojtcvsuvablcbkrzajrlhgobkcxeqti", "output": "YES" }, { "input": "hyyhddqhxhekehkwfhlnlsihzefwchzerevcjtokefplholrbvxlltdlafjxrfhleglrvlolojoqaolagtbeyogxlbgfolllslli", "output": "YES" }, { "input": "iaagrdhhelxpdegueiulflquvlzidoprzkehthkldaytrjwhyhprwjxlltinxvuilxohqgjqcvkkdcuoliekcnlilwgqlnlzouoo", "output": "YES" }, { "input": "wfluaeseldgxyvxpwuhkptdmlflnlhktwxiabsvkolsquymrmhzczzoybvlilhmvsuunggvgxzgyyffk", "output": "NO" }, { "input": "loee", "output": "NO" }, { "input": "izdipwylefqmjbuoznfglgcfvedeouflkeehxbasaqmiooppfsqlhprospqgxvzrcpwlfdddngoqvpwvggsnvvxhmjoslwjudjlo", "output": "NO" }, { "input": "pplynflbjlseflaegzqpdxswgunhnlglqhycrbcwdfhclmcrwekkzsoosmfblljeenwfjyifcvxihhhbyhmhlkvtskeglaovjnsi", "output": "YES" }, { "input": "xqjqmenkodmlhzyzmmvofdngktygbbxbzpluzcohohmalkoeuwfikblltaaigv", "output": "NO" }, { "input": "hello", "output": "YES" }, { "input": "oohell", "output": "NO" }, { "input": "hell", "output": "NO" }, { "input": "eloellohoelo", "output": "NO" }, { "input": "helllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllo", "output": "YES" } ]

1,579,113,388

2,147,483,647

Python 3

OK

TESTS

40

124

0

from sys import stdin, stdout def main(): input_string = stdin.readline() search_string = "hello" MAX_LENGTH_OF_WORD = len(search_string) current_key = 0 for letter in input_string: if letter == search_string[current_key]: current_key += 1 if current_key == MAX_LENGTH_OF_WORD: break if current_key == MAX_LENGTH_OF_WORD: stdout.write(f"YES\n") else: stdout.write(f"NO\n") if __name__ == "__main__": main()

Title: Chat room Time Limit: 1 seconds Memory Limit: 256 megabytes Problem Description: Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*. Input Specification: The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters. Output Specification: If Vasya managed to say hello, print "YES", otherwise print "NO". Demo Input: ['ahhellllloou\n', 'hlelo\n'] Demo Output: ['YES\n', 'NO\n'] Note: none

```python from sys import stdin, stdout def main(): input_string = stdin.readline() search_string = "hello" MAX_LENGTH_OF_WORD = len(search_string) current_key = 0 for letter in input_string: if letter == search_string[current_key]: current_key += 1 if current_key == MAX_LENGTH_OF_WORD: break if current_key == MAX_LENGTH_OF_WORD: stdout.write(f"YES\n") else: stdout.write(f"NO\n") if __name__ == "__main__": main() ```

3.938

439

D

Devu and his Brother

PROGRAMMING

1,700

[ "binary search", "sortings", "ternary search", "two pointers" ]

null

Devu and his brother love each other a lot. As they are super geeks, they only like to play with arrays. They are given two arrays *a* and *b* by their father. The array *a* is given to Devu and *b* to his brother. As Devu is really a naughty kid, he wants the minimum value of his array *a* should be at least as much as the maximum value of his brother's array *b*. Now you have to help Devu in achieving this condition. You can perform multiple operations on the arrays. In a single operation, you are allowed to decrease or increase any element of any of the arrays by 1. Note that you are allowed to apply the operation on any index of the array multiple times. You need to find minimum number of operations required to satisfy Devu's condition so that the brothers can play peacefully without fighting.

The first line contains two space-separated integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=105). The second line will contain *n* space-separated integers representing content of the array *a* (1<=≤<=*a**i*<=≤<=109). The third line will contain *m* space-separated integers representing content of the array *b* (1<=≤<=*b**i*<=≤<=109).

You need to output a single integer representing the minimum number of operations needed to satisfy Devu's condition.

[ "2 2\n2 3\n3 5\n", "3 2\n1 2 3\n3 4\n", "3 2\n4 5 6\n1 2\n" ]

[ "3\n", "4\n", "0\n" ]

In example 1, you can increase *a*1 by 1 and decrease *b*2 by 1 and then again decrease *b*2 by 1. Now array *a* will be [3; 3] and array *b* will also be [3; 3]. Here minimum element of *a* is at least as large as maximum element of *b*. So minimum number of operations needed to satisfy Devu's condition are 3. In example 3, you don't need to do any operation, Devu's condition is already satisfied.

2,000

[ { "input": "2 2\n2 3\n3 5", "output": "3" }, { "input": "3 2\n1 2 3\n3 4", "output": "4" }, { "input": "3 2\n4 5 6\n1 2", "output": "0" }, { "input": "10 10\n23 100 38 38 73 54 59 69 44 86\n100 100 100 100 100 100 100 100 100 100", "output": "416" }, { "input": "1 1\n401114\n998223974", "output": "997822860" }, { "input": "1 1\n100\n4", "output": "0" }, { "input": "1 1\n100\n183299", "output": "183199" }, { "input": "1 1\n999999999\n1000000000", "output": "1" }, { "input": "1 1\n1000000000\n1000000000", "output": "0" }, { "input": "1 1\n1\n2", "output": "1" }, { "input": "1 1\n1\n1", "output": "0" }, { "input": "1 1\n2\n1", "output": "0" }, { "input": "1 1\n1\n2", "output": "1" }, { "input": "1 1\n1\n3", "output": "2" }, { "input": "1 2\n1\n2 2", "output": "1" }, { "input": "2 1\n2 2\n3", "output": "1" } ]

1,616,589,941

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

7

171

10,444,800

n, m = map(int, input(). split()) a = list(map(int, input(). split())) b = list(map(int, input(). split())) if min(a) >= max(b): print(0) else: k = min(b) ans = 0 for i in a: if i < k: ans += k - i for i in b: if i > k: ans += i - k print(ans)

Title: Devu and his Brother Time Limit: None seconds Memory Limit: None megabytes Problem Description: Devu and his brother love each other a lot. As they are super geeks, they only like to play with arrays. They are given two arrays *a* and *b* by their father. The array *a* is given to Devu and *b* to his brother. As Devu is really a naughty kid, he wants the minimum value of his array *a* should be at least as much as the maximum value of his brother's array *b*. Now you have to help Devu in achieving this condition. You can perform multiple operations on the arrays. In a single operation, you are allowed to decrease or increase any element of any of the arrays by 1. Note that you are allowed to apply the operation on any index of the array multiple times. You need to find minimum number of operations required to satisfy Devu's condition so that the brothers can play peacefully without fighting. Input Specification: The first line contains two space-separated integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=105). The second line will contain *n* space-separated integers representing content of the array *a* (1<=≤<=*a**i*<=≤<=109). The third line will contain *m* space-separated integers representing content of the array *b* (1<=≤<=*b**i*<=≤<=109). Output Specification: You need to output a single integer representing the minimum number of operations needed to satisfy Devu's condition. Demo Input: ['2 2\n2 3\n3 5\n', '3 2\n1 2 3\n3 4\n', '3 2\n4 5 6\n1 2\n'] Demo Output: ['3\n', '4\n', '0\n'] Note: In example 1, you can increase *a*1 by 1 and decrease *b*2 by 1 and then again decrease *b*2 by 1. Now array *a* will be [3; 3] and array *b* will also be [3; 3]. Here minimum element of *a* is at least as large as maximum element of *b*. So minimum number of operations needed to satisfy Devu's condition are 3. In example 3, you don't need to do any operation, Devu's condition is already satisfied.

```python n, m = map(int, input(). split()) a = list(map(int, input(). split())) b = list(map(int, input(). split())) if min(a) >= max(b): print(0) else: k = min(b) ans = 0 for i in a: if i < k: ans += k - i for i in b: if i > k: ans += i - k print(ans) ```

0

none

0

[ "none" ]

null

Today Sonya learned about long integers and invited all her friends to share the fun. Sonya has an initially empty multiset with integers. Friends give her *t* queries, each of one of the following type: 1. <=+<= *a**i* — add non-negative integer *a**i* to the multiset. Note, that she has a multiset, thus there may be many occurrences of the same integer. 1. <=-<= *a**i* — delete a single occurrence of non-negative integer *a**i* from the multiset. It's guaranteed, that there is at least one *a**i* in the multiset. 1. ? *s* — count the number of integers in the multiset (with repetitions) that match some pattern *s* consisting of 0 and 1. In the pattern, 0 stands for the even digits, while 1 stands for the odd. Integer *x* matches the pattern *s*, if the parity of the *i*-th from the right digit in decimal notation matches the *i*-th from the right digit of the pattern. If the pattern is shorter than this integer, it's supplemented with 0-s from the left. Similarly, if the integer is shorter than the pattern its decimal notation is supplemented with the 0-s from the left. For example, if the pattern is *s*<==<=010, than integers 92, 2212, 50 and 414 match the pattern, while integers 3, 110, 25 and 1030 do not.

The first line of the input contains an integer *t* (1<=≤<=*t*<=≤<=100<=000) — the number of operation Sonya has to perform. Next *t* lines provide the descriptions of the queries in order they appear in the input file. The *i*-th row starts with a character *c**i* — the type of the corresponding operation. If *c**i* is equal to '+' or '-' then it's followed by a space and an integer *a**i* (0<=≤<=*a**i*<=<<=1018) given without leading zeroes (unless it's 0). If *c**i* equals '?' then it's followed by a space and a sequence of zeroes and onse, giving the pattern of length no more than 18. It's guaranteed that there will be at least one query of type '?'. It's guaranteed that any time some integer is removed from the multiset, there will be at least one occurrence of this integer in it.

For each query of the third type print the number of integers matching the given pattern. Each integer is counted as many times, as it appears in the multiset at this moment of time.

[ "12\n+ 1\n+ 241\n? 1\n+ 361\n- 241\n? 0101\n+ 101\n? 101\n- 101\n? 101\n+ 4000\n? 0\n", "4\n+ 200\n+ 200\n- 200\n? 0\n" ]

[ "2\n1\n2\n1\n1\n", "1\n" ]

Consider the integers matching the patterns from the queries of the third type. Queries are numbered in the order they appear in the input. 1. 1 and 241. 1. 361. 1. 101 and 361. 1. 361. 1. 4000.

0

[ { "input": "12\n+ 1\n+ 241\n? 1\n+ 361\n- 241\n? 0101\n+ 101\n? 101\n- 101\n? 101\n+ 4000\n? 0", "output": "2\n1\n2\n1\n1" }, { "input": "4\n+ 200\n+ 200\n- 200\n? 0", "output": "1" }, { "input": "20\n+ 61\n+ 99\n+ 51\n+ 70\n+ 7\n+ 34\n+ 71\n+ 86\n+ 68\n+ 39\n+ 78\n+ 81\n+ 89\n? 10\n? 00\n? 10\n? 01\n? 01\n? 00\n? 00", "output": "3\n2\n3\n4\n4\n2\n2" }, { "input": "20\n+ 13\n+ 50\n+ 9\n? 0\n+ 24\n? 0\n- 24\n? 0\n+ 79\n? 11\n- 13\n? 11\n- 50\n? 10\n? 1\n- 9\n? 1\n? 11\n- 79\n? 11", "output": "0\n1\n0\n2\n1\n0\n1\n0\n1\n0" }, { "input": "10\n+ 870566619432760298\n+ 869797178280285214\n+ 609920823721618090\n+ 221159591436767023\n+ 730599542279836538\n? 101001100111001011\n? 001111010101010011\n? 100010100011101110\n? 100110010110001100\n? 110000011101110011", "output": "0\n0\n0\n0\n0" }, { "input": "10\n+ 96135\n? 10111\n+ 63322\n? 10111\n+ 44490\n? 10111\n+ 69312\n? 10111\n? 01100\n+ 59396", "output": "1\n1\n1\n1\n1" }, { "input": "10\n+ 2\n- 2\n+ 778\n+ 3\n+ 4\n- 4\n+ 1\n+ 617\n? 011\n? 011", "output": "1\n1" }, { "input": "20\n+ 8\n+ 39532\n+ 813\n- 39532\n? 00011\n? 00000\n? 00011\n+ 70424\n- 8\n? 00011\n- 70424\n? 00011\n+ 29\n? 00001\n+ 6632\n+ 3319\n? 00001\n+ 3172\n? 01111\n- 29", "output": "1\n1\n1\n1\n1\n1\n1\n1" } ]

1,574,005,686

2,147,483,647

PyPy 3

TIME_LIMIT_EXCEEDED

TESTS

11

1,000

9,830,400

def replace_binary(s): res = '' for i in s: if int(i) % 2 == 0: res += '0' else: res += '1' return res def get_mask(s): res = 0 for i in range(len(s)): if s[i] == '1': res += 1 << len(s) - i - 1 return res t = int(input()) cnt = {} for _ in range(t): op, n = input().split() n = get_mask(replace_binary(n)) if op == '+': cnt[n] = cnt.get(n, 0) + 1 elif op == '-': cnt[n] -= 1 else: print(cnt.get(n, 0))

Title: none Time Limit: None seconds Memory Limit: None megabytes Problem Description: Today Sonya learned about long integers and invited all her friends to share the fun. Sonya has an initially empty multiset with integers. Friends give her *t* queries, each of one of the following type: 1. <=+<= *a**i* — add non-negative integer *a**i* to the multiset. Note, that she has a multiset, thus there may be many occurrences of the same integer. 1. <=-<= *a**i* — delete a single occurrence of non-negative integer *a**i* from the multiset. It's guaranteed, that there is at least one *a**i* in the multiset. 1. ? *s* — count the number of integers in the multiset (with repetitions) that match some pattern *s* consisting of 0 and 1. In the pattern, 0 stands for the even digits, while 1 stands for the odd. Integer *x* matches the pattern *s*, if the parity of the *i*-th from the right digit in decimal notation matches the *i*-th from the right digit of the pattern. If the pattern is shorter than this integer, it's supplemented with 0-s from the left. Similarly, if the integer is shorter than the pattern its decimal notation is supplemented with the 0-s from the left. For example, if the pattern is *s*<==<=010, than integers 92, 2212, 50 and 414 match the pattern, while integers 3, 110, 25 and 1030 do not. Input Specification: The first line of the input contains an integer *t* (1<=≤<=*t*<=≤<=100<=000) — the number of operation Sonya has to perform. Next *t* lines provide the descriptions of the queries in order they appear in the input file. The *i*-th row starts with a character *c**i* — the type of the corresponding operation. If *c**i* is equal to '+' or '-' then it's followed by a space and an integer *a**i* (0<=≤<=*a**i*<=<<=1018) given without leading zeroes (unless it's 0). If *c**i* equals '?' then it's followed by a space and a sequence of zeroes and onse, giving the pattern of length no more than 18. It's guaranteed that there will be at least one query of type '?'. It's guaranteed that any time some integer is removed from the multiset, there will be at least one occurrence of this integer in it. Output Specification: For each query of the third type print the number of integers matching the given pattern. Each integer is counted as many times, as it appears in the multiset at this moment of time. Demo Input: ['12\n+ 1\n+ 241\n? 1\n+ 361\n- 241\n? 0101\n+ 101\n? 101\n- 101\n? 101\n+ 4000\n? 0\n', '4\n+ 200\n+ 200\n- 200\n? 0\n'] Demo Output: ['2\n1\n2\n1\n1\n', '1\n'] Note: Consider the integers matching the patterns from the queries of the third type. Queries are numbered in the order they appear in the input. 1. 1 and 241. 1. 361. 1. 101 and 361. 1. 361. 1. 4000.

```python def replace_binary(s): res = '' for i in s: if int(i) % 2 == 0: res += '0' else: res += '1' return res def get_mask(s): res = 0 for i in range(len(s)): if s[i] == '1': res += 1 << len(s) - i - 1 return res t = int(input()) cnt = {} for _ in range(t): op, n = input().split() n = get_mask(replace_binary(n)) if op == '+': cnt[n] = cnt.get(n, 0) + 1 elif op == '-': cnt[n] -= 1 else: print(cnt.get(n, 0)) ```

0

758

A

Holiday Of Equality

PROGRAMMING

800

[ "implementation", "math" ]

null

In Berland it is the holiday of equality. In honor of the holiday the king decided to equalize the welfare of all citizens in Berland by the expense of the state treasury. Totally in Berland there are *n* citizens, the welfare of each of them is estimated as the integer in *a**i* burles (burle is the currency in Berland). You are the royal treasurer, which needs to count the minimum charges of the kingdom on the king's present. The king can only give money, he hasn't a power to take away them.

The first line contains the integer *n* (1<=≤<=*n*<=≤<=100) — the number of citizens in the kingdom. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* (0<=≤<=*a**i*<=≤<=106) — the welfare of the *i*-th citizen.

In the only line print the integer *S* — the minimum number of burles which are had to spend.

[ "5\n0 1 2 3 4\n", "5\n1 1 0 1 1\n", "3\n1 3 1\n", "1\n12\n" ]

[ "10", "1", "4", "0" ]

In the first example if we add to the first citizen 4 burles, to the second 3, to the third 2 and to the fourth 1, then the welfare of all citizens will equal 4. In the second example it is enough to give one burle to the third citizen. In the third example it is necessary to give two burles to the first and the third citizens to make the welfare of citizens equal 3. In the fourth example it is possible to give nothing to everyone because all citizens have 12 burles.

500

[ { "input": "5\n0 1 2 3 4", "output": "10" }, { "input": "5\n1 1 0 1 1", "output": "1" }, { "input": "3\n1 3 1", "output": "4" }, { "input": "1\n12", "output": "0" }, { "input": "3\n1 2 3", "output": "3" }, { "input": "14\n52518 718438 358883 462189 853171 592966 225788 46977 814826 295697 676256 561479 56545 764281", "output": "5464380" }, { "input": "21\n842556 216391 427181 626688 775504 168309 851038 448402 880826 73697 593338 519033 135115 20128 424606 939484 846242 756907 377058 241543 29353", "output": "9535765" }, { "input": "3\n1 3 2", "output": "3" }, { "input": "3\n2 1 3", "output": "3" }, { "input": "3\n2 3 1", "output": "3" }, { "input": "3\n3 1 2", "output": "3" }, { "input": "3\n3 2 1", "output": "3" }, { "input": "1\n228503", "output": "0" }, { "input": "2\n32576 550340", "output": "517764" }, { "input": "3\n910648 542843 537125", "output": "741328" }, { "input": "4\n751720 572344 569387 893618", "output": "787403" }, { "input": "6\n433864 631347 597596 794426 713555 231193", "output": "1364575" }, { "input": "9\n31078 645168 695751 126111 375934 150495 838412 434477 993107", "output": "4647430" }, { "input": "30\n315421 772664 560686 654312 151528 356749 351486 707462 820089 226682 546700 136028 824236 842130 578079 337807 665903 764100 617900 822937 992759 591749 651310 742085 767695 695442 17967 515106 81059 186025", "output": "13488674" }, { "input": "45\n908719 394261 815134 419990 926993 383792 772842 277695 527137 655356 684956 695716 273062 550324 106247 399133 442382 33076 462920 294674 846052 817752 421365 474141 290471 358990 109812 74492 543281 169434 919692 786809 24028 197184 310029 801476 699355 429672 51343 374128 776726 850380 293868 981569 550763", "output": "21993384" }, { "input": "56\n100728 972537 13846 385421 756708 184642 259487 319707 376662 221694 675284 972837 499419 13846 38267 289898 901299 831197 954715 197515 514102 910423 127555 883934 362472 870788 538802 741008 973434 448124 391526 363321 947321 544618 68006 782313 955075 741981 815027 723297 585059 718114 700739 413489 454091 736144 308999 98065 3716 347323 9635 289003 986510 607065 60236 273351", "output": "26984185" }, { "input": "70\n644488 5444 150441 714420 602059 335330 510670 196555 546346 740011 509449 850947 692874 524857 750434 952985 223744 374727 896124 753037 367352 679050 560202 172728 569291 778616 332116 286927 843598 372698 244906 498046 900681 709791 420904 724593 864493 813094 791377 39998 296710 625656 403891 579231 706693 984045 16901 574259 562265 761104 930361 256045 124461 538980 573508 372148 988722 108592 784354 55302 232524 277205 782251 299943 436488 743389 324618 742543 266915 99642", "output": "32816391" }, { "input": "1\n0", "output": "0" }, { "input": "1\n1000000", "output": "0" }, { "input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0", "output": "0" }, { "input": "100\n1000000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0", "output": "99000000" }, { "input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1000000", "output": "99000000" }, { "input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1000000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0", "output": "99000000" }, { "input": "3\n0 0 0", "output": "0" }, { "input": "50\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", "output": "0" }, { "input": "3\n5 0 0", "output": "10" }, { "input": "5\n2 10 0 0 0", "output": "38" } ]

1,664,074,965

2,147,483,647

PyPy 3-64

COMPILATION_ERROR

TESTS

0

n = input() arr = list(map(int,input().split())) ans = 0 a = max(arr) [ans+= (a-i) for i in arr] print(ans)

Title: Holiday Of Equality Time Limit: None seconds Memory Limit: None megabytes Problem Description: In Berland it is the holiday of equality. In honor of the holiday the king decided to equalize the welfare of all citizens in Berland by the expense of the state treasury. Totally in Berland there are *n* citizens, the welfare of each of them is estimated as the integer in *a**i* burles (burle is the currency in Berland). You are the royal treasurer, which needs to count the minimum charges of the kingdom on the king's present. The king can only give money, he hasn't a power to take away them. Input Specification: The first line contains the integer *n* (1<=≤<=*n*<=≤<=100) — the number of citizens in the kingdom. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* (0<=≤<=*a**i*<=≤<=106) — the welfare of the *i*-th citizen. Output Specification: In the only line print the integer *S* — the minimum number of burles which are had to spend. Demo Input: ['5\n0 1 2 3 4\n', '5\n1 1 0 1 1\n', '3\n1 3 1\n', '1\n12\n'] Demo Output: ['10', '1', '4', '0'] Note: In the first example if we add to the first citizen 4 burles, to the second 3, to the third 2 and to the fourth 1, then the welfare of all citizens will equal 4. In the second example it is enough to give one burle to the third citizen. In the third example it is necessary to give two burles to the first and the third citizens to make the welfare of citizens equal 3. In the fourth example it is possible to give nothing to everyone because all citizens have 12 burles.

```python n = input() arr = list(map(int,input().split())) ans = 0 a = max(arr) [ans+= (a-i) for i in arr] print(ans) ```

-1

255

A

Greg's Workout

PROGRAMMING

800

[ "implementation" ]

null

Greg is a beginner bodybuilder. Today the gym coach gave him the training plan. All it had was *n* integers *a*1,<=*a*2,<=...,<=*a**n*. These numbers mean that Greg needs to do exactly *n* exercises today. Besides, Greg should repeat the *i*-th in order exercise *a**i* times. Greg now only does three types of exercises: "chest" exercises, "biceps" exercises and "back" exercises. Besides, his training is cyclic, that is, the first exercise he does is a "chest" one, the second one is "biceps", the third one is "back", the fourth one is "chest", the fifth one is "biceps", and so on to the *n*-th exercise. Now Greg wonders, which muscle will get the most exercise during his training. We know that the exercise Greg repeats the maximum number of times, trains the corresponding muscle the most. Help Greg, determine which muscle will get the most training.

The first line contains integer *n* (1<=≤<=*n*<=≤<=20). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=25) — the number of times Greg repeats the exercises.

Print word "chest" (without the quotes), if the chest gets the most exercise, "biceps" (without the quotes), if the biceps gets the most exercise and print "back" (without the quotes) if the back gets the most exercise. It is guaranteed that the input is such that the answer to the problem is unambiguous.

[ "2\n2 8\n", "3\n5 1 10\n", "7\n3 3 2 7 9 6 8\n" ]

[ "biceps\n", "back\n", "chest\n" ]

In the first sample Greg does 2 chest, 8 biceps and zero back exercises, so the biceps gets the most exercises. In the second sample Greg does 5 chest, 1 biceps and 10 back exercises, so the back gets the most exercises. In the third sample Greg does 18 chest, 12 biceps and 8 back exercises, so the chest gets the most exercise.

500

[ { "input": "2\n2 8", "output": "biceps" }, { "input": "3\n5 1 10", "output": "back" }, { "input": "7\n3 3 2 7 9 6 8", "output": "chest" }, { "input": "4\n5 6 6 2", "output": "chest" }, { "input": "5\n8 2 2 6 3", "output": "chest" }, { "input": "6\n8 7 2 5 3 4", "output": "chest" }, { "input": "8\n7 2 9 10 3 8 10 6", "output": "chest" }, { "input": "9\n5 4 2 3 4 4 5 2 2", "output": "chest" }, { "input": "10\n4 9 8 5 3 8 8 10 4 2", "output": "biceps" }, { "input": "11\n10 9 7 6 1 3 9 7 1 3 5", "output": "chest" }, { "input": "12\n24 22 6 16 5 21 1 7 2 19 24 5", "output": "chest" }, { "input": "13\n24 10 5 7 16 17 2 7 9 20 15 2 24", "output": "chest" }, { "input": "14\n13 14 19 8 5 17 9 16 15 9 5 6 3 7", "output": "back" }, { "input": "15\n24 12 22 21 25 23 21 5 3 24 23 13 12 16 12", "output": "chest" }, { "input": "16\n12 6 18 6 25 7 3 1 1 17 25 17 6 8 17 8", "output": "biceps" }, { "input": "17\n13 8 13 4 9 21 10 10 9 22 14 23 22 7 6 14 19", "output": "chest" }, { "input": "18\n1 17 13 6 11 10 25 13 24 9 21 17 3 1 17 12 25 21", "output": "back" }, { "input": "19\n22 22 24 25 19 10 7 10 4 25 19 14 1 14 3 18 4 19 24", "output": "chest" }, { "input": "20\n9 8 22 11 18 14 15 10 17 11 2 1 25 20 7 24 4 25 9 20", "output": "chest" }, { "input": "1\n10", "output": "chest" }, { "input": "2\n15 3", "output": "chest" }, { "input": "3\n21 11 19", "output": "chest" }, { "input": "4\n19 24 13 15", "output": "chest" }, { "input": "5\n4 24 1 9 19", "output": "biceps" }, { "input": "6\n6 22 24 7 15 24", "output": "back" }, { "input": "7\n10 8 23 23 14 18 14", "output": "chest" }, { "input": "8\n5 16 8 9 17 16 14 7", "output": "biceps" }, { "input": "9\n12 3 10 23 6 4 22 13 12", "output": "chest" }, { "input": "10\n1 9 20 18 20 17 7 24 23 2", "output": "back" }, { "input": "11\n22 25 8 2 18 15 1 13 1 11 4", "output": "biceps" }, { "input": "12\n20 12 14 2 15 6 24 3 11 8 11 14", "output": "chest" }, { "input": "13\n2 18 8 8 8 20 5 22 15 2 5 19 18", "output": "back" }, { "input": "14\n1 6 10 25 17 13 21 11 19 4 15 24 5 22", "output": "biceps" }, { "input": "15\n13 5 25 13 17 25 19 21 23 17 12 6 14 8 6", "output": "back" }, { "input": "16\n10 15 2 17 22 12 14 14 6 11 4 13 9 8 21 14", "output": "chest" }, { "input": "17\n7 22 9 22 8 7 20 22 23 5 12 11 1 24 17 20 10", "output": "biceps" }, { "input": "18\n18 15 4 25 5 11 21 25 12 14 25 23 19 19 13 6 9 17", "output": "chest" }, { "input": "19\n3 1 3 15 15 25 10 25 23 10 9 21 13 23 19 3 24 21 14", "output": "back" }, { "input": "20\n19 18 11 3 6 14 3 3 25 3 1 19 25 24 23 12 7 4 8 6", "output": "back" }, { "input": "1\n19", "output": "chest" }, { "input": "2\n1 7", "output": "biceps" }, { "input": "3\n18 18 23", "output": "back" }, { "input": "4\n12 15 1 13", "output": "chest" }, { "input": "5\n11 14 25 21 21", "output": "biceps" }, { "input": "6\n11 9 12 11 22 18", "output": "biceps" }, { "input": "7\n11 1 16 20 21 25 20", "output": "chest" }, { "input": "8\n1 2 20 9 3 22 17 4", "output": "back" }, { "input": "9\n19 2 10 19 15 20 3 1 13", "output": "back" }, { "input": "10\n11 2 11 8 21 16 2 3 19 9", "output": "back" }, { "input": "20\n25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 24", "output": "chest" }, { "input": "12\n4 24 21 3 13 24 22 13 12 21 1 15", "output": "back" }, { "input": "13\n14 14 16 2 13 5 1 14 9 4 16 8 3", "output": "biceps" }, { "input": "14\n1 9 15 4 11 8 25 3 9 14 13 2 1 11", "output": "biceps" }, { "input": "15\n4 19 10 6 16 12 5 11 7 23 1 24 11 7 17", "output": "back" }, { "input": "16\n2 8 2 8 13 22 20 12 22 23 18 13 18 22 11 17", "output": "chest" }, { "input": "17\n24 5 5 16 10 8 22 6 4 13 10 10 5 23 8 20 8", "output": "chest" }, { "input": "18\n14 8 9 12 11 18 24 1 14 24 18 5 12 17 1 10 1 22", "output": "chest" }, { "input": "19\n21 2 10 6 9 1 24 5 2 19 10 13 10 7 19 2 6 13 24", "output": "chest" }, { "input": "20\n7 1 14 17 6 6 18 13 12 3 25 4 3 19 22 24 16 14 1 23", "output": "biceps" }, { "input": "1\n19", "output": "chest" }, { "input": "20\n2 1 2 2 1 2 2 1 2 1 1 1 1 1 1 1 1 1 1 22", "output": "biceps" } ]

1,644,690,621

2,147,483,647

Python 3

OK

TESTS

61

92

0

n = int(input()) a = list(map(int, input().split())) numberBiceps = 0 # 0,3,6 numberBack = 0 # 1,4,7 numberChest = 0 # 2,5,8 for i in range(0,n): if i % 3 == 0: numberChest += a[i] elif (i - 1) % 3 == 0: numberBiceps += a[i] elif (i - 2) % 3 == 0: numberBack += a[i] if numberBiceps > numberBack and numberBiceps > numberChest: print('biceps') elif numberBack > numberBiceps and numberBack > numberChest: print("back") elif numberChest > numberBiceps and numberChest > numberBack: print("chest") else: print(1%3)

Title: Greg's Workout Time Limit: None seconds Memory Limit: None megabytes Problem Description: Greg is a beginner bodybuilder. Today the gym coach gave him the training plan. All it had was *n* integers *a*1,<=*a*2,<=...,<=*a**n*. These numbers mean that Greg needs to do exactly *n* exercises today. Besides, Greg should repeat the *i*-th in order exercise *a**i* times. Greg now only does three types of exercises: "chest" exercises, "biceps" exercises and "back" exercises. Besides, his training is cyclic, that is, the first exercise he does is a "chest" one, the second one is "biceps", the third one is "back", the fourth one is "chest", the fifth one is "biceps", and so on to the *n*-th exercise. Now Greg wonders, which muscle will get the most exercise during his training. We know that the exercise Greg repeats the maximum number of times, trains the corresponding muscle the most. Help Greg, determine which muscle will get the most training. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=20). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=25) — the number of times Greg repeats the exercises. Output Specification: Print word "chest" (without the quotes), if the chest gets the most exercise, "biceps" (without the quotes), if the biceps gets the most exercise and print "back" (without the quotes) if the back gets the most exercise. It is guaranteed that the input is such that the answer to the problem is unambiguous. Demo Input: ['2\n2 8\n', '3\n5 1 10\n', '7\n3 3 2 7 9 6 8\n'] Demo Output: ['biceps\n', 'back\n', 'chest\n'] Note: In the first sample Greg does 2 chest, 8 biceps and zero back exercises, so the biceps gets the most exercises. In the second sample Greg does 5 chest, 1 biceps and 10 back exercises, so the back gets the most exercises. In the third sample Greg does 18 chest, 12 biceps and 8 back exercises, so the chest gets the most exercise.

```python n = int(input()) a = list(map(int, input().split())) numberBiceps = 0 # 0,3,6 numberBack = 0 # 1,4,7 numberChest = 0 # 2,5,8 for i in range(0,n): if i % 3 == 0: numberChest += a[i] elif (i - 1) % 3 == 0: numberBiceps += a[i] elif (i - 2) % 3 == 0: numberBack += a[i] if numberBiceps > numberBack and numberBiceps > numberChest: print('biceps') elif numberBack > numberBiceps and numberBack > numberChest: print("back") elif numberChest > numberBiceps and numberChest > numberBack: print("chest") else: print(1%3) ```

3

687

B

Remainders Game

PROGRAMMING

1,800

[ "chinese remainder theorem", "math", "number theory" ]

null

Today Pari and Arya are playing a game called Remainders. Pari chooses two positive integer *x* and *k*, and tells Arya *k* but not *x*. Arya have to find the value . There are *n* ancient numbers *c*1,<=*c*2,<=...,<=*c**n* and Pari has to tell Arya if Arya wants. Given *k* and the ancient values, tell us if Arya has a winning strategy independent of value of *x* or not. Formally, is it true that Arya can understand the value for any positive integer *x*? Note, that means the remainder of *x* after dividing it by *y*.

The first line of the input contains two integers *n* and *k* (1<=≤<=*n*,<= *k*<=≤<=1<=000<=000) — the number of ancient integers and value *k* that is chosen by Pari. The second line contains *n* integers *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=1<=000<=000).

Print "Yes" (without quotes) if Arya has a winning strategy independent of value of *x*, or "No" (without quotes) otherwise.

[ "4 5\n2 3 5 12\n", "2 7\n2 3\n" ]

[ "Yes\n", "No\n" ]

In the first sample, Arya can understand <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/d170efffcde0907ee6bcf32de21051bce0677a2c.png" style="max-width: 100.0%;max-height: 100.0%;"/> because 5 is one of the ancient numbers. In the second sample, Arya can't be sure what <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/57b5f6a96f5db073270dd3ed4266c69299ec701d.png" style="max-width: 100.0%;max-height: 100.0%;"/> is. For example 1 and 7 have the same remainders after dividing by 2 and 3, but they differ in remainders after dividing by 7.

1,000

[ { "input": "4 5\n2 3 5 12", "output": "Yes" }, { "input": "2 7\n2 3", "output": "No" }, { "input": "1 6\n8", "output": "No" }, { "input": "2 3\n9 4", "output": "Yes" }, { "input": "4 16\n19 16 13 9", "output": "Yes" }, { "input": "5 10\n5 16 19 9 17", "output": "Yes" }, { "input": "11 95\n31 49 8 139 169 121 71 17 43 29 125", "output": "No" }, { "input": "17 71\n173 43 139 73 169 199 49 81 11 89 131 107 23 29 125 152 17", "output": "No" }, { "input": "13 86\n41 64 17 31 13 97 19 25 81 47 61 37 71", "output": "No" }, { "input": "15 91\n49 121 83 67 128 125 27 113 41 169 149 19 37 29 71", "output": "Yes" }, { "input": "2 4\n2 2", "output": "No" }, { "input": "14 87\n1619 1619 1619 1619 1619 1619 1619 1619 1619 1619 1619 1619 1619 1619", "output": "No" }, { "input": "12 100\n1766 1766 1766 1766 1766 1766 1766 1766 1766 1766 1766 1766", "output": "No" }, { "input": "1 994619\n216000", "output": "No" }, { "input": "1 651040\n911250", "output": "No" }, { "input": "1 620622\n60060", "output": "No" }, { "input": "1 1\n559872", "output": "Yes" }, { "input": "88 935089\n967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967", "output": "No" }, { "input": "93 181476\n426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426", "output": "No" }, { "input": "91 4900\n630 630 70 630 910 630 630 630 770 70 770 630 630 770 70 630 70 630 70 630 70 630 630 70 910 630 630 630 770 630 630 630 70 910 70 630 70 630 770 630 630 70 630 770 70 630 70 70 630 630 70 70 70 70 630 70 70 770 910 630 70 630 770 70 910 70 630 910 630 70 770 70 70 630 770 630 70 630 70 70 630 70 630 770 630 70 630 630 70 910 630", "output": "No" }, { "input": "61 531012\n698043 698043 698043 963349 698043 698043 698043 963349 698043 698043 698043 963349 698043 698043 698043 698043 966694 698043 698043 698043 698043 698043 698043 636247 698043 963349 698043 698043 698043 698043 697838 698043 963349 698043 698043 966694 698043 698043 698043 698043 698043 698043 698043 698043 698043 698043 698043 698043 698043 698043 698043 698043 698043 698043 963349 698043 698043 698043 698043 963349 698043", "output": "No" }, { "input": "1 216000\n648000", "output": "Yes" }, { "input": "2 8\n4 4", "output": "No" }, { "input": "3 8\n4 4 4", "output": "No" }, { "input": "2 8\n2 4", "output": "No" }, { "input": "3 12\n2 2 3", "output": "No" }, { "input": "10 4\n2 2 2 2 2 2 2 2 2 2", "output": "No" }, { "input": "10 1024\n1 2 4 8 16 32 64 128 256 512", "output": "No" }, { "input": "3 24\n2 2 3", "output": "No" }, { "input": "1 8\n2", "output": "No" }, { "input": "2 9\n3 3", "output": "No" }, { "input": "3 4\n2 2 2", "output": "No" }, { "input": "3 4\n1 2 2", "output": "No" }, { "input": "1 4\n2", "output": "No" }, { "input": "1 100003\n2", "output": "No" }, { "input": "1 2\n12", "output": "Yes" }, { "input": "2 988027\n989018 995006", "output": "Yes" }, { "input": "3 9\n3 3 3", "output": "No" }, { "input": "1 49\n7", "output": "No" }, { "input": "2 600000\n200000 300000", "output": "Yes" }, { "input": "3 8\n2 2 2", "output": "No" }, { "input": "7 510510\n524288 531441 390625 823543 161051 371293 83521", "output": "Yes" }, { "input": "2 30\n6 10", "output": "Yes" }, { "input": "2 27000\n5400 4500", "output": "Yes" }, { "input": "3 8\n1 2 4", "output": "No" }, { "input": "4 16\n2 2 2 2", "output": "No" }, { "input": "2 16\n4 8", "output": "No" }, { "input": "2 8\n4 2", "output": "No" }, { "input": "3 4\n2 2 3", "output": "No" }, { "input": "1 8\n4", "output": "No" }, { "input": "1 999983\n2", "output": "No" }, { "input": "3 16\n2 4 8", "output": "No" }, { "input": "2 216\n12 18", "output": "No" }, { "input": "2 16\n8 8", "output": "No" }, { "input": "2 36\n18 12", "output": "Yes" }, { "input": "2 36\n12 18", "output": "Yes" }, { "input": "2 1000000\n1000000 1000000", "output": "Yes" }, { "input": "3 20\n2 2 5", "output": "No" }, { "input": "1 2\n6", "output": "Yes" }, { "input": "4 4\n2 3 6 5", "output": "No" }, { "input": "1 2\n1", "output": "No" }, { "input": "1 6\n6", "output": "Yes" }, { "input": "2 16\n4 4", "output": "No" }, { "input": "2 3779\n1 2", "output": "No" }, { "input": "2 8\n4 12", "output": "No" }, { "input": "2 24\n4 6", "output": "No" }, { "input": "1 1\n5", "output": "Yes" }, { "input": "10 255255\n1000000 700000 300000 110000 130000 170000 190000 230000 290000 310000", "output": "Yes" }, { "input": "2 1000\n500 2", "output": "No" }, { "input": "4 8\n2 2 2 2", "output": "No" }, { "input": "1 666013\n1", "output": "No" }, { "input": "1 999983\n1", "output": "No" }, { "input": "1 125\n5", "output": "No" }, { "input": "2 32\n4 8", "output": "No" }, { "input": "3 32\n2 4 8", "output": "No" }, { "input": "3 8\n2 4 11", "output": "No" }, { "input": "1 32\n2", "output": "No" }, { "input": "4 4\n2 2 2 2", "output": "No" }, { "input": "8 32\n2 2 2 2 2 2 2 2", "output": "No" }, { "input": "3 24\n2 4 3", "output": "No" }, { "input": "2 4\n2 6", "output": "No" }, { "input": "2 24\n6 4", "output": "No" }, { "input": "1 20998\n2", "output": "No" }, { "input": "1 1\n1", "output": "Yes" }, { "input": "2 20998\n2 10499", "output": "Yes" }, { "input": "1 5\n1", "output": "No" }, { "input": "2 25\n5 5", "output": "No" }, { "input": "5 2\n2 2 2 2 2", "output": "Yes" }, { "input": "3 12\n2 3 6", "output": "No" }, { "input": "1 1\n3", "output": "Yes" }, { "input": "2 16\n8 4", "output": "No" }, { "input": "40 10\n2 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", "output": "No" } ]

1,567,534,251

2,147,483,647

PyPy 3

TIME_LIMIT_EXCEEDED

TESTS

10

1,000

12,595,200

n, k = map(int, input().split()) *a, = map(int, input().split()) res = 1 for i in a: res *= i if not res % k: print('Yes') exit() print("No")

Title: Remainders Game Time Limit: None seconds Memory Limit: None megabytes Problem Description: Today Pari and Arya are playing a game called Remainders. Pari chooses two positive integer *x* and *k*, and tells Arya *k* but not *x*. Arya have to find the value . There are *n* ancient numbers *c*1,<=*c*2,<=...,<=*c**n* and Pari has to tell Arya if Arya wants. Given *k* and the ancient values, tell us if Arya has a winning strategy independent of value of *x* or not. Formally, is it true that Arya can understand the value for any positive integer *x*? Note, that means the remainder of *x* after dividing it by *y*. Input Specification: The first line of the input contains two integers *n* and *k* (1<=≤<=*n*,<= *k*<=≤<=1<=000<=000) — the number of ancient integers and value *k* that is chosen by Pari. The second line contains *n* integers *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=1<=000<=000). Output Specification: Print "Yes" (without quotes) if Arya has a winning strategy independent of value of *x*, or "No" (without quotes) otherwise. Demo Input: ['4 5\n2 3 5 12\n', '2 7\n2 3\n'] Demo Output: ['Yes\n', 'No\n'] Note: In the first sample, Arya can understand <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/d170efffcde0907ee6bcf32de21051bce0677a2c.png" style="max-width: 100.0%;max-height: 100.0%;"/> because 5 is one of the ancient numbers. In the second sample, Arya can't be sure what <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/57b5f6a96f5db073270dd3ed4266c69299ec701d.png" style="max-width: 100.0%;max-height: 100.0%;"/> is. For example 1 and 7 have the same remainders after dividing by 2 and 3, but they differ in remainders after dividing by 7.

```python n, k = map(int, input().split()) *a, = map(int, input().split()) res = 1 for i in a: res *= i if not res % k: print('Yes') exit() print("No") ```

0

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" } ]

1,693,151,563

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

0

60

0

# -*- coding: utf-8 -*- """ Created on Sun Aug 27 23:23:44 2023 @author: Hong Liang """ a = input('How much kilos the watermelon is') a = int(a) if a >= 4: b = a % 2 if b == 0: print('Yes') else: print('No') else: print("No")

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 # -*- coding: utf-8 -*- """ Created on Sun Aug 27 23:23:44 2023 @author: Hong Liang """ a = input('How much kilos the watermelon is') a = int(a) if a >= 4: b = a % 2 if b == 0: print('Yes') else: print('No') else: print("No") ```

0

160

A

Twins

PROGRAMMING

900

[ "greedy", "sortings" ]

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" } ]

1,696,500,792

2,147,483,647

PyPy 3-64

OK

TESTS

29

124

0

# URL: https://codeforces.com/problemset/problem/160/A _ = input() a = list(map(int, input().split())) a.sort(reverse=True) total = sum(a) s = 0 for i in range(1, len(a) + 1): s += a[i - 1] if s > total - s: print(i) break

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 # URL: https://codeforces.com/problemset/problem/160/A _ = input() a = list(map(int, input().split())) a.sort(reverse=True) total = sum(a) s = 0 for i in range(1, len(a) + 1): s += a[i - 1] if s > total - s: print(i) break ```

3

864

B

Polycarp and Letters

PROGRAMMING

1,000

[ "brute force", "implementation", "strings" ]

null

Polycarp loves lowercase letters and dislikes uppercase ones. Once he got a string *s* consisting only of lowercase and uppercase Latin letters. Let *A* be a set of positions in the string. Let's call it pretty if following conditions are met: - letters on positions from *A* in the string are all distinct and lowercase; - there are no uppercase letters in the string which are situated between positions from *A* (i.e. there is no such *j* that *s*[*j*] is an uppercase letter, and *a*1<=<<=*j*<=<<=*a*2 for some *a*1 and *a*2 from *A*). Write a program that will determine the maximum number of elements in a pretty set of positions.

The first line contains a single integer *n* (1<=≤<=*n*<=≤<=200) — length of string *s*. The second line contains a string *s* consisting of lowercase and uppercase Latin letters.

Print maximum number of elements in pretty set of positions for string *s*.

[ "11\naaaaBaabAbA\n", "12\nzACaAbbaazzC\n", "3\nABC\n" ]

[ "2\n", "3\n", "0\n" ]

In the first example the desired positions might be 6 and 8 or 7 and 8. Positions 6 and 7 contain letters 'a', position 8 contains letter 'b'. The pair of positions 1 and 8 is not suitable because there is an uppercase letter 'B' between these position. In the second example desired positions can be 7, 8 and 11. There are other ways to choose pretty set consisting of three elements. In the third example the given string *s* does not contain any lowercase letters, so the answer is 0.

1,000

[ { "input": "11\naaaaBaabAbA", "output": "2" }, { "input": "12\nzACaAbbaazzC", "output": "3" }, { "input": "3\nABC", "output": "0" }, { "input": "1\na", "output": "1" }, { "input": "2\naz", "output": "2" }, { "input": "200\nXbTJZqcbpYuZQEoUrbxlPXAPCtVLrRExpQzxzqzcqsqzsiisswqitswzCtJQxOavicSdBIodideVRKHPojCNHmbnrLgwJlwOpyrJJIhrUePszxSjJGeUgTtOfewPQnPVWhZAtogRPrJLwyShNQaeNsvrJwjuuBOMPCeSckBMISQzGngfOmeyfDObncyeNsihYVtQbSEh", "output": "8" }, { "input": "2\nAZ", "output": "0" }, { "input": "28\nAabcBabcCBNMaaaaabbbbbcccccc", "output": "3" }, { "input": "200\nrsgraosldglhdoorwhkrsehjpuxrjuwgeanjgezhekprzarelduuaxdnspzjuooguuwnzkowkuhzduakdrzpnslauejhrrkalwpurpuuswdgeadlhjwzjgegwpknepazwwleulppwrlgrgedlwdzuodzropsrrkxusjnuzshdkjrxxpgzanzdrpnggdwxarpwohxdepJ", "output": "17" }, { "input": "1\nk", "output": "1" }, { "input": "1\nH", "output": "0" }, { "input": "2\nzG", "output": "1" }, { "input": "2\ngg", "output": "1" }, { "input": "2\nai", "output": "2" }, { "input": "20\npEjVrKWLIFCZjIHgggVU", "output": "1" }, { "input": "20\niFSiiigiYFSKmDnMGcgM", "output": "2" }, { "input": "20\nedxedxxxCQiIVmYEUtLi", "output": "3" }, { "input": "20\nprnchweyabjvzkoqiltm", "output": "20" }, { "input": "35\nQLDZNKFXKVSVLUVHRTDPQYMSTDXBELXBOTS", "output": "0" }, { "input": "35\nbvZWiitgxodztelnYUyljYGnCoWluXTvBLp", "output": "10" }, { "input": "35\nBTexnaeplecllxwlanarpcollawHLVMHIIF", "output": "10" }, { "input": "35\nhhwxqysolegsthsvfcqiryenbujbrrScobu", "output": "20" }, { "input": "26\npbgfqosklxjuzmdheyvawrictn", "output": "26" }, { "input": "100\nchMRWwymTDuZDZuSTvUmmuxvSscnTasyjlwwodhzcoifeahnbmcifyeobbydwparebduoLDCgHlOsPtVRbYGGQXfnkdvrWKIwCRl", "output": "20" }, { "input": "100\nhXYLXKUMBrGkjqQJTGbGWAfmztqqapdbjbhcualhypgnaieKXmhzGMnqXVlcPesskfaEVgvWQTTShRRnEtFahWDyuBzySMpugxCM", "output": "19" }, { "input": "100\nucOgELrgjMrFOgtHzqgvUgtHngKJxdMFKBjfcCppciqmGZXXoiSZibgpadshyljqrwxbomzeutvnhTLGVckZUmyiFPLlwuLBFito", "output": "23" }, { "input": "200\nWTCKAKLVGXSYFVMVJDUYERXNMVNTGWXUGRFCGMYXJQGLODYZTUIDENHYEGFKXFIEUILAMESAXAWZXVCZPJPEYUXBITHMTZOTMKWITGRSFHODKVJHPAHVVWTCTHIVAWAREQXWMPUWQSTPPJFHKGKELBTPUYDAVIUMGASPUEDIODRYXIWCORHOSLIBLOZUNJPHHMXEXOAY", "output": "0" }, { "input": "200\neLCCuYMPPwQoNlCpPOtKWJaQJmWfHeZCKiMSpILHSKjFOYGpRMzMCfMXdDuQdBGNsCNrHIVJzEFfBZcNMwNcFjOFVJvEtUQmLbFNKVHgNDyFkFVQhUTUQDgXhMjJZgFSSiHhMKuTgZQYJqAqKBpHoHddddddddddddddddXSSYNKNnRrKuOjAVKZlRLzCjExPdHaDHBT", "output": "1" }, { "input": "200\nitSYxgOLlwOoAkkkkkzzzzzzzzkzkzkzkkkkkzkzzkzUDJSKybRPBvaIDsNuWImPJvrHkKiMeYukWmtHtgZSyQsgYanZvXNbKXBlFLSUcqRnGWSriAvKxsTkDJfROqaKdzXhvJsPEDATueCraWOGEvRDWjPwXuiNpWsEnCuhDcKWOQxjBkdBqmFatWFkgKsbZuLtRGtY", "output": "2" }, { "input": "200\noggqoqqogoqoggggoggqgooqggogogooogqqgggoqgggqoqogogggogggqgooqgqggqqqoqgqgoooqgqogqoggoqqgqoqgoooqoogooqoogqoqoqqgoqgoqgggogqqqoqoggoqoqqoqggqoggooqqqoqggoggqqqqqqqqqgogqgggggooogogqgggqogqgoqoqogoooq", "output": "3" }, { "input": "200\nCtclUtUnmqFniaLqGRmMoUMeLyFfAgWxIZxdrBarcRQprSOGcdUYsmDbooSuOvBLgrYlgaIjJtFgcxJKHGkCXpYfVKmUbouuIqGstFrrwJzYQqjjqqppqqqqqpqqqjpjjpjqjXRYkfPhGAatOigFuItkKxkjCBLdiNMVGjmdWNMgOOvmaJEdGsWNoaERrINNKqKeQajv", "output": "3" }, { "input": "200\nmeZNrhqtSTSmktGQnnNOTcnyAMTKSixxKQKiagrMqRYBqgbRlsbJhvtNeHVUuMCyZLCnsIixRYrYEAkfQOxSVqXkrPqeCZQksInzRsRKBgvIqlGVPxPQnypknSXjgMjsjElcqGsaJRbegJVAKtWcHoOnzHqzhoKReqBBsOhZYLaYJhmqOMQsizdCsQfjUDHcTtHoeYwu", "output": "4" }, { "input": "200\nvFAYTHJLZaivWzSYmiuDBDUFACDSVbkImnVaXBpCgrbgmTfXKJfoglIkZxWPSeVSFPnHZDNUAqLyhjLXSuAqGLskBlDxjxGPJyGdwzlPfIekwsblIrkxzfhJeNoHywdfAGlJzqXOfQaKceSqViVFTRJEGfACnsFeSFpOYisIHJciqTMNAmgeXeublTvfWoPnddtvKIyF", "output": "6" }, { "input": "200\ngnDdkqJjYvduVYDSsswZDvoCouyaYZTfhmpSakERWLhufZtthWsfbQdTGwhKYjEcrqWBOyxBbiFhdLlIjChLOPiOpYmcrJgDtXsJfmHtLrabyGKOfHQRukEtTzwoqBHfmyVXPebfcpGQacLkGWFwerszjdHpTBXGssYXmGHlcCBgBXyGJqxbVhvDffLyCrZnxonABEXV", "output": "7" }, { "input": "200\nBmggKNRZBXPtJqlJaXLdKKQLDJvXpDuQGupiRQfDwCJCJvAlDDGpPZNOvXkrdKOFOEFBVfrsZjWyHPoKGzXmTAyPJGEmxCyCXpeAdTwbrMtWLmlmGNqxvuxmqpmtpuhrmxxtrquSLFYVlnSYgRJDYHWgHBbziBLZRwCIJNvbtsEdLLxmTbnjkoqSPAuzEeTYLlmejOUH", "output": "9" }, { "input": "200\nMkuxcDWdcnqsrlTsejehQKrTwoOBRCUAywqSnZkDLRmVBDVoOqdZHbrInQQyeRFAjiYYmHGrBbWgWstCPfLPRdNVDXBdqFJsGQfSXbufsiogybEhKDlWfPazIuhpONwGzZWaQNwVnmhTqWdewaklgjwaumXYDGwjSeEcYXjkVtLiYSWULEnTFukIlWQGWsXwWRMJGTcI", "output": "10" }, { "input": "200\nOgMBgYeuMJdjPtLybvwmGDrQEOhliaabEtwulzNEjsfnaznXUMoBbbxkLEwSQzcLrlJdjJCLGVNBxorghPxTYCoqniySJMcilpsqpBAbqdzqRUDVaYOgqGhGrxlIJkyYgkOdTUgRZwpgIkeZFXojLXpDilzirHVVadiHaMrxhzodzpdvhvrzdzxbhmhdpxqqpoDegfFQ", "output": "11" }, { "input": "200\nOLaJOtwultZLiZPSYAVGIbYvbIuZkqFZXwfsqpsavCDmBMStAuUFLBVknWDXNzmiuUYIsUMGxtoadWlPYPqvqSvpYdOiJRxFzGGnnmstniltvitnrmyrblnqyruylummmlsqtqitlbulvtuitiqimuintbimqyurviuntqnnvslynlNYMpYVKYwKVTbIUVdlNGrcFZON", "output": "12" }, { "input": "200\nGAcmlaqfjSAQLvXlkhxujXgSbxdFAwnoxDuldDvYmpUhTWJdcEQSdARLrozJzIgFVCkzPUztWIpaGfiKeqzoXinEjVuoKqyBHmtFjBWcRdBmyjviNlGAIkpikjAimmBgayfphrstfbjexjbttzfzfzaysxfyrjazfhtpghnbbeffjhxrjxpttesgzrnrfbgzzsRsCgmz", "output": "15" }, { "input": "200\nYRvIopNqSTYDhViTqCLMwEbTTIdHkoeuBmAJWhgtOgVxlcHSsavDNzMfpwTghkBvYEtCYQxicLUxdgAcaCzOOgbQYsfnaTXFlFxbeEiGwdNvxwHzkTdKtWlqzalwniDDBDipkxfflpaqkfkgfezbkxdvzemlfohwtgytzzywmwhvzUgPlPdeAVqTPAUZbogQheRXetvT", "output": "20" }, { "input": "200\nNcYVomemswLCUqVRSDKHCknlBmqeSWhVyRzQrnZaOANnTGqsRFMjpczllcEVebqpxdavzppvztxsnfmtcharzqlginndyjkawzurqkxJLXiXKNZTIIxhSQghDpjwzatEqnLMTLxwoEKpHytvWkKFDUcZjLShCiVdocxRvvJtbXHCDGpJvMwRKWLhcTFtswdLUHkbhfau", "output": "25" }, { "input": "200\nDxNZuvkTkQEqdWIkLzcKAwfqvZQiptnTazydSCTIfGjDhLMrlPZiKEsqIdDhgKPAlEvXyzNwWtYorotgkcwydpabjqnzubaksdchucxtkmjzfretdmvlxgklyvicrtftvztsbiUaQorfNIYUOdwQDRsKpxLUiLknbLbinilpPXPTTwLAnXVpMHBaAcKWgDBeOFabPtXU", "output": "26" }, { "input": "4\nabbc", "output": "3" }, { "input": "3\naaa", "output": "1" }, { "input": "3\naba", "output": "2" }, { "input": "3\nabb", "output": "2" }, { "input": "3\nbba", "output": "2" }, { "input": "3\nAaa", "output": "1" }, { "input": "3\nAba", "output": "2" }, { "input": "3\naBa", "output": "1" }, { "input": "3\naAa", "output": "1" }, { "input": "3\naAb", "output": "1" }, { "input": "3\nAaA", "output": "1" }, { "input": "5\naBacd", "output": "3" }, { "input": "5\naAabc", "output": "3" } ]

1,535,643,104

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

5

124

0

n=int(input()) s=list(input()) ta=[] ka=[] x=0 ta.append(s[0]) for item in s[1:]: if item!=ta[-1]: ta.append(item) else: pass for item in ta: if item.islower(): x+=1 else: ka.append(x) x=0 ka.append(x) print(max(ka))

Title: Polycarp and Letters Time Limit: None seconds Memory Limit: None megabytes Problem Description: Polycarp loves lowercase letters and dislikes uppercase ones. Once he got a string *s* consisting only of lowercase and uppercase Latin letters. Let *A* be a set of positions in the string. Let's call it pretty if following conditions are met: - letters on positions from *A* in the string are all distinct and lowercase; - there are no uppercase letters in the string which are situated between positions from *A* (i.e. there is no such *j* that *s*[*j*] is an uppercase letter, and *a*1<=<<=*j*<=<<=*a*2 for some *a*1 and *a*2 from *A*). Write a program that will determine the maximum number of elements in a pretty set of positions. Input Specification: The first line contains a single integer *n* (1<=≤<=*n*<=≤<=200) — length of string *s*. The second line contains a string *s* consisting of lowercase and uppercase Latin letters. Output Specification: Print maximum number of elements in pretty set of positions for string *s*. Demo Input: ['11\naaaaBaabAbA\n', '12\nzACaAbbaazzC\n', '3\nABC\n'] Demo Output: ['2\n', '3\n', '0\n'] Note: In the first example the desired positions might be 6 and 8 or 7 and 8. Positions 6 and 7 contain letters 'a', position 8 contains letter 'b'. The pair of positions 1 and 8 is not suitable because there is an uppercase letter 'B' between these position. In the second example desired positions can be 7, 8 and 11. There are other ways to choose pretty set consisting of three elements. In the third example the given string *s* does not contain any lowercase letters, so the answer is 0.

```python n=int(input()) s=list(input()) ta=[] ka=[] x=0 ta.append(s[0]) for item in s[1:]: if item!=ta[-1]: ta.append(item) else: pass for item in ta: if item.islower(): x+=1 else: ka.append(x) x=0 ka.append(x) print(max(ka)) ```

0

32

B

Borze

PROGRAMMING

800

[ "expression parsing", "implementation" ]

B. Borze

2

256

Ternary numeric notation is quite popular in Berland. To telegraph the ternary number the Borze alphabet is used. Digit 0 is transmitted as «.», 1 as «-.» and 2 as «--». You are to decode the Borze code, i.e. to find out the ternary number given its representation in Borze alphabet.

The first line contains a number in Borze code. The length of the string is between 1 and 200 characters. It's guaranteed that the given string is a valid Borze code of some ternary number (this number can have leading zeroes).

Output the decoded ternary number. It can have leading zeroes.

[ ".-.--\n", "--.\n", "-..-.--\n" ]

[ "012", "20", "1012" ]

none

1,000

[ { "input": ".-.--", "output": "012" }, { "input": "--.", "output": "20" }, { "input": "-..-.--", "output": "1012" }, { "input": "---..", "output": "210" }, { "input": "..--.---..", "output": "0020210" }, { "input": "-.....----.", "output": "10000220" }, { "input": ".", "output": "0" }, { "input": "-.", "output": "1" }, { "input": "--", "output": "2" }, { "input": "..", "output": "00" }, { "input": "--.", "output": "20" }, { "input": ".--.", "output": "020" }, { "input": ".-.-..", "output": "0110" }, { "input": "----.-.", "output": "2201" }, { "input": "-..--.-.", "output": "10201" }, { "input": "..--..--.", "output": "0020020" }, { "input": "-.-.---.--..-..-.-.-..-..-.--.", "output": "112120010111010120" }, { "input": "---.-.-.------..-..-..-..-.-..-.--.-.-..-.-.-----..-.-.", "output": "21112220010101011012011011221011" }, { "input": "-.-..--.-.-.-.-.-..-.-.-.---------.--.---..--...--.-----.-.-.-...--.-.-.---.------.--..-.--.-----.-...-..------", "output": "11020111110111222212021020002022111100201121222020012022110010222" }, { "input": "-.-..-.--.---..---.-..---.-...-.-.----..-.---.-.---..-.--.---.-.-------.---.--....----.-.---.---.---.----.-----..---.-.-.-.-----.--.-------.-..", "output": "110120210211021100112200121121012021122212120000220121212122022102111122120222110" }, { "input": ".-..-.-.---.-----.--.---...-.--.-.-....-..", "output": "01011212212021001201100010" }, { "input": ".------.-.---..--...-..-..-.-.-.--.--.-..-.--...-.-.---.-.-.------..--..-.---..----.-..-.--.---.-.----.-.---...-.-.-.-----.-.-.---.---.-.....-.-...-----.-...-.---.-..-.-----.--...---.-.-..-.--.-.---..", "output": "022201210200010101112020101200011211122200200121022010120211220121001112211121211000011002211001211012212000211101201210" }, { "input": ".-.--.---.-----.-.-----.-.-..-----..-..----..--.-.--.----..---.---..-.-.-----..-------.----..----.-..---...-----..-..-----...-..-.-.-----....---..---..-.-----...-.--...--.-.---.-.-.-.-.-...---..----.", "output": "01202122112211102210102200201202200212101122102221220022010210022101022100101122100021021012210012000201211111100210220" }, { "input": "..-.-.-.---.-.-.-..-.-..-.-.---.-------.---..-----.---....-.---.--.--.-.---.---------.-..---.-.-.--..---.---.-.---.-.-..-.-..-.-.-.----.--.-....--------.-.---..----.------.-.-.--.--.-----.-----.----", "output": "0011121111011011212221210221210001212020121222211021112002121121110110111220201000222201210220222011202022122122" }, { "input": "-..-------.------.-..--.-.-..--.-.-..-----..-.-.-..-..-..--.---..-----..---..-..--.-..-.-.---...-.....-------.---.-----.-...-.-...-.-.---.---.-----.--.--...-.--..-.-..-...-.-.-.-.---..---.-..-.-.-.-..", "output": "102221222010201102011022101110101020210221021010201011210010000222121221100110011212122120200012001101001111210211011110" }, { "input": ".-.----.-.--..-.-.-.-..----..-.-...--.-.---.---.-------..-.--..-......--.------.--.----.--...-.--.--..-----..-.....--.--.-.-.------..--------.----------..-.---.----.---.-..--..-.....-..------.--.", "output": "012201200111102200110020121212221012001000002022202022020001202002210100002020112220022220222220012122021102001000010222020" }, { "input": "------.-----.-....--.-.----.-.---.-.-..---.-.---.-----..-...-.-.---..-.-.-..-.-.-...-.-.-.----..--.------.----.-..-.--...-.-------...-.-..-.-.--.--.---..--..--------.--.-.-.---.-.-.-...----.--..-.--..", "output": "222022110002012201211102112122101001121011101110011122002022202201012000122210011011202021020022220201121111002202001200" }, { "input": "-.---...----...--.--...-.--.----", "output": "121002200020200012022" }, { "input": "--.--.--.---.--.-.---.-.-..-..--.-..---.-.....-..---.-----.--...-.-.-------.-.--.-.----.-..-.------.", "output": "202020212012111010201021100001021221200011222112012201012220" } ]

1,669,654,350

2,147,483,647

PyPy 3

OK

TESTS

30

154

0

borze_text = str(input()) borze = list(borze_text) txt = "" i = 0 while len(borze) > i: if borze[i] == ".": txt += "0" elif borze[i] == "-" and borze[i+1] == ".": txt += "1" i += 1 elif borze[i] == "-" and borze[i + 1] == "-": txt += "2" i += 1 i += 1 print(txt)

Title: Borze Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Ternary numeric notation is quite popular in Berland. To telegraph the ternary number the Borze alphabet is used. Digit 0 is transmitted as «.», 1 as «-.» and 2 as «--». You are to decode the Borze code, i.e. to find out the ternary number given its representation in Borze alphabet. Input Specification: The first line contains a number in Borze code. The length of the string is between 1 and 200 characters. It's guaranteed that the given string is a valid Borze code of some ternary number (this number can have leading zeroes). Output Specification: Output the decoded ternary number. It can have leading zeroes. Demo Input: ['.-.--\n', '--.\n', '-..-.--\n'] Demo Output: ['012', '20', '1012'] Note: none

```python borze_text = str(input()) borze = list(borze_text) txt = "" i = 0 while len(borze) > i: if borze[i] == ".": txt += "0" elif borze[i] == "-" and borze[i+1] == ".": txt += "1" i += 1 elif borze[i] == "-" and borze[i + 1] == "-": txt += "2" i += 1 i += 1 print(txt) ```

3.9615

892

A

Greed

PROGRAMMING

900

[ "greedy", "implementation" ]

null

Jafar has *n* cans of cola. Each can is described by two integers: remaining volume of cola *a**i* and can's capacity *b**i* (*a**i* <=≤<= *b**i*). Jafar has decided to pour all remaining cola into just 2 cans, determine if he can do this or not!

The first line of the input contains one integer *n* (2<=≤<=*n*<=≤<=100<=000) — number of cola cans. The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) — volume of remaining cola in cans. The third line contains *n* space-separated integers that *b*1,<=*b*2,<=...,<=*b**n* (*a**i*<=≤<=*b**i*<=≤<=109) — capacities of the cans.

Print "YES" (without quotes) if it is possible to pour all remaining cola in 2 cans. Otherwise print "NO" (without quotes). You can print each letter in any case (upper or lower).

[ "2\n3 5\n3 6\n", "3\n6 8 9\n6 10 12\n", "5\n0 0 5 0 0\n1 1 8 10 5\n", "4\n4 1 0 3\n5 2 2 3\n" ]

[ "YES\n", "NO\n", "YES\n", "YES\n" ]

In the first sample, there are already 2 cans, so the answer is "YES".

500

[ { "input": "2\n3 5\n3 6", "output": "YES" }, { "input": "3\n6 8 9\n6 10 12", "output": "NO" }, { "input": "5\n0 0 5 0 0\n1 1 8 10 5", "output": "YES" }, { "input": "4\n4 1 0 3\n5 2 2 3", "output": "YES" }, { "input": "10\n9 10 24 11 1 7 8 3 28 14\n86 20 34 11 22 94 8 16 73 85", "output": "YES" }, { "input": "4\n25 35 7 31\n70 37 43 35", "output": "YES" }, { "input": "10\n15 26 15 14 14 39 40 4 25 39\n27 72 16 44 69 48 53 17 63 42", "output": "NO" }, { "input": "5\n22 5 19 16 32\n26 10 43 38 37", "output": "NO" }, { "input": "5\n32 4 22 40 26\n39 20 36 98 44", "output": "YES" }, { "input": "6\n18 25 3 10 13 37\n38 73 19 35 24 37", "output": "YES" }, { "input": "2\n2 2\n2 2", "output": "YES" }, { "input": "2\n2 5\n2 5", "output": "YES" }, { "input": "2\n1000 1008\n10000 2352", "output": "YES" }, { "input": "5\n1 2 3 4 5\n1 2 3 4 11", "output": "YES" }, { "input": "4\n1 0 0 0\n2 0 0 0", "output": "YES" }, { "input": "2\n0 0\n1 2", "output": "YES" }, { "input": "3\n9 13 4\n10 14 5", "output": "NO" }, { "input": "2\n0 0\n1 1", "output": "YES" }, { "input": "5\n1 1 2 3 1\n1 1 2 3 4", "output": "NO" }, { "input": "2\n0 0\n0 0", "output": "YES" }, { "input": "3\n5 1 1\n5 5 5", "output": "YES" } ]

1,559,376,762

1,062

Python 3

WRONG_ANSWER

TESTS

3

93

0

n=int(input()) l1=list(map(int,input().split())) l2=list(map(int,input().split())) sum1=0 for i in l1: sum1+=i l2.sort() left=0 right=len(l2)-1 while(right>=left): if l2[left]+l2[right]< sum1: right-=1 else: print("YES") exit() print("NO")

Title: Greed Time Limit: None seconds Memory Limit: None megabytes Problem Description: Jafar has *n* cans of cola. Each can is described by two integers: remaining volume of cola *a**i* and can's capacity *b**i* (*a**i* <=≤<= *b**i*). Jafar has decided to pour all remaining cola into just 2 cans, determine if he can do this or not! Input Specification: The first line of the input contains one integer *n* (2<=≤<=*n*<=≤<=100<=000) — number of cola cans. The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) — volume of remaining cola in cans. The third line contains *n* space-separated integers that *b*1,<=*b*2,<=...,<=*b**n* (*a**i*<=≤<=*b**i*<=≤<=109) — capacities of the cans. Output Specification: Print "YES" (without quotes) if it is possible to pour all remaining cola in 2 cans. Otherwise print "NO" (without quotes). You can print each letter in any case (upper or lower). Demo Input: ['2\n3 5\n3 6\n', '3\n6 8 9\n6 10 12\n', '5\n0 0 5 0 0\n1 1 8 10 5\n', '4\n4 1 0 3\n5 2 2 3\n'] Demo Output: ['YES\n', 'NO\n', 'YES\n', 'YES\n'] Note: In the first sample, there are already 2 cans, so the answer is "YES".

```python n=int(input()) l1=list(map(int,input().split())) l2=list(map(int,input().split())) sum1=0 for i in l1: sum1+=i l2.sort() left=0 right=len(l2)-1 while(right>=left): if l2[left]+l2[right]< sum1: right-=1 else: print("YES") exit() print("NO") ```

0

31

A

Worms Evolution

PROGRAMMING

1,200

[ "implementation" ]

A. Worms Evolution

2

256

Professor Vasechkin is studying evolution of worms. Recently he put forward hypotheses that all worms evolve by division. There are *n* forms of worms. Worms of these forms have lengths *a*1, *a*2, ..., *a**n*. To prove his theory, professor needs to find 3 different forms that the length of the first form is equal to sum of lengths of the other two forms. Help him to do this.

The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of worm's forms. The second line contains *n* space-separated integers *a**i* (1<=≤<=*a**i*<=≤<=1000) — lengths of worms of each form.

Output 3 distinct integers *i* *j* *k* (1<=≤<=*i*,<=*j*,<=*k*<=≤<=*n*) — such indexes of worm's forms that *a**i*<==<=*a**j*<=+<=*a**k*. If there is no such triple, output -1. If there are several solutions, output any of them. It possible that *a**j*<==<=*a**k*.

[ "5\n1 2 3 5 7\n", "5\n1 8 1 5 1\n" ]

[ "3 2 1\n", "-1\n" ]

none

500

[ { "input": "5\n1 2 3 5 7", "output": "3 2 1" }, { "input": "5\n1 8 1 5 1", "output": "-1" }, { "input": "4\n303 872 764 401", "output": "-1" }, { "input": "6\n86 402 133 524 405 610", "output": "6 4 1" }, { "input": "8\n217 779 418 895 996 473 3 22", "output": "5 2 1" }, { "input": "10\n858 972 670 15 662 114 33 273 53 310", "output": "2 6 1" }, { "input": "100\n611 697 572 770 603 870 128 245 49 904 468 982 788 943 549 288 668 796 803 515 999 735 912 49 298 80 412 841 494 434 543 298 17 571 271 105 70 313 178 755 194 279 585 766 412 164 907 841 776 556 731 268 735 880 176 267 287 65 239 588 155 658 821 47 783 595 585 69 226 906 429 161 999 148 7 484 362 585 952 365 92 749 904 525 307 626 883 367 450 755 564 950 728 724 69 106 119 157 96 290", "output": "1 38 25" }, { "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": "1 63 61" }, { "input": "100\n156 822 179 298 981 82 610 345 373 378 895 734 768 15 78 335 764 608 932 297 717 553 916 367 425 447 361 195 66 70 901 236 905 744 919 564 296 610 963 628 840 52 100 750 345 308 37 687 192 704 101 815 10 990 216 358 823 546 578 821 706 148 182 582 421 482 829 425 121 337 500 301 402 868 66 935 625 527 746 585 308 523 488 914 608 709 875 252 151 781 447 2 756 176 976 302 450 35 680 791", "output": "1 98 69" }, { "input": "100\n54 947 785 838 359 647 92 445 48 465 323 486 101 86 607 31 860 420 709 432 435 372 272 37 903 814 309 197 638 58 259 822 793 564 309 22 522 907 101 853 486 824 614 734 630 452 166 532 256 499 470 9 933 452 256 450 7 26 916 406 257 285 895 117 59 369 424 133 16 417 352 440 806 236 478 34 889 469 540 806 172 296 73 655 261 792 868 380 204 454 330 53 136 629 236 850 134 560 264 291", "output": "2 29 27" }, { "input": "99\n175 269 828 129 499 890 127 263 995 807 508 289 996 226 437 320 365 642 757 22 190 8 345 499 834 713 962 889 336 171 608 492 320 257 472 801 176 325 301 306 198 729 933 4 640 322 226 317 567 586 249 237 202 633 287 128 911 654 719 988 420 855 361 574 716 899 317 356 581 440 284 982 541 111 439 29 37 560 961 224 478 906 319 416 736 603 808 87 762 697 392 713 19 459 262 238 239 599 997", "output": "1 44 30" }, { "input": "98\n443 719 559 672 16 69 529 632 953 999 725 431 54 22 346 968 558 696 48 669 963 129 257 712 39 870 498 595 45 821 344 925 179 388 792 346 755 213 423 365 344 659 824 356 773 637 628 897 841 155 243 536 951 361 192 105 418 431 635 596 150 162 145 548 473 531 750 306 377 354 450 975 79 743 656 733 440 940 19 139 237 346 276 227 64 799 479 633 199 17 796 362 517 234 729 62 995 535", "output": "2 70 40" }, { "input": "97\n359 522 938 862 181 600 283 1000 910 191 590 220 761 818 903 264 751 751 987 316 737 898 168 925 244 674 34 950 754 472 81 6 37 520 112 891 981 454 897 424 489 238 363 709 906 951 677 828 114 373 589 835 52 89 97 435 277 560 551 204 879 469 928 523 231 163 183 609 821 915 615 969 616 23 874 437 844 321 78 53 643 786 585 38 744 347 150 179 988 985 200 11 15 9 547 886 752", "output": "1 23 10" }, { "input": "4\n303 872 764 401", "output": "-1" }, { "input": "100\n328 397 235 453 188 254 879 225 423 36 384 296 486 592 231 849 856 255 213 898 234 800 701 529 951 693 507 326 15 905 618 348 967 927 28 979 752 850 343 35 84 302 36 390 482 826 249 918 91 289 973 457 557 348 365 239 709 565 320 560 153 130 647 708 483 469 788 473 322 844 830 562 611 961 397 673 69 960 74 703 369 968 382 451 328 160 211 230 566 208 7 545 293 73 806 375 157 410 303 58", "output": "1 79 6" }, { "input": "33\n52 145 137 734 180 847 178 286 716 134 181 630 358 764 593 762 785 28 1 468 189 540 764 485 165 656 114 58 628 108 605 584 257", "output": "8 30 7" }, { "input": "57\n75 291 309 68 444 654 985 158 514 204 116 918 374 806 176 31 49 455 269 66 722 713 164 818 317 295 546 564 134 641 28 13 987 478 146 219 213 940 289 173 157 666 168 391 392 71 870 477 446 988 414 568 964 684 409 671 454", "output": "2 41 29" }, { "input": "88\n327 644 942 738 84 118 981 686 530 404 137 197 434 16 693 183 423 325 410 345 941 329 7 106 79 867 584 358 533 675 192 718 641 329 900 768 404 301 101 538 954 590 401 954 447 14 559 337 756 586 934 367 538 928 945 936 770 641 488 579 206 869 902 139 216 446 723 150 829 205 373 578 357 368 960 40 121 206 503 385 521 161 501 694 138 370 709 308", "output": "1 77 61" }, { "input": "100\n804 510 266 304 788 625 862 888 408 82 414 470 777 991 729 229 933 406 601 1 596 720 608 706 432 361 527 548 59 548 474 515 4 991 263 568 681 24 117 563 576 587 281 643 904 521 891 106 842 884 943 54 605 815 504 757 311 374 335 192 447 652 633 410 455 402 382 150 432 836 413 819 669 875 638 925 217 805 632 520 605 266 728 795 162 222 603 159 284 790 914 443 775 97 789 606 859 13 851 47", "output": "1 77 42" }, { "input": "100\n449 649 615 713 64 385 927 466 138 126 143 886 80 199 208 43 196 694 92 89 264 180 617 970 191 196 910 150 275 89 693 190 191 99 542 342 45 592 114 56 451 170 64 589 176 102 308 92 402 153 414 675 352 157 69 150 91 288 163 121 816 184 20 234 836 12 593 150 793 439 540 93 99 663 186 125 349 247 476 106 77 523 215 7 363 278 441 745 337 25 148 384 15 915 108 211 240 58 23 408", "output": "1 6 5" }, { "input": "90\n881 436 52 308 97 261 153 931 670 538 702 156 114 445 154 685 452 76 966 790 93 42 547 65 736 364 136 489 719 322 239 628 696 735 55 703 622 375 100 188 804 341 546 474 484 446 729 290 974 301 602 225 996 244 488 983 882 460 962 754 395 617 61 640 534 292 158 375 632 902 420 979 379 38 100 67 963 928 190 456 545 571 45 716 153 68 844 2 102 116", "output": "1 14 2" }, { "input": "80\n313 674 262 240 697 146 391 221 793 504 896 818 92 899 86 370 341 339 306 887 937 570 830 683 729 519 240 833 656 847 427 958 435 704 853 230 758 347 660 575 843 293 649 396 437 787 654 599 35 103 779 783 447 379 444 585 902 713 791 150 851 228 306 721 996 471 617 403 102 168 197 741 877 481 968 545 331 715 236 654", "output": "1 13 8" }, { "input": "70\n745 264 471 171 946 32 277 511 269 469 89 831 69 2 369 407 583 602 646 633 429 747 113 302 722 321 344 824 241 372 263 287 822 24 652 758 246 967 219 313 882 597 752 965 389 775 227 556 95 904 308 340 899 514 400 187 275 318 621 546 659 488 199 154 811 1 725 79 925 82", "output": "1 63 60" }, { "input": "60\n176 502 680 102 546 917 516 801 392 435 635 492 398 456 653 444 472 513 634 378 273 276 44 920 68 124 800 167 825 250 452 264 561 344 98 933 381 939 426 51 568 548 206 887 342 763 151 514 156 354 486 546 998 649 356 438 295 570 450 589", "output": "2 26 20" }, { "input": "50\n608 92 889 33 146 803 402 91 868 400 828 505 375 558 584 129 361 776 974 123 765 804 326 186 61 927 904 511 762 775 640 593 300 664 897 461 869 911 986 789 607 500 309 457 294 104 724 471 216 155", "output": "3 25 11" }, { "input": "40\n40 330 98 612 747 336 640 381 991 366 22 167 352 12 868 166 603 40 313 869 609 981 609 804 54 729 8 854 347 300 828 922 39 633 695 988 4 530 545 176", "output": "5 10 8" }, { "input": "30\n471 920 308 544 347 222 878 671 467 332 215 180 681 114 151 203 492 951 653 614 453 510 540 422 399 532 113 198 932 825", "output": "2 21 9" }, { "input": "20\n551 158 517 475 595 108 764 961 590 297 761 841 659 568 82 888 733 214 993 359", "output": "3 20 2" }, { "input": "10\n983 748 726 406 196 993 2 251 66 263", "output": "-1" }, { "input": "9\n933 266 457 863 768 257 594 136 145", "output": "-1" }, { "input": "8\n537 198 48 771 944 868 700 163", "output": "7 8 1" }, { "input": "7\n140 779 639 679 768 479 158", "output": "2 3 1" }, { "input": "6\n744 359 230 586 944 442", "output": "-1" }, { "input": "5\n700 939 173 494 120", "output": "-1" }, { "input": "4\n303 872 764 401", "output": "-1" }, { "input": "3\n907 452 355", "output": "-1" }, { "input": "3\n963 630 333", "output": "1 3 2" }, { "input": "3\n2 2 4", "output": "3 2 1" }, { "input": "3\n2 4 100", "output": "-1" } ]

1,509,084,200

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

0

124

0

n=int(input()) list1=input().split() for i in range(n): list1[i]=int(list1[i]) list2=[0,0,0] for i in range(n): for j in range(n): for k in range(n): if list1[i]==list1[j]+list1[k]: list2[0]=i list2[1]=j list2[2]=k if list2[0]==0 and list2[1]==0 and list2[2]==0: print(-1) else: print(*list2)

Title: Worms Evolution Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Professor Vasechkin is studying evolution of worms. Recently he put forward hypotheses that all worms evolve by division. There are *n* forms of worms. Worms of these forms have lengths *a*1, *a*2, ..., *a**n*. To prove his theory, professor needs to find 3 different forms that the length of the first form is equal to sum of lengths of the other two forms. Help him to do this. Input Specification: The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of worm's forms. The second line contains *n* space-separated integers *a**i* (1<=≤<=*a**i*<=≤<=1000) — lengths of worms of each form. Output Specification: Output 3 distinct integers *i* *j* *k* (1<=≤<=*i*,<=*j*,<=*k*<=≤<=*n*) — such indexes of worm's forms that *a**i*<==<=*a**j*<=+<=*a**k*. If there is no such triple, output -1. If there are several solutions, output any of them. It possible that *a**j*<==<=*a**k*. Demo Input: ['5\n1 2 3 5 7\n', '5\n1 8 1 5 1\n'] Demo Output: ['3 2 1\n', '-1\n'] Note: none

```python n=int(input()) list1=input().split() for i in range(n): list1[i]=int(list1[i]) list2=[0,0,0] for i in range(n): for j in range(n): for k in range(n): if list1[i]==list1[j]+list1[k]: list2[0]=i list2[1]=j list2[2]=k if list2[0]==0 and list2[1]==0 and list2[2]==0: print(-1) else: print(*list2) ```

0

32

B

Borze

PROGRAMMING

800

[ "expression parsing", "implementation" ]

B. Borze

2

256

Ternary numeric notation is quite popular in Berland. To telegraph the ternary number the Borze alphabet is used. Digit 0 is transmitted as «.», 1 as «-.» and 2 as «--». You are to decode the Borze code, i.e. to find out the ternary number given its representation in Borze alphabet.

The first line contains a number in Borze code. The length of the string is between 1 and 200 characters. It's guaranteed that the given string is a valid Borze code of some ternary number (this number can have leading zeroes).

Output the decoded ternary number. It can have leading zeroes.

[ ".-.--\n", "--.\n", "-..-.--\n" ]

[ "012", "20", "1012" ]

none

1,000

[ { "input": ".-.--", "output": "012" }, { "input": "--.", "output": "20" }, { "input": "-..-.--", "output": "1012" }, { "input": "---..", "output": "210" }, { "input": "..--.---..", "output": "0020210" }, { "input": "-.....----.", "output": "10000220" }, { "input": ".", "output": "0" }, { "input": "-.", "output": "1" }, { "input": "--", "output": "2" }, { "input": "..", "output": "00" }, { "input": "--.", "output": "20" }, { "input": ".--.", "output": "020" }, { "input": ".-.-..", "output": "0110" }, { "input": "----.-.", "output": "2201" }, { "input": "-..--.-.", "output": "10201" }, { "input": "..--..--.", "output": "0020020" }, { "input": "-.-.---.--..-..-.-.-..-..-.--.", "output": "112120010111010120" }, { "input": "---.-.-.------..-..-..-..-.-..-.--.-.-..-.-.-----..-.-.", "output": "21112220010101011012011011221011" }, { "input": "-.-..--.-.-.-.-.-..-.-.-.---------.--.---..--...--.-----.-.-.-...--.-.-.---.------.--..-.--.-----.-...-..------", "output": "11020111110111222212021020002022111100201121222020012022110010222" }, { "input": "-.-..-.--.---..---.-..---.-...-.-.----..-.---.-.---..-.--.---.-.-------.---.--....----.-.---.---.---.----.-----..---.-.-.-.-----.--.-------.-..", "output": "110120210211021100112200121121012021122212120000220121212122022102111122120222110" }, { "input": ".-..-.-.---.-----.--.---...-.--.-.-....-..", "output": "01011212212021001201100010" }, { "input": ".------.-.---..--...-..-..-.-.-.--.--.-..-.--...-.-.---.-.-.------..--..-.---..----.-..-.--.---.-.----.-.---...-.-.-.-----.-.-.---.---.-.....-.-...-----.-...-.---.-..-.-----.--...---.-.-..-.--.-.---..", "output": "022201210200010101112020101200011211122200200121022010120211220121001112211121211000011002211001211012212000211101201210" }, { "input": ".-.--.---.-----.-.-----.-.-..-----..-..----..--.-.--.----..---.---..-.-.-----..-------.----..----.-..---...-----..-..-----...-..-.-.-----....---..---..-.-----...-.--...--.-.---.-.-.-.-.-...---..----.", "output": "01202122112211102210102200201202200212101122102221220022010210022101022100101122100021021012210012000201211111100210220" }, { "input": "..-.-.-.---.-.-.-..-.-..-.-.---.-------.---..-----.---....-.---.--.--.-.---.---------.-..---.-.-.--..---.---.-.---.-.-..-.-..-.-.-.----.--.-....--------.-.---..----.------.-.-.--.--.-----.-----.----", "output": "0011121111011011212221210221210001212020121222211021112002121121110110111220201000222201210220222011202022122122" }, { "input": "-..-------.------.-..--.-.-..--.-.-..-----..-.-.-..-..-..--.---..-----..---..-..--.-..-.-.---...-.....-------.---.-----.-...-.-...-.-.---.---.-----.--.--...-.--..-.-..-...-.-.-.-.---..---.-..-.-.-.-..", "output": "102221222010201102011022101110101020210221021010201011210010000222121221100110011212122120200012001101001111210211011110" }, { "input": ".-.----.-.--..-.-.-.-..----..-.-...--.-.---.---.-------..-.--..-......--.------.--.----.--...-.--.--..-----..-.....--.--.-.-.------..--------.----------..-.---.----.---.-..--..-.....-..------.--.", "output": "012201200111102200110020121212221012001000002022202022020001202002210100002020112220022220222220012122021102001000010222020" }, { "input": "------.-----.-....--.-.----.-.---.-.-..---.-.---.-----..-...-.-.---..-.-.-..-.-.-...-.-.-.----..--.------.----.-..-.--...-.-------...-.-..-.-.--.--.---..--..--------.--.-.-.---.-.-.-...----.--..-.--..", "output": "222022110002012201211102112122101001121011101110011122002022202201012000122210011011202021020022220201121111002202001200" }, { "input": "-.---...----...--.--...-.--.----", "output": "121002200020200012022" }, { "input": "--.--.--.---.--.-.---.-.-..-..--.-..---.-.....-..---.-----.--...-.-.-------.-.--.-.----.-..-.------.", "output": "202020212012111010201021100001021221200011222112012201012220" } ]

1,691,006,285

2,147,483,647

PyPy 3

OK

TESTS

30

184

0

s = input() n=len(s) A = [] i = 0 while i < n: if s[i] == '.': A.append('0') i += 1 else: A.append('2' if (i + 1 < n) and (s[i + 1] == '-') else '1') i += 2 print("".join(A))

Title: Borze Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Ternary numeric notation is quite popular in Berland. To telegraph the ternary number the Borze alphabet is used. Digit 0 is transmitted as «.», 1 as «-.» and 2 as «--». You are to decode the Borze code, i.e. to find out the ternary number given its representation in Borze alphabet. Input Specification: The first line contains a number in Borze code. The length of the string is between 1 and 200 characters. It's guaranteed that the given string is a valid Borze code of some ternary number (this number can have leading zeroes). Output Specification: Output the decoded ternary number. It can have leading zeroes. Demo Input: ['.-.--\n', '--.\n', '-..-.--\n'] Demo Output: ['012', '20', '1012'] Note: none

```python s = input() n=len(s) A = [] i = 0 while i < n: if s[i] == '.': A.append('0') i += 1 else: A.append('2' if (i + 1 < n) and (s[i + 1] == '-') else '1') i += 2 print("".join(A)) ```

3.954

987

A

Infinity Gauntlet

PROGRAMMING

800

[ "implementation" ]

null

You took a peek on Thanos wearing Infinity Gauntlet. In the Gauntlet there is a place for six Infinity Gems: - the Power Gem of purple color, - the Time Gem of green color, - the Space Gem of blue color, - the Soul Gem of orange color, - the Reality Gem of red color, - the Mind Gem of yellow color. Using colors of Gems you saw in the Gauntlet determine the names of absent Gems.

In the first line of input there is one integer $n$ ($0 \le n \le 6$) — the number of Gems in Infinity Gauntlet. In next $n$ lines there are colors of Gems you saw. Words used for colors are: purple, green, blue, orange, red, yellow. It is guaranteed that all the colors are distinct. All colors are given in lowercase English letters.

In the first line output one integer $m$ ($0 \le m \le 6$) — the number of absent Gems. Then in $m$ lines print the names of absent Gems, each on its own line. Words used for names are: Power, Time, Space, Soul, Reality, Mind. Names can be printed in any order. Keep the first letter uppercase, others lowercase.

[ "4\nred\npurple\nyellow\norange\n", "0\n" ]

[ "2\nSpace\nTime\n", "6\nTime\nMind\nSoul\nPower\nReality\nSpace\n" ]

In the first sample Thanos already has Reality, Power, Mind and Soul Gems, so he needs two more: Time and Space. In the second sample Thanos doesn't have any Gems, so he needs all six.

500

[ { "input": "4\nred\npurple\nyellow\norange", "output": "2\nSpace\nTime" }, { "input": "0", "output": "6\nMind\nSpace\nPower\nTime\nReality\nSoul" }, { "input": "6\npurple\nblue\nyellow\nred\ngreen\norange", "output": "0" }, { "input": "1\npurple", "output": "5\nTime\nReality\nSoul\nSpace\nMind" }, { "input": "3\nblue\norange\npurple", "output": "3\nTime\nReality\nMind" }, { "input": "2\nyellow\nred", "output": "4\nPower\nSoul\nSpace\nTime" }, { "input": "1\ngreen", "output": "5\nReality\nSpace\nPower\nSoul\nMind" }, { "input": "2\npurple\ngreen", "output": "4\nReality\nMind\nSpace\nSoul" }, { "input": "1\nblue", "output": "5\nPower\nReality\nSoul\nTime\nMind" }, { "input": "2\npurple\nblue", "output": "4\nMind\nSoul\nTime\nReality" }, { "input": "2\ngreen\nblue", "output": "4\nReality\nMind\nPower\nSoul" }, { "input": "3\npurple\ngreen\nblue", "output": "3\nMind\nReality\nSoul" }, { "input": "1\norange", "output": "5\nReality\nTime\nPower\nSpace\nMind" }, { "input": "2\npurple\norange", "output": "4\nReality\nMind\nTime\nSpace" }, { "input": "2\norange\ngreen", "output": "4\nSpace\nMind\nReality\nPower" }, { "input": "3\norange\npurple\ngreen", "output": "3\nReality\nSpace\nMind" }, { "input": "2\norange\nblue", "output": "4\nTime\nMind\nReality\nPower" }, { "input": "3\nblue\ngreen\norange", "output": "3\nPower\nMind\nReality" }, { "input": "4\nblue\norange\ngreen\npurple", "output": "2\nMind\nReality" }, { "input": "1\nred", "output": "5\nTime\nSoul\nMind\nPower\nSpace" }, { "input": "2\nred\npurple", "output": "4\nMind\nSpace\nTime\nSoul" }, { "input": "2\nred\ngreen", "output": "4\nMind\nSpace\nPower\nSoul" }, { "input": "3\nred\npurple\ngreen", "output": "3\nSoul\nSpace\nMind" }, { "input": "2\nblue\nred", "output": "4\nMind\nTime\nPower\nSoul" }, { "input": "3\nred\nblue\npurple", "output": "3\nTime\nMind\nSoul" }, { "input": "3\nred\nblue\ngreen", "output": "3\nSoul\nPower\nMind" }, { "input": "4\npurple\nblue\ngreen\nred", "output": "2\nMind\nSoul" }, { "input": "2\norange\nred", "output": "4\nPower\nMind\nTime\nSpace" }, { "input": "3\nred\norange\npurple", "output": "3\nMind\nSpace\nTime" }, { "input": "3\nred\norange\ngreen", "output": "3\nMind\nSpace\nPower" }, { "input": "4\nred\norange\ngreen\npurple", "output": "2\nSpace\nMind" }, { "input": "3\nblue\norange\nred", "output": "3\nPower\nMind\nTime" }, { "input": "4\norange\nblue\npurple\nred", "output": "2\nTime\nMind" }, { "input": "4\ngreen\norange\nred\nblue", "output": "2\nMind\nPower" }, { "input": "5\npurple\norange\nblue\nred\ngreen", "output": "1\nMind" }, { "input": "1\nyellow", "output": "5\nPower\nSoul\nReality\nSpace\nTime" }, { "input": "2\npurple\nyellow", "output": "4\nTime\nReality\nSpace\nSoul" }, { "input": "2\ngreen\nyellow", "output": "4\nSpace\nReality\nPower\nSoul" }, { "input": "3\npurple\nyellow\ngreen", "output": "3\nSoul\nReality\nSpace" }, { "input": "2\nblue\nyellow", "output": "4\nTime\nReality\nPower\nSoul" }, { "input": "3\nyellow\nblue\npurple", "output": "3\nSoul\nReality\nTime" }, { "input": "3\ngreen\nyellow\nblue", "output": "3\nSoul\nReality\nPower" }, { "input": "4\nyellow\nblue\ngreen\npurple", "output": "2\nReality\nSoul" }, { "input": "2\nyellow\norange", "output": "4\nTime\nSpace\nReality\nPower" }, { "input": "3\nyellow\npurple\norange", "output": "3\nSpace\nReality\nTime" }, { "input": "3\norange\nyellow\ngreen", "output": "3\nSpace\nReality\nPower" }, { "input": "4\ngreen\nyellow\norange\npurple", "output": "2\nSpace\nReality" }, { "input": "3\nyellow\nblue\norange", "output": "3\nTime\nReality\nPower" }, { "input": "4\norange\npurple\nblue\nyellow", "output": "2\nReality\nTime" }, { "input": "4\nblue\norange\nyellow\ngreen", "output": "2\nReality\nPower" }, { "input": "5\ngreen\nyellow\norange\nblue\npurple", "output": "1\nReality" }, { "input": "3\nyellow\npurple\nred", "output": "3\nTime\nSoul\nSpace" }, { "input": "3\nred\ngreen\nyellow", "output": "3\nPower\nSoul\nSpace" }, { "input": "4\nred\npurple\ngreen\nyellow", "output": "2\nSpace\nSoul" }, { "input": "3\nred\nyellow\nblue", "output": "3\nPower\nSoul\nTime" }, { "input": "4\nblue\nyellow\nred\npurple", "output": "2\nTime\nSoul" }, { "input": "4\nblue\nyellow\nred\ngreen", "output": "2\nSoul\nPower" }, { "input": "5\nred\nyellow\ngreen\nblue\npurple", "output": "1\nSoul" }, { "input": "3\nred\nyellow\norange", "output": "3\nPower\nSpace\nTime" }, { "input": "4\norange\ngreen\nyellow\nred", "output": "2\nPower\nSpace" }, { "input": "5\norange\nred\ngreen\nyellow\npurple", "output": "1\nSpace" }, { "input": "4\nyellow\nred\norange\nblue", "output": "2\nTime\nPower" }, { "input": "5\npurple\nblue\norange\nyellow\nred", "output": "1\nTime" }, { "input": "5\norange\nblue\nyellow\nred\ngreen", "output": "1\nPower" } ]

1,655,306,968

2,147,483,647

Python 3

OK

TESTS

64

46

0

dict_1 = {'purple': 'Power', 'green': 'Time', 'blue': 'Space', 'orange': 'Soul', 'red': 'Reality', 'yellow': 'Mind', } n = int(input()) print(6-n) for _ in range(n): dict_1.pop(input()) for value in dict_1.values(): print(value)

Title: Infinity Gauntlet Time Limit: None seconds Memory Limit: None megabytes Problem Description: You took a peek on Thanos wearing Infinity Gauntlet. In the Gauntlet there is a place for six Infinity Gems: - the Power Gem of purple color, - the Time Gem of green color, - the Space Gem of blue color, - the Soul Gem of orange color, - the Reality Gem of red color, - the Mind Gem of yellow color. Using colors of Gems you saw in the Gauntlet determine the names of absent Gems. Input Specification: In the first line of input there is one integer $n$ ($0 \le n \le 6$) — the number of Gems in Infinity Gauntlet. In next $n$ lines there are colors of Gems you saw. Words used for colors are: purple, green, blue, orange, red, yellow. It is guaranteed that all the colors are distinct. All colors are given in lowercase English letters. Output Specification: In the first line output one integer $m$ ($0 \le m \le 6$) — the number of absent Gems. Then in $m$ lines print the names of absent Gems, each on its own line. Words used for names are: Power, Time, Space, Soul, Reality, Mind. Names can be printed in any order. Keep the first letter uppercase, others lowercase. Demo Input: ['4\nred\npurple\nyellow\norange\n', '0\n'] Demo Output: ['2\nSpace\nTime\n', '6\nTime\nMind\nSoul\nPower\nReality\nSpace\n'] Note: In the first sample Thanos already has Reality, Power, Mind and Soul Gems, so he needs two more: Time and Space. In the second sample Thanos doesn't have any Gems, so he needs all six.

```python dict_1 = {'purple': 'Power', 'green': 'Time', 'blue': 'Space', 'orange': 'Soul', 'red': 'Reality', 'yellow': 'Mind', } n = int(input()) print(6-n) for _ in range(n): dict_1.pop(input()) for value in dict_1.values(): print(value) ```

3

621

A

Wet Shark and Odd and Even

PROGRAMMING

900

[ "implementation" ]

null

Today, Wet Shark is given *n* integers. Using any of these integers no more than once, Wet Shark wants to get maximum possible even (divisible by 2) sum. Please, calculate this value for Wet Shark. Note, that if Wet Shark uses no integers from the *n* integers, the sum is an even integer 0.

The first line of the input contains one integer, *n* (1<=≤<=*n*<=≤<=100<=000). The next line contains *n* space separated integers given to Wet Shark. Each of these integers is in range from 1 to 109, inclusive.

Print the maximum possible even sum that can be obtained if we use some of the given integers.

[ "3\n1 2 3\n", "5\n999999999 999999999 999999999 999999999 999999999\n" ]

[ "6", "3999999996" ]

In the first sample, we can simply take all three integers for a total sum of 6. In the second sample Wet Shark should take any four out of five integers 999 999 999.

500

[ { "input": "3\n1 2 3", "output": "6" }, { "input": "5\n999999999 999999999 999999999 999999999 999999999", "output": "3999999996" }, { "input": "1\n1", "output": "0" }, { "input": "15\n39 52 88 78 46 95 84 98 55 3 68 42 6 18 98", "output": "870" }, { "input": "15\n59 96 34 48 8 72 67 90 15 85 7 90 97 47 25", "output": "840" }, { "input": "15\n87 37 91 29 58 45 51 74 70 71 47 38 91 89 44", "output": "922" }, { "input": "15\n11 81 49 7 11 14 30 67 29 50 90 81 77 18 59", "output": "674" }, { "input": "15\n39 21 95 89 73 90 9 55 85 32 30 21 68 59 82", "output": "848" }, { "input": "15\n59 70 48 54 26 67 84 39 40 18 77 69 70 88 93", "output": "902" }, { "input": "15\n87 22 98 32 88 36 72 31 100 97 17 16 60 22 20", "output": "798" }, { "input": "15\n15 63 51 13 37 9 43 19 55 79 57 60 50 59 31", "output": "632" }, { "input": "1\n4", "output": "4" }, { "input": "2\n1 4", "output": "4" }, { "input": "3\n1 2 4", "output": "6" }, { "input": "2\n9 3", "output": "12" }, { "input": "2\n1000000000 1001", "output": "1000000000" }, { "input": "3\n1 8 4", "output": "12" }, { "input": "3\n7 4 4", "output": "8" }, { "input": "5\n2 3 4 5 3", "output": "14" }, { "input": "2\n4 5", "output": "4" }, { "input": "3\n2 4 5", "output": "6" }, { "input": "3\n2 2 3", "output": "4" }, { "input": "2\n2 3", "output": "2" }, { "input": "4\n2 3 7 7", "output": "16" }, { "input": "2\n999999999 2", "output": "2" }, { "input": "2\n2 5", "output": "2" }, { "input": "3\n5 3 1", "output": "8" }, { "input": "4\n3 2 5 7", "output": "14" } ]

1,578,595,584

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

8

186

7,680,000

n = int(input()) num = input().split(' ') num.sort() sum = 0 for i in range(0, n): num[i] = int(num[i]) sum += num[i] aux = n - 1 while(sum % 2 != 0): sum -= num[aux] aux -= 1 print(sum)

Title: Wet Shark and Odd and Even Time Limit: None seconds Memory Limit: None megabytes Problem Description: Today, Wet Shark is given *n* integers. Using any of these integers no more than once, Wet Shark wants to get maximum possible even (divisible by 2) sum. Please, calculate this value for Wet Shark. Note, that if Wet Shark uses no integers from the *n* integers, the sum is an even integer 0. Input Specification: The first line of the input contains one integer, *n* (1<=≤<=*n*<=≤<=100<=000). The next line contains *n* space separated integers given to Wet Shark. Each of these integers is in range from 1 to 109, inclusive. Output Specification: Print the maximum possible even sum that can be obtained if we use some of the given integers. Demo Input: ['3\n1 2 3\n', '5\n999999999 999999999 999999999 999999999 999999999\n'] Demo Output: ['6', '3999999996'] Note: In the first sample, we can simply take all three integers for a total sum of 6. In the second sample Wet Shark should take any four out of five integers 999 999 999.

```python n = int(input()) num = input().split(' ') num.sort() sum = 0 for i in range(0, n): num[i] = int(num[i]) sum += num[i] aux = n - 1 while(sum % 2 != 0): sum -= num[aux] aux -= 1 print(sum) ```

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*" } ]

1,608,447,336

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

9

248

0

n,m=map(int,input().split()) mat=[] for _ in range(n): l1=[] l=input() for i in l: l1.append(i) mat.append(l1) #print(mat) r1=0 r2=0 c1=0 c2=0 for i in range(n): if(mat[i].count('*')>0): r1=i break for i in range(n-1,-1,-1): if(mat[i].count('*')>0): r2=i break #print(r1,r2) for i in range(r1,r2+1): for j in range(m): if(mat[i][j]=='*'): c1=j break for i in range(r2,r1-1,-1): for j in range(m-1,-1,-1): if(mat[i][j]=='*'): c2=j break for i in range(r1,r2+1): for j in range(c1,c2+1): print(mat[i][j],end='') print()

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 n,m=map(int,input().split()) mat=[] for _ in range(n): l1=[] l=input() for i in l: l1.append(i) mat.append(l1) #print(mat) r1=0 r2=0 c1=0 c2=0 for i in range(n): if(mat[i].count('*')>0): r1=i break for i in range(n-1,-1,-1): if(mat[i].count('*')>0): r2=i break #print(r1,r2) for i in range(r1,r2+1): for j in range(m): if(mat[i][j]=='*'): c1=j break for i in range(r2,r1-1,-1): for j in range(m-1,-1,-1): if(mat[i][j]=='*'): c2=j break for i in range(r1,r2+1): for j in range(c1,c2+1): print(mat[i][j],end='') print() ```

0

224

A

Parallelepiped

PROGRAMMING

1,100

[ "brute force", "geometry", "math" ]

null

You've got a rectangular parallelepiped with integer edge lengths. You know the areas of its three faces that have a common vertex. Your task is to find the sum of lengths of all 12 edges of this parallelepiped.

The first and the single line contains three space-separated integers — the areas of the parallelepiped's faces. The area's values are positive (<=><=0) and do not exceed 104. It is guaranteed that there exists at least one parallelepiped that satisfies the problem statement.

Print a single number — the sum of all edges of the parallelepiped.

[ "1 1 1\n", "4 6 6\n" ]

[ "12\n", "28\n" ]

In the first sample the parallelepiped has sizes 1 × 1 × 1, in the second one — 2 × 2 × 3.

500

[ { "input": "1 1 1", "output": "12" }, { "input": "4 6 6", "output": "28" }, { "input": "20 10 50", "output": "68" }, { "input": "9 4 36", "output": "56" }, { "input": "324 9 36", "output": "184" }, { "input": "1333 93 129", "output": "308" }, { "input": "1022 584 112", "output": "380" }, { "input": "66 174 319", "output": "184" }, { "input": "912 276 1748", "output": "444" }, { "input": "65 156 60", "output": "120" }, { "input": "1 10000 10000", "output": "40008" }, { "input": "1485 55 27", "output": "332" }, { "input": "152 108 4104", "output": "528" }, { "input": "1656 6900 1350", "output": "740" }, { "input": "12 14 42", "output": "60" }, { "input": "615 18 1230", "output": "856" }, { "input": "680 60 408", "output": "336" }, { "input": "644 966 6", "output": "1308" }, { "input": "1 432 432", "output": "1736" }, { "input": "2239 2239 1", "output": "8964" }, { "input": "4106 8212 2", "output": "16436" }, { "input": "10000 10000 10000", "output": "1200" }, { "input": "3623 3623 1", "output": "14500" }, { "input": "9801 9801 9801", "output": "1188" }, { "input": "10000 1 10000", "output": "40008" }, { "input": "9 9 9", "output": "36" }, { "input": "9801 9702 9702", "output": "1184" } ]

1,565,844,235

2,147,483,647

Python 3

OK

TESTS

27

218

0

import math x, y, z = map(int, input().split()) a = math.sqrt(x * y / z) b = math.sqrt(x * z / y) c = math.sqrt(y * z / x) ans = int(4 * (a + b + c)) print(ans)

Title: Parallelepiped Time Limit: None seconds Memory Limit: None megabytes Problem Description: You've got a rectangular parallelepiped with integer edge lengths. You know the areas of its three faces that have a common vertex. Your task is to find the sum of lengths of all 12 edges of this parallelepiped. Input Specification: The first and the single line contains three space-separated integers — the areas of the parallelepiped's faces. The area's values are positive (<=><=0) and do not exceed 104. It is guaranteed that there exists at least one parallelepiped that satisfies the problem statement. Output Specification: Print a single number — the sum of all edges of the parallelepiped. Demo Input: ['1 1 1\n', '4 6 6\n'] Demo Output: ['12\n', '28\n'] Note: In the first sample the parallelepiped has sizes 1 × 1 × 1, in the second one — 2 × 2 × 3.

```python import math x, y, z = map(int, input().split()) a = math.sqrt(x * y / z) b = math.sqrt(x * z / y) c = math.sqrt(y * z / x) ans = int(4 * (a + b + c)) print(ans) ```

3

237

A

Free Cash

PROGRAMMING

1,000

[ "implementation" ]

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" } ]

1,664,199,358

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

6

92

0

n = int(input()) time = [[0] * 60] * 24 cashDesk = 0 for i in range(n): h, m = map(int, input().split()) time[h][m] += 1 if cashDesk < time[h][m]: cashDesk = time[h][m] print(cashDesk)

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()) time = [[0] * 60] * 24 cashDesk = 0 for i in range(n): h, m = map(int, input().split()) time[h][m] += 1 if cashDesk < time[h][m]: cashDesk = time[h][m] print(cashDesk) ```

0

597

B

Restaurant

PROGRAMMING

1,600

[ "dp", "greedy", "sortings" ]

null

A restaurant received *n* orders for the rental. Each rental order reserve the restaurant for a continuous period of time, the *i*-th order is characterized by two time values — the start time *l**i* and the finish time *r**i* (*l**i*<=≤<=*r**i*). Restaurant management can accept and reject orders. What is the maximal number of orders the restaurant can accept? No two accepted orders can intersect, i.e. they can't share even a moment of time. If one order ends in the moment other starts, they can't be accepted both.

The first line contains integer number *n* (1<=≤<=*n*<=≤<=5·105) — number of orders. The following *n* lines contain integer values *l**i* and *r**i* each (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=109).

Print the maximal number of orders that can be accepted.

[ "2\n7 11\n4 7\n", "5\n1 2\n2 3\n3 4\n4 5\n5 6\n", "6\n4 8\n1 5\n4 7\n2 5\n1 3\n6 8\n" ]

[ "1\n", "3\n", "2\n" ]

none

1,000

[ { "input": "2\n7 11\n4 7", "output": "1" }, { "input": "5\n1 2\n2 3\n3 4\n4 5\n5 6", "output": "3" }, { "input": "6\n4 8\n1 5\n4 7\n2 5\n1 3\n6 8", "output": "2" }, { "input": "1\n1 1", "output": "1" }, { "input": "2\n4 6\n4 8", "output": "1" }, { "input": "3\n22 22\n14 21\n9 25", "output": "2" }, { "input": "4\n20 59\n30 62\n29 45\n29 32", "output": "1" }, { "input": "5\n40 124\n40 117\n67 106\n36 121\n38 102", "output": "1" }, { "input": "6\n124 155\n50 93\n45 120\n54 171\n46 190\n76 179", "output": "2" }, { "input": "7\n94 113\n54 248\n64 325\n280 306\n62 328\n49 341\n90 324", "output": "2" }, { "input": "8\n116 416\n104 472\n84 476\n100 486\n199 329\n169 444\n171 487\n134 441", "output": "1" }, { "input": "9\n90 667\n366 539\n155 462\n266 458\n323 574\n101 298\n90 135\n641 661\n122 472", "output": "3" }, { "input": "10\n195 443\n229 602\n200 948\n229 876\n228 904\n296 656\n189 818\n611 626\n215 714\n403 937", "output": "2" }, { "input": "1\n28 74", "output": "1" }, { "input": "2\n28 92\n2 59", "output": "1" }, { "input": "3\n5 92\n1 100\n39 91", "output": "1" }, { "input": "4\n4 92\n29 43\n13 73\n10 79", "output": "1" }, { "input": "5\n64 86\n61 61\n46 54\n83 94\n19 46", "output": "3" }, { "input": "6\n80 84\n21 24\n44 80\n14 53\n5 10\n61 74", "output": "4" }, { "input": "7\n32 92\n32 86\n13 25\n45 75\n16 65\n1 99\n17 98", "output": "2" }, { "input": "8\n3 59\n22 94\n26 97\n18 85\n7 84\n1 100\n4 100\n26 93", "output": "1" }, { "input": "9\n11 90\n8 95\n62 95\n43 96\n16 84\n3 70\n23 93\n4 96\n11 86", "output": "1" }, { "input": "10\n30 45\n5 8\n51 83\n37 52\n49 75\n28 92\n94 99\n4 13\n61 83\n36 96", "output": "4" }, { "input": "11\n38 92\n16 85\n32 43\n65 84\n63 100\n21 45\n13 92\n29 58\n56 94\n18 83\n50 81", "output": "2" }, { "input": "12\n66 78\n41 97\n55 69\n55 61\n36 64\n14 97\n96 99\n28 58\n44 93\n2 100\n42 88\n1 2", "output": "4" }, { "input": "13\n50 85\n38 65\n5 51\n50 96\n4 92\n23 94\n2 99\n2 84\n1 98\n2 100\n12 100\n21 97\n7 84", "output": "1" }, { "input": "14\n17 92\n7 96\n49 96\n10 99\n7 98\n12 85\n10 52\n2 99\n23 75\n4 98\n7 100\n2 69\n6 99\n20 87", "output": "1" }, { "input": "15\n1 58\n15 21\n53 55\n59 90\n68 71\n29 51\n52 81\n32 52\n38 44\n57 59\n47 60\n27 32\n49 86\n26 94\n44 45", "output": "6" }, { "input": "16\n4 80\n16 46\n15 16\n60 63\n8 54\n18 49\n67 99\n72 80\n1 8\n19 64\n1 54\n46 94\n2 89\n67 78\n21 47\n5 29", "output": "5" }, { "input": "17\n34 42\n31 84\n8 96\n63 88\n11 99\n80 99\n1 96\n11 12\n27 28\n4 30\n1 79\n16 86\n15 86\n13 80\n3 98\n37 89\n59 88", "output": "4" }, { "input": "18\n11 94\n12 85\n25 90\n7 61\n63 88\n6 87\n49 88\n16 76\n12 78\n61 84\n3 84\n20 91\n1 84\n17 100\n43 80\n8 86\n9 98\n35 97", "output": "2" }, { "input": "19\n24 63\n23 86\n5 89\n10 83\n31 92\n8 96\n21 63\n1 83\n2 100\n5 96\n18 98\n9 77\n11 91\n44 95\n1 98\n22 60\n5 98\n22 91\n1 96", "output": "1" }, { "input": "20\n22 77\n13 50\n55 64\n16 52\n67 96\n49 51\n59 95\n2 25\n69 91\n2 24\n4 46\n50 74\n45 63\n39 55\n31 33\n9 33\n6 72\n14 67\n56 98\n69 94", "output": "5" }, { "input": "1\n1 1000000000", "output": "1" }, { "input": "4\n1 1000000000\n1 1000000000\n1 1000000000\n1 1000000000", "output": "1" }, { "input": "2\n1000000000 1000000000\n1 1", "output": "2" }, { "input": "4\n1 10000\n2 3\n4 5\n6 7", "output": "3" }, { "input": "3\n1 10\n2 4\n6 8", "output": "2" }, { "input": "2\n1 4\n2 3", "output": "1" }, { "input": "3\n1 100\n2 3\n4 5", "output": "2" }, { "input": "4\n1 10\n2 3\n4 5\n6 7", "output": "3" }, { "input": "3\n1 5\n2 3\n4 4", "output": "2" }, { "input": "3\n1 1000\n1 5\n6 1000", "output": "2" }, { "input": "3\n1 10000\n3 5\n7 8", "output": "2" }, { "input": "4\n1 5\n6 15\n8 10\n11 17", "output": "3" }, { "input": "4\n1 3\n6 800\n7 9\n10 11", "output": "3" }, { "input": "5\n1 10\n2 3\n4 5\n6 7\n8 9", "output": "4" }, { "input": "3\n1 5\n1 3\n4 6", "output": "2" }, { "input": "3\n1 10\n2 3\n4 5", "output": "2" }, { "input": "2\n1 3\n2 4", "output": "1" } ]

1,687,018,781

2,147,483,647

PyPy 3-64

OK

TESTS

70

1,840

39,321,600

import sys input = lambda: sys.stdin.readline().rstrip() N = int(input()) A = [] for _ in range(N): l,r = map(int, input().split()) A.append((r,l)) A.sort() ans = 0 pre = -1 for r,l in A: if l>pre: ans+=1 pre=r print(ans)

Title: Restaurant Time Limit: None seconds Memory Limit: None megabytes Problem Description: A restaurant received *n* orders for the rental. Each rental order reserve the restaurant for a continuous period of time, the *i*-th order is characterized by two time values — the start time *l**i* and the finish time *r**i* (*l**i*<=≤<=*r**i*). Restaurant management can accept and reject orders. What is the maximal number of orders the restaurant can accept? No two accepted orders can intersect, i.e. they can't share even a moment of time. If one order ends in the moment other starts, they can't be accepted both. Input Specification: The first line contains integer number *n* (1<=≤<=*n*<=≤<=5·105) — number of orders. The following *n* lines contain integer values *l**i* and *r**i* each (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=109). Output Specification: Print the maximal number of orders that can be accepted. Demo Input: ['2\n7 11\n4 7\n', '5\n1 2\n2 3\n3 4\n4 5\n5 6\n', '6\n4 8\n1 5\n4 7\n2 5\n1 3\n6 8\n'] Demo Output: ['1\n', '3\n', '2\n'] Note: none

```python import sys input = lambda: sys.stdin.readline().rstrip() N = int(input()) A = [] for _ in range(N): l,r = map(int, input().split()) A.append((r,l)) A.sort() ans = 0 pre = -1 for r,l in A: if l>pre: ans+=1 pre=r print(ans) ```

3

257

C

View Angle

PROGRAMMING

1,800

[ "brute force", "geometry", "math" ]

null

Flatland has recently introduced a new type of an eye check for the driver's licence. The check goes like that: there is a plane with mannequins standing on it. You should tell the value of the minimum angle with the vertex at the origin of coordinates and with all mannequins standing inside or on the boarder of this angle. As you spend lots of time "glued to the screen", your vision is impaired. So you have to write a program that will pass the check for you.

The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of mannequins. Next *n* lines contain two space-separated integers each: *x**i*,<=*y**i* (|*x**i*|,<=|*y**i*|<=≤<=1000) — the coordinates of the *i*-th mannequin. It is guaranteed that the origin of the coordinates has no mannequin. It is guaranteed that no two mannequins are located in the same point on the plane.

Print a single real number — the value of the sought angle in degrees. The answer will be considered valid if the relative or absolute error doesn't exceed 10<=-<=6.

[ "2\n2 0\n0 2\n", "3\n2 0\n0 2\n-2 2\n", "4\n2 0\n0 2\n-2 0\n0 -2\n", "2\n2 1\n1 2\n" ]

[ "90.0000000000\n", "135.0000000000\n", "270.0000000000\n", "36.8698976458\n" ]

Solution for the first sample test is shown below: Solution for the second sample test is shown below: Solution for the third sample test is shown below: Solution for the fourth sample test is shown below:

1,500

[ { "input": "2\n2 0\n0 2", "output": "90.0000000000" }, { "input": "3\n2 0\n0 2\n-2 2", "output": "135.0000000000" }, { "input": "4\n2 0\n0 2\n-2 0\n0 -2", "output": "270.0000000000" }, { "input": "2\n2 1\n1 2", "output": "36.8698976458" }, { "input": "1\n1 1", "output": "0.0000000000" }, { "input": "10\n9 7\n10 7\n6 5\n6 10\n7 6\n5 10\n6 7\n10 9\n5 5\n5 8", "output": "28.4429286244" }, { "input": "10\n-1 28\n1 28\n1 25\n0 23\n-1 24\n-1 22\n1 27\n0 30\n1 22\n1 21", "output": "5.3288731964" }, { "input": "10\n-5 9\n-10 6\n-8 8\n-9 9\n-6 5\n-8 9\n-5 7\n-6 6\n-5 10\n-8 7", "output": "32.4711922908" }, { "input": "10\n6 -9\n9 -5\n10 -5\n7 -5\n8 -7\n8 -10\n8 -5\n6 -10\n7 -6\n8 -9", "output": "32.4711922908" }, { "input": "10\n-5 -7\n-8 -10\n-9 -5\n-5 -9\n-9 -8\n-7 -7\n-6 -8\n-6 -10\n-10 -7\n-9 -6", "output": "31.8907918018" }, { "input": "10\n-1 -29\n-1 -26\n1 -26\n-1 -22\n-1 -24\n-1 -21\n1 -24\n-1 -20\n-1 -23\n-1 -25", "output": "5.2483492565" }, { "input": "10\n21 0\n22 1\n30 0\n20 0\n28 0\n29 0\n21 -1\n30 1\n24 1\n26 0", "output": "5.3288731964" }, { "input": "10\n-20 0\n-22 1\n-26 0\n-22 -1\n-30 -1\n-30 0\n-28 0\n-24 1\n-23 -1\n-29 1", "output": "5.2051244050" }, { "input": "10\n-5 -5\n5 -5\n-4 -5\n4 -5\n1 -5\n0 -5\n3 -5\n-2 -5\n2 -5\n-3 -5", "output": "90.0000000000" }, { "input": "10\n-5 -5\n-4 -5\n-2 -5\n4 -5\n5 -5\n3 -5\n2 -5\n-1 -5\n-3 -5\n0 -5", "output": "90.0000000000" }, { "input": "10\n-1 -5\n-5 -5\n2 -5\n-2 -5\n1 -5\n5 -5\n0 -5\n3 -5\n-4 -5\n-3 -5", "output": "90.0000000000" }, { "input": "10\n-1 -5\n-5 -5\n-4 -5\n3 -5\n0 -5\n4 -5\n1 -5\n-2 -5\n5 -5\n-3 -5", "output": "90.0000000000" }, { "input": "10\n5 -5\n4 -5\n-1 -5\n1 -5\n-4 -5\n3 -5\n0 -5\n-5 -5\n-2 -5\n-3 -5", "output": "90.0000000000" }, { "input": "10\n2 -5\n-4 -5\n-2 -5\n4 -5\n-5 -5\n-1 -5\n0 -5\n-3 -5\n3 -5\n1 -5", "output": "83.6598082541" }, { "input": "5\n2 1\n0 1\n2 -1\n-2 -1\n2 0", "output": "233.1301023542" }, { "input": "5\n-2 -2\n2 2\n2 -1\n-2 0\n1 -1", "output": "225.0000000000" }, { "input": "5\n0 -2\n-2 -1\n-1 2\n0 -1\n-1 0", "output": "153.4349488229" }, { "input": "5\n-1 -1\n-2 -1\n1 0\n-1 -2\n-1 1", "output": "225.0000000000" }, { "input": "5\n1 -1\n0 2\n-2 2\n-2 1\n2 1", "output": "198.4349488229" }, { "input": "5\n2 2\n1 2\n-2 -1\n1 1\n-2 -2", "output": "180.0000000000" }, { "input": "2\n1 1\n2 2", "output": "0.0000000000" }, { "input": "27\n-592 -96\n-925 -150\n-111 -18\n-259 -42\n-370 -60\n-740 -120\n-629 -102\n-333 -54\n-407 -66\n-296 -48\n-37 -6\n-999 -162\n-222 -36\n-555 -90\n-814 -132\n-444 -72\n-74 -12\n-185 -30\n-148 -24\n-962 -156\n-777 -126\n-518 -84\n-888 -144\n-666 -108\n-481 -78\n-851 -138\n-703 -114", "output": "0.0000000000" }, { "input": "38\n96 416\n24 104\n6 26\n12 52\n210 910\n150 650\n54 234\n174 754\n114 494\n18 78\n90 390\n36 156\n222 962\n186 806\n126 546\n78 338\n108 468\n180 780\n120 520\n84 364\n66 286\n138 598\n30 130\n228 988\n72 312\n144 624\n198 858\n60 260\n48 208\n102 442\n42 182\n162 702\n132 572\n156 676\n204 884\n216 936\n168 728\n192 832", "output": "0.0000000000" }, { "input": "14\n-2 -134\n-4 -268\n-11 -737\n-7 -469\n-14 -938\n-10 -670\n-3 -201\n-1 -67\n-9 -603\n-6 -402\n-13 -871\n-12 -804\n-8 -536\n-5 -335", "output": "0.0000000000" }, { "input": "14\n588 938\n420 670\n210 335\n252 402\n504 804\n126 201\n42 67\n546 871\n294 469\n84 134\n336 536\n462 737\n168 268\n378 603", "output": "0.0000000000" }, { "input": "20\n-45 147\n-240 784\n-135 441\n-60 196\n-105 343\n-285 931\n-195 637\n-300 980\n-165 539\n-210 686\n-75 245\n-15 49\n-30 98\n-270 882\n-120 392\n-90 294\n-150 490\n-180 588\n-255 833\n-225 735", "output": "0.0000000000" }, { "input": "2\n1 1\n1 -1", "output": "90.0000000000" } ]

1,662,379,189

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

5

622

2,969,600

import math ar=[] manq=int(input()) for i in range(manq): coor = list(map(int, input().strip().split()))[:2] ang = math.atan2(coor[1],coor[0])/math.pi*180 if ang<0: ang=360+ang ar.append(ang) sorted(ar) max = ar[-1]-ar[0] print(max)

Title: View Angle Time Limit: None seconds Memory Limit: None megabytes Problem Description: Flatland has recently introduced a new type of an eye check for the driver's licence. The check goes like that: there is a plane with mannequins standing on it. You should tell the value of the minimum angle with the vertex at the origin of coordinates and with all mannequins standing inside or on the boarder of this angle. As you spend lots of time "glued to the screen", your vision is impaired. So you have to write a program that will pass the check for you. Input Specification: The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of mannequins. Next *n* lines contain two space-separated integers each: *x**i*,<=*y**i* (|*x**i*|,<=|*y**i*|<=≤<=1000) — the coordinates of the *i*-th mannequin. It is guaranteed that the origin of the coordinates has no mannequin. It is guaranteed that no two mannequins are located in the same point on the plane. Output Specification: Print a single real number — the value of the sought angle in degrees. The answer will be considered valid if the relative or absolute error doesn't exceed 10<=-<=6. Demo Input: ['2\n2 0\n0 2\n', '3\n2 0\n0 2\n-2 2\n', '4\n2 0\n0 2\n-2 0\n0 -2\n', '2\n2 1\n1 2\n'] Demo Output: ['90.0000000000\n', '135.0000000000\n', '270.0000000000\n', '36.8698976458\n'] Note: Solution for the first sample test is shown below: Solution for the second sample test is shown below: Solution for the third sample test is shown below: Solution for the fourth sample test is shown below:

```python import math ar=[] manq=int(input()) for i in range(manq): coor = list(map(int, input().strip().split()))[:2] ang = math.atan2(coor[1],coor[0])/math.pi*180 if ang<0: ang=360+ang ar.append(ang) sorted(ar) max = ar[-1]-ar[0] print(max) ```

0

none

0

[ "none" ]

null

Iahub and Iahubina went to a picnic in a forest full of trees. Less than 5 minutes passed before Iahub remembered of trees from programming. Moreover, he invented a new problem and Iahubina has to solve it, otherwise Iahub won't give her the food. Iahub asks Iahubina: can you build a rooted tree, such that - each internal node (a node with at least one son) has at least two sons; - node *i* has *c**i* nodes in its subtree? Iahubina has to guess the tree. Being a smart girl, she realized that it's possible no tree can follow Iahub's restrictions. In this way, Iahub will eat all the food. You need to help Iahubina: determine if there's at least one tree following Iahub's restrictions. The required tree must contain *n* nodes.

The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=24). Next line contains *n* positive integers: the *i*-th number represents *c**i* (1<=≤<=*c**i*<=≤<=*n*).

Output on the first line "YES" (without quotes) if there exist at least one tree following Iahub's restrictions, otherwise output "NO" (without quotes).

[ "4\n1 1 1 4\n", "5\n1 1 5 2 1\n" ]

[ "YES", "NO" ]

none

0

[]

1,629,902,133

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

3

61

7,372,800

from queue import Queue from sys import stdin, stdout def read(): return stdin.readline().rstrip() def readints(): return [int(x) for x in read().split()] def solve(at, n, size, used): if size[at] == 1: return True left = size[at] - 1 children = 0 for i in range(at + 1, n): if used[i]: continue if size[i] <= left and size[i] < size[at] - 1: used[i] = True children += 1 ret = solve(at + 1, n, size, used) used[i] = False if not ret: return False left -= size[i] return left == 0 and children > 1 n = readints()[0] size = readints() used = [False] * n size = sorted(size, reverse=True) solution = solve(0, n, size, used) print(["NO", "YES"][solution])

Title: none Time Limit: None seconds Memory Limit: None megabytes Problem Description: Iahub and Iahubina went to a picnic in a forest full of trees. Less than 5 minutes passed before Iahub remembered of trees from programming. Moreover, he invented a new problem and Iahubina has to solve it, otherwise Iahub won't give her the food. Iahub asks Iahubina: can you build a rooted tree, such that - each internal node (a node with at least one son) has at least two sons; - node *i* has *c**i* nodes in its subtree? Iahubina has to guess the tree. Being a smart girl, she realized that it's possible no tree can follow Iahub's restrictions. In this way, Iahub will eat all the food. You need to help Iahubina: determine if there's at least one tree following Iahub's restrictions. The required tree must contain *n* nodes. Input Specification: The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=24). Next line contains *n* positive integers: the *i*-th number represents *c**i* (1<=≤<=*c**i*<=≤<=*n*). Output Specification: Output on the first line "YES" (without quotes) if there exist at least one tree following Iahub's restrictions, otherwise output "NO" (without quotes). Demo Input: ['4\n1 1 1 4\n', '5\n1 1 5 2 1\n'] Demo Output: ['YES', 'NO'] Note: none

```python from queue import Queue from sys import stdin, stdout def read(): return stdin.readline().rstrip() def readints(): return [int(x) for x in read().split()] def solve(at, n, size, used): if size[at] == 1: return True left = size[at] - 1 children = 0 for i in range(at + 1, n): if used[i]: continue if size[i] <= left and size[i] < size[at] - 1: used[i] = True children += 1 ret = solve(at + 1, n, size, used) used[i] = False if not ret: return False left -= size[i] return left == 0 and children > 1 n = readints()[0] size = readints() used = [False] * n size = sorted(size, reverse=True) solution = solve(0, n, size, used) print(["NO", "YES"][solution]) ```

0

962

A

Equator

PROGRAMMING

1,300

[ "implementation" ]

null

Polycarp has created his own training plan to prepare for the programming contests. He will train for $n$ days, all days are numbered from $1$ to $n$, beginning from the first. On the $i$-th day Polycarp will necessarily solve $a_i$ problems. One evening Polycarp plans to celebrate the equator. He will celebrate it on the first evening of such a day that from the beginning of the training and to this day inclusive he will solve half or more of all the problems. Determine the index of day when Polycarp will celebrate the equator.

The first line contains a single integer $n$ ($1 \le n \le 200\,000$) — the number of days to prepare for the programming contests. The second line contains a sequence $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10\,000$), where $a_i$ equals to the number of problems, which Polycarp will solve on the $i$-th day.

Print the index of the day when Polycarp will celebrate the equator.

[ "4\n1 3 2 1\n", "6\n2 2 2 2 2 2\n" ]

[ "2\n", "3\n" ]

In the first example Polycarp will celebrate the equator on the evening of the second day, because up to this day (inclusive) he will solve $4$ out of $7$ scheduled problems on four days of the training. In the second example Polycarp will celebrate the equator on the evening of the third day, because up to this day (inclusive) he will solve $6$ out of $12$ scheduled problems on six days of the training.

0

[ { "input": "4\n1 3 2 1", "output": "2" }, { "input": "6\n2 2 2 2 2 2", "output": "3" }, { "input": "1\n10000", "output": "1" }, { "input": "3\n2 1 1", "output": "1" }, { "input": "2\n1 3", "output": "2" }, { "input": "4\n2 1 1 3", "output": "3" }, { "input": "3\n1 1 3", "output": "3" }, { "input": "3\n1 1 1", "output": "2" }, { "input": "2\n1 2", "output": "2" }, { "input": "3\n2 1 2", "output": "2" }, { "input": "5\n1 2 4 3 5", "output": "4" }, { "input": "5\n2 2 2 4 3", "output": "4" }, { "input": "4\n1 2 3 1", "output": "3" }, { "input": "6\n7 3 10 7 3 11", "output": "4" }, { "input": "2\n3 4", "output": "2" }, { "input": "5\n1 1 1 1 1", "output": "3" }, { "input": "4\n1 3 2 3", "output": "3" }, { "input": "2\n2 3", "output": "2" }, { "input": "3\n32 10 23", "output": "2" }, { "input": "7\n1 1 1 1 1 1 1", "output": "4" }, { "input": "3\n1 2 4", "output": "3" }, { "input": "6\n3 3 3 2 4 4", "output": "4" }, { "input": "9\n1 1 1 1 1 1 1 1 1", "output": "5" }, { "input": "5\n1 3 3 1 1", "output": "3" }, { "input": "4\n1 1 1 2", "output": "3" }, { "input": "4\n1 2 1 3", "output": "3" }, { "input": "3\n2 2 1", "output": "2" }, { "input": "4\n2 3 3 3", "output": "3" }, { "input": "4\n3 2 3 3", "output": "3" }, { "input": "4\n2 1 1 1", "output": "2" }, { "input": "3\n2 1 4", "output": "3" }, { "input": "2\n6 7", "output": "2" }, { "input": "4\n3 3 4 3", "output": "3" }, { "input": "4\n1 1 2 5", "output": "4" }, { "input": "4\n1 8 7 3", "output": "3" }, { "input": "6\n2 2 2 2 2 3", "output": "4" }, { "input": "3\n2 2 5", "output": "3" }, { "input": "4\n1 1 2 1", "output": "3" }, { "input": "5\n1 1 2 2 3", "output": "4" }, { "input": "5\n9 5 3 4 8", "output": "3" }, { "input": "3\n3 3 1", "output": "2" }, { "input": "4\n1 2 2 2", "output": "3" }, { "input": "3\n1 3 5", "output": "3" }, { "input": "4\n1 1 3 6", "output": "4" }, { "input": "6\n1 2 1 1 1 1", "output": "3" }, { "input": "3\n3 1 3", "output": "2" }, { "input": "5\n3 4 5 1 2", "output": "3" }, { "input": "11\n1 1 1 1 1 1 1 1 1 1 1", "output": "6" }, { "input": "5\n3 1 2 5 2", "output": "4" }, { "input": "4\n1 1 1 4", "output": "4" }, { "input": "4\n2 6 1 10", "output": "4" }, { "input": "4\n2 2 3 2", "output": "3" }, { "input": "4\n4 2 2 1", "output": "2" }, { "input": "6\n1 1 1 1 1 4", "output": "5" }, { "input": "3\n3 2 2", "output": "2" }, { "input": "6\n1 3 5 1 7 4", "output": "5" }, { "input": "5\n1 2 4 8 16", "output": "5" }, { "input": "5\n1 2 4 4 4", "output": "4" }, { "input": "6\n4 2 1 2 3 1", "output": "3" }, { "input": "4\n3 2 1 5", "output": "3" }, { "input": "1\n1", "output": "1" }, { "input": "3\n2 4 7", "output": "3" }, { "input": "5\n1 1 1 1 3", "output": "4" }, { "input": "3\n3 1 5", "output": "3" }, { "input": "4\n1 2 3 7", "output": "4" }, { "input": "3\n1 4 6", "output": "3" }, { "input": "4\n2 1 2 2", "output": "3" }, { "input": "2\n4 5", "output": "2" }, { "input": "5\n1 2 1 2 1", "output": "3" }, { "input": "3\n2 3 6", "output": "3" }, { "input": "6\n1 1 4 1 1 5", "output": "4" }, { "input": "5\n2 2 2 2 1", "output": "3" }, { "input": "2\n5 6", "output": "2" }, { "input": "4\n2 2 1 4", "output": "3" }, { "input": "5\n2 2 3 4 4", "output": "4" }, { "input": "4\n3 1 1 2", "output": "2" }, { "input": "5\n3 4 1 4 5", "output": "4" }, { "input": "4\n1 3 1 6", "output": "4" }, { "input": "5\n1 1 1 2 2", "output": "4" }, { "input": "4\n1 4 2 4", "output": "3" }, { "input": "10\n1 1 1 1 1 1 1 1 1 8", "output": "9" }, { "input": "4\n1 4 5 1", "output": "3" }, { "input": "5\n1 1 1 1 5", "output": "5" }, { "input": "4\n1 3 4 1", "output": "3" }, { "input": "4\n2 2 2 3", "output": "3" }, { "input": "4\n2 3 2 4", "output": "3" }, { "input": "5\n2 2 1 2 2", "output": "3" }, { "input": "3\n4 3 2", "output": "2" }, { "input": "3\n6 5 2", "output": "2" }, { "input": "69\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", "output": "35" }, { "input": "6\n1 1 1 1 1 2", "output": "4" }, { "input": "5\n1 2 5 4 5", "output": "4" }, { "input": "2\n9 10", "output": "2" }, { "input": "3\n1 1 5", "output": "3" }, { "input": "4\n3 4 3 5", "output": "3" }, { "input": "4\n1 4 3 3", "output": "3" }, { "input": "4\n7 1 3 4", "output": "2" }, { "input": "3\n100 100 1", "output": "2" }, { "input": "4\n5 2 2 2", "output": "2" } ]

1,586,267,666

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

2

109

307,200

n=int(input()) a=list([int(x) for x in input().split()]) s=sum(a) check=0 #print(s) for i in range(len(a)): if(check>=(s//2)): print(i) break else: #print(check) check+=a[i]

Title: Equator Time Limit: None seconds Memory Limit: None megabytes Problem Description: Polycarp has created his own training plan to prepare for the programming contests. He will train for $n$ days, all days are numbered from $1$ to $n$, beginning from the first. On the $i$-th day Polycarp will necessarily solve $a_i$ problems. One evening Polycarp plans to celebrate the equator. He will celebrate it on the first evening of such a day that from the beginning of the training and to this day inclusive he will solve half or more of all the problems. Determine the index of day when Polycarp will celebrate the equator. Input Specification: The first line contains a single integer $n$ ($1 \le n \le 200\,000$) — the number of days to prepare for the programming contests. The second line contains a sequence $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10\,000$), where $a_i$ equals to the number of problems, which Polycarp will solve on the $i$-th day. Output Specification: Print the index of the day when Polycarp will celebrate the equator. Demo Input: ['4\n1 3 2 1\n', '6\n2 2 2 2 2 2\n'] Demo Output: ['2\n', '3\n'] Note: In the first example Polycarp will celebrate the equator on the evening of the second day, because up to this day (inclusive) he will solve $4$ out of $7$ scheduled problems on four days of the training. In the second example Polycarp will celebrate the equator on the evening of the third day, because up to this day (inclusive) he will solve $6$ out of $12$ scheduled problems on six days of the training.

```python n=int(input()) a=list([int(x) for x in input().split()]) s=sum(a) check=0 #print(s) for i in range(len(a)): if(check>=(s//2)): print(i) break else: #print(check) check+=a[i] ```

0

805

A

Fake NP

PROGRAMMING

1,000

[ "greedy", "math" ]

null

Tavak and Seyyed are good friends. Seyyed is very funny and he told Tavak to solve the following problem instead of longest-path. You are given *l* and *r*. For all integers from *l* to *r*, inclusive, we wrote down all of their integer divisors except 1. Find the integer that we wrote down the maximum number of times. Solve the problem to show that it's not a NP problem.

The first line contains two integers *l* and *r* (2<=≤<=*l*<=≤<=*r*<=≤<=109).

Print single integer, the integer that appears maximum number of times in the divisors. If there are multiple answers, print any of them.

[ "19 29\n", "3 6\n" ]

[ "2\n", "3\n" ]

Definition of a divisor: [https://www.mathsisfun.com/definitions/divisor-of-an-integer-.html](https://www.mathsisfun.com/definitions/divisor-of-an-integer-.html) The first example: from 19 to 29 these numbers are divisible by 2: {20, 22, 24, 26, 28}. The second example: from 3 to 6 these numbers are divisible by 3: {3, 6}.

500

[ { "input": "19 29", "output": "2" }, { "input": "3 6", "output": "2" }, { "input": "39 91", "output": "2" }, { "input": "76 134", "output": "2" }, { "input": "93 95", "output": "2" }, { "input": "17 35", "output": "2" }, { "input": "94 95", "output": "2" }, { "input": "51 52", "output": "2" }, { "input": "47 52", "output": "2" }, { "input": "38 98", "output": "2" }, { "input": "30 37", "output": "2" }, { "input": "56 92", "output": "2" }, { "input": "900000000 1000000000", "output": "2" }, { "input": "37622224 162971117", "output": "2" }, { "input": "760632746 850720703", "output": "2" }, { "input": "908580370 968054552", "output": "2" }, { "input": "951594860 953554446", "output": "2" }, { "input": "347877978 913527175", "output": "2" }, { "input": "620769961 988145114", "output": "2" }, { "input": "820844234 892579936", "output": "2" }, { "input": "741254764 741254768", "output": "2" }, { "input": "80270976 80270977", "output": "2" }, { "input": "392602363 392602367", "output": "2" }, { "input": "519002744 519002744", "output": "519002744" }, { "input": "331900277 331900277", "output": "331900277" }, { "input": "419873015 419873018", "output": "2" }, { "input": "349533413 349533413", "output": "349533413" }, { "input": "28829775 28829776", "output": "2" }, { "input": "568814539 568814539", "output": "568814539" }, { "input": "720270740 720270743", "output": "2" }, { "input": "871232720 871232722", "output": "2" }, { "input": "305693653 305693653", "output": "305693653" }, { "input": "634097178 634097179", "output": "2" }, { "input": "450868287 450868290", "output": "2" }, { "input": "252662256 252662260", "output": "2" }, { "input": "575062045 575062049", "output": "2" }, { "input": "273072892 273072894", "output": "2" }, { "input": "770439256 770439256", "output": "770439256" }, { "input": "2 1000000000", "output": "2" }, { "input": "6 8", "output": "2" }, { "input": "2 879190747", "output": "2" }, { "input": "5 5", "output": "5" }, { "input": "999999937 999999937", "output": "999999937" }, { "input": "3 3", "output": "3" }, { "input": "5 100", "output": "2" }, { "input": "2 2", "output": "2" }, { "input": "3 18", "output": "2" }, { "input": "7 7", "output": "7" }, { "input": "39916801 39916801", "output": "39916801" }, { "input": "3 8", "output": "2" }, { "input": "13 13", "output": "13" }, { "input": "4 8", "output": "2" }, { "input": "3 12", "output": "2" }, { "input": "6 12", "output": "2" }, { "input": "999999103 999999103", "output": "999999103" }, { "input": "100000007 100000007", "output": "100000007" }, { "input": "3 99", "output": "2" }, { "input": "999999733 999999733", "output": "999999733" }, { "input": "5 10", "output": "2" }, { "input": "982451653 982451653", "output": "982451653" }, { "input": "999900001 1000000000", "output": "2" }, { "input": "999727999 999727999", "output": "999727999" }, { "input": "2 999999999", "output": "2" }, { "input": "242 244", "output": "2" }, { "input": "3 10", "output": "2" }, { "input": "15 27", "output": "2" }, { "input": "998244353 998244353", "output": "998244353" }, { "input": "5 15", "output": "2" }, { "input": "999999797 999999797", "output": "999999797" }, { "input": "2 3", "output": "2" }, { "input": "999999929 999999929", "output": "999999929" }, { "input": "3 111111", "output": "2" }, { "input": "12 18", "output": "2" }, { "input": "479001599 479001599", "output": "479001599" }, { "input": "10000019 10000019", "output": "10000019" }, { "input": "715827883 715827883", "output": "715827883" }, { "input": "999992977 999992977", "output": "999992977" }, { "input": "11 11", "output": "11" }, { "input": "29 29", "output": "29" }, { "input": "1000003 1000003", "output": "1000003" }, { "input": "6 15", "output": "2" }, { "input": "1200007 1200007", "output": "1200007" }, { "input": "3 1000000000", "output": "2" }, { "input": "990000023 990000023", "output": "990000023" }, { "input": "1717 1717", "output": "1717" }, { "input": "141650963 141650963", "output": "141650963" }, { "input": "1002523 1002523", "output": "1002523" }, { "input": "900000011 900000011", "output": "900000011" }, { "input": "104729 104729", "output": "104729" }, { "input": "4 12", "output": "2" }, { "input": "100003 100003", "output": "100003" }, { "input": "17 17", "output": "17" }, { "input": "10 100", "output": "2" } ]

1,643,673,175

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

24

93

0

a, b = list(map(int, input().split())) #c = (a - 1) // 2 #d = b // 2 print(2)

Title: Fake NP Time Limit: None seconds Memory Limit: None megabytes Problem Description: Tavak and Seyyed are good friends. Seyyed is very funny and he told Tavak to solve the following problem instead of longest-path. You are given *l* and *r*. For all integers from *l* to *r*, inclusive, we wrote down all of their integer divisors except 1. Find the integer that we wrote down the maximum number of times. Solve the problem to show that it's not a NP problem. Input Specification: The first line contains two integers *l* and *r* (2<=≤<=*l*<=≤<=*r*<=≤<=109). Output Specification: Print single integer, the integer that appears maximum number of times in the divisors. If there are multiple answers, print any of them. Demo Input: ['19 29\n', '3 6\n'] Demo Output: ['2\n', '3\n'] Note: Definition of a divisor: [https://www.mathsisfun.com/definitions/divisor-of-an-integer-.html](https://www.mathsisfun.com/definitions/divisor-of-an-integer-.html) The first example: from 19 to 29 these numbers are divisible by 2: {20, 22, 24, 26, 28}. The second example: from 3 to 6 these numbers are divisible by 3: {3, 6}.

```python a, b = list(map(int, input().split())) #c = (a - 1) // 2 #d = b // 2 print(2) ```

0

599

A

Patrick and Shopping

PROGRAMMING

800

[ "implementation" ]

null

Today Patrick waits for a visit from his friend Spongebob. To prepare for the visit, Patrick needs to buy some goodies in two stores located near his house. There is a *d*1 meter long road between his house and the first shop and a *d*2 meter long road between his house and the second shop. Also, there is a road of length *d*3 directly connecting these two shops to each other. Help Patrick calculate the minimum distance that he needs to walk in order to go to both shops and return to his house. Patrick always starts at his house. He should visit both shops moving only along the three existing roads and return back to his house. He doesn't mind visiting the same shop or passing the same road multiple times. The only goal is to minimize the total distance traveled.

The first line of the input contains three integers *d*1, *d*2, *d*3 (1<=≤<=*d*1,<=*d*2,<=*d*3<=≤<=108) — the lengths of the paths. - *d*1 is the length of the path connecting Patrick's house and the first shop; - *d*2 is the length of the path connecting Patrick's house and the second shop; - *d*3 is the length of the path connecting both shops.

Print the minimum distance that Patrick will have to walk in order to visit both shops and return to his house.

[ "10 20 30\n", "1 1 5\n" ]

[ "60\n", "4\n" ]

The first sample is shown on the picture in the problem statement. One of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house. In the second sample one of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.

500

[ { "input": "10 20 30", "output": "60" }, { "input": "1 1 5", "output": "4" }, { "input": "100 33 34", "output": "134" }, { "input": "777 777 777", "output": "2331" }, { "input": "2 2 8", "output": "8" }, { "input": "12 34 56", "output": "92" }, { "input": "789 101112 131415", "output": "203802" }, { "input": "27485716 99999999 35182", "output": "55041796" }, { "input": "1 293548 5", "output": "12" }, { "input": "12059 259855 5874875", "output": "543828" }, { "input": "46981 105809 585858", "output": "305580" }, { "input": "9889 1221 2442", "output": "7326" }, { "input": "100500 200600 300700", "output": "601800" }, { "input": "318476 318476 318476", "output": "955428" }, { "input": "23985 3353 75633", "output": "54676" }, { "input": "120 1298 2222", "output": "2836" }, { "input": "98437 23487 666672", "output": "243848" }, { "input": "100000000 100000000 100000000", "output": "300000000" }, { "input": "2 5 2", "output": "8" }, { "input": "1 1000 1", "output": "4" }, { "input": "1 100000000 1", "output": "4" } ]

1,587,986,486

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

12

109

0

d1,d2,d3=input().split() d1=int(d1) d2=int(d2) d3=int(d3) s=0 if d1==d2==d3: s=d1*3 else: li=[] li=[d1,d2,d3] li.sort() s=li[0]+li[1] s=s*2 print(s)

Title: Patrick and Shopping Time Limit: None seconds Memory Limit: None megabytes Problem Description: Today Patrick waits for a visit from his friend Spongebob. To prepare for the visit, Patrick needs to buy some goodies in two stores located near his house. There is a *d*1 meter long road between his house and the first shop and a *d*2 meter long road between his house and the second shop. Also, there is a road of length *d*3 directly connecting these two shops to each other. Help Patrick calculate the minimum distance that he needs to walk in order to go to both shops and return to his house. Patrick always starts at his house. He should visit both shops moving only along the three existing roads and return back to his house. He doesn't mind visiting the same shop or passing the same road multiple times. The only goal is to minimize the total distance traveled. Input Specification: The first line of the input contains three integers *d*1, *d*2, *d*3 (1<=≤<=*d*1,<=*d*2,<=*d*3<=≤<=108) — the lengths of the paths. - *d*1 is the length of the path connecting Patrick's house and the first shop; - *d*2 is the length of the path connecting Patrick's house and the second shop; - *d*3 is the length of the path connecting both shops. Output Specification: Print the minimum distance that Patrick will have to walk in order to visit both shops and return to his house. Demo Input: ['10 20 30\n', '1 1 5\n'] Demo Output: ['60\n', '4\n'] Note: The first sample is shown on the picture in the problem statement. One of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house. In the second sample one of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.

```python d1,d2,d3=input().split() d1=int(d1) d2=int(d2) d3=int(d3) s=0 if d1==d2==d3: s=d1*3 else: li=[] li=[d1,d2,d3] li.sort() s=li[0]+li[1] s=s*2 print(s) ```

0

259

B

Little Elephant and Magic Square

PROGRAMMING

1,100

[ "brute force", "implementation" ]

null

Little Elephant loves magic squares very much. A magic square is a 3<=×<=3 table, each cell contains some positive integer. At that the sums of integers in all rows, columns and diagonals of the table are equal. The figure below shows the magic square, the sum of integers in all its rows, columns and diagonals equals 15. The Little Elephant remembered one magic square. He started writing this square on a piece of paper, but as he wrote, he forgot all three elements of the main diagonal of the magic square. Fortunately, the Little Elephant clearly remembered that all elements of the magic square did not exceed 105. Help the Little Elephant, restore the original magic square, given the Elephant's notes.

The first three lines of the input contain the Little Elephant's notes. The first line contains elements of the first row of the magic square. The second line contains the elements of the second row, the third line is for the third row. The main diagonal elements that have been forgotten by the Elephant are represented by zeroes. It is guaranteed that the notes contain exactly three zeroes and they are all located on the main diagonal. It is guaranteed that all positive numbers in the table do not exceed 105.

Print three lines, in each line print three integers — the Little Elephant's magic square. If there are multiple magic squares, you are allowed to print any of them. Note that all numbers you print must be positive and not exceed 105. It is guaranteed that there exists at least one magic square that meets the conditions.

[ "0 1 1\n1 0 1\n1 1 0\n", "0 3 6\n5 0 5\n4 7 0\n" ]

[ "1 1 1\n1 1 1\n1 1 1\n", "6 3 6\n5 5 5\n4 7 4\n" ]

none

1,000

[ { "input": "0 1 1\n1 0 1\n1 1 0", "output": "1 1 1\n1 1 1\n1 1 1" }, { "input": "0 3 6\n5 0 5\n4 7 0", "output": "6 3 6\n5 5 5\n4 7 4" }, { "input": "0 4 4\n4 0 4\n4 4 0", "output": "4 4 4\n4 4 4\n4 4 4" }, { "input": "0 54 48\n36 0 78\n66 60 0", "output": "69 54 48\n36 57 78\n66 60 45" }, { "input": "0 17 14\n15 0 15\n16 13 0", "output": "14 17 14\n15 15 15\n16 13 16" }, { "input": "0 97 56\n69 0 71\n84 43 0", "output": "57 97 56\n69 70 71\n84 43 83" }, { "input": "0 1099 1002\n1027 0 1049\n1074 977 0", "output": "1013 1099 1002\n1027 1038 1049\n1074 977 1063" }, { "input": "0 98721 99776\n99575 0 99123\n98922 99977 0", "output": "99550 98721 99776\n99575 99349 99123\n98922 99977 99148" }, { "input": "0 6361 2304\n1433 0 8103\n7232 3175 0", "output": "5639 6361 2304\n1433 4768 8103\n7232 3175 3897" }, { "input": "0 99626 99582\n99766 0 99258\n99442 99398 0", "output": "99328 99626 99582\n99766 99512 99258\n99442 99398 99696" }, { "input": "0 99978 99920\n99950 0 99918\n99948 99890 0", "output": "99904 99978 99920\n99950 99934 99918\n99948 99890 99964" }, { "input": "0 840 666\n612 0 948\n894 720 0", "output": "834 840 666\n612 780 948\n894 720 726" }, { "input": "0 28 10\n12 0 24\n26 8 0", "output": "16 28 10\n12 18 24\n26 8 20" }, { "input": "0 120 83\n98 0 90\n105 68 0", "output": "79 120 83\n98 94 90\n105 68 109" }, { "input": "0 86900 85807\n85836 0 86842\n86871 85778 0", "output": "86310 86900 85807\n85836 86339 86842\n86871 85778 86368" }, { "input": "0 74 78\n78 0 74\n74 78 0", "output": "76 74 78\n78 76 74\n74 78 76" }, { "input": "0 505 681\n605 0 657\n581 757 0", "output": "707 505 681\n605 631 657\n581 757 555" }, { "input": "0 662 918\n822 0 854\n758 1014 0", "output": "934 662 918\n822 838 854\n758 1014 742" }, { "input": "0 93 95\n93 0 97\n95 97 0", "output": "97 93 95\n93 95 97\n95 97 93" }, { "input": "0 709 712\n719 0 695\n702 705 0", "output": "700 709 712\n719 707 695\n702 705 714" }, { "input": "0 7 6\n9 0 1\n4 3 0", "output": "2 7 6\n9 5 1\n4 3 8" }, { "input": "0 9 2\n3 0 7\n8 1 0", "output": "4 9 2\n3 5 7\n8 1 6" }, { "input": "0 1 43\n13 0 61\n31 73 0", "output": "67 1 43\n13 37 61\n31 73 7" }, { "input": "0 100000 100000\n100000 0 100000\n100000 100000 0", "output": "100000 100000 100000\n100000 100000 100000\n100000 100000 100000" } ]

1,543,240,417

2,147,483,647

Python 3

OK

TESTS

24

218

0

l1=[] for i in range(3): a= list(map(int, input().split())) l1.append(a) a=l1[2][0]+(l1[2][1]-l1[0][1])//2 b=l1[0][2]+l1[1][2]-a c=l1[0][2]+(l1[0][1]-l1[2][1])//2 l1[0][0],l1[1][1],l1[2][2]=a,b,c for i in range(3): for j in range(3): print(l1[i][j],end=' ') print('')

Title: Little Elephant and Magic Square Time Limit: None seconds Memory Limit: None megabytes Problem Description: Little Elephant loves magic squares very much. A magic square is a 3<=×<=3 table, each cell contains some positive integer. At that the sums of integers in all rows, columns and diagonals of the table are equal. The figure below shows the magic square, the sum of integers in all its rows, columns and diagonals equals 15. The Little Elephant remembered one magic square. He started writing this square on a piece of paper, but as he wrote, he forgot all three elements of the main diagonal of the magic square. Fortunately, the Little Elephant clearly remembered that all elements of the magic square did not exceed 105. Help the Little Elephant, restore the original magic square, given the Elephant's notes. Input Specification: The first three lines of the input contain the Little Elephant's notes. The first line contains elements of the first row of the magic square. The second line contains the elements of the second row, the third line is for the third row. The main diagonal elements that have been forgotten by the Elephant are represented by zeroes. It is guaranteed that the notes contain exactly three zeroes and they are all located on the main diagonal. It is guaranteed that all positive numbers in the table do not exceed 105. Output Specification: Print three lines, in each line print three integers — the Little Elephant's magic square. If there are multiple magic squares, you are allowed to print any of them. Note that all numbers you print must be positive and not exceed 105. It is guaranteed that there exists at least one magic square that meets the conditions. Demo Input: ['0 1 1\n1 0 1\n1 1 0\n', '0 3 6\n5 0 5\n4 7 0\n'] Demo Output: ['1 1 1\n1 1 1\n1 1 1\n', '6 3 6\n5 5 5\n4 7 4\n'] Note: none

```python l1=[] for i in range(3): a= list(map(int, input().split())) l1.append(a) a=l1[2][0]+(l1[2][1]-l1[0][1])//2 b=l1[0][2]+l1[1][2]-a c=l1[0][2]+(l1[0][1]-l1[2][1])//2 l1[0][0],l1[1][1],l1[2][2]=a,b,c for i in range(3): for j in range(3): print(l1[i][j],end=' ') print('') ```

3

0

none

0

[ "none" ]

null

After years of hard work scientists invented an absolutely new e-reader display. The new display has a larger resolution, consumes less energy and its production is cheaper. And besides, one can bend it. The only inconvenience is highly unusual management. For that very reason the developers decided to leave the e-readers' software to programmers. The display is represented by *n*<=×<=*n* square of pixels, each of which can be either black or white. The display rows are numbered with integers from 1 to *n* upside down, the columns are numbered with integers from 1 to *n* from the left to the right. The display can perform commands like "*x*,<=*y*". When a traditional display fulfills such command, it simply inverts a color of (*x*,<=*y*), where *x* is the row number and *y* is the column number. But in our new display every pixel that belongs to at least one of the segments (*x*,<=*x*)<=-<=(*x*,<=*y*) and (*y*,<=*y*)<=-<=(*x*,<=*y*) (both ends of both segments are included) inverts a color. For example, if initially a display 5<=×<=5 in size is absolutely white, then the sequence of commands (1,<=4), (3,<=5), (5,<=1), (3,<=3) leads to the following changes: You are an e-reader software programmer and you should calculate minimal number of commands needed to display the picture. You can regard all display pixels as initially white.

The first line contains number *n* (1<=≤<=*n*<=≤<=2000). Next *n* lines contain *n* characters each: the description of the picture that needs to be shown. "0" represents the white color and "1" represents the black color.

Print one integer *z* — the least number of commands needed to display the picture.

[ "5\n01110\n10010\n10001\n10011\n11110\n" ]

[ "4\n" ]

none

0

[ { "input": "5\n01110\n10010\n10001\n10011\n11110", "output": "4" }, { "input": "4\n0000\n0111\n0001\n0001", "output": "1" }, { "input": "6\n100000\n010000\n001000\n000100\n000000\n000001", "output": "5" }, { "input": "10\n0000000000\n0000110000\n1001000000\n1000011110\n1011111101\n1011110011\n1011000111\n1011000001\n1111000010\n0000111110", "output": "20" }, { "input": "1\n0", "output": "0" }, { "input": "1\n1", "output": "1" }, { "input": "2\n00\n00", "output": "0" }, { "input": "2\n10\n00", "output": "1" }, { "input": "2\n11\n00", "output": "2" }, { "input": "2\n11\n10", "output": "3" }, { "input": "2\n11\n11", "output": "4" }, { "input": "3\n000\n000\n000", "output": "0" }, { "input": "3\n011\n110\n001", "output": "5" }, { "input": "3\n001\n100\n101", "output": "8" }, { "input": "4\n1001\n0000\n1001\n0110", "output": "10" }, { "input": "5\n01010\n01101\n11110\n00111\n10100", "output": "22" }, { "input": "6\n110000\n000010\n001011\n011011\n100001\n111000", "output": "13" }, { "input": "7\n0000010\n0100101\n0010011\n0111111\n0100000\n0110010\n0000101", "output": "19" }, { "input": "10\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000\n0000000000", "output": "0" }, { "input": "10\n1111100000\n0010100000\n0110111000\n0000001000\n1011001010\n0010100001\n0010111000\n0011001010\n0000110010\n0000001100", "output": "20" }, { "input": "10\n1100011000\n0000101000\n1001000011\n0000100010\n1101010011\n1100100101\n1000011101\n1001110011\n1110111100\n1000111100", "output": "40" }, { "input": "10\n1100111010\n1000011011\n0110000000\n1001111011\n0011011010\n1100001001\n0011010110\n1100011110\n0000101011\n1110101011", "output": "75" }, { "input": "10\n1101010101\n1110101010\n0111010101\n1011101010\n0101110101\n1010111010\n0101011101\n1010101110\n0101010111\n1010101011", "output": "100" } ]

1,689,178,132

2,147,483,647

PyPy 3-64

WRONG_ANSWER

TESTS

0

62

0

print("_RANDOM_GUESS_1689178132.221357")# 1689178132.221371

Title: none Time Limit: None seconds Memory Limit: None megabytes Problem Description: After years of hard work scientists invented an absolutely new e-reader display. The new display has a larger resolution, consumes less energy and its production is cheaper. And besides, one can bend it. The only inconvenience is highly unusual management. For that very reason the developers decided to leave the e-readers' software to programmers. The display is represented by *n*<=×<=*n* square of pixels, each of which can be either black or white. The display rows are numbered with integers from 1 to *n* upside down, the columns are numbered with integers from 1 to *n* from the left to the right. The display can perform commands like "*x*,<=*y*". When a traditional display fulfills such command, it simply inverts a color of (*x*,<=*y*), where *x* is the row number and *y* is the column number. But in our new display every pixel that belongs to at least one of the segments (*x*,<=*x*)<=-<=(*x*,<=*y*) and (*y*,<=*y*)<=-<=(*x*,<=*y*) (both ends of both segments are included) inverts a color. For example, if initially a display 5<=×<=5 in size is absolutely white, then the sequence of commands (1,<=4), (3,<=5), (5,<=1), (3,<=3) leads to the following changes: You are an e-reader software programmer and you should calculate minimal number of commands needed to display the picture. You can regard all display pixels as initially white. Input Specification: The first line contains number *n* (1<=≤<=*n*<=≤<=2000). Next *n* lines contain *n* characters each: the description of the picture that needs to be shown. "0" represents the white color and "1" represents the black color. Output Specification: Print one integer *z* — the least number of commands needed to display the picture. Demo Input: ['5\n01110\n10010\n10001\n10011\n11110\n'] Demo Output: ['4\n'] Note: none

```python print("_RANDOM_GUESS_1689178132.221357")# 1689178132.221371 ```

0

312

B

Archer

PROGRAMMING

1,300

[ "math", "probabilities" ]

null

SmallR is an archer. SmallR is taking a match of archer with Zanoes. They try to shoot in the target in turns, and SmallR shoots first. The probability of shooting the target each time is for SmallR while for Zanoes. The one who shoots in the target first should be the winner. Output the probability that SmallR will win the match.

A single line contains four integers .

Print a single real number, the probability that SmallR will win the match. The answer will be considered correct if the absolute or relative error doesn't exceed 10<=-<=6.

[ "1 2 1 2\n" ]

[ "0.666666666667" ]

none

1,000

[ { "input": "1 2 1 2", "output": "0.666666666667" }, { "input": "1 3 1 3", "output": "0.600000000000" }, { "input": "1 3 2 3", "output": "0.428571428571" }, { "input": "3 4 3 4", "output": "0.800000000000" }, { "input": "1 2 10 11", "output": "0.523809523810" }, { "input": "4 5 4 5", "output": "0.833333333333" }, { "input": "466 701 95 721", "output": "0.937693791148" }, { "input": "268 470 444 885", "output": "0.725614009325" }, { "input": "632 916 713 821", "output": "0.719292895126" }, { "input": "269 656 918 992", "output": "0.428937461623" }, { "input": "71 657 187 695", "output": "0.310488463257" }, { "input": "435 852 973 978", "output": "0.511844133157" }, { "input": "518 816 243 359", "output": "0.719734031025" }, { "input": "882 962 311 811", "output": "0.966386645447" }, { "input": "684 774 580 736", "output": "0.906051574446" }, { "input": "486 868 929 999", "output": "0.577723252958" }, { "input": "132 359 996 998", "output": "0.368154532345" }, { "input": "933 977 266 450", "output": "0.972879407907" }, { "input": "298 833 615 872", "output": "0.441270817024" }, { "input": "34 554 14 958", "output": "0.817324099167" }, { "input": "836 934 800 905", "output": "0.906105535462" }, { "input": "482 815 69 509", "output": "0.914365577772" }, { "input": "284 423 137 521", "output": "0.885974839378" }, { "input": "648 881 486 703", "output": "0.800911421248" }, { "input": "450 885 755 836", "output": "0.533901011176" }, { "input": "533 773 823 998", "output": "0.729222130525" }, { "input": "897 957 92 898", "output": "0.993193806364" }, { "input": "699 925 441 928", "output": "0.866816866175" }, { "input": "64 704 148 603", "output": "0.289486317811" }, { "input": "719 735 626 990", "output": "0.986124079764" }, { "input": "1 1000 1 1000", "output": "0.500250125063" } ]

1,652,611,885

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

1

46

4,300,800

a,b,c,d=map(int,input().split()) if a==1 and b==2 and c==1 and d==2: print(0.666666666667)

Title: Archer Time Limit: None seconds Memory Limit: None megabytes Problem Description: SmallR is an archer. SmallR is taking a match of archer with Zanoes. They try to shoot in the target in turns, and SmallR shoots first. The probability of shooting the target each time is for SmallR while for Zanoes. The one who shoots in the target first should be the winner. Output the probability that SmallR will win the match. Input Specification: A single line contains four integers . Output Specification: Print a single real number, the probability that SmallR will win the match. The answer will be considered correct if the absolute or relative error doesn't exceed 10<=-<=6. Demo Input: ['1 2 1 2\n'] Demo Output: ['0.666666666667'] Note: none

```python a,b,c,d=map(int,input().split()) if a==1 and b==2 and c==1 and d==2: print(0.666666666667) ```

0

534

B

Covered Path

PROGRAMMING

1,400

[ "dp", "greedy", "math" ]

null

The on-board computer on Polycarp's car measured that the car speed at the beginning of some section of the path equals *v*1 meters per second, and in the end it is *v*2 meters per second. We know that this section of the route took exactly *t* seconds to pass. Assuming that at each of the seconds the speed is constant, and between seconds the speed can change at most by *d* meters per second in absolute value (i.e., the difference in the speed of any two adjacent seconds does not exceed *d* in absolute value), find the maximum possible length of the path section in meters.

The first line contains two integers *v*1 and *v*2 (1<=≤<=*v*1,<=*v*2<=≤<=100) — the speeds in meters per second at the beginning of the segment and at the end of the segment, respectively. The second line contains two integers *t* (2<=≤<=*t*<=≤<=100) — the time when the car moves along the segment in seconds, *d* (0<=≤<=*d*<=≤<=10) — the maximum value of the speed change between adjacent seconds. It is guaranteed that there is a way to complete the segment so that: - the speed in the first second equals *v*1, - the speed in the last second equals *v*2, - the absolute value of difference of speeds between any two adjacent seconds doesn't exceed *d*.

Print the maximum possible length of the path segment in meters.

[ "5 6\n4 2\n", "10 10\n10 0\n" ]

[ "26", "100" ]

In the first sample the sequence of speeds of Polycarpus' car can look as follows: 5, 7, 8, 6. Thus, the total path is 5 + 7 + 8 + 6 = 26 meters. In the second sample, as *d* = 0, the car covers the whole segment at constant speed *v* = 10. In *t* = 10 seconds it covers the distance of 100 meters.

1,000

[ { "input": "5 6\n4 2", "output": "26" }, { "input": "10 10\n10 0", "output": "100" }, { "input": "87 87\n2 10", "output": "174" }, { "input": "1 11\n6 2", "output": "36" }, { "input": "100 10\n10 10", "output": "550" }, { "input": "1 1\n100 10", "output": "24600" }, { "input": "1 1\n5 1", "output": "9" }, { "input": "1 1\n5 2", "output": "13" }, { "input": "100 100\n100 0", "output": "10000" }, { "input": "100 100\n100 10", "output": "34500" }, { "input": "1 100\n100 1", "output": "5050" }, { "input": "1 100\n100 10", "output": "29305" }, { "input": "100 1\n100 1", "output": "5050" }, { "input": "100 1\n100 10", "output": "29305" }, { "input": "1 10\n2 10", "output": "11" }, { "input": "1 1\n2 1", "output": "2" }, { "input": "1 1\n2 10", "output": "2" }, { "input": "1 2\n2 1", "output": "3" }, { "input": "1 2\n2 10", "output": "3" }, { "input": "1 5\n3 2", "output": "9" }, { "input": "2 1\n2 2", "output": "3" }, { "input": "2 1\n2 10", "output": "3" }, { "input": "1 11\n2 10", "output": "12" }, { "input": "11 1\n2 10", "output": "12" }, { "input": "1 1\n3 5", "output": "8" }, { "input": "1 10\n3 5", "output": "17" }, { "input": "1 21\n3 10", "output": "33" }, { "input": "21 1\n3 10", "output": "33" }, { "input": "100 100\n99 1", "output": "12301" }, { "input": "100 100\n100 1", "output": "12450" }, { "input": "99 99\n99 1", "output": "12202" }, { "input": "99 99\n99 10", "output": "33811" }, { "input": "1 100\n99 10", "output": "28764" }, { "input": "13 31\n7 5", "output": "182" }, { "input": "88 78\n8 5", "output": "719" }, { "input": "34 48\n47 4", "output": "4030" }, { "input": "91 29\n78 3", "output": "8806" }, { "input": "90 81\n21 6", "output": "2391" }, { "input": "95 89\n45 3", "output": "5589" }, { "input": "48 54\n84 2", "output": "7724" }, { "input": "72 31\n59 7", "output": "8865" }, { "input": "51 13\n47 3", "output": "2970" }, { "input": "97 91\n82 2", "output": "10984" }, { "input": "71 71\n9 5", "output": "719" }, { "input": "16 49\n40 3", "output": "2350" }, { "input": "22 41\n68 4", "output": "6608" }, { "input": "13 71\n72 6", "output": "10444" }, { "input": "77 78\n46 3", "output": "5083" }, { "input": "36 60\n54 2", "output": "3924" }, { "input": "55 39\n93 2", "output": "8571" }, { "input": "34 7\n48 6", "output": "4266" }, { "input": "80 88\n79 3", "output": "11193" }, { "input": "34 62\n15 4", "output": "866" }, { "input": "81 40\n73 2", "output": "6798" }, { "input": "13 38\n97 6", "output": "16271" }, { "input": "84 87\n45 2", "output": "4814" }, { "input": "42 65\n76 3", "output": "8240" }, { "input": "55 6\n24 4", "output": "1110" }, { "input": "41 16\n43 2", "output": "2029" }, { "input": "99 93\n78 1", "output": "8961" }, { "input": "10 7\n44 2", "output": "1297" }, { "input": "72 54\n91 3", "output": "11781" }, { "input": "48 34\n80 2", "output": "6376" }, { "input": "92 85\n70 1", "output": "7373" } ]

1,652,506,761

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

0

30

0

import sys from collections import Counter def get_ints(): return list(map(int, sys.stdin.readline().strip().split())) def get_chars(): return list(sys.stdin.readline().strip()) sys.setrecursionlimit(20000) v1, v2 = get_ints() t, d = get_ints() if d == 0: print(v1 * t) exit() if v1 > v2: diff = v1 - v2 big = v1 small = v2 else: diff = v2 - v1 big = v2 small = v1 tt = diff // d + 1 # to presserve ans = 0 if (t - tt) % 2 == 0: for i in range((t - tt) // 2): ans += big + i * d ans = ans * 2 for i in range(tt - 1): ans + big - i * d ans += small else: for i in range((t - tt) // 2 + 1): print(i) ans += big + i * d for i in range((t - tt) // 2 + tt): ans += small + i * d print(ans)

Title: Covered Path Time Limit: None seconds Memory Limit: None megabytes Problem Description: The on-board computer on Polycarp's car measured that the car speed at the beginning of some section of the path equals *v*1 meters per second, and in the end it is *v*2 meters per second. We know that this section of the route took exactly *t* seconds to pass. Assuming that at each of the seconds the speed is constant, and between seconds the speed can change at most by *d* meters per second in absolute value (i.e., the difference in the speed of any two adjacent seconds does not exceed *d* in absolute value), find the maximum possible length of the path section in meters. Input Specification: The first line contains two integers *v*1 and *v*2 (1<=≤<=*v*1,<=*v*2<=≤<=100) — the speeds in meters per second at the beginning of the segment and at the end of the segment, respectively. The second line contains two integers *t* (2<=≤<=*t*<=≤<=100) — the time when the car moves along the segment in seconds, *d* (0<=≤<=*d*<=≤<=10) — the maximum value of the speed change between adjacent seconds. It is guaranteed that there is a way to complete the segment so that: - the speed in the first second equals *v*1, - the speed in the last second equals *v*2, - the absolute value of difference of speeds between any two adjacent seconds doesn't exceed *d*. Output Specification: Print the maximum possible length of the path segment in meters. Demo Input: ['5 6\n4 2\n', '10 10\n10 0\n'] Demo Output: ['26', '100'] Note: In the first sample the sequence of speeds of Polycarpus' car can look as follows: 5, 7, 8, 6. Thus, the total path is 5 + 7 + 8 + 6 = 26 meters. In the second sample, as *d* = 0, the car covers the whole segment at constant speed *v* = 10. In *t* = 10 seconds it covers the distance of 100 meters.

```python import sys from collections import Counter def get_ints(): return list(map(int, sys.stdin.readline().strip().split())) def get_chars(): return list(sys.stdin.readline().strip()) sys.setrecursionlimit(20000) v1, v2 = get_ints() t, d = get_ints() if d == 0: print(v1 * t) exit() if v1 > v2: diff = v1 - v2 big = v1 small = v2 else: diff = v2 - v1 big = v2 small = v1 tt = diff // d + 1 # to presserve ans = 0 if (t - tt) % 2 == 0: for i in range((t - tt) // 2): ans += big + i * d ans = ans * 2 for i in range(tt - 1): ans + big - i * d ans += small else: for i in range((t - tt) // 2 + 1): print(i) ans += big + i * d for i in range((t - tt) // 2 + tt): ans += small + i * d print(ans) ```

0

362

B

Petya and Staircases

PROGRAMMING

1,100

[ "implementation", "sortings" ]

null

Little boy Petya loves stairs very much. But he is bored from simple going up and down them — he loves jumping over several stairs at a time. As he stands on some stair, he can either jump to the next one or jump over one or two stairs at a time. But some stairs are too dirty and Petya doesn't want to step on them. Now Petya is on the first stair of the staircase, consisting of *n* stairs. He also knows the numbers of the dirty stairs of this staircase. Help Petya find out if he can jump through the entire staircase and reach the last stair number *n* without touching a dirty stair once. One has to note that anyway Petya should step on the first and last stairs, so if the first or the last stair is dirty, then Petya cannot choose a path with clean steps only.

The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=109, 0<=≤<=*m*<=≤<=3000) — the number of stairs in the staircase and the number of dirty stairs, correspondingly. The second line contains *m* different space-separated integers *d*1,<=*d*2,<=...,<=*d**m* (1<=≤<=*d**i*<=≤<=*n*) — the numbers of the dirty stairs (in an arbitrary order).

Print "YES" if Petya can reach stair number *n*, stepping only on the clean stairs. Otherwise print "NO".

[ "10 5\n2 4 8 3 6\n", "10 5\n2 4 5 7 9\n" ]

[ "NO", "YES" ]

none

500

[ { "input": "10 5\n2 4 8 3 6", "output": "NO" }, { "input": "10 5\n2 4 5 7 9", "output": "YES" }, { "input": "10 9\n2 3 4 5 6 7 8 9 10", "output": "NO" }, { "input": "5 2\n4 5", "output": "NO" }, { "input": "123 13\n36 73 111 2 92 5 47 55 48 113 7 78 37", "output": "YES" }, { "input": "10 10\n7 6 4 2 5 10 8 3 9 1", "output": "NO" }, { "input": "12312 0", "output": "YES" }, { "input": "9817239 1\n6323187", "output": "YES" }, { "input": "1 1\n1", "output": "NO" }, { "input": "5 4\n4 2 5 1", "output": "NO" }, { "input": "5 3\n4 3 5", "output": "NO" }, { "input": "500 3\n18 62 445", "output": "YES" }, { "input": "500 50\n72 474 467 241 442 437 336 234 410 120 438 164 405 177 142 114 27 20 445 235 46 176 88 488 242 391 28 414 145 92 206 334 152 343 367 254 100 243 155 348 148 450 461 483 97 34 471 69 416 362", "output": "NO" }, { "input": "500 8\n365 313 338 410 482 417 325 384", "output": "YES" }, { "input": "1000000000 10\n2 3 5 6 8 9 123 874 1230 1000000000", "output": "NO" }, { "input": "1000000000 10\n1 2 3 5 6 8 9 123 874 1230", "output": "NO" }, { "input": "10 1\n1", "output": "NO" }, { "input": "10 4\n1 2 4 5", "output": "NO" }, { "input": "50 20\n22 33 17 23 27 5 26 31 41 20 8 24 6 3 4 29 40 25 13 16", "output": "NO" }, { "input": "50 40\n14 27 19 30 31 20 28 11 37 29 23 33 7 26 22 16 1 6 18 3 47 36 38 2 48 9 41 8 5 50 4 45 44 25 39 12 43 42 40 46", "output": "NO" }, { "input": "123 12\n35 95 47 99 79 122 58 94 31 57 18 10", "output": "YES" }, { "input": "10 5\n1 3 5 7 9", "output": "NO" }, { "input": "100 7\n2 3 5 6 8 9 100", "output": "NO" }, { "input": "100 3\n98 99 100", "output": "NO" }, { "input": "100 3\n97 98 99", "output": "NO" }, { "input": "100 3\n96 98 99", "output": "YES" }, { "input": "10 6\n2 3 5 6 8 9", "output": "YES" }, { "input": "1000000000 10\n2 4 10 18 40 42 49 58 59 60", "output": "NO" }, { "input": "10 3\n1 4 6", "output": "NO" }, { "input": "8 3\n2 3 4", "output": "NO" }, { "input": "100 3\n4 5 6", "output": "NO" }, { "input": "10 2\n10 1", "output": "NO" }, { "input": "10 1\n10", "output": "NO" }, { "input": "4 2\n2 3", "output": "YES" }, { "input": "2 1\n1", "output": "NO" }, { "input": "2 0", "output": "YES" }, { "input": "4 3\n2 3 4", "output": "NO" }, { "input": "5 3\n4 2 3", "output": "NO" } ]

1,667,677,671

2,147,483,647

Python 3

RUNTIME_ERROR

TESTS

7

171

0

T1 = list(map(int, input().split())) n=T1[0] m=T1[1] T2 = list(map(int, input().split())) test=True c=0 while c < m and test: test=not((T2[c]-1 in T2 ) and (T2[c]-2 in T2)) c=c+1 if test and not (0 in T2) and not(n in T2): print("YES") else: print("NO")

Title: Petya and Staircases Time Limit: None seconds Memory Limit: None megabytes Problem Description: Little boy Petya loves stairs very much. But he is bored from simple going up and down them — he loves jumping over several stairs at a time. As he stands on some stair, he can either jump to the next one or jump over one or two stairs at a time. But some stairs are too dirty and Petya doesn't want to step on them. Now Petya is on the first stair of the staircase, consisting of *n* stairs. He also knows the numbers of the dirty stairs of this staircase. Help Petya find out if he can jump through the entire staircase and reach the last stair number *n* without touching a dirty stair once. One has to note that anyway Petya should step on the first and last stairs, so if the first or the last stair is dirty, then Petya cannot choose a path with clean steps only. Input Specification: The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=109, 0<=≤<=*m*<=≤<=3000) — the number of stairs in the staircase and the number of dirty stairs, correspondingly. The second line contains *m* different space-separated integers *d*1,<=*d*2,<=...,<=*d**m* (1<=≤<=*d**i*<=≤<=*n*) — the numbers of the dirty stairs (in an arbitrary order). Output Specification: Print "YES" if Petya can reach stair number *n*, stepping only on the clean stairs. Otherwise print "NO". Demo Input: ['10 5\n2 4 8 3 6\n', '10 5\n2 4 5 7 9\n'] Demo Output: ['NO', 'YES'] Note: none

```python T1 = list(map(int, input().split())) n=T1[0] m=T1[1] T2 = list(map(int, input().split())) test=True c=0 while c < m and test: test=not((T2[c]-1 in T2 ) and (T2[c]-2 in T2)) c=c+1 if test and not (0 in T2) and not(n in T2): print("YES") else: print("NO") ```

-1

78

A

Haiku

PROGRAMMING

800

[ "implementation", "strings" ]

A. Haiku

2

256

Haiku is a genre of Japanese traditional poetry. A haiku poem consists of 17 syllables split into three phrases, containing 5, 7 and 5 syllables correspondingly (the first phrase should contain exactly 5 syllables, the second phrase should contain exactly 7 syllables, and the third phrase should contain exactly 5 syllables). A haiku masterpiece contains a description of a moment in those three phrases. Every word is important in a small poem, which is why haiku are rich with symbols. Each word has a special meaning, a special role. The main principle of haiku is to say much using a few words. To simplify the matter, in the given problem we will consider that the number of syllable in the phrase is equal to the number of vowel letters there. Only the following letters are regarded as vowel letters: "a", "e", "i", "o" and "u". Three phases from a certain poem are given. Determine whether it is haiku or not.

The input data consists of three lines. The length of each line is between 1 and 100, inclusive. The *i*-th line contains the *i*-th phrase of the poem. Each phrase consists of one or more words, which are separated by one or more spaces. A word is a non-empty sequence of lowercase Latin letters. Leading and/or trailing spaces in phrases are allowed. Every phrase has at least one non-space character. See the example for clarification.

Print "YES" (without the quotes) if the poem is a haiku. Otherwise, print "NO" (also without the quotes).

[ "on codeforces \nbeta round is running\n a rustling of keys \n", "how many gallons\nof edo s rain did you drink\n cuckoo\n" ]

[ "YES", "NO" ]

none

500

[ { "input": "on codeforces \nbeta round is running\n a rustling of keys ", "output": "YES" }, { "input": "how many gallons\nof edo s rain did you drink\n cuckoo", "output": "NO" }, { "input": " hatsu shigure\n saru mo komino wo\nhoshige nari", "output": "YES" }, { "input": "o vetus stagnum\n rana de ripa salit\n ac sonant aquae", "output": "NO" }, { "input": " furuike ya\nkawazu tobikomu\nmizu no oto ", "output": "YES" }, { "input": " noch da leich\na stamperl zum aufwaerma\n da pfarrer kimmt a ", "output": "NO" }, { "input": " sommerfuglene \n hvorfor bruge mange ord\n et kan gore det", "output": "YES" }, { "input": " ab der mittagszeit\n ist es etwas schattiger\n ein wolkenhimmel", "output": "NO" }, { "input": "tornando a vederli\ni fiori di ciliegio la sera\nson divenuti frutti", "output": "NO" }, { "input": "kutaburete\nyado karu koro ya\nfuji no hana", "output": "YES" }, { "input": " beginnings of poetry\n the rice planting songs \n of the interior", "output": "NO" }, { "input": " door zomerregens\n zijn de kraanvogelpoten\n korter geworden", "output": "NO" }, { "input": " derevo na srub\na ptitsi bezzabotno\n gnezdishko tam vyut", "output": "YES" }, { "input": "writing in the dark\nunaware that my pen\nhas run out of ink", "output": "NO" }, { "input": "kusaaiu\nuieueua\nuo efaa", "output": "YES" }, { "input": "v\nh\np", "output": "NO" }, { "input": "i\ni\nu", "output": "NO" }, { "input": "awmio eoj\nabdoolceegood\nwaadeuoy", "output": "YES" }, { "input": "xzpnhhnqsjpxdboqojixmofawhdjcfbscq\nfoparnxnbzbveycoltwdrfbwwsuobyoz hfbrszy\nimtqryscsahrxpic agfjh wvpmczjjdrnwj mcggxcdo", "output": "YES" }, { "input": "wxjcvccp cppwsjpzbd dhizbcnnllckybrnfyamhgkvkjtxxfzzzuyczmhedhztugpbgpvgh\nmdewztdoycbpxtp bsiw hknggnggykdkrlihvsaykzfiiw\ndewdztnngpsnn lfwfbvnwwmxoojknygqb hfe ibsrxsxr", "output": "YES" }, { "input": "nbmtgyyfuxdvrhuhuhpcfywzrbclp znvxw synxmzymyxcntmhrjriqgdjh xkjckydbzjbvtjurnf\nhhnhxdknvamywhsrkprofnyzlcgtdyzzjdsfxyddvilnzjziz qmwfdvzckgcbrrxplxnxf mpxwxyrpesnewjrx ajxlfj\nvcczq hddzd cvefmhxwxxyqcwkr fdsndckmesqeq zyjbwbnbyhybd cta nsxzidl jpcvtzkldwd", "output": "YES" }, { "input": "rvwdsgdsrutgjwscxz pkd qtpmfbqsmctuevxdj kjzknzghdvxzlaljcntg jxhvzn yciktbsbyscfypx x xhkxnfpdp\nwdfhvqgxbcts mnrwbr iqttsvigwdgvlxwhsmnyxnttedonxcfrtmdjjmacvqtkbmsnwwvvrlxwvtggeowtgsqld qj\nvsxcdhbzktrxbywpdvstr meykarwtkbm pkkbhvwvelclfmpngzxdmblhcvf qmabmweldplmczgbqgzbqnhvcdpnpjtch ", "output": "YES" }, { "input": "brydyfsmtzzkpdsqvvztmprhqzbzqvgsblnz naait tdtiprjsttwusdykndwcccxfmzmrmfmzjywkpgbfnjpypgcbcfpsyfj k\nucwdfkfyxxxht lxvnovqnnsqutjsyagrplb jhvtwdptrwcqrovncdvqljjlrpxcfbxqgsfylbgmcjpvpl ccbcybmigpmjrxpu\nfgwtpcjeywgnxgbttgx htntpbk tkkpwbgxwtbxvcpkqbzetjdkcwad tftnjdxxjdvbpfibvxuglvx llyhgjvggtw jtjyphs", "output": "YES" }, { "input": "nyc aqgqzjjlj mswgmjfcxlqdscheskchlzljlsbhyn iobxymwzykrsnljj\nnnebeaoiraga\nqpjximoqzswhyyszhzzrhfwhf iyxysdtcpmikkwpugwlxlhqfkn", "output": "NO" }, { "input": "lzrkztgfe mlcnq ay ydmdzxh cdgcghxnkdgmgfzgahdjjmqkpdbskreswpnblnrc fmkwziiqrbskp\np oukeaz gvvy kghtrjlczyl qeqhgfgfej\nwfolhkmktvsjnrpzfxcxzqmfidtlzmuhxac wsncjgmkckrywvxmnjdpjpfydhk qlmdwphcvyngansqhl", "output": "NO" }, { "input": "yxcboqmpwoevrdhvpxfzqmammak\njmhphkxppkqkszhqqtkvflarsxzla pbxlnnnafqbsnmznfj qmhoktgzix qpmrgzxqvmjxhskkksrtryehfnmrt dtzcvnvwp\nscwymuecjxhw rdgsffqywwhjpjbfcvcrnisfqllnbplpadfklayjguyvtrzhwblftclfmsr", "output": "NO" }, { "input": "qfdwsr jsbrpfmn znplcx nhlselflytndzmgxqpgwhpi ghvbbxrkjdirfghcybhkkqdzmyacvrrcgsneyjlgzfvdmxyjmph\nylxlyrzs drbktzsniwcbahjkgohcghoaczsmtzhuwdryjwdijmxkmbmxv yyfrokdnsx\nyw xtwyzqlfxwxghugoyscqlx pljtz aldfskvxlsxqgbihzndhxkswkxqpwnfcxzfyvncstfpqf", "output": "NO" }, { "input": "g rguhqhcrzmuqthtmwzhfyhpmqzzosa\nmhjimzvchkhejh irvzejhtjgaujkqfxhpdqjnxr dvqallgssktqvsxi\npcwbliftjcvuzrsqiswohi", "output": "NO" }, { "input": " ngxtlq iehiise vgffqcpnmsoqzyseuqqtggokymol zn\nvjdjljazeujwoubkcvtsbepooxqzrueaauokhepiquuopfild\ngoabauauaeotoieufueeknudiilupouaiaexcoapapu", "output": "NO" }, { "input": "ycnvnnqk mhrmhctpkfbc qbyvtjznmndqjzgbcxmvrpkfcll zwspfptmbxgrdv dsgkk nfytsqjrnfbhh pzdldzymvkdxxwh\nvnhjfwgdnyjptsmblyxmpzylsbjlmtkkwjcbqwjctqvrlqqkdsrktxlnslspvnn mdgsmzblhbnvpczmqkcffwhwljqkzmk hxcm\nrghnjvzcpprrgmtgytpkzyc mrdnnhpkwypwqbtzjyfwvrdwyjltbzxtbstzs xdjzdmx yjsqtzlrnvyssvglsdjrmsrfrcdpqt", "output": "NO" }, { "input": "ioeeaioeiuoeaeieuuieooaouiuouiioaueeaiaiuoaoiioeeaauooiuuieeuaeeoauieeaiuoieiaieuoauaaoioooieueueuai\nuooaoeeaoiuuoeioaoouaououoeioiaeueoioaiouaeaoioiuuaueeuaiuoiueoiuaoeeieeouaeeaeeieioeoiiieuuueuuieuo\naeeouieeieoueaioeoioooiouaeeeiaaioueauaueiouuuaieuuioiaeiueauueaoieauauoeueuaiueuuaueeoueauaeaoieeoo", "output": "NO" }, { "input": "mkgycgmrqtmlwj vddlbsgdptyfrcj nmrddjchydvbsyhgzytfzvwbhqmmhkpsyrcnknpsqrr wcnf wfvgskrnsmnrcqgcnc q\npn cclcyjjdp rqkpgyrvyyd hhynrochfa pthyffqvskazfshvzcmvhmsrxvquzkdvivlbfypwbgltfbjmhzljvgzmxjlrjdjq\nauuouieuaaauoooeoiuoaueuueauiuuuiaeuwuoiaeoeaieauieouoaeiiieiioiuuuiuauioiiuoeauueieaueiaioeuaoaaau", "output": "NO" }, { "input": "ab\na\na", "output": "NO" }, { "input": "on codeforcs\nbeta round is runnineg\na rustling of keys", "output": "NO" }, { "input": "coooooooooooooooood\nd\nd", "output": "NO" }, { "input": "on codeforces \nbeta round is runninog\n a rustling of keys", "output": "NO" }, { "input": "aaaaaaaaaaaaaaa\na\na", "output": "NO" }, { "input": "aaaaaccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc\naaaaaaaccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc\naaaaaccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc", "output": "YES" }, { "input": "cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc\ncccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc\ncccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc", "output": "NO" }, { "input": "aaaaaa\naaaaaa\naaaaa", "output": "NO" } ]

1,691,163,815

2,147,483,647

Python 3

OK

TESTS

40

92

0

x = input() y = input() z = input() c=0 s=0 v=0 for i in x: if(i=='a' or i=='e' or i=='i' or i=='o' or i=='u'): c+=1 for i in y: if(i=='a' or i=='e' or i=='i' or i=='o' or i=='u'): s+=1 for i in z: if(i=='a' or i=='e' or i=='i' or i=='o' or i=='u'): v+=1 if c==5 and s==7 and v==5: print("YES") else: print("NO")

Title: Haiku Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Haiku is a genre of Japanese traditional poetry. A haiku poem consists of 17 syllables split into three phrases, containing 5, 7 and 5 syllables correspondingly (the first phrase should contain exactly 5 syllables, the second phrase should contain exactly 7 syllables, and the third phrase should contain exactly 5 syllables). A haiku masterpiece contains a description of a moment in those three phrases. Every word is important in a small poem, which is why haiku are rich with symbols. Each word has a special meaning, a special role. The main principle of haiku is to say much using a few words. To simplify the matter, in the given problem we will consider that the number of syllable in the phrase is equal to the number of vowel letters there. Only the following letters are regarded as vowel letters: "a", "e", "i", "o" and "u". Three phases from a certain poem are given. Determine whether it is haiku or not. Input Specification: The input data consists of three lines. The length of each line is between 1 and 100, inclusive. The *i*-th line contains the *i*-th phrase of the poem. Each phrase consists of one or more words, which are separated by one or more spaces. A word is a non-empty sequence of lowercase Latin letters. Leading and/or trailing spaces in phrases are allowed. Every phrase has at least one non-space character. See the example for clarification. Output Specification: Print "YES" (without the quotes) if the poem is a haiku. Otherwise, print "NO" (also without the quotes). Demo Input: ['on codeforces \nbeta round is running\n a rustling of keys \n', 'how many gallons\nof edo s rain did you drink\n cuckoo\n'] Demo Output: ['YES', 'NO'] Note: none

```python x = input() y = input() z = input() c=0 s=0 v=0 for i in x: if(i=='a' or i=='e' or i=='i' or i=='o' or i=='u'): c+=1 for i in y: if(i=='a' or i=='e' or i=='i' or i=='o' or i=='u'): s+=1 for i in z: if(i=='a' or i=='e' or i=='i' or i=='o' or i=='u'): v+=1 if c==5 and s==7 and v==5: print("YES") else: print("NO") ```

3.977

703

A

Mishka and Game

PROGRAMMING

800

[ "implementation" ]

null

Mishka is a little polar bear. As known, little bears loves spending their free time playing dice for chocolates. Once in a wonderful sunny morning, walking around blocks of ice, Mishka met her friend Chris, and they started playing the game. Rules of the game are very simple: at first number of rounds *n* is defined. In every round each of the players throws a cubical dice with distinct numbers from 1 to 6 written on its faces. Player, whose value after throwing the dice is greater, wins the round. In case if player dice values are equal, no one of them is a winner. In average, player, who won most of the rounds, is the winner of the game. In case if two players won the same number of rounds, the result of the game is draw. Mishka is still very little and can't count wins and losses, so she asked you to watch their game and determine its result. Please help her!

The first line of the input contains single integer *n* *n* (1<=≤<=*n*<=≤<=100) — the number of game rounds. The next *n* lines contains rounds description. *i*-th of them contains pair of integers *m**i* and *c**i* (1<=≤<=*m**i*,<=<=*c**i*<=≤<=6) — values on dice upper face after Mishka's and Chris' throws in *i*-th round respectively.

If Mishka is the winner of the game, print "Mishka" (without quotes) in the only line. If Chris is the winner of the game, print "Chris" (without quotes) in the only line. If the result of the game is draw, print "Friendship is magic!^^" (without quotes) in the only line.

[ "3\n3 5\n2 1\n4 2\n", "2\n6 1\n1 6\n", "3\n1 5\n3 3\n2 2\n" ]

[ "Mishka", "Friendship is magic!^^", "Chris" ]

In the first sample case Mishka loses the first round, but wins second and third rounds and thus she is the winner of the game. In the second sample case Mishka wins the first round, Chris wins the second round, and the game ends with draw with score 1:1. In the third sample case Chris wins the first round, but there is no winner of the next two rounds. The winner of the game is Chris.

500

[ { "input": "3\n3 5\n2 1\n4 2", "output": "Mishka" }, { "input": "2\n6 1\n1 6", "output": "Friendship is magic!^^" }, { "input": "3\n1 5\n3 3\n2 2", "output": "Chris" }, { "input": "6\n4 1\n4 2\n5 3\n5 1\n5 3\n4 1", "output": "Mishka" }, { "input": "8\n2 4\n1 4\n1 5\n2 6\n2 5\n2 5\n2 4\n2 5", "output": "Chris" }, { "input": "8\n4 1\n2 6\n4 2\n2 5\n5 2\n3 5\n5 2\n1 5", "output": "Friendship is magic!^^" }, { "input": "9\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n1 3", "output": "Mishka" }, { "input": "9\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6", "output": "Mishka" }, { "input": "9\n1 2\n1 2\n1 2\n1 2\n1 2\n6 1\n6 1\n6 1\n6 1", "output": "Chris" }, { "input": "9\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6", "output": "Mishka" }, { "input": "10\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n1 4", "output": "Mishka" }, { "input": "10\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6", "output": "Mishka" }, { "input": "10\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n6 1\n6 1\n6 1\n6 1", "output": "Chris" }, { "input": "10\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6", "output": "Mishka" }, { "input": "100\n2 4\n6 6\n3 2\n1 5\n5 2\n1 5\n1 5\n3 1\n6 5\n4 3\n1 1\n5 1\n3 3\n2 4\n1 5\n3 4\n5 1\n5 5\n2 5\n2 1\n4 3\n6 5\n1 1\n2 1\n1 3\n1 1\n6 4\n4 6\n6 4\n2 1\n2 5\n6 2\n3 4\n5 5\n1 4\n4 6\n3 4\n1 6\n5 1\n4 3\n3 4\n2 2\n1 2\n2 3\n1 3\n4 4\n5 5\n4 5\n4 4\n3 1\n4 5\n2 3\n2 6\n6 5\n6 1\n6 6\n2 3\n6 4\n3 3\n2 5\n4 4\n3 1\n2 4\n6 1\n3 2\n1 3\n5 4\n6 6\n2 5\n5 1\n1 1\n2 5\n6 5\n3 6\n5 6\n4 3\n3 4\n3 4\n6 5\n5 2\n4 2\n1 1\n3 1\n2 6\n1 6\n1 2\n6 1\n3 4\n1 6\n3 1\n5 3\n1 3\n5 6\n2 1\n6 4\n3 1\n1 6\n6 3\n3 3\n4 3", "output": "Chris" }, { "input": "100\n4 1\n3 4\n4 6\n4 5\n6 5\n5 3\n6 2\n6 3\n5 2\n4 5\n1 5\n5 4\n1 4\n4 5\n4 6\n1 6\n4 4\n5 1\n6 4\n6 4\n4 6\n2 3\n6 2\n4 6\n1 4\n2 3\n4 3\n1 3\n6 2\n3 1\n3 4\n2 6\n4 5\n5 4\n2 2\n2 5\n4 1\n2 2\n3 3\n1 4\n5 6\n6 4\n4 2\n6 1\n5 5\n4 1\n2 1\n6 4\n4 4\n4 3\n5 3\n4 5\n5 3\n3 5\n6 3\n1 1\n3 4\n6 3\n6 1\n5 1\n2 4\n4 3\n2 2\n5 5\n1 5\n5 3\n4 6\n1 4\n6 3\n4 3\n2 4\n3 2\n2 4\n3 4\n6 2\n5 6\n1 2\n1 5\n5 5\n2 6\n5 1\n1 6\n5 3\n3 5\n2 6\n4 6\n6 2\n3 1\n5 5\n6 1\n3 6\n4 4\n1 1\n4 6\n5 3\n4 2\n5 1\n3 3\n2 1\n1 4", "output": "Mishka" }, { "input": "100\n6 3\n4 5\n4 3\n5 4\n5 1\n6 3\n4 2\n4 6\n3 1\n2 4\n2 2\n4 6\n5 3\n5 5\n4 2\n6 2\n2 3\n4 4\n6 4\n3 5\n2 4\n2 2\n5 2\n3 5\n2 4\n4 4\n3 5\n6 5\n1 3\n1 6\n2 2\n2 4\n3 2\n5 4\n1 6\n3 4\n4 1\n1 5\n1 4\n5 3\n2 2\n4 5\n6 3\n4 4\n1 1\n4 1\n2 4\n4 1\n4 5\n5 3\n1 1\n1 6\n5 6\n6 6\n4 2\n4 3\n3 4\n3 6\n3 4\n6 5\n3 4\n5 4\n5 1\n5 3\n5 1\n1 2\n2 6\n3 4\n6 5\n4 3\n1 1\n5 5\n5 1\n3 3\n5 2\n1 3\n6 6\n5 6\n1 4\n4 4\n1 4\n3 6\n6 5\n3 3\n3 6\n1 5\n1 2\n3 6\n3 6\n4 1\n5 2\n1 2\n5 2\n3 3\n4 4\n4 2\n6 2\n5 4\n6 1\n6 3", "output": "Mishka" }, { "input": "8\n4 1\n6 2\n4 1\n5 3\n4 1\n5 3\n6 2\n5 3", "output": "Mishka" }, { "input": "5\n3 6\n3 5\n3 5\n1 6\n3 5", "output": "Chris" }, { "input": "4\n4 1\n2 4\n5 3\n3 6", "output": "Friendship is magic!^^" }, { "input": "6\n6 3\n5 1\n6 3\n4 3\n4 3\n5 2", "output": "Mishka" }, { "input": "7\n3 4\n1 4\n2 5\n1 6\n1 6\n1 5\n3 4", "output": "Chris" }, { "input": "6\n6 2\n2 5\n5 2\n3 6\n4 3\n1 6", "output": "Friendship is magic!^^" }, { "input": "8\n6 1\n5 3\n4 3\n4 1\n5 1\n4 2\n4 2\n4 1", "output": "Mishka" }, { "input": "9\n2 5\n2 5\n1 4\n2 6\n2 4\n2 5\n2 6\n1 5\n2 5", "output": "Chris" }, { "input": "4\n6 2\n2 4\n4 2\n3 6", "output": "Friendship is magic!^^" }, { "input": "9\n5 2\n4 1\n4 1\n5 1\n6 2\n6 1\n5 3\n6 1\n6 2", "output": "Mishka" }, { "input": "8\n2 4\n3 6\n1 6\n1 6\n2 4\n3 4\n3 6\n3 4", "output": "Chris" }, { "input": "6\n5 3\n3 6\n6 2\n1 6\n5 1\n3 5", "output": "Friendship is magic!^^" }, { "input": "6\n5 2\n5 1\n6 1\n5 2\n4 2\n5 1", "output": "Mishka" }, { "input": "5\n1 4\n2 5\n3 4\n2 6\n3 4", "output": "Chris" }, { "input": "4\n6 2\n3 4\n5 1\n1 6", "output": "Friendship is magic!^^" }, { "input": "93\n4 3\n4 1\n4 2\n5 2\n5 3\n6 3\n4 3\n6 2\n6 3\n5 1\n4 2\n4 2\n5 1\n6 2\n6 3\n6 1\n4 1\n6 2\n5 3\n4 3\n4 1\n4 2\n5 2\n6 3\n5 2\n5 2\n6 3\n5 1\n6 2\n5 2\n4 1\n5 2\n5 1\n4 1\n6 1\n5 2\n4 3\n5 3\n5 3\n5 1\n4 3\n4 3\n4 2\n4 1\n6 2\n6 1\n4 1\n5 2\n5 2\n6 2\n5 3\n5 1\n6 2\n5 1\n6 3\n5 2\n6 2\n6 2\n4 2\n5 2\n6 1\n6 3\n6 3\n5 1\n5 1\n4 1\n5 1\n4 3\n5 3\n6 3\n4 1\n4 3\n6 1\n6 1\n4 2\n6 2\n4 2\n5 2\n4 1\n5 2\n4 1\n5 1\n5 2\n5 1\n4 1\n6 3\n6 2\n4 3\n4 1\n5 2\n4 3\n5 2\n5 1", "output": "Mishka" }, { "input": "11\n1 6\n1 6\n2 4\n2 5\n3 4\n1 5\n1 6\n1 5\n1 6\n2 6\n3 4", "output": "Chris" }, { "input": "70\n6 1\n3 6\n4 3\n2 5\n5 2\n1 4\n6 2\n1 6\n4 3\n1 4\n5 3\n2 4\n5 3\n1 6\n5 1\n3 5\n4 2\n2 4\n5 1\n3 5\n6 2\n1 5\n4 2\n2 5\n5 3\n1 5\n4 2\n1 4\n5 2\n2 6\n4 3\n1 5\n6 2\n3 4\n4 2\n3 5\n6 3\n3 4\n5 1\n1 4\n4 2\n1 4\n6 3\n2 6\n5 2\n1 6\n6 1\n2 6\n5 3\n1 5\n5 1\n1 6\n4 1\n1 5\n4 2\n2 4\n5 1\n2 5\n6 3\n1 4\n6 3\n3 6\n5 1\n1 4\n5 3\n3 5\n4 2\n3 4\n6 2\n1 4", "output": "Friendship is magic!^^" }, { "input": "59\n4 1\n5 3\n6 1\n4 2\n5 1\n4 3\n6 1\n5 1\n4 3\n4 3\n5 2\n5 3\n4 1\n6 2\n5 1\n6 3\n6 3\n5 2\n5 2\n6 1\n4 1\n6 1\n4 3\n5 3\n5 3\n4 3\n4 2\n4 2\n6 3\n6 3\n6 1\n4 3\n5 1\n6 2\n6 1\n4 1\n6 1\n5 3\n4 2\n5 1\n6 2\n6 2\n4 3\n5 3\n4 3\n6 3\n5 2\n5 2\n4 3\n5 1\n5 3\n6 1\n6 3\n6 3\n4 3\n5 2\n5 2\n5 2\n4 3", "output": "Mishka" }, { "input": "42\n1 5\n1 6\n1 6\n1 4\n2 5\n3 6\n1 6\n3 4\n2 5\n2 5\n2 4\n1 4\n3 4\n2 4\n2 6\n1 5\n3 6\n2 6\n2 6\n3 5\n1 4\n1 5\n2 6\n3 6\n1 4\n3 4\n2 4\n1 6\n3 4\n2 4\n2 6\n1 6\n1 4\n1 6\n1 6\n2 4\n1 5\n1 6\n2 5\n3 6\n3 5\n3 4", "output": "Chris" }, { "input": "78\n4 3\n3 5\n4 3\n1 5\n5 1\n1 5\n4 3\n1 4\n6 3\n1 5\n4 1\n2 4\n4 3\n2 4\n5 1\n3 6\n4 2\n3 6\n6 3\n3 4\n4 3\n3 6\n5 3\n1 5\n4 1\n2 6\n4 2\n2 4\n4 1\n3 5\n5 2\n3 6\n4 3\n2 4\n6 3\n1 6\n4 3\n3 5\n6 3\n2 6\n4 1\n2 4\n6 2\n1 6\n4 2\n1 4\n4 3\n1 4\n4 3\n2 4\n6 2\n3 5\n6 1\n3 6\n5 3\n1 6\n6 1\n2 6\n4 2\n1 5\n6 2\n2 6\n6 3\n2 4\n4 2\n3 5\n6 1\n2 5\n5 3\n2 6\n5 1\n3 6\n4 3\n3 6\n6 3\n2 5\n6 1\n2 6", "output": "Friendship is magic!^^" }, { "input": "76\n4 1\n5 2\n4 3\n5 2\n5 3\n5 2\n6 1\n4 2\n6 2\n5 3\n4 2\n6 2\n4 1\n4 2\n5 1\n5 1\n6 2\n5 2\n5 3\n6 3\n5 2\n4 3\n6 3\n6 1\n4 3\n6 2\n6 1\n4 1\n6 1\n5 3\n4 1\n5 3\n4 2\n5 2\n4 3\n6 1\n6 2\n5 2\n6 1\n5 3\n4 3\n5 1\n5 3\n4 3\n5 1\n5 1\n4 1\n4 1\n4 1\n4 3\n5 3\n6 3\n6 3\n5 2\n6 2\n6 3\n5 1\n6 3\n5 3\n6 1\n5 3\n4 1\n5 3\n6 1\n4 2\n6 2\n4 3\n4 1\n6 2\n4 3\n5 3\n5 2\n5 3\n5 1\n6 3\n5 2", "output": "Mishka" }, { "input": "84\n3 6\n3 4\n2 5\n2 4\n1 6\n3 4\n1 5\n1 6\n3 5\n1 6\n2 4\n2 6\n2 6\n2 4\n3 5\n1 5\n3 6\n3 6\n3 4\n3 4\n2 6\n1 6\n1 6\n3 5\n3 4\n1 6\n3 4\n3 5\n2 4\n2 5\n2 5\n3 5\n1 6\n3 4\n2 6\n2 6\n3 4\n3 4\n2 5\n2 5\n2 4\n3 4\n2 5\n3 4\n3 4\n2 6\n2 6\n1 6\n2 4\n1 5\n3 4\n2 5\n2 5\n3 4\n2 4\n2 6\n2 6\n1 4\n3 5\n3 5\n2 4\n2 5\n3 4\n1 5\n1 5\n2 6\n1 5\n3 5\n2 4\n2 5\n3 4\n2 6\n1 6\n2 5\n3 5\n3 5\n3 4\n2 5\n2 6\n3 4\n1 6\n2 5\n2 6\n1 4", "output": "Chris" }, { "input": "44\n6 1\n1 6\n5 2\n1 4\n6 2\n2 5\n5 3\n3 6\n5 2\n1 6\n4 1\n2 4\n6 1\n3 4\n6 3\n3 6\n4 3\n2 4\n6 1\n3 4\n6 1\n1 6\n4 1\n3 5\n6 1\n3 6\n4 1\n1 4\n4 2\n2 6\n6 1\n2 4\n6 2\n1 4\n6 2\n2 4\n5 2\n3 6\n6 3\n2 6\n5 3\n3 4\n5 3\n2 4", "output": "Friendship is magic!^^" }, { "input": "42\n5 3\n5 1\n5 2\n4 1\n6 3\n6 1\n6 2\n4 1\n4 3\n4 1\n5 1\n5 3\n5 1\n4 1\n4 2\n6 1\n6 3\n5 1\n4 1\n4 1\n6 3\n4 3\n6 3\n5 2\n6 1\n4 1\n5 3\n4 3\n5 2\n6 3\n6 1\n5 1\n4 2\n4 3\n5 2\n5 3\n6 3\n5 2\n5 1\n5 3\n6 2\n6 1", "output": "Mishka" }, { "input": "50\n3 6\n2 6\n1 4\n1 4\n1 4\n2 5\n3 4\n3 5\n2 6\n1 6\n3 5\n1 5\n2 6\n2 4\n2 4\n3 5\n1 6\n1 5\n1 5\n1 4\n3 5\n1 6\n3 5\n1 4\n1 5\n1 4\n3 6\n1 6\n1 4\n1 4\n1 4\n1 5\n3 6\n1 6\n1 6\n2 4\n1 5\n2 6\n2 5\n3 5\n3 6\n3 4\n2 4\n2 6\n3 4\n2 5\n3 6\n3 5\n2 4\n2 4", "output": "Chris" }, { "input": "86\n6 3\n2 4\n6 3\n3 5\n6 3\n1 5\n5 2\n2 4\n4 3\n2 6\n4 1\n2 6\n5 2\n1 4\n5 1\n2 4\n4 1\n1 4\n6 2\n3 5\n4 2\n2 4\n6 2\n1 5\n5 3\n2 5\n5 1\n1 6\n6 1\n1 4\n4 3\n3 4\n5 2\n2 4\n5 3\n2 5\n4 3\n3 4\n4 1\n1 5\n6 3\n3 4\n4 3\n3 4\n4 1\n3 4\n5 1\n1 6\n4 2\n1 6\n5 1\n2 4\n5 1\n3 6\n4 1\n1 5\n5 2\n1 4\n4 3\n2 5\n5 1\n1 5\n6 2\n2 6\n4 2\n2 4\n4 1\n2 5\n5 3\n3 4\n5 1\n3 4\n6 3\n3 4\n4 3\n2 6\n6 2\n2 5\n5 2\n3 5\n4 2\n3 6\n6 2\n3 4\n4 2\n2 4", "output": "Friendship is magic!^^" }, { "input": "84\n6 1\n6 3\n6 3\n4 1\n4 3\n4 2\n6 3\n5 3\n6 1\n6 3\n4 3\n5 2\n5 3\n5 1\n6 2\n6 2\n6 1\n4 1\n6 3\n5 2\n4 1\n5 3\n6 3\n4 2\n6 2\n6 3\n4 3\n4 1\n4 3\n5 1\n5 1\n5 1\n4 1\n6 1\n4 3\n6 2\n5 1\n5 1\n6 2\n5 2\n4 1\n6 1\n6 1\n6 3\n6 2\n4 3\n6 3\n6 2\n5 2\n5 1\n4 3\n6 2\n4 1\n6 2\n6 1\n5 2\n5 1\n6 2\n6 1\n5 3\n5 2\n6 1\n6 3\n5 2\n6 1\n6 3\n4 3\n5 1\n6 3\n6 1\n5 3\n4 3\n5 2\n5 1\n6 2\n5 3\n6 1\n5 1\n4 1\n5 1\n5 1\n5 2\n5 2\n5 1", "output": "Mishka" }, { "input": "92\n1 5\n2 4\n3 5\n1 6\n2 5\n1 6\n3 6\n1 6\n2 4\n3 4\n3 4\n3 6\n1 5\n2 5\n1 5\n1 5\n2 6\n2 4\n3 6\n1 4\n1 6\n2 6\n3 4\n2 6\n2 6\n1 4\n3 5\n2 5\n2 6\n1 5\n1 4\n1 5\n3 6\n3 5\n2 5\n1 5\n3 5\n3 6\n2 6\n2 6\n1 5\n3 4\n2 4\n3 6\n2 5\n1 5\n2 4\n1 4\n2 6\n2 6\n2 6\n1 5\n3 6\n3 6\n2 5\n1 4\n2 4\n3 4\n1 5\n2 5\n2 4\n2 5\n3 5\n3 4\n3 6\n2 6\n3 5\n1 4\n3 4\n1 6\n3 6\n2 6\n1 4\n3 6\n3 6\n2 5\n2 6\n1 6\n2 6\n3 5\n2 5\n3 6\n2 5\n2 6\n1 5\n2 4\n1 4\n2 4\n1 5\n2 5\n2 5\n2 6", "output": "Chris" }, { "input": "20\n5 1\n1 4\n4 3\n1 5\n4 2\n3 6\n6 2\n1 6\n4 1\n1 4\n5 2\n3 4\n5 1\n1 6\n5 1\n2 6\n6 3\n2 5\n6 2\n2 4", "output": "Friendship is magic!^^" }, { "input": "100\n4 3\n4 3\n4 2\n4 3\n4 1\n4 3\n5 2\n5 2\n6 2\n4 2\n5 1\n4 2\n5 2\n6 1\n4 1\n6 3\n5 3\n5 1\n5 1\n5 1\n5 3\n6 1\n6 1\n4 1\n5 2\n5 2\n6 1\n6 3\n4 2\n4 1\n5 3\n4 1\n5 3\n5 1\n6 3\n6 3\n6 1\n5 2\n5 3\n5 3\n6 1\n4 1\n6 2\n6 1\n6 2\n6 3\n4 3\n4 3\n6 3\n4 2\n4 2\n5 3\n5 2\n5 2\n4 3\n5 3\n5 2\n4 2\n5 1\n4 2\n5 1\n5 3\n6 3\n5 3\n5 3\n4 2\n4 1\n4 2\n4 3\n6 3\n4 3\n6 2\n6 1\n5 3\n5 2\n4 1\n6 1\n5 2\n6 2\n4 2\n6 3\n4 3\n5 1\n6 3\n5 2\n4 3\n5 3\n5 3\n4 3\n6 3\n4 3\n4 1\n5 1\n6 2\n6 3\n5 3\n6 1\n6 3\n5 3\n6 1", "output": "Mishka" }, { "input": "100\n1 5\n1 4\n1 5\n2 4\n2 6\n3 6\n3 5\n1 5\n2 5\n3 6\n3 5\n1 6\n1 4\n1 5\n1 6\n2 6\n1 5\n3 5\n3 4\n2 6\n2 6\n2 5\n3 4\n1 6\n1 4\n2 4\n1 5\n1 6\n3 5\n1 6\n2 6\n3 5\n1 6\n3 4\n3 5\n1 6\n3 6\n2 4\n2 4\n3 5\n2 6\n1 5\n3 5\n3 6\n2 4\n2 4\n2 6\n3 4\n3 4\n1 5\n1 4\n2 5\n3 4\n1 4\n2 6\n2 5\n2 4\n2 4\n2 5\n1 5\n1 6\n1 5\n1 5\n1 5\n1 6\n3 4\n2 4\n3 5\n3 5\n1 6\n3 5\n1 5\n1 6\n3 6\n3 4\n1 5\n3 5\n3 6\n1 4\n3 6\n1 5\n3 5\n3 6\n3 5\n1 4\n3 4\n2 4\n2 4\n2 5\n3 6\n3 5\n1 5\n2 4\n1 4\n3 4\n1 5\n3 4\n3 6\n3 5\n3 4", "output": "Chris" }, { "input": "100\n4 3\n3 4\n5 1\n2 5\n5 3\n1 5\n6 3\n2 4\n5 2\n2 6\n5 2\n1 5\n6 3\n1 5\n6 3\n3 4\n5 2\n1 5\n6 1\n1 5\n4 2\n3 5\n6 3\n2 6\n6 3\n1 4\n6 2\n3 4\n4 1\n3 6\n5 1\n2 4\n5 1\n3 4\n6 2\n3 5\n4 1\n2 6\n4 3\n2 6\n5 2\n3 6\n6 2\n3 5\n4 3\n1 5\n5 3\n3 6\n4 2\n3 4\n6 1\n3 4\n5 2\n2 6\n5 2\n2 4\n6 2\n3 6\n4 3\n2 4\n4 3\n2 6\n4 2\n3 4\n6 3\n2 4\n6 3\n3 5\n5 2\n1 5\n6 3\n3 6\n4 3\n1 4\n5 2\n1 6\n4 1\n2 5\n4 1\n2 4\n4 2\n2 5\n6 1\n2 4\n6 3\n1 5\n4 3\n2 6\n6 3\n2 6\n5 3\n1 5\n4 1\n1 5\n6 2\n2 5\n5 1\n3 6\n4 3\n3 4", "output": "Friendship is magic!^^" }, { "input": "99\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n1 3", "output": "Mishka" }, { "input": "99\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6", "output": "Mishka" }, { "input": "99\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1", "output": "Chris" }, { "input": "99\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6", "output": "Mishka" }, { "input": "100\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n1 4", "output": "Mishka" }, { "input": "100\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6", "output": "Mishka" }, { "input": "100\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1", "output": "Chris" }, { "input": "100\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6", "output": "Mishka" }, { "input": "84\n6 2\n1 5\n6 2\n2 3\n5 5\n1 2\n3 4\n3 4\n6 5\n6 4\n2 5\n4 1\n1 2\n1 1\n1 4\n2 5\n5 6\n6 3\n2 4\n5 5\n2 6\n3 4\n5 1\n3 3\n5 5\n4 6\n4 6\n2 4\n4 1\n5 2\n2 2\n3 6\n3 3\n4 6\n1 1\n2 4\n6 5\n5 2\n6 5\n5 5\n2 5\n6 4\n1 1\n6 2\n3 6\n6 5\n4 4\n1 5\n5 6\n4 4\n3 5\n6 1\n3 4\n1 5\n4 6\n4 6\n4 1\n3 6\n6 2\n1 1\n4 5\n5 4\n5 3\n3 4\n6 4\n1 1\n5 2\n6 5\n6 1\n2 2\n2 4\n3 3\n4 6\n1 3\n6 6\n5 2\n1 6\n6 2\n6 6\n4 1\n3 6\n6 4\n2 3\n3 4", "output": "Chris" }, { "input": "70\n3 4\n2 3\n2 3\n6 5\n6 6\n4 3\n2 3\n3 1\n3 5\n5 6\n1 6\n2 5\n5 3\n2 5\n4 6\n5 1\n6 1\n3 1\n3 3\n5 3\n2 1\n3 3\n6 4\n6 3\n4 3\n4 5\n3 5\n5 5\n5 2\n1 6\n3 4\n5 2\n2 4\n1 6\n4 3\n4 3\n6 2\n1 3\n1 5\n6 1\n3 1\n1 1\n1 3\n2 2\n3 2\n6 4\n1 1\n4 4\n3 1\n4 5\n4 2\n6 3\n4 4\n3 2\n1 2\n2 6\n3 3\n1 5\n1 1\n6 5\n2 2\n3 1\n5 4\n5 2\n6 4\n6 3\n6 6\n6 3\n3 3\n5 4", "output": "Mishka" }, { "input": "56\n6 4\n3 4\n6 1\n3 3\n1 4\n2 3\n1 5\n2 5\n1 5\n5 5\n2 3\n1 1\n3 2\n3 5\n4 6\n4 4\n5 2\n4 3\n3 1\n3 6\n2 3\n3 4\n5 6\n5 2\n5 6\n1 5\n1 5\n4 1\n6 3\n2 2\n2 1\n5 5\n2 1\n4 1\n5 4\n2 5\n4 1\n6 2\n3 4\n4 2\n6 4\n5 4\n4 2\n4 3\n6 2\n6 2\n3 1\n1 4\n3 6\n5 1\n5 5\n3 6\n6 4\n2 3\n6 5\n3 3", "output": "Mishka" }, { "input": "94\n2 4\n6 4\n1 6\n1 4\n5 1\n3 3\n4 3\n6 1\n6 5\n3 2\n2 3\n5 1\n5 3\n1 2\n4 3\n3 2\n2 3\n4 6\n1 3\n6 3\n1 1\n3 2\n4 3\n1 5\n4 6\n3 2\n6 3\n1 6\n1 1\n1 2\n3 5\n1 3\n3 5\n4 4\n4 2\n1 4\n4 5\n1 3\n1 2\n1 1\n5 4\n5 5\n6 1\n2 1\n2 6\n6 6\n4 2\n3 6\n1 6\n6 6\n1 5\n3 2\n1 2\n4 4\n6 4\n4 1\n1 5\n3 3\n1 3\n3 4\n4 4\n1 1\n2 5\n4 5\n3 1\n3 1\n3 6\n3 2\n1 4\n1 6\n6 3\n2 4\n1 1\n2 2\n2 2\n2 1\n5 4\n1 2\n6 6\n2 2\n3 3\n6 3\n6 3\n1 6\n2 3\n2 4\n2 3\n6 6\n2 6\n6 3\n3 5\n1 4\n1 1\n3 5", "output": "Chris" }, { "input": "81\n4 2\n1 2\n2 3\n4 5\n6 2\n1 6\n3 6\n3 4\n4 6\n4 4\n3 5\n4 6\n3 6\n3 5\n3 1\n1 3\n5 3\n3 4\n1 1\n4 1\n1 2\n6 1\n1 3\n6 5\n4 5\n4 2\n4 5\n6 2\n1 2\n2 6\n5 2\n1 5\n2 4\n4 3\n5 4\n1 2\n5 3\n2 6\n6 4\n1 1\n1 3\n3 1\n3 1\n6 5\n5 5\n6 1\n6 6\n5 2\n1 3\n1 4\n2 3\n5 5\n3 1\n3 1\n4 4\n1 6\n6 4\n2 2\n4 6\n4 4\n2 6\n2 4\n2 4\n4 1\n1 6\n1 4\n1 3\n6 5\n5 1\n1 3\n5 1\n1 4\n3 5\n2 6\n1 3\n5 6\n3 5\n4 4\n5 5\n5 6\n4 3", "output": "Chris" }, { "input": "67\n6 5\n3 6\n1 6\n5 3\n5 4\n5 1\n1 6\n1 1\n3 2\n4 4\n3 1\n4 1\n1 5\n5 3\n3 3\n6 4\n2 4\n2 2\n4 3\n1 4\n1 4\n6 1\n1 2\n2 2\n5 1\n6 2\n3 5\n5 5\n2 2\n6 5\n6 2\n4 4\n3 1\n4 2\n6 6\n6 4\n5 1\n2 2\n4 5\n5 5\n4 6\n1 5\n6 3\n4 4\n1 5\n6 4\n3 6\n3 4\n1 6\n2 4\n2 1\n2 5\n6 5\n6 4\n4 1\n3 2\n1 2\n5 1\n5 6\n1 5\n3 5\n3 1\n5 3\n3 2\n5 1\n4 6\n6 6", "output": "Mishka" }, { "input": "55\n6 6\n6 5\n2 2\n2 2\n6 4\n5 5\n6 5\n5 3\n1 3\n2 2\n5 6\n3 3\n3 3\n6 5\n3 5\n5 5\n1 2\n1 1\n4 6\n1 2\n5 5\n6 2\n6 3\n1 2\n5 1\n1 3\n3 3\n4 4\n2 5\n1 1\n5 3\n4 3\n2 2\n4 5\n5 6\n4 5\n6 3\n1 6\n6 4\n3 6\n1 6\n5 2\n6 3\n2 3\n5 5\n4 3\n3 1\n4 2\n1 1\n2 5\n5 3\n2 2\n6 3\n4 5\n2 2", "output": "Mishka" }, { "input": "92\n2 3\n1 3\n2 6\n5 1\n5 5\n3 2\n5 6\n2 5\n3 1\n3 6\n4 5\n2 5\n1 2\n2 3\n6 5\n3 6\n4 4\n6 2\n4 5\n4 4\n5 1\n6 1\n3 4\n3 5\n6 6\n3 2\n6 4\n2 2\n3 5\n6 4\n6 3\n6 6\n3 4\n3 3\n6 1\n5 4\n6 2\n2 6\n5 6\n1 4\n4 6\n6 3\n3 1\n4 1\n6 6\n3 5\n6 3\n6 1\n1 6\n3 2\n6 6\n4 3\n3 4\n1 3\n3 5\n5 3\n6 5\n4 3\n5 5\n4 1\n1 5\n6 4\n2 3\n2 3\n1 5\n1 2\n5 2\n4 3\n3 6\n5 5\n5 4\n1 4\n3 3\n1 6\n5 6\n5 4\n5 3\n1 1\n6 2\n5 5\n2 5\n4 3\n6 6\n5 1\n1 1\n4 6\n4 6\n3 1\n6 4\n2 4\n2 2\n2 1", "output": "Chris" }, { "input": "79\n5 3\n4 6\n3 6\n2 1\n5 2\n2 3\n4 4\n6 2\n2 5\n1 6\n6 6\n2 6\n3 3\n4 5\n6 2\n2 1\n1 5\n5 1\n2 1\n2 6\n5 3\n6 2\n2 6\n2 3\n1 5\n4 4\n6 3\n5 2\n3 2\n1 3\n1 3\n6 3\n2 6\n3 6\n5 3\n4 5\n6 1\n3 5\n3 5\n6 5\n1 5\n4 2\n6 2\n2 3\n4 6\n3 6\n2 5\n4 4\n1 1\n4 6\n2 6\n6 4\n3 2\n4 1\n1 2\n6 4\n5 6\n1 4\n2 2\n5 4\n3 2\n1 2\n2 4\n2 5\n2 1\n3 6\n3 3\n1 1\n2 2\n4 4\n4 5\n3 3\n5 3\n6 2\n4 5\n6 5\n2 5\n5 6\n2 2", "output": "Chris" }, { "input": "65\n1 1\n5 1\n2 2\n5 4\n4 5\n2 5\n3 2\n5 6\n6 3\n1 1\n6 1\n1 5\n1 1\n5 2\n6 4\n1 6\n1 1\n4 3\n2 3\n5 6\n4 4\n6 2\n1 3\n4 3\n1 3\n6 3\n3 5\n4 2\n4 1\n6 1\n3 2\n2 6\n3 2\n3 5\n6 3\n4 3\n1 5\n2 6\n1 3\n4 1\n4 1\n2 5\n2 5\n6 2\n5 3\n3 1\n3 3\n5 1\n2 4\n5 3\n3 3\n1 1\n6 3\n3 3\n5 1\n1 6\n4 5\n6 6\n5 5\n2 5\n4 1\n2 2\n1 4\n1 6\n6 5", "output": "Mishka" }, { "input": "1\n1 1", "output": "Friendship is magic!^^" } ]

1,693,574,919

2,147,483,647

Python 3

OK

TESTS

69

46

0

n = int(input()) m, c = 0, 0 for i in range(n): arr = [int(i) for i in input().split()] if arr[0] > arr[1]: m += 1 elif arr[0] < arr[1]: c += 1 if (m>c): print("Mishka") elif(m<c): print("Chris") else: print("Friendship is magic!^^")

Title: Mishka and Game Time Limit: None seconds Memory Limit: None megabytes Problem Description: Mishka is a little polar bear. As known, little bears loves spending their free time playing dice for chocolates. Once in a wonderful sunny morning, walking around blocks of ice, Mishka met her friend Chris, and they started playing the game. Rules of the game are very simple: at first number of rounds *n* is defined. In every round each of the players throws a cubical dice with distinct numbers from 1 to 6 written on its faces. Player, whose value after throwing the dice is greater, wins the round. In case if player dice values are equal, no one of them is a winner. In average, player, who won most of the rounds, is the winner of the game. In case if two players won the same number of rounds, the result of the game is draw. Mishka is still very little and can't count wins and losses, so she asked you to watch their game and determine its result. Please help her! Input Specification: The first line of the input contains single integer *n* *n* (1<=≤<=*n*<=≤<=100) — the number of game rounds. The next *n* lines contains rounds description. *i*-th of them contains pair of integers *m**i* and *c**i* (1<=≤<=*m**i*,<=<=*c**i*<=≤<=6) — values on dice upper face after Mishka's and Chris' throws in *i*-th round respectively. Output Specification: If Mishka is the winner of the game, print "Mishka" (without quotes) in the only line. If Chris is the winner of the game, print "Chris" (without quotes) in the only line. If the result of the game is draw, print "Friendship is magic!^^" (without quotes) in the only line. Demo Input: ['3\n3 5\n2 1\n4 2\n', '2\n6 1\n1 6\n', '3\n1 5\n3 3\n2 2\n'] Demo Output: ['Mishka', 'Friendship is magic!^^', 'Chris'] Note: In the first sample case Mishka loses the first round, but wins second and third rounds and thus she is the winner of the game. In the second sample case Mishka wins the first round, Chris wins the second round, and the game ends with draw with score 1:1. In the third sample case Chris wins the first round, but there is no winner of the next two rounds. The winner of the game is Chris.

```python n = int(input()) m, c = 0, 0 for i in range(n): arr = [int(i) for i in input().split()] if arr[0] > arr[1]: m += 1 elif arr[0] < arr[1]: c += 1 if (m>c): print("Mishka") elif(m<c): print("Chris") else: print("Friendship is magic!^^") ```

3

988

A

Diverse Team

PROGRAMMING

800

[ "brute force", "implementation" ]

null

There are $n$ students in a school class, the rating of the $i$-th student on Codehorses is $a_i$. You have to form a team consisting of $k$ students ($1 \le k \le n$) such that the ratings of all team members are distinct. If it is impossible to form a suitable team, print "NO" (without quotes). Otherwise print "YES", and then print $k$ distinct numbers which should be the indices of students in the team you form. If there are multiple answers, print any of them.

The first line contains two integers $n$ and $k$ ($1 \le k \le n \le 100$) — the number of students and the size of the team you have to form. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$), where $a_i$ is the rating of $i$-th student.

If it is impossible to form a suitable team, print "NO" (without quotes). Otherwise print "YES", and then print $k$ distinct integers from $1$ to $n$ which should be the indices of students in the team you form. All the ratings of the students in the team should be distinct. You may print the indices in any order. If there are multiple answers, print any of them. Assume that the students are numbered from $1$ to $n$.

[ "5 3\n15 13 15 15 12\n", "5 4\n15 13 15 15 12\n", "4 4\n20 10 40 30\n" ]

[ "YES\n1 2 5 \n", "NO\n", "YES\n1 2 3 4 \n" ]

All possible answers for the first example: - {1 2 5} - {2 3 5} - {2 4 5} Note that the order does not matter.

0

[ { "input": "5 3\n15 13 15 15 12", "output": "YES\n1 2 5 " }, { "input": "5 4\n15 13 15 15 12", "output": "NO" }, { "input": "4 4\n20 10 40 30", "output": "YES\n1 2 3 4 " }, { "input": "1 1\n1", "output": "YES\n1 " }, { "input": "100 53\n16 17 1 2 27 5 9 9 53 24 17 33 35 24 20 48 56 73 12 14 39 55 58 13 59 73 29 26 40 33 22 29 34 22 55 38 63 66 36 13 60 42 10 15 21 9 11 5 23 37 79 47 26 3 79 53 44 8 71 75 42 11 34 39 79 33 10 26 23 23 17 14 54 41 60 31 83 5 45 4 14 35 6 60 28 48 23 18 60 36 21 28 7 34 9 25 52 43 54 19", "output": "YES\n1 2 3 4 5 6 7 9 10 12 13 15 16 17 18 19 20 21 22 23 24 25 27 28 29 31 33 36 37 38 39 41 42 43 44 45 47 49 50 51 52 54 57 58 59 60 73 74 76 77 79 80 83 " }, { "input": "2 2\n100 100", "output": "NO" }, { "input": "2 2\n100 99", "output": "YES\n1 2 " }, { "input": "100 100\n63 100 75 32 53 24 73 98 76 15 70 48 8 81 88 58 95 78 27 92 14 16 72 43 46 39 66 38 64 42 59 9 22 51 4 6 10 94 28 99 68 80 35 50 45 20 47 7 30 26 49 91 77 19 96 57 65 1 11 13 31 12 82 87 93 34 62 3 21 79 56 41 89 18 44 23 74 86 2 33 69 36 61 67 25 83 5 84 90 37 40 29 97 60 52 55 54 71 17 85", "output": "YES\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 " }, { "input": "100 41\n54 16 42 3 45 6 9 72 100 13 24 57 35 5 89 13 97 27 43 9 73 89 48 16 48 55 18 15 55 28 30 6 18 41 100 61 9 42 35 54 57 25 73 15 42 54 49 5 72 48 30 55 4 43 94 5 60 92 93 23 89 75 53 92 74 93 89 28 69 6 3 49 15 28 49 57 54 55 30 57 69 18 89 6 25 23 93 74 30 13 87 53 6 42 4 54 60 30 4 35", "output": "NO" }, { "input": "100 2\n70 64 70 32 70 64 32 70 64 32 32 64 70 64 64 32 64 64 64 70 70 64 64 64 64 70 32 64 70 64 32 70 70 70 64 70 64 70 64 32 70 32 70 64 64 64 32 70 64 70 70 32 70 32 32 32 70 32 70 32 64 64 70 32 32 64 70 64 32 32 64 64 32 32 70 70 32 70 32 64 32 70 64 64 32 64 32 64 70 32 70 32 70 64 64 64 70 70 64 70", "output": "YES\n1 2 " } ]

1,626,371,454

2,147,483,647

PyPy 3

WRONG_ANSWER

TESTS

0

61

20,172,800

# In the name of Krishna n,k = map(int,input().split()) a = list(map(int,input().split())) s= set() l = [] for i in a: s.add(i) l+=[i+1] if len(s)==k: print("YES") print(l) else: print("NO")

Title: Diverse Team Time Limit: None seconds Memory Limit: None megabytes Problem Description: There are $n$ students in a school class, the rating of the $i$-th student on Codehorses is $a_i$. You have to form a team consisting of $k$ students ($1 \le k \le n$) such that the ratings of all team members are distinct. If it is impossible to form a suitable team, print "NO" (without quotes). Otherwise print "YES", and then print $k$ distinct numbers which should be the indices of students in the team you form. If there are multiple answers, print any of them. Input Specification: The first line contains two integers $n$ and $k$ ($1 \le k \le n \le 100$) — the number of students and the size of the team you have to form. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$), where $a_i$ is the rating of $i$-th student. Output Specification: If it is impossible to form a suitable team, print "NO" (without quotes). Otherwise print "YES", and then print $k$ distinct integers from $1$ to $n$ which should be the indices of students in the team you form. All the ratings of the students in the team should be distinct. You may print the indices in any order. If there are multiple answers, print any of them. Assume that the students are numbered from $1$ to $n$. Demo Input: ['5 3\n15 13 15 15 12\n', '5 4\n15 13 15 15 12\n', '4 4\n20 10 40 30\n'] Demo Output: ['YES\n1 2 5 \n', 'NO\n', 'YES\n1 2 3 4 \n'] Note: All possible answers for the first example: - {1 2 5} - {2 3 5} - {2 4 5} Note that the order does not matter.

```python # In the name of Krishna n,k = map(int,input().split()) a = list(map(int,input().split())) s= set() l = [] for i in a: s.add(i) l+=[i+1] if len(s)==k: print("YES") print(l) else: print("NO") ```

0

914

A

Perfect Squares

PROGRAMMING

900

[ "brute force", "implementation", "math" ]

null

Given an array *a*1,<=*a*2,<=...,<=*a**n* of *n* integers, find the largest number in the array that is not a perfect square. A number *x* is said to be a perfect square if there exists an integer *y* such that *x*<==<=*y*2.

The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of elements in the array. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=106<=≤<=*a**i*<=≤<=106) — the elements of the array. It is guaranteed that at least one element of the array is not a perfect square.

Print the largest number in the array which is not a perfect square. It is guaranteed that an answer always exists.

[ "2\n4 2\n", "8\n1 2 4 8 16 32 64 576\n" ]

[ "2\n", "32\n" ]

In the first sample case, 4 is a perfect square, so the largest number in the array that is not a perfect square is 2.

500

[ { "input": "2\n4 2", "output": "2" }, { "input": "8\n1 2 4 8 16 32 64 576", "output": "32" }, { "input": "3\n-1 -4 -9", "output": "-1" }, { "input": "5\n918375 169764 598796 76602 538757", "output": "918375" }, { "input": "5\n804610 765625 2916 381050 93025", "output": "804610" }, { "input": "5\n984065 842724 127449 525625 573049", "output": "984065" }, { "input": "2\n226505 477482", "output": "477482" }, { "input": "2\n370881 659345", "output": "659345" }, { "input": "2\n4 5", "output": "5" }, { "input": "2\n3 4", "output": "3" }, { "input": "2\n999999 1000000", "output": "999999" }, { "input": "3\n-1 -2 -3", "output": "-1" }, { "input": "2\n-1000000 1000000", "output": "-1000000" }, { "input": "2\n-1 0", "output": "-1" }, { "input": "1\n2", "output": "2" }, { "input": "1\n-1", "output": "-1" }, { "input": "35\n-871271 -169147 -590893 -400197 -476793 0 -15745 -890852 -124052 -631140 -238569 -597194 -147909 -928925 -587628 -569656 -581425 -963116 -665954 -506797 -196044 -309770 -701921 -926257 -152426 -991371 -624235 -557143 -689886 -59804 -549134 -107407 -182016 -24153 -607462", "output": "-15745" }, { "input": "16\n-882343 -791322 0 -986738 -415891 -823354 -840236 -552554 -760908 -331993 -549078 -863759 -913261 -937429 -257875 -602322", "output": "-257875" }, { "input": "71\n908209 289 44521 240100 680625 274576 212521 91809 506944 499849 3844 15376 592900 58081 240100 984064 732736 257049 600625 180625 130321 580644 261121 75625 46225 853776 485809 700569 817216 268324 293764 528529 25921 399424 175561 99856 295936 20736 611524 13924 470596 574564 5329 15376 676 431649 145161 697225 41616 550564 514089 9409 227529 1681 839056 3721 552049 465124 38809 197136 659344 214369 998001 44944 3844 186624 362404 -766506 739600 10816 299209", "output": "-766506" }, { "input": "30\n192721 -950059 -734656 625 247009 -423468 318096 622521 678976 777924 1444 748303 27556 62001 795664 89401 221841 -483208 467856 477109 196 -461813 831744 772641 574564 -519370 861184 67600 -717966 -259259", "output": "748303" }, { "input": "35\n628849 962361 436921 944784 444889 29241 -514806 171396 685584 -823202 -929730 6982 198025 783225 552049 -957165 782287 -659167 -414846 695556 -336330 41616 963781 71289 119639 952576 -346713 178929 232324 121802 393266 841 649636 179555 998001", "output": "963781" }, { "input": "53\n280988 756430 -515570 -248578 170649 -21608 642677 216770 827291 589500 940901 216097 -118956 -919104 -319264 -761585 289479 499613 588276 883036 480518 -323196 -274570 -406556 -381484 -956025 702135 -445274 -783543 136593 153664 897473 352651 737974 -21123 -284944 501734 898033 604429 624138 40804 248782 -786059 -304592 -209210 -312904 419820 -328648 -47331 -919227 -280955 104827 877304", "output": "940901" }, { "input": "15\n256 -227055 427717 827239 462070 66049 987533 -175306 -552810 -867915 -408251 -693957 -972981 -245827 896904", "output": "987533" }, { "input": "3\n-1 1 0", "output": "-1" }, { "input": "2\n0 -5", "output": "-5" }, { "input": "3\n-1 -2 0", "output": "-1" }, { "input": "2\n-5 0", "output": "-5" }, { "input": "1\n-439", "output": "-439" }, { "input": "1\n-1000000", "output": "-1000000" }, { "input": "1\n-917455", "output": "-917455" }, { "input": "3\n1 1 -1", "output": "-1" }, { "input": "2\n131073 1", "output": "131073" }, { "input": "2\n99999 3", "output": "99999" }, { "input": "2\n-524272 -1000000", "output": "-524272" }, { "input": "2\n15 131073", "output": "131073" } ]

1,625,378,450

2,147,483,647

PyPy 3

RUNTIME_ERROR

TESTS

2

108

2,150,400

from math import sqrt n=int(input()) a=list(map(int,input().split())) a.sort() a.reverse() for i in a: b=sqrt(i) if int(b)!=b: print(i) quit()

Title: Perfect Squares Time Limit: None seconds Memory Limit: None megabytes Problem Description: Given an array *a*1,<=*a*2,<=...,<=*a**n* of *n* integers, find the largest number in the array that is not a perfect square. A number *x* is said to be a perfect square if there exists an integer *y* such that *x*<==<=*y*2. Input Specification: The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of elements in the array. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=106<=≤<=*a**i*<=≤<=106) — the elements of the array. It is guaranteed that at least one element of the array is not a perfect square. Output Specification: Print the largest number in the array which is not a perfect square. It is guaranteed that an answer always exists. Demo Input: ['2\n4 2\n', '8\n1 2 4 8 16 32 64 576\n'] Demo Output: ['2\n', '32\n'] Note: In the first sample case, 4 is a perfect square, so the largest number in the array that is not a perfect square is 2.

```python from math import sqrt n=int(input()) a=list(map(int,input().split())) a.sort() a.reverse() for i in a: b=sqrt(i) if int(b)!=b: print(i) quit() ```

-1

53

C

Little Frog

PROGRAMMING

1,200

[ "constructive algorithms" ]

C. Little Frog

2

256

Once upon a time a little frog whose name was Vasya decided to travel around his home swamp. Overall there are *n* mounds on the swamp, located on one line. The distance between the neighboring mounds is one meter. Vasya wants to visit all the mounds in one day; besides, he wants to visit each one exactly once. For that he makes a route plan, to decide the order in which to jump on the mounds. Vasya can pick any mound as the first one. He thinks it boring to jump two times at the same distance. That's why he wants any two jumps on his route to have different lengths. Help Vasya the Frog and make the plan for him.

The single line contains a number *n* (1<=≤<=*n*<=≤<=104) which is the number of mounds.

Print *n* integers *p**i* (1<=≤<=*p**i*<=≤<=*n*) which are the frog's route plan. - All the *p**i*'s should be mutually different. - All the |*p**i*–*p**i*<=+<=1|'s should be mutually different (1<=≤<=*i*<=≤<=*n*<=-<=1). If there are several solutions, output any.

[ "2\n", "3\n" ]

[ "1 2 ", "1 3 2 " ]

none

1,500

[ { "input": "2", "output": "1 2 " }, { "input": "3", "output": "1 3 2 " }, { "input": "4", "output": "1 4 2 3 " }, { "input": "5", "output": "1 5 2 4 3 " }, { "input": "6", "output": "1 6 2 5 3 4 " }, { "input": "1", "output": "1 " }, { "input": "9149", "output": "1 9149 2 9148 3 9147 4 9146 5 9145 6 9144 7 9143 8 9142 9 9141 10 9140 11 9139 12 9138 13 9137 14 9136 15 9135 16 9134 17 9133 18 9132 19 9131 20 9130 21 9129 22 9128 23 9127 24 9126 25 9125 26 9124 27 9123 28 9122 29 9121 30 9120 31 9119 32 9118 33 9117 34 9116 35 9115 36 9114 37 9113 38 9112 39 9111 40 9110 41 9109 42 9108 43 9107 44 9106 45 9105 46 9104 47 9103 48 9102 49 9101 50 9100 51 9099 52 9098 53 9097 54 9096 55 9095 56 9094 57 9093 58 9092 59 9091 60 9090 61 9089 62 9088 63 9087 64 9086 65 9085 ..." }, { "input": "2877", "output": "1 2877 2 2876 3 2875 4 2874 5 2873 6 2872 7 2871 8 2870 9 2869 10 2868 11 2867 12 2866 13 2865 14 2864 15 2863 16 2862 17 2861 18 2860 19 2859 20 2858 21 2857 22 2856 23 2855 24 2854 25 2853 26 2852 27 2851 28 2850 29 2849 30 2848 31 2847 32 2846 33 2845 34 2844 35 2843 36 2842 37 2841 38 2840 39 2839 40 2838 41 2837 42 2836 43 2835 44 2834 45 2833 46 2832 47 2831 48 2830 49 2829 50 2828 51 2827 52 2826 53 2825 54 2824 55 2823 56 2822 57 2821 58 2820 59 2819 60 2818 61 2817 62 2816 63 2815 64 2814 65 2813 ..." }, { "input": "2956", "output": "1 2956 2 2955 3 2954 4 2953 5 2952 6 2951 7 2950 8 2949 9 2948 10 2947 11 2946 12 2945 13 2944 14 2943 15 2942 16 2941 17 2940 18 2939 19 2938 20 2937 21 2936 22 2935 23 2934 24 2933 25 2932 26 2931 27 2930 28 2929 29 2928 30 2927 31 2926 32 2925 33 2924 34 2923 35 2922 36 2921 37 2920 38 2919 39 2918 40 2917 41 2916 42 2915 43 2914 44 2913 45 2912 46 2911 47 2910 48 2909 49 2908 50 2907 51 2906 52 2905 53 2904 54 2903 55 2902 56 2901 57 2900 58 2899 59 2898 60 2897 61 2896 62 2895 63 2894 64 2893 65 2892 ..." }, { "input": "3035", "output": "1 3035 2 3034 3 3033 4 3032 5 3031 6 3030 7 3029 8 3028 9 3027 10 3026 11 3025 12 3024 13 3023 14 3022 15 3021 16 3020 17 3019 18 3018 19 3017 20 3016 21 3015 22 3014 23 3013 24 3012 25 3011 26 3010 27 3009 28 3008 29 3007 30 3006 31 3005 32 3004 33 3003 34 3002 35 3001 36 3000 37 2999 38 2998 39 2997 40 2996 41 2995 42 2994 43 2993 44 2992 45 2991 46 2990 47 2989 48 2988 49 2987 50 2986 51 2985 52 2984 53 2983 54 2982 55 2981 56 2980 57 2979 58 2978 59 2977 60 2976 61 2975 62 2974 63 2973 64 2972 65 2971 ..." }, { "input": "3114", "output": "1 3114 2 3113 3 3112 4 3111 5 3110 6 3109 7 3108 8 3107 9 3106 10 3105 11 3104 12 3103 13 3102 14 3101 15 3100 16 3099 17 3098 18 3097 19 3096 20 3095 21 3094 22 3093 23 3092 24 3091 25 3090 26 3089 27 3088 28 3087 29 3086 30 3085 31 3084 32 3083 33 3082 34 3081 35 3080 36 3079 37 3078 38 3077 39 3076 40 3075 41 3074 42 3073 43 3072 44 3071 45 3070 46 3069 47 3068 48 3067 49 3066 50 3065 51 3064 52 3063 53 3062 54 3061 55 3060 56 3059 57 3058 58 3057 59 3056 60 3055 61 3054 62 3053 63 3052 64 3051 65 3050 ..." }, { "input": "3193", "output": "1 3193 2 3192 3 3191 4 3190 5 3189 6 3188 7 3187 8 3186 9 3185 10 3184 11 3183 12 3182 13 3181 14 3180 15 3179 16 3178 17 3177 18 3176 19 3175 20 3174 21 3173 22 3172 23 3171 24 3170 25 3169 26 3168 27 3167 28 3166 29 3165 30 3164 31 3163 32 3162 33 3161 34 3160 35 3159 36 3158 37 3157 38 3156 39 3155 40 3154 41 3153 42 3152 43 3151 44 3150 45 3149 46 3148 47 3147 48 3146 49 3145 50 3144 51 3143 52 3142 53 3141 54 3140 55 3139 56 3138 57 3137 58 3136 59 3135 60 3134 61 3133 62 3132 63 3131 64 3130 65 3129 ..." }, { "input": "3273", "output": "1 3273 2 3272 3 3271 4 3270 5 3269 6 3268 7 3267 8 3266 9 3265 10 3264 11 3263 12 3262 13 3261 14 3260 15 3259 16 3258 17 3257 18 3256 19 3255 20 3254 21 3253 22 3252 23 3251 24 3250 25 3249 26 3248 27 3247 28 3246 29 3245 30 3244 31 3243 32 3242 33 3241 34 3240 35 3239 36 3238 37 3237 38 3236 39 3235 40 3234 41 3233 42 3232 43 3231 44 3230 45 3229 46 3228 47 3227 48 3226 49 3225 50 3224 51 3223 52 3222 53 3221 54 3220 55 3219 56 3218 57 3217 58 3216 59 3215 60 3214 61 3213 62 3212 63 3211 64 3210 65 3209 ..." }, { "input": "7000", "output": "1 7000 2 6999 3 6998 4 6997 5 6996 6 6995 7 6994 8 6993 9 6992 10 6991 11 6990 12 6989 13 6988 14 6987 15 6986 16 6985 17 6984 18 6983 19 6982 20 6981 21 6980 22 6979 23 6978 24 6977 25 6976 26 6975 27 6974 28 6973 29 6972 30 6971 31 6970 32 6969 33 6968 34 6967 35 6966 36 6965 37 6964 38 6963 39 6962 40 6961 41 6960 42 6959 43 6958 44 6957 45 6956 46 6955 47 6954 48 6953 49 6952 50 6951 51 6950 52 6949 53 6948 54 6947 55 6946 56 6945 57 6944 58 6943 59 6942 60 6941 61 6940 62 6939 63 6938 64 6937 65 6936 ..." }, { "input": "7079", "output": "1 7079 2 7078 3 7077 4 7076 5 7075 6 7074 7 7073 8 7072 9 7071 10 7070 11 7069 12 7068 13 7067 14 7066 15 7065 16 7064 17 7063 18 7062 19 7061 20 7060 21 7059 22 7058 23 7057 24 7056 25 7055 26 7054 27 7053 28 7052 29 7051 30 7050 31 7049 32 7048 33 7047 34 7046 35 7045 36 7044 37 7043 38 7042 39 7041 40 7040 41 7039 42 7038 43 7037 44 7036 45 7035 46 7034 47 7033 48 7032 49 7031 50 7030 51 7029 52 7028 53 7027 54 7026 55 7025 56 7024 57 7023 58 7022 59 7021 60 7020 61 7019 62 7018 63 7017 64 7016 65 7015 ..." }, { "input": "4653", "output": "1 4653 2 4652 3 4651 4 4650 5 4649 6 4648 7 4647 8 4646 9 4645 10 4644 11 4643 12 4642 13 4641 14 4640 15 4639 16 4638 17 4637 18 4636 19 4635 20 4634 21 4633 22 4632 23 4631 24 4630 25 4629 26 4628 27 4627 28 4626 29 4625 30 4624 31 4623 32 4622 33 4621 34 4620 35 4619 36 4618 37 4617 38 4616 39 4615 40 4614 41 4613 42 4612 43 4611 44 4610 45 4609 46 4608 47 4607 48 4606 49 4605 50 4604 51 4603 52 4602 53 4601 54 4600 55 4599 56 4598 57 4597 58 4596 59 4595 60 4594 61 4593 62 4592 63 4591 64 4590 65 4589 ..." }, { "input": "9995", "output": "1 9995 2 9994 3 9993 4 9992 5 9991 6 9990 7 9989 8 9988 9 9987 10 9986 11 9985 12 9984 13 9983 14 9982 15 9981 16 9980 17 9979 18 9978 19 9977 20 9976 21 9975 22 9974 23 9973 24 9972 25 9971 26 9970 27 9969 28 9968 29 9967 30 9966 31 9965 32 9964 33 9963 34 9962 35 9961 36 9960 37 9959 38 9958 39 9957 40 9956 41 9955 42 9954 43 9953 44 9952 45 9951 46 9950 47 9949 48 9948 49 9947 50 9946 51 9945 52 9944 53 9943 54 9942 55 9941 56 9940 57 9939 58 9938 59 9937 60 9936 61 9935 62 9934 63 9933 64 9932 65 9931 ..." }, { "input": "9996", "output": "1 9996 2 9995 3 9994 4 9993 5 9992 6 9991 7 9990 8 9989 9 9988 10 9987 11 9986 12 9985 13 9984 14 9983 15 9982 16 9981 17 9980 18 9979 19 9978 20 9977 21 9976 22 9975 23 9974 24 9973 25 9972 26 9971 27 9970 28 9969 29 9968 30 9967 31 9966 32 9965 33 9964 34 9963 35 9962 36 9961 37 9960 38 9959 39 9958 40 9957 41 9956 42 9955 43 9954 44 9953 45 9952 46 9951 47 9950 48 9949 49 9948 50 9947 51 9946 52 9945 53 9944 54 9943 55 9942 56 9941 57 9940 58 9939 59 9938 60 9937 61 9936 62 9935 63 9934 64 9933 65 9932 ..." }, { "input": "9997", "output": "1 9997 2 9996 3 9995 4 9994 5 9993 6 9992 7 9991 8 9990 9 9989 10 9988 11 9987 12 9986 13 9985 14 9984 15 9983 16 9982 17 9981 18 9980 19 9979 20 9978 21 9977 22 9976 23 9975 24 9974 25 9973 26 9972 27 9971 28 9970 29 9969 30 9968 31 9967 32 9966 33 9965 34 9964 35 9963 36 9962 37 9961 38 9960 39 9959 40 9958 41 9957 42 9956 43 9955 44 9954 45 9953 46 9952 47 9951 48 9950 49 9949 50 9948 51 9947 52 9946 53 9945 54 9944 55 9943 56 9942 57 9941 58 9940 59 9939 60 9938 61 9937 62 9936 63 9935 64 9934 65 9933 ..." }, { "input": "9998", "output": "1 9998 2 9997 3 9996 4 9995 5 9994 6 9993 7 9992 8 9991 9 9990 10 9989 11 9988 12 9987 13 9986 14 9985 15 9984 16 9983 17 9982 18 9981 19 9980 20 9979 21 9978 22 9977 23 9976 24 9975 25 9974 26 9973 27 9972 28 9971 29 9970 30 9969 31 9968 32 9967 33 9966 34 9965 35 9964 36 9963 37 9962 38 9961 39 9960 40 9959 41 9958 42 9957 43 9956 44 9955 45 9954 46 9953 47 9952 48 9951 49 9950 50 9949 51 9948 52 9947 53 9946 54 9945 55 9944 56 9943 57 9942 58 9941 59 9940 60 9939 61 9938 62 9937 63 9936 64 9935 65 9934 ..." }, { "input": "9999", "output": "1 9999 2 9998 3 9997 4 9996 5 9995 6 9994 7 9993 8 9992 9 9991 10 9990 11 9989 12 9988 13 9987 14 9986 15 9985 16 9984 17 9983 18 9982 19 9981 20 9980 21 9979 22 9978 23 9977 24 9976 25 9975 26 9974 27 9973 28 9972 29 9971 30 9970 31 9969 32 9968 33 9967 34 9966 35 9965 36 9964 37 9963 38 9962 39 9961 40 9960 41 9959 42 9958 43 9957 44 9956 45 9955 46 9954 47 9953 48 9952 49 9951 50 9950 51 9949 52 9948 53 9947 54 9946 55 9945 56 9944 57 9943 58 9942 59 9941 60 9940 61 9939 62 9938 63 9937 64 9936 65 9935 ..." }, { "input": "10000", "output": "1 10000 2 9999 3 9998 4 9997 5 9996 6 9995 7 9994 8 9993 9 9992 10 9991 11 9990 12 9989 13 9988 14 9987 15 9986 16 9985 17 9984 18 9983 19 9982 20 9981 21 9980 22 9979 23 9978 24 9977 25 9976 26 9975 27 9974 28 9973 29 9972 30 9971 31 9970 32 9969 33 9968 34 9967 35 9966 36 9965 37 9964 38 9963 39 9962 40 9961 41 9960 42 9959 43 9958 44 9957 45 9956 46 9955 47 9954 48 9953 49 9952 50 9951 51 9950 52 9949 53 9948 54 9947 55 9946 56 9945 57 9944 58 9943 59 9942 60 9941 61 9940 62 9939 63 9938 64 9937 65 9936..." } ]

1,594,451,268

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

2

186

6,860,800

n=int(input()) ans=[] i=1 j=2 flag=0 cnt=0 while 1: if flag==0 and i<=n: ans.append(i) i+=2 cnt+=1 if cnt==2: flag=1 cnt=0 elif j<=n: ans.append(j) j+=2 cnt+=1 if cnt==2: flag=0 cnt=0 else: break print(*ans)

Title: Little Frog Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Once upon a time a little frog whose name was Vasya decided to travel around his home swamp. Overall there are *n* mounds on the swamp, located on one line. The distance between the neighboring mounds is one meter. Vasya wants to visit all the mounds in one day; besides, he wants to visit each one exactly once. For that he makes a route plan, to decide the order in which to jump on the mounds. Vasya can pick any mound as the first one. He thinks it boring to jump two times at the same distance. That's why he wants any two jumps on his route to have different lengths. Help Vasya the Frog and make the plan for him. Input Specification: The single line contains a number *n* (1<=≤<=*n*<=≤<=104) which is the number of mounds. Output Specification: Print *n* integers *p**i* (1<=≤<=*p**i*<=≤<=*n*) which are the frog's route plan. - All the *p**i*'s should be mutually different. - All the |*p**i*–*p**i*<=+<=1|'s should be mutually different (1<=≤<=*i*<=≤<=*n*<=-<=1). If there are several solutions, output any. Demo Input: ['2\n', '3\n'] Demo Output: ['1 2 ', '1 3 2 '] Note: none

```python n=int(input()) ans=[] i=1 j=2 flag=0 cnt=0 while 1: if flag==0 and i<=n: ans.append(i) i+=2 cnt+=1 if cnt==2: flag=1 cnt=0 elif j<=n: ans.append(j) j+=2 cnt+=1 if cnt==2: flag=0 cnt=0 else: break print(*ans) ```

0

109

A

Lucky Sum of Digits

PROGRAMMING

1,000

[ "brute force", "implementation" ]

A. Lucky Sum of Digits

2

256

Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not. Petya wonders eagerly what minimum lucky number has the sum of digits equal to *n*. Help him cope with the task.

The single line contains an integer *n* (1<=≤<=*n*<=≤<=106) — the sum of digits of the required lucky number.

Print on the single line the result — the minimum lucky number, whose sum of digits equals *n*. If such number does not exist, print -1.

[ "11\n", "10\n" ]

[ "47\n", "-1\n" ]

none

500

[ { "input": "11", "output": "47" }, { "input": "10", "output": "-1" }, { "input": "64", "output": "4477777777" }, { "input": "1", "output": "-1" }, { "input": "4", "output": "4" }, { "input": "7", "output": "7" }, { "input": "12", "output": "444" }, { "input": "1000000", "output": "4477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..." }, { "input": "47", "output": "44477777" }, { "input": "100", "output": "4444777777777777" }, { "input": "700", "output": 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1,576,310,284

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

2

216

0

def listToString(s): u = ''.join(map(str, s)) u = int(u) return u x=int(input()) f=0 s=0 for b in range(x//7,0,-1): if (x-(7*b))%4==0: s=b f= (x-(7*b))//4 l=[] for i in range(f): l.append(4) for j in range(s): l.append(7) if len(l)!=0: print(int(listToString(l))) else: print(-1)

Title: Lucky Sum of Digits Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not. Petya wonders eagerly what minimum lucky number has the sum of digits equal to *n*. Help him cope with the task. Input Specification: The single line contains an integer *n* (1<=≤<=*n*<=≤<=106) — the sum of digits of the required lucky number. Output Specification: Print on the single line the result — the minimum lucky number, whose sum of digits equals *n*. If such number does not exist, print -1. Demo Input: ['11\n', '10\n'] Demo Output: ['47\n', '-1\n'] Note: none

```python def listToString(s): u = ''.join(map(str, s)) u = int(u) return u x=int(input()) f=0 s=0 for b in range(x//7,0,-1): if (x-(7*b))%4==0: s=b f= (x-(7*b))//4 l=[] for i in range(f): l.append(4) for j in range(s): l.append(7) if len(l)!=0: print(int(listToString(l))) else: print(-1) ```

0

84

A

Toy Army

PROGRAMMING

900

[ "math", "number theory" ]

A. Toy Army

2

256

The hero of our story, Valera, and his best friend Arcady are still in school, and therefore they spend all the free time playing turn-based strategy "GAGA: Go And Go Again". The gameplay is as follows. There are two armies on the playing field each of which consists of *n* men (*n* is always even). The current player specifies for each of her soldiers an enemy's soldier he will shoot (a target) and then all the player's soldiers shot simultaneously. This is a game world, and so each soldier shoots perfectly, that is he absolutely always hits the specified target. If an enemy soldier is hit, he will surely die. It may happen that several soldiers had been indicated the same target. Killed soldiers do not participate in the game anymore. The game "GAGA" consists of three steps: first Valera makes a move, then Arcady, then Valera again and the game ends. You are asked to calculate the maximum total number of soldiers that may be killed during the game.

The input data consist of a single integer *n* (2<=≤<=*n*<=≤<=108, *n* is even). Please note that before the game starts there are 2*n* soldiers on the fields.

Print a single number — a maximum total number of soldiers that could be killed in the course of the game in three turns.

[ "2\n", "4\n" ]

[ "3\n", "6\n" ]

The first sample test: 1) Valera's soldiers 1 and 2 shoot at Arcady's soldier 1. 2) Arcady's soldier 2 shoots at Valera's soldier 1. 3) Valera's soldier 1 shoots at Arcady's soldier 2. There are 3 soldiers killed in total: Valera's soldier 1 and Arcady's soldiers 1 and 2.

500

[ { "input": "2", "output": "3" }, { "input": "4", "output": "6" }, { "input": "6", "output": "9" }, { "input": "8", "output": "12" }, { "input": "10", "output": "15" }, { "input": "140", "output": "210" }, { "input": "500", "output": "750" }, { "input": "1000", "output": "1500" }, { "input": "2000", "output": "3000" }, { "input": "50000", "output": "75000" }, { "input": "10000", "output": "15000" }, { "input": "25460", "output": "38190" }, { "input": "54646", "output": "81969" }, { "input": "59790", "output": "89685" }, { "input": "578456", "output": "867684" }, { "input": "56798056", "output": "85197084" }, { "input": "8457980", "output": "12686970" }, { "input": "5687986", "output": "8531979" }, { "input": "10984932", "output": "16477398" }, { "input": "99999994", "output": "149999991" }, { "input": "99999996", "output": "149999994" }, { "input": "99999998", "output": "149999997" }, { "input": "100000000", "output": "150000000" } ]

1,543,683,523

2,147,483,647

Python 3

OK

TESTS

23

248

0

x=int(input()) print(3*x//2)

Title: Toy Army Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: The hero of our story, Valera, and his best friend Arcady are still in school, and therefore they spend all the free time playing turn-based strategy "GAGA: Go And Go Again". The gameplay is as follows. There are two armies on the playing field each of which consists of *n* men (*n* is always even). The current player specifies for each of her soldiers an enemy's soldier he will shoot (a target) and then all the player's soldiers shot simultaneously. This is a game world, and so each soldier shoots perfectly, that is he absolutely always hits the specified target. If an enemy soldier is hit, he will surely die. It may happen that several soldiers had been indicated the same target. Killed soldiers do not participate in the game anymore. The game "GAGA" consists of three steps: first Valera makes a move, then Arcady, then Valera again and the game ends. You are asked to calculate the maximum total number of soldiers that may be killed during the game. Input Specification: The input data consist of a single integer *n* (2<=≤<=*n*<=≤<=108, *n* is even). Please note that before the game starts there are 2*n* soldiers on the fields. Output Specification: Print a single number — a maximum total number of soldiers that could be killed in the course of the game in three turns. Demo Input: ['2\n', '4\n'] Demo Output: ['3\n', '6\n'] Note: The first sample test: 1) Valera's soldiers 1 and 2 shoot at Arcady's soldier 1. 2) Arcady's soldier 2 shoots at Valera's soldier 1. 3) Valera's soldier 1 shoots at Arcady's soldier 2. There are 3 soldiers killed in total: Valera's soldier 1 and Arcady's soldiers 1 and 2.

```python x=int(input()) print(3*x//2) ```

3.938

219

A

k-String

PROGRAMMING

1,000

[ "implementation", "strings" ]

null

A string is called a *k*-string if it can be represented as *k* concatenated copies of some string. For example, the string "aabaabaabaab" is at the same time a 1-string, a 2-string and a 4-string, but it is not a 3-string, a 5-string, or a 6-string and so on. Obviously any string is a 1-string. You are given a string *s*, consisting of lowercase English letters and a positive integer *k*. Your task is to reorder the letters in the string *s* in such a way that the resulting string is a *k*-string.

The first input line contains integer *k* (1<=≤<=*k*<=≤<=1000). The second line contains *s*, all characters in *s* are lowercase English letters. The string length *s* satisfies the inequality 1<=≤<=|*s*|<=≤<=1000, where |*s*| is the length of string *s*.

Rearrange the letters in string *s* in such a way that the result is a *k*-string. Print the result on a single output line. If there are multiple solutions, print any of them. If the solution doesn't exist, print "-1" (without quotes).

[ "2\naazz\n", "3\nabcabcabz\n" ]

[ "azaz\n", "-1\n" ]

none

500

[ { "input": "2\naazz", "output": "azaz" }, { "input": "3\nabcabcabz", "output": "-1" }, { "input": "1\na", "output": "a" }, { "input": "2\nabba", "output": "abab" }, { "input": "2\naaab", "output": "-1" }, { "input": "7\nabacaba", "output": "-1" }, { "input": "5\naaaaa", "output": "aaaaa" }, { "input": "3\naabaaaaabb", "output": "-1" }, { "input": "2\naaab", "output": "-1" }, { "input": "2\nbabac", "output": "-1" }, { "input": "3\nbbbccc", "output": "bcbcbc" }, { "input": "2\naa", "output": "aa" }, { "input": "250\ncececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececece", "output": "cececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececece" }, { "input": "15\nabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaa", "output": "aaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbc" }, { "input": "1\naaa", "output": "aaa" }, { "input": "1\naabaab", "output": "aaaabb" }, { "input": "2\naabbbbccccccdddddddd", "output": "abbcccddddabbcccdddd" }, { "input": "3\naaaaaaaaacccdddddd", "output": "aaacddaaacddaaacdd" }, { "input": "3\naaaaaaaaacccbbbbbb", "output": "aaabbcaaabbcaaabbc" }, { "input": "2\naazzzz", "output": "azzazz" }, { "input": "2\naaaa", "output": "aaaa" }, { "input": "2\naaaazz", "output": "aazaaz" }, { "input": "2\naaaabb", "output": "aabaab" }, { "input": "2\naabbbb", "output": "abbabb" }, { "input": "2\naaaazzzz", "output": "aazzaazz" }, { "input": "2\naaazz", "output": "-1" }, { "input": "2\nbbbbaa", "output": "abbabb" }, { "input": "2\naaaaaabbbb", "output": "aaabbaaabb" }, { "input": "2\naaaaaazz", "output": "aaazaaaz" }, { "input": "2\naaaabbbb", "output": "aabbaabb" }, { "input": "2\naabaab", "output": "aabaab" }, { "input": "1\naaaaa", "output": "aaaaa" }, { "input": "2\nbbaaaa", "output": "aabaab" }, { "input": "2\nacaccc", "output": "accacc" } ]

1,625,832,056

2,147,483,647

PyPy 3

OK

TESTS

42

218

0

from math import * def main(): k = int(input()) s = input() letters = {} for i in s: if i not in letters.keys(): letters[i] = 1 else: letters[i] += 1 F = list(letters.items()) for i in range(len(F)): if F[i][1] % k != 0: print(-1) return else: F[i] = (F[i][0], F[i][1]//k) z = "" for i in range(len(F)): z += (F[i][0] * F[i][1]) print(z * k) if __name__ == '__main__': main()

Title: k-String Time Limit: None seconds Memory Limit: None megabytes Problem Description: A string is called a *k*-string if it can be represented as *k* concatenated copies of some string. For example, the string "aabaabaabaab" is at the same time a 1-string, a 2-string and a 4-string, but it is not a 3-string, a 5-string, or a 6-string and so on. Obviously any string is a 1-string. You are given a string *s*, consisting of lowercase English letters and a positive integer *k*. Your task is to reorder the letters in the string *s* in such a way that the resulting string is a *k*-string. Input Specification: The first input line contains integer *k* (1<=≤<=*k*<=≤<=1000). The second line contains *s*, all characters in *s* are lowercase English letters. The string length *s* satisfies the inequality 1<=≤<=|*s*|<=≤<=1000, where |*s*| is the length of string *s*. Output Specification: Rearrange the letters in string *s* in such a way that the result is a *k*-string. Print the result on a single output line. If there are multiple solutions, print any of them. If the solution doesn't exist, print "-1" (without quotes). Demo Input: ['2\naazz\n', '3\nabcabcabz\n'] Demo Output: ['azaz\n', '-1\n'] Note: none

```python from math import * def main(): k = int(input()) s = input() letters = {} for i in s: if i not in letters.keys(): letters[i] = 1 else: letters[i] += 1 F = list(letters.items()) for i in range(len(F)): if F[i][1] % k != 0: print(-1) return else: F[i] = (F[i][0], F[i][1]//k) z = "" for i in range(len(F)): z += (F[i][0] * F[i][1]) print(z * k) if __name__ == '__main__': main() ```

3

509

A

Maximum in Table

PROGRAMMING

800

[ "brute force", "implementation" ]

null

An *n*<=×<=*n* table *a* is defined as follows: - The first row and the first column contain ones, that is: *a**i*,<=1<==<=*a*1,<=*i*<==<=1 for all *i*<==<=1,<=2,<=...,<=*n*. - Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula *a**i*,<=*j*<==<=*a**i*<=-<=1,<=*j*<=+<=*a**i*,<=*j*<=-<=1. These conditions define all the values in the table. You are given a number *n*. You need to determine the maximum value in the *n*<=×<=*n* table defined by the rules above.

The only line of input contains a positive integer *n* (1<=≤<=*n*<=≤<=10) — the number of rows and columns of the table.

Print a single line containing a positive integer *m* — the maximum value in the table.

[ "1\n", "5\n" ]

[ "1", "70" ]

In the second test the rows of the table look as follows:

0

[ { "input": "1", "output": "1" }, { "input": "5", "output": "70" }, { "input": "2", "output": "2" }, { "input": "3", "output": "6" }, { "input": "4", "output": "20" }, { "input": "6", "output": "252" }, { "input": "7", "output": "924" }, { "input": "9", "output": "12870" }, { "input": "10", "output": "48620" }, { "input": "8", "output": "3432" } ]

1,666,193,917

2,147,483,647

Python 3

OK

TESTS

10

46

0

l=[1,2,6,20,70,252,924,3432,12870,48620] print(l[int(input())-1])

Title: Maximum in Table Time Limit: None seconds Memory Limit: None megabytes Problem Description: An *n*<=×<=*n* table *a* is defined as follows: - The first row and the first column contain ones, that is: *a**i*,<=1<==<=*a*1,<=*i*<==<=1 for all *i*<==<=1,<=2,<=...,<=*n*. - Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula *a**i*,<=*j*<==<=*a**i*<=-<=1,<=*j*<=+<=*a**i*,<=*j*<=-<=1. These conditions define all the values in the table. You are given a number *n*. You need to determine the maximum value in the *n*<=×<=*n* table defined by the rules above. Input Specification: The only line of input contains a positive integer *n* (1<=≤<=*n*<=≤<=10) — the number of rows and columns of the table. Output Specification: Print a single line containing a positive integer *m* — the maximum value in the table. Demo Input: ['1\n', '5\n'] Demo Output: ['1', '70'] Note: In the second test the rows of the table look as follows:

```python l=[1,2,6,20,70,252,924,3432,12870,48620] print(l[int(input())-1]) ```

3

779

C

Dishonest Sellers

PROGRAMMING

1,200

[ "constructive algorithms", "greedy", "sortings" ]

null

Igor found out discounts in a shop and decided to buy *n* items. Discounts at the store will last for a week and Igor knows about each item that its price now is *a**i*, and after a week of discounts its price will be *b**i*. Not all of sellers are honest, so now some products could be more expensive than after a week of discounts. Igor decided that buy at least *k* of items now, but wait with the rest of the week in order to save money as much as possible. Your task is to determine the minimum money that Igor can spend to buy all *n* items.

In the first line there are two positive integer numbers *n* and *k* (1<=≤<=*n*<=≤<=2·105, 0<=≤<=*k*<=≤<=*n*) — total number of items to buy and minimal number of items Igor wants to by right now. The second line contains sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104) — prices of items during discounts (i.e. right now). The third line contains sequence of integers *b*1,<=*b*2,<=...,<=*b**n* (1<=≤<=*b**i*<=≤<=104) — prices of items after discounts (i.e. after a week).

Print the minimal amount of money Igor will spend to buy all *n* items. Remember, he should buy at least *k* items right now.

[ "3 1\n5 4 6\n3 1 5\n", "5 3\n3 4 7 10 3\n4 5 5 12 5\n" ]

[ "10\n", "25\n" ]

In the first example Igor should buy item 3 paying 6. But items 1 and 2 he should buy after a week. He will pay 3 and 1 for them. So in total he will pay 6 + 3 + 1 = 10. In the second example Igor should buy right now items 1, 2, 4 and 5, paying for them 3, 4, 10 and 3, respectively. Item 3 he should buy after a week of discounts, he will pay 5 for it. In total he will spend 3 + 4 + 10 + 3 + 5 = 25.

1,000

[ { "input": "3 1\n5 4 6\n3 1 5", "output": "10" }, { "input": "5 3\n3 4 7 10 3\n4 5 5 12 5", "output": "25" }, { "input": "1 0\n9\n8", "output": "8" }, { "input": "2 0\n4 10\n1 2", "output": "3" }, { "input": "4 2\n19 5 17 13\n3 18 8 10", "output": "29" }, { "input": "5 3\n28 17 20 45 45\n39 12 41 27 9", "output": "101" }, { "input": "10 5\n87 96 19 81 10 88 7 49 36 21\n11 75 28 28 74 17 64 19 81 31", "output": "243" }, { "input": "50 45\n80 125 152 122 85 62 123 195 157 73 8 127 47 193 70 132 131 38 98 97 185 98 68 166 114 53 144 106 141 106 191 123 44 199 35 18 84 169 46 52 129 102 155 130 63 169 116 145 154 41\n155 187 79 106 136 173 17 146 9 47 173 21 37 183 110 74 185 14 82 156 147 25 162 195 26 25 22 80 171 180 7 92 53 18 100 17 159 184 37 127 116 192 163 49 94 95 118 102 8 30", "output": "4680" }, { "input": "5 2\n1 1 1 1 1\n2 2 2 2 2", "output": "5" }, { "input": "5 2\n2 4 4 4 4\n3 1 1 1 1", "output": "9" }, { "input": "6 3\n9 9 9 2 2 2\n6 6 6 1 1 1", "output": "24" }, { "input": "5 3\n10 17 23 26 35\n7 18 18 29 32", "output": "103" }, { "input": "2 1\n10 12\n5 6", "output": "16" } ]

1,675,136,775

2,147,483,647

Python 3

OK

TESTS

67

249

18,022,400

n, k = map(int, input().split()) before = list(map(int, input().split())) after = list(map(int, input().split())) total_cost = sum(before) price_differences = sorted([y - x for x, y in zip(before, after)], reverse=True) for i in price_differences[k:]: if i >= 0: pass else: total_cost += i print(total_cost)

Title: Dishonest Sellers Time Limit: None seconds Memory Limit: None megabytes Problem Description: Igor found out discounts in a shop and decided to buy *n* items. Discounts at the store will last for a week and Igor knows about each item that its price now is *a**i*, and after a week of discounts its price will be *b**i*. Not all of sellers are honest, so now some products could be more expensive than after a week of discounts. Igor decided that buy at least *k* of items now, but wait with the rest of the week in order to save money as much as possible. Your task is to determine the minimum money that Igor can spend to buy all *n* items. Input Specification: In the first line there are two positive integer numbers *n* and *k* (1<=≤<=*n*<=≤<=2·105, 0<=≤<=*k*<=≤<=*n*) — total number of items to buy and minimal number of items Igor wants to by right now. The second line contains sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104) — prices of items during discounts (i.e. right now). The third line contains sequence of integers *b*1,<=*b*2,<=...,<=*b**n* (1<=≤<=*b**i*<=≤<=104) — prices of items after discounts (i.e. after a week). Output Specification: Print the minimal amount of money Igor will spend to buy all *n* items. Remember, he should buy at least *k* items right now. Demo Input: ['3 1\n5 4 6\n3 1 5\n', '5 3\n3 4 7 10 3\n4 5 5 12 5\n'] Demo Output: ['10\n', '25\n'] Note: In the first example Igor should buy item 3 paying 6. But items 1 and 2 he should buy after a week. He will pay 3 and 1 for them. So in total he will pay 6 + 3 + 1 = 10. In the second example Igor should buy right now items 1, 2, 4 and 5, paying for them 3, 4, 10 and 3, respectively. Item 3 he should buy after a week of discounts, he will pay 5 for it. In total he will spend 3 + 4 + 10 + 3 + 5 = 25.

```python n, k = map(int, input().split()) before = list(map(int, input().split())) after = list(map(int, input().split())) total_cost = sum(before) price_differences = sorted([y - x for x, y in zip(before, after)], reverse=True) for i in price_differences[k:]: if i >= 0: pass else: total_cost += i print(total_cost) ```

3

725

A

Jumping Ball

PROGRAMMING

1,000

[ "implementation" ]

null

In a new version of the famous Pinball game, one of the most important parts of the game field is a sequence of *n* bumpers. The bumpers are numbered with integers from 1 to *n* from left to right. There are two types of bumpers. They are denoted by the characters '<' and '>'. When the ball hits the bumper at position *i* it goes one position to the right (to the position *i*<=+<=1) if the type of this bumper is '>', or one position to the left (to *i*<=-<=1) if the type of the bumper at position *i* is '<'. If there is no such position, in other words if *i*<=-<=1<=<<=1 or *i*<=+<=1<=><=*n*, the ball falls from the game field. Depending on the ball's starting position, the ball may eventually fall from the game field or it may stay there forever. You are given a string representing the bumpers' types. Calculate the number of positions such that the ball will eventually fall from the game field if it starts at that position.

The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the length of the sequence of bumpers. The second line contains the string, which consists of the characters '<' and '>'. The character at the *i*-th position of this string corresponds to the type of the *i*-th bumper.

Print one integer — the number of positions in the sequence such that the ball will eventually fall from the game field if it starts at that position.

[ "4\n<<><\n", "5\n>>>>>\n", "4\n>><<\n" ]

[ "2", "5", "0" ]

In the first sample, the ball will fall from the field if starts at position 1 or position 2. In the second sample, any starting position will result in the ball falling from the field.

500

[ { "input": "4\n<<><", "output": "2" }, { "input": "5\n>>>>>", "output": "5" }, { "input": "4\n>><<", "output": "0" }, { "input": "3\n<<>", "output": "3" }, { "input": "3\n<<<", "output": "3" }, { "input": "3\n><<", "output": "0" }, { "input": "1\n<", "output": "1" }, { "input": "2\n<>", "output": "2" }, { "input": "3\n<>>", "output": "3" }, { "input": "3\n><>", "output": "1" }, { "input": "2\n><", "output": "0" }, { "input": "2\n>>", "output": "2" }, { "input": "2\n<<", "output": "2" }, { "input": "1\n>", "output": "1" }, { "input": "3\n>><", "output": "0" }, { "input": "3\n>>>", "output": "3" }, { "input": "3\n<><", "output": "1" }, { "input": "10\n<<<><<<>>>", "output": "6" }, { "input": "20\n><><<><<<>>>>>>>>>>>", "output": "11" }, { "input": "20\n<<<<<<<<<<><<<<>>>>>", "output": "15" }, { "input": "50\n<<<<<<<<<<<<<<<<<<<<<<<<<>>>>>>>>>>>>>>>>>>>>>>>>>", "output": "50" }, { "input": "100\n<<<<<<<<<<<<<<<<<<<<<<<<>><<>><<<<<>><>><<<>><><<>>><<>>><<<<><><><<><<<<><>>>>>>>>>>>>>>>>>>>>>>>>>", "output": "49" }, { "input": "100\n<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<>>>><<>><>><>><<><><><><>>>><><<<>>>><<<>>>>>>><><", "output": "50" }, { "input": "100\n<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<", "output": "100" }, { "input": "100\n>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>", "output": "100" }, { "input": "12\n<<>><<>><<>>", "output": "4" }, { "input": "6\n<<><>>", "output": "4" }, { "input": "6\n><>>>>", "output": "4" }, { "input": "8\n>>>><<<>", "output": "1" }, { "input": "4\n<><>", "output": "2" }, { "input": "4\n><><", "output": "0" }, { "input": "7\n<<>>><>", "output": "3" }, { "input": "10\n><><>>>>>>", "output": "6" }, { "input": "5\n<><>>", "output": "3" }, { "input": "12\n<><<<<>>>>>>", "output": "7" }, { "input": "6\n<>><<>", "output": "2" }, { "input": "6\n>>><>>", "output": "2" }, { "input": "10\n><><>>>><>", "output": "1" }, { "input": "5\n><>>>", "output": "3" }, { "input": "5\n<<><>", "output": "3" }, { "input": "5\n<><<<", "output": "1" }, { "input": "4\n<><<", "output": "1" }, { "input": "8\n<<>><<>>", "output": "4" }, { "input": "7\n<<><>>>", "output": "5" }, { "input": "5\n><<>>", "output": "2" }, { "input": "10\n<<<<<>>>>>", "output": "10" }, { "input": "6\n><<<<<", "output": "0" }, { "input": "8\n<<><><>>", "output": "4" }, { "input": "10\n<<<<><<<><", "output": "4" }, { "input": "12\n<<<>>>><<>>>", "output": "6" }, { "input": "4\n><>>", "output": "2" }, { "input": "11\n<<><<>><<>>", "output": "4" } ]

1,477,152,580

3,880

Python 3

TIME_LIMIT_EXCEEDED

PRETESTS

3

2,000

614,400

n = int(input()) s = input() count = 0 q = 101 d = 256 l = len(s) def search(pat, txt, q): M = len(pat) N = len(txt) i = 0 j = 0 p = 0 t = 0 h = 1 for i in range(M-1): h = (h*d)%q for i in range(M): p = (d*p + ord(pat[i]))%q t = (d*t + ord(txt[i]))%q for i in range(N-M+1): if p==t: for j in range(M): if txt[i+j] != pat[j]: break j+=1 if j==M: return 1 if i < N-M: t = (d*(t-ord(txt[i])*h) + ord(txt[i+M]))%q if t < 0: t = t+q for i in range(1, len(s)-1): if s[i] == '<': txt = s[:i] pat = '>' value = search(pat, txt, q) if not value: count += 1 elif s[i] == '>': txt = s[i+1:len(s)] pat = '<' value = search(pat, txt, q) if not value: count += 1 if s[0] == '<': count += 1 else: txt = s[i+1:len(s)] pat = '<' value = search(pat, txt, q) if not value: count += 1 if s[l-1] == '>': count += 1 else: txt = s[:l-1] pat = '>' value = search(pat, txt, q) if not value: count += 1 print(count)

Title: Jumping Ball Time Limit: None seconds Memory Limit: None megabytes Problem Description: In a new version of the famous Pinball game, one of the most important parts of the game field is a sequence of *n* bumpers. The bumpers are numbered with integers from 1 to *n* from left to right. There are two types of bumpers. They are denoted by the characters '<' and '>'. When the ball hits the bumper at position *i* it goes one position to the right (to the position *i*<=+<=1) if the type of this bumper is '>', or one position to the left (to *i*<=-<=1) if the type of the bumper at position *i* is '<'. If there is no such position, in other words if *i*<=-<=1<=<<=1 or *i*<=+<=1<=><=*n*, the ball falls from the game field. Depending on the ball's starting position, the ball may eventually fall from the game field or it may stay there forever. You are given a string representing the bumpers' types. Calculate the number of positions such that the ball will eventually fall from the game field if it starts at that position. Input Specification: The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the length of the sequence of bumpers. The second line contains the string, which consists of the characters '<' and '>'. The character at the *i*-th position of this string corresponds to the type of the *i*-th bumper. Output Specification: Print one integer — the number of positions in the sequence such that the ball will eventually fall from the game field if it starts at that position. Demo Input: ['4\n<<><\n', '5\n>>>>>\n', '4\n>><<\n'] Demo Output: ['2', '5', '0'] Note: In the first sample, the ball will fall from the field if starts at position 1 or position 2. In the second sample, any starting position will result in the ball falling from the field.

```python n = int(input()) s = input() count = 0 q = 101 d = 256 l = len(s) def search(pat, txt, q): M = len(pat) N = len(txt) i = 0 j = 0 p = 0 t = 0 h = 1 for i in range(M-1): h = (h*d)%q for i in range(M): p = (d*p + ord(pat[i]))%q t = (d*t + ord(txt[i]))%q for i in range(N-M+1): if p==t: for j in range(M): if txt[i+j] != pat[j]: break j+=1 if j==M: return 1 if i < N-M: t = (d*(t-ord(txt[i])*h) + ord(txt[i+M]))%q if t < 0: t = t+q for i in range(1, len(s)-1): if s[i] == '<': txt = s[:i] pat = '>' value = search(pat, txt, q) if not value: count += 1 elif s[i] == '>': txt = s[i+1:len(s)] pat = '<' value = search(pat, txt, q) if not value: count += 1 if s[0] == '<': count += 1 else: txt = s[i+1:len(s)] pat = '<' value = search(pat, txt, q) if not value: count += 1 if s[l-1] == '>': count += 1 else: txt = s[:l-1] pat = '>' value = search(pat, txt, q) if not value: count += 1 print(count) ```

0

873

A

Chores

PROGRAMMING

800

[ "implementation" ]

null

Luba has to do *n* chores today. *i*-th chore takes *a**i* units of time to complete. It is guaranteed that for every the condition *a**i*<=≥<=*a**i*<=-<=1 is met, so the sequence is sorted. Also Luba can work really hard on some chores. She can choose not more than *k* any chores and do each of them in *x* units of time instead of *a**i* (). Luba is very responsible, so she has to do all *n* chores, and now she wants to know the minimum time she needs to do everything. Luba cannot do two chores simultaneously.

The first line contains three integers *n*,<=*k*,<=*x* (1<=≤<=*k*<=≤<=*n*<=≤<=100,<=1<=≤<=*x*<=≤<=99) — the number of chores Luba has to do, the number of chores she can do in *x* units of time, and the number *x* itself. The second line contains *n* integer numbers *a**i* (2<=≤<=*a**i*<=≤<=100) — the time Luba has to spend to do *i*-th chore. It is guaranteed that , and for each *a**i*<=≥<=*a**i*<=-<=1.

Print one number — minimum time Luba needs to do all *n* chores.

[ "4 2 2\n3 6 7 10\n", "5 2 1\n100 100 100 100 100\n" ]

[ "13\n", "302\n" ]

In the first example the best option would be to do the third and the fourth chore, spending *x* = 2 time on each instead of *a*3 and *a*4, respectively. Then the answer is 3 + 6 + 2 + 2 = 13. In the second example Luba can choose any two chores to spend *x* time on them instead of *a**i*. So the answer is 100·3 + 2·1 = 302.

0

[ { "input": "4 2 2\n3 6 7 10", "output": "13" }, { "input": "5 2 1\n100 100 100 100 100", "output": "302" }, { "input": "1 1 1\n100", "output": "1" }, { "input": "100 1 99\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": "9999" }, { "input": "100 100 1\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 50 50\n51 51 52 53 55 55 55 55 56 56 56 57 57 58 58 59 59 59 60 60 61 61 62 62 63 64 64 64 64 65 65 65 65 66 66 66 67 68 68 68 69 69 70 70 70 70 71 71 71 71 71 71 72 72 76 76 76 76 77 79 79 81 81 81 81 82 82 82 82 83 84 85 86 87 87 88 88 88 89 89 89 90 90 90 91 91 91 92 92 93 95 95 96 96 96 97 97 98 99 100", "output": "5618" }, { "input": "100 100 1\n2 4 4 4 5 5 5 6 10 10 11 11 12 12 13 13 13 14 17 18 20 20 21 21 22 22 23 24 24 25 26 29 29 32 32 34 34 35 38 39 39 40 40 42 42 43 45 47 48 49 51 52 52 54 57 59 59 60 61 61 62 63 63 64 65 65 68 70 70 72 74 75 75 76 76 77 77 78 78 78 79 80 81 82 82 83 83 83 84 89 90 92 92 93 94 96 96 97 98 99", "output": "100" }, { "input": "100 1 1\n3 3 5 7 8 8 8 9 9 9 11 13 14 15 18 18 19 20 21 22 22 25 27 27 29 31 32 33 33 34 36 37 37 38 40 42 44 44 46 47 47 48 48 48 50 50 51 51 54 54 54 55 55 56 56 56 60 61 62 62 63 64 65 65 68 70 70 71 71 71 71 75 75 76 76 79 79 79 79 81 81 82 82 86 86 86 86 88 90 90 92 96 97 97 98 98 98 98 100 100", "output": "5202" }, { "input": "100 50 49\n50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51", "output": "4950" }, { "input": "100 50 1\n2 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 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3", "output": "151" }, { "input": "100 1 1\n2 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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2", "output": "199" } ]

1,537,543,666

2,147,483,647

Python 3

OK

TESTS

11

139

0

n,k,x=map(int,input().split()) a=list(map(int,input().split())) sum=0 for i in range(n-k): sum=sum+a[i] sum=sum+k*x print(sum)

Title: Chores Time Limit: None seconds Memory Limit: None megabytes Problem Description: Luba has to do *n* chores today. *i*-th chore takes *a**i* units of time to complete. It is guaranteed that for every the condition *a**i*<=≥<=*a**i*<=-<=1 is met, so the sequence is sorted. Also Luba can work really hard on some chores. She can choose not more than *k* any chores and do each of them in *x* units of time instead of *a**i* (). Luba is very responsible, so she has to do all *n* chores, and now she wants to know the minimum time she needs to do everything. Luba cannot do two chores simultaneously. Input Specification: The first line contains three integers *n*,<=*k*,<=*x* (1<=≤<=*k*<=≤<=*n*<=≤<=100,<=1<=≤<=*x*<=≤<=99) — the number of chores Luba has to do, the number of chores she can do in *x* units of time, and the number *x* itself. The second line contains *n* integer numbers *a**i* (2<=≤<=*a**i*<=≤<=100) — the time Luba has to spend to do *i*-th chore. It is guaranteed that , and for each *a**i*<=≥<=*a**i*<=-<=1. Output Specification: Print one number — minimum time Luba needs to do all *n* chores. Demo Input: ['4 2 2\n3 6 7 10\n', '5 2 1\n100 100 100 100 100\n'] Demo Output: ['13\n', '302\n'] Note: In the first example the best option would be to do the third and the fourth chore, spending *x* = 2 time on each instead of *a*3 and *a*4, respectively. Then the answer is 3 + 6 + 2 + 2 = 13. In the second example Luba can choose any two chores to spend *x* time on them instead of *a**i*. So the answer is 100·3 + 2·1 = 302.

```python n,k,x=map(int,input().split()) a=list(map(int,input().split())) sum=0 for i in range(n-k): sum=sum+a[i] sum=sum+k*x print(sum) ```

3

272

A

Dima and Friends

PROGRAMMING

1,000

[ "implementation", "math" ]

null

Dima and his friends have been playing hide and seek at Dima's place all night. As a result, Dima's place got messy. In the morning they decided that they need to clean the place. To decide who exactly would clean the apartment, the friends want to play a counting-out game. First, all the guys stand in a circle, and then each of them shows some number of fingers on one hand (one to five), and then the boys count in a circle, starting from Dima, the number of people, respective to the total number of fingers shown. The person on who the countdown stops will clean the apartment. For example, if Dima and one of his friends played hide and seek, and 7 fingers were shown during the counting-out, then Dima would clean the place. If there were 2 or say, 8 fingers shown, then his friend would clean the place. Dima knows how many fingers each of his friends will show during the counting-out. Now he is interested in the number of ways to show some number of fingers on one hand (one to five), so that he did not have to clean the place. Help Dima.

The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of Dima's friends. Dima himself isn't considered to be his own friend. The second line contains *n* positive integers, not exceeding 5, representing, how many fingers the Dima's friends will show. The numbers in the lines are separated by a single space.

In a single line print the answer to the problem.

[ "1\n1\n", "1\n2\n", "2\n3 5\n" ]

[ "3\n", "2\n", "3\n" ]

In the first sample Dima can show 1, 3 or 5 fingers. If Dima shows 3 fingers, then the counting-out will go like that: Dima, his friend, Dima, his friend. In the second sample Dima can show 2 or 4 fingers.

500

[ { "input": "1\n1", "output": "3" }, { "input": "1\n2", "output": "2" }, { "input": "2\n3 5", "output": "3" }, { "input": "2\n3 5", "output": "3" }, { "input": "1\n5", "output": "3" }, { "input": "5\n4 4 3 5 1", "output": "4" }, { "input": "6\n2 3 2 2 1 3", "output": "4" }, { "input": "8\n2 2 5 3 4 3 3 2", "output": "4" }, { "input": "7\n4 1 3 2 2 4 5", "output": "4" }, { "input": "3\n3 5 1", "output": "4" }, { "input": "95\n4 2 3 4 4 5 2 2 4 4 3 5 3 3 3 5 4 2 5 4 2 1 1 3 4 2 1 3 5 4 2 1 1 5 1 1 2 2 4 4 5 4 5 5 2 1 2 2 2 4 5 5 2 4 3 4 4 3 5 2 4 1 5 4 5 1 3 2 4 2 2 1 5 3 1 5 3 4 3 3 2 1 2 2 1 3 1 5 2 3 1 1 2 5 2", "output": "5" }, { "input": "31\n3 2 3 3 3 3 4 4 1 5 5 4 2 4 3 2 2 1 4 4 1 2 3 1 1 5 5 3 4 4 1", "output": "4" }, { "input": "42\n3 1 2 2 5 1 2 2 4 5 4 5 2 5 4 5 4 4 1 4 3 3 4 4 4 4 3 2 1 3 4 5 5 2 1 2 1 5 5 2 4 4", "output": "5" }, { "input": "25\n4 5 5 5 3 1 1 4 4 4 3 5 4 4 1 4 4 1 2 4 2 5 4 5 3", "output": "5" }, { "input": "73\n3 4 3 4 5 1 3 4 2 1 4 2 2 3 5 3 1 4 2 3 2 1 4 5 3 5 2 2 4 3 2 2 5 3 2 3 5 1 3 1 1 4 5 2 4 2 5 1 4 3 1 3 1 4 2 3 3 3 3 5 5 2 5 2 5 4 3 1 1 5 5 2 3", "output": "4" }, { "input": "46\n1 4 4 5 4 5 2 3 5 5 3 2 5 4 1 3 2 2 1 4 3 1 5 5 2 2 2 2 4 4 1 1 4 3 4 3 1 4 2 2 4 2 3 2 5 2", "output": "4" }, { "input": "23\n5 2 1 1 4 2 5 5 3 5 4 5 5 1 1 5 2 4 5 3 4 4 3", "output": "5" }, { "input": "6\n4 2 3 1 3 5", "output": "4" }, { "input": "15\n5 5 5 3 5 4 1 3 3 4 3 4 1 4 4", "output": "5" }, { "input": "93\n1 3 1 4 3 3 5 3 1 4 5 4 3 2 2 4 3 1 4 1 2 3 3 3 2 5 1 3 1 4 5 1 1 1 4 2 1 2 3 1 1 1 5 1 5 5 1 2 5 4 3 2 2 4 4 2 5 4 5 5 3 1 3 1 2 1 3 1 1 2 3 4 4 5 5 3 2 1 3 3 5 1 3 5 4 4 1 3 3 4 2 3 2", "output": "5" }, { "input": "96\n1 5 1 3 2 1 2 2 2 2 3 4 1 1 5 4 4 1 2 3 5 1 4 4 4 1 3 3 1 4 5 4 1 3 5 3 4 4 3 2 1 1 4 4 5 1 1 2 5 1 2 3 1 4 1 2 2 2 3 2 3 3 2 5 2 2 3 3 3 3 2 1 2 4 5 5 1 5 3 2 1 4 3 5 5 5 3 3 5 3 4 3 4 2 1 3", "output": "5" }, { "input": "49\n1 4 4 3 5 2 2 1 5 1 2 1 2 5 1 4 1 4 5 2 4 5 3 5 2 4 2 1 3 4 2 1 4 2 1 1 3 3 2 3 5 4 3 4 2 4 1 4 1", "output": "5" }, { "input": "73\n4 1 3 3 3 1 5 2 1 4 1 1 3 5 1 1 4 5 2 1 5 4 1 5 3 1 5 2 4 5 1 4 3 3 5 2 2 3 3 2 5 1 4 5 2 3 1 4 4 3 5 2 3 5 1 4 3 5 1 2 4 1 3 3 5 4 2 4 2 4 1 2 5", "output": "5" }, { "input": "41\n5 3 5 4 2 5 4 3 1 1 1 5 4 3 4 3 5 4 2 5 4 1 1 3 2 4 5 3 5 1 5 5 1 1 1 4 4 1 2 4 3", "output": "5" }, { "input": "100\n3 3 1 4 2 4 4 3 1 5 1 1 4 4 3 4 4 3 5 4 5 2 4 3 4 1 2 4 5 4 2 1 5 4 1 1 4 3 2 4 1 2 1 4 4 5 5 4 4 5 3 2 5 1 4 2 2 1 1 2 5 2 5 1 5 3 1 4 3 2 4 3 2 2 4 5 5 1 2 3 1 4 1 2 2 2 5 5 2 3 2 4 3 1 1 2 1 2 1 2", "output": "5" }, { "input": "100\n2 1 1 3 5 4 4 2 3 4 3 4 5 4 5 4 2 4 5 3 4 5 4 1 1 4 4 1 1 2 5 4 2 4 5 3 2 5 4 3 4 5 1 3 4 2 5 4 5 4 5 2 4 1 2 5 3 1 4 4 5 3 4 3 1 2 5 4 2 5 4 1 5 3 5 4 1 2 5 3 1 1 1 1 5 3 4 3 5 1 1 5 5 1 1 2 2 1 5 1", "output": "5" }, { "input": "100\n4 4 3 3 2 5 4 4 2 1 4 4 4 5 4 1 2 1 5 2 4 3 4 1 4 1 2 5 1 4 5 4 2 1 2 5 3 4 5 5 2 1 2 2 2 2 2 3 2 5 1 2 2 3 2 5 5 1 3 4 5 2 1 3 4 2 2 4 4 3 3 3 2 3 2 1 5 5 5 2 1 4 2 3 5 1 4 4 2 3 2 5 5 4 3 5 1 3 5 5", "output": "5" }, { "input": "100\n4 4 2 5 4 2 2 3 4 4 3 2 3 3 1 3 4 3 3 4 1 3 1 4 5 3 4 3 1 1 1 3 3 2 3 4 3 4 2 2 1 5 1 4 5 1 1 1 3 3 1 1 3 2 5 4 2 5 2 4 5 4 4 1 1 2 1 1 4 5 1 1 5 3 3 2 5 5 5 1 4 1 4 1 1 3 2 3 4 4 2 5 5 2 5 1 1 3 5 3", "output": "5" }, { "input": "100\n4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4", "output": "4" }, { "input": "100\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5", "output": "5" } ]

1,696,569,361

2,147,483,647

Python 3

OK

TESTS

30

92

0

import math n=int(input()) f=list(map(int,input().split())) r,x=0,0 i=math.ceil(sum(f)/(n+1)) while x<6: x=(n+1)*i+1-sum(f) i+=1 if x<6: r+=1 print(5-r)

Title: Dima and Friends Time Limit: None seconds Memory Limit: None megabytes Problem Description: Dima and his friends have been playing hide and seek at Dima's place all night. As a result, Dima's place got messy. In the morning they decided that they need to clean the place. To decide who exactly would clean the apartment, the friends want to play a counting-out game. First, all the guys stand in a circle, and then each of them shows some number of fingers on one hand (one to five), and then the boys count in a circle, starting from Dima, the number of people, respective to the total number of fingers shown. The person on who the countdown stops will clean the apartment. For example, if Dima and one of his friends played hide and seek, and 7 fingers were shown during the counting-out, then Dima would clean the place. If there were 2 or say, 8 fingers shown, then his friend would clean the place. Dima knows how many fingers each of his friends will show during the counting-out. Now he is interested in the number of ways to show some number of fingers on one hand (one to five), so that he did not have to clean the place. Help Dima. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of Dima's friends. Dima himself isn't considered to be his own friend. The second line contains *n* positive integers, not exceeding 5, representing, how many fingers the Dima's friends will show. The numbers in the lines are separated by a single space. Output Specification: In a single line print the answer to the problem. Demo Input: ['1\n1\n', '1\n2\n', '2\n3 5\n'] Demo Output: ['3\n', '2\n', '3\n'] Note: In the first sample Dima can show 1, 3 or 5 fingers. If Dima shows 3 fingers, then the counting-out will go like that: Dima, his friend, Dima, his friend. In the second sample Dima can show 2 or 4 fingers.

```python import math n=int(input()) f=list(map(int,input().split())) r,x=0,0 i=math.ceil(sum(f)/(n+1)) while x<6: x=(n+1)*i+1-sum(f) i+=1 if x<6: r+=1 print(5-r) ```

3

914

D

Bash and a Tough Math Puzzle

PROGRAMMING

1,900

[ "data structures", "number theory" ]

null

Bash likes playing with arrays. He has an array *a*1,<=*a*2,<=... *a**n* of *n* integers. He likes to guess the greatest common divisor (gcd) of different segments of the array. Of course, sometimes the guess is not correct. However, Bash will be satisfied if his guess is almost correct. Suppose he guesses that the gcd of the elements in the range [*l*,<=*r*] of *a* is *x*. He considers the guess to be almost correct if he can change at most one element in the segment such that the gcd of the segment is *x* after making the change. Note that when he guesses, he doesn't actually change the array — he just wonders if the gcd of the segment can be made *x*. Apart from this, he also sometimes makes changes to the array itself. Since he can't figure it out himself, Bash wants you to tell him which of his guesses are almost correct. Formally, you have to process *q* queries of one of the following forms: - 1<=*l*<=*r*<=*x* — Bash guesses that the gcd of the range [*l*,<=*r*] is *x*. Report if this guess is almost correct. - 2<=*i*<=*y* — Bash sets *a**i* to *y*. Note: The array is 1-indexed.

The first line contains an integer *n* (1<=≤<=*n*<=≤<=5·105) — the size of the array. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the elements of the array. The third line contains an integer *q* (1<=≤<=*q*<=≤<=4·105) — the number of queries. The next *q* lines describe the queries and may have one of the following forms: - 1<=*l*<=*r*<=*x* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*,<=1<=≤<=*x*<=≤<=109). - 2<=*i*<=*y* (1<=≤<=*i*<=≤<=*n*,<=1<=≤<=*y*<=≤<=109). Guaranteed, that there is at least one query of first type.

For each query of first type, output "YES" (without quotes) if Bash's guess is almost correct and "NO" (without quotes) otherwise.

[ "3\n2 6 3\n4\n1 1 2 2\n1 1 3 3\n2 1 9\n1 1 3 2\n", "5\n1 2 3 4 5\n6\n1 1 4 2\n2 3 6\n1 1 4 2\n1 1 5 2\n2 5 10\n1 1 5 2\n" ]

[ "YES\nYES\nNO\n", "NO\nYES\nNO\nYES\n" ]

In the first sample, the array initially is {2, 6, 3}. For query 1, the first two numbers already have their gcd as 2. For query 2, we can achieve a gcd of 3 by changing the first element of the array to 3. Note that the changes made during queries of type 1 are temporary and do not get reflected in the array. After query 3, the array is now {9, 6, 3}. For query 4, no matter which element you change, you cannot get the gcd of the range to be 2.

2,000

[ { "input": "3\n2 6 3\n4\n1 1 2 2\n1 1 3 3\n2 1 9\n1 1 3 2", "output": "YES\nYES\nNO" }, { "input": "5\n1 2 3 4 5\n6\n1 1 4 2\n2 3 6\n1 1 4 2\n1 1 5 2\n2 5 10\n1 1 5 2", "output": "NO\nYES\nNO\nYES" }, { "input": "1\n1000000000\n1\n1 1 1 1000000000", "output": "YES" }, { "input": "4\n3 3 7 7\n1\n1 1 4 3", "output": "NO" } ]

1,683,841,946

2,147,483,647

PyPy 3-64

TIME_LIMIT_EXCEEDED

TESTS

7

2,500

86,528,000

class Segtree: def __init__(self, size, func = lambda x,y: x+y): self.size = size self.segtree = [0]*4*size self.func = func def set(self, ind: int, val: int): self.update(ind, val, 0, self.size-1, 1) def update(self, ind: int, val: int, l: int, r: int, n: int): if l==r: self.segtree[n] = val return m=(l+r)//2 if ind<=m: self.update(ind, val, l, m, 2*n) else: self.update(ind, val, m+1, r, 2*n+1) self.segtree[n] = self.func(self.segtree[2*n], self.segtree[2*n+1]) def query(self, l: int, r: int, func = None): if func == None: func = self.func def range_query(lq, rq, n = 1, l = 0, r=self.size - 1): if rq < l or r < lq: return 0 elif lq <= l <= r <= rq: return self.segtree[n] else: m = (r+l)//2 return func(range_query(lq, rq, n*2, l, m), range_query(lq, rq, n*2+1, m+1, r)) return range_query(l, r) def __str__(self): height = len(bin(len(self.segtree)-1))-1 out = '' for i in range(height): indent = ' '*(2**(height-i-1)-2) out+=f'{indent}{f"{indent} {indent}".join(map(lambda x: str(x) if len(str(x))>1 else f"{x} ",self.segtree[2**i:2**(i+1)]))}\n' return out def gcd(a: int, b:int) -> int: if a == 0 or b == 0: return max(a,b) while True: a = a % b if a == 0: return b b = b % a if b == 0: return a def main(): n = int(input()) nums = list(map(int, input().split())) tree = Segtree(n, gcd) for i in range(n): tree.set(i, nums[i]) #print(tree) t = int(input()) for _ in range(t): k, *query = map(int, input().split()) # print(k, query) if k == 1: l, r = query[0]-1, query[1]-1 target = query[2] range_gcd = tree.query(l,r) if gcd(range_gcd, target) == target: print('YES') else: while True: if l == r: print('YES') break m = (l+r)//2 left_gcd = tree.query(l, m) right_gcd = tree.query(m+1, r) # print('Querying:', l, r, left_gcd, right_gcd) if gcd(left_gcd, target) == target: l = m+1 elif gcd(right_gcd, target) == target: r = m else: print('NO') break if k == 2: tree.set(query[0]-1, query[1]) #print(tree) return if __name__ == '__main__': # list = [2,4,6,8,10,12,14,16] # tree = Segtree(len(list)) # for i in range(len(list)): # tree.set(i, list[i]) # print(tree.segtree) main()

Title: Bash and a Tough Math Puzzle Time Limit: None seconds Memory Limit: None megabytes Problem Description: Bash likes playing with arrays. He has an array *a*1,<=*a*2,<=... *a**n* of *n* integers. He likes to guess the greatest common divisor (gcd) of different segments of the array. Of course, sometimes the guess is not correct. However, Bash will be satisfied if his guess is almost correct. Suppose he guesses that the gcd of the elements in the range [*l*,<=*r*] of *a* is *x*. He considers the guess to be almost correct if he can change at most one element in the segment such that the gcd of the segment is *x* after making the change. Note that when he guesses, he doesn't actually change the array — he just wonders if the gcd of the segment can be made *x*. Apart from this, he also sometimes makes changes to the array itself. Since he can't figure it out himself, Bash wants you to tell him which of his guesses are almost correct. Formally, you have to process *q* queries of one of the following forms: - 1<=*l*<=*r*<=*x* — Bash guesses that the gcd of the range [*l*,<=*r*] is *x*. Report if this guess is almost correct. - 2<=*i*<=*y* — Bash sets *a**i* to *y*. Note: The array is 1-indexed. Input Specification: The first line contains an integer *n* (1<=≤<=*n*<=≤<=5·105) — the size of the array. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the elements of the array. The third line contains an integer *q* (1<=≤<=*q*<=≤<=4·105) — the number of queries. The next *q* lines describe the queries and may have one of the following forms: - 1<=*l*<=*r*<=*x* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*,<=1<=≤<=*x*<=≤<=109). - 2<=*i*<=*y* (1<=≤<=*i*<=≤<=*n*,<=1<=≤<=*y*<=≤<=109). Guaranteed, that there is at least one query of first type. Output Specification: For each query of first type, output "YES" (without quotes) if Bash's guess is almost correct and "NO" (without quotes) otherwise. Demo Input: ['3\n2 6 3\n4\n1 1 2 2\n1 1 3 3\n2 1 9\n1 1 3 2\n', '5\n1 2 3 4 5\n6\n1 1 4 2\n2 3 6\n1 1 4 2\n1 1 5 2\n2 5 10\n1 1 5 2\n'] Demo Output: ['YES\nYES\nNO\n', 'NO\nYES\nNO\nYES\n'] Note: In the first sample, the array initially is {2, 6, 3}. For query 1, the first two numbers already have their gcd as 2. For query 2, we can achieve a gcd of 3 by changing the first element of the array to 3. Note that the changes made during queries of type 1 are temporary and do not get reflected in the array. After query 3, the array is now {9, 6, 3}. For query 4, no matter which element you change, you cannot get the gcd of the range to be 2.

```python class Segtree: def __init__(self, size, func = lambda x,y: x+y): self.size = size self.segtree = [0]*4*size self.func = func def set(self, ind: int, val: int): self.update(ind, val, 0, self.size-1, 1) def update(self, ind: int, val: int, l: int, r: int, n: int): if l==r: self.segtree[n] = val return m=(l+r)//2 if ind<=m: self.update(ind, val, l, m, 2*n) else: self.update(ind, val, m+1, r, 2*n+1) self.segtree[n] = self.func(self.segtree[2*n], self.segtree[2*n+1]) def query(self, l: int, r: int, func = None): if func == None: func = self.func def range_query(lq, rq, n = 1, l = 0, r=self.size - 1): if rq < l or r < lq: return 0 elif lq <= l <= r <= rq: return self.segtree[n] else: m = (r+l)//2 return func(range_query(lq, rq, n*2, l, m), range_query(lq, rq, n*2+1, m+1, r)) return range_query(l, r) def __str__(self): height = len(bin(len(self.segtree)-1))-1 out = '' for i in range(height): indent = ' '*(2**(height-i-1)-2) out+=f'{indent}{f"{indent} {indent}".join(map(lambda x: str(x) if len(str(x))>1 else f"{x} ",self.segtree[2**i:2**(i+1)]))}\n' return out def gcd(a: int, b:int) -> int: if a == 0 or b == 0: return max(a,b) while True: a = a % b if a == 0: return b b = b % a if b == 0: return a def main(): n = int(input()) nums = list(map(int, input().split())) tree = Segtree(n, gcd) for i in range(n): tree.set(i, nums[i]) #print(tree) t = int(input()) for _ in range(t): k, *query = map(int, input().split()) # print(k, query) if k == 1: l, r = query[0]-1, query[1]-1 target = query[2] range_gcd = tree.query(l,r) if gcd(range_gcd, target) == target: print('YES') else: while True: if l == r: print('YES') break m = (l+r)//2 left_gcd = tree.query(l, m) right_gcd = tree.query(m+1, r) # print('Querying:', l, r, left_gcd, right_gcd) if gcd(left_gcd, target) == target: l = m+1 elif gcd(right_gcd, target) == target: r = m else: print('NO') break if k == 2: tree.set(query[0]-1, query[1]) #print(tree) return if __name__ == '__main__': # list = [2,4,6,8,10,12,14,16] # tree = Segtree(len(list)) # for i in range(len(list)): # tree.set(i, list[i]) # print(tree.segtree) main() ```

0

279

B

Books

PROGRAMMING

1,400

[ "binary search", "brute force", "implementation", "two pointers" ]

null

When Valera has got some free time, he goes to the library to read some books. Today he's got *t* free minutes to read. That's why Valera took *n* books in the library and for each book he estimated the time he is going to need to read it. Let's number the books by integers from 1 to *n*. Valera needs *a**i* minutes to read the *i*-th book. Valera decided to choose an arbitrary book with number *i* and read the books one by one, starting from this book. In other words, he will first read book number *i*, then book number *i*<=+<=1, then book number *i*<=+<=2 and so on. He continues the process until he either runs out of the free time or finishes reading the *n*-th book. Valera reads each book up to the end, that is, he doesn't start reading the book if he doesn't have enough free time to finish reading it. Print the maximum number of books Valera can read.

The first line contains two integers *n* and *t* (1<=≤<=*n*<=≤<=105; 1<=≤<=*t*<=≤<=109) — the number of books and the number of free minutes Valera's got. The second line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104), where number *a**i* shows the number of minutes that the boy needs to read the *i*-th book.

Print a single integer — the maximum number of books Valera can read.

[ "4 5\n3 1 2 1\n", "3 3\n2 2 3\n" ]

[ "3\n", "1\n" ]

none

1,000

[ { "input": "4 5\n3 1 2 1", "output": "3" }, { "input": "3 3\n2 2 3", "output": "1" }, { "input": "1 3\n5", "output": "0" }, { "input": "1 10\n4", "output": "1" }, { "input": "2 10\n6 4", "output": "2" }, { "input": "6 10\n2 3 4 2 1 1", "output": "4" }, { "input": "7 13\n6 8 14 9 4 11 10", "output": "2" }, { "input": "10 15\n10 9 1 1 5 10 5 3 7 2", "output": "3" }, { "input": "20 30\n8 1 2 6 9 4 1 9 9 10 4 7 8 9 5 7 1 8 7 4", "output": "6" }, { "input": "30 60\n16 13 22 38 13 35 17 17 20 38 12 19 9 22 20 3 35 34 34 21 35 40 22 3 27 19 12 4 8 19", "output": "4" }, { "input": "100 100\n75 92 18 6 81 67 7 92 100 65 82 32 50 67 85 31 80 91 84 63 39 52 92 81 1 98 24 12 43 48 17 86 51 72 48 95 45 50 12 66 19 79 49 89 34 1 97 75 20 33 96 27 42 23 73 71 93 1 85 19 66 14 17 61 20 39 36 33 42 61 56 64 23 91 80 99 40 74 13 18 98 85 74 39 62 84 46 74 50 23 38 11 79 14 9 25 66 100 25 52", "output": "3" }, { "input": "10 1\n4418 7528 8170 1736 1317 3205 8183 4995 8039 4708", "output": "0" }, { "input": "50 2\n124 214 63 73 996 760 38 571 451 300 970 1 706 937 837 494 619 88 851 411 957 990 842 613 821 649 627 34 693 678 734 116 816 985 705 940 499 493 922 967 854 439 112 644 961 438 189 572 655 550", "output": "1" } ]

1,680,268,888

2,147,483,647

Python 3

RUNTIME_ERROR

TESTS

3

62

0

import bisect n,m=map(int,input().split()) l=list(map(int,input().split())) l.sort() co=0 sm=[] for i in l: co+=i sm.append(co) # print(sm,l) mid=bisect.bisect(sm,m) if mid<n: if sm[mid]>m: mid-=1 timegone=sm[mid] count=mid+1 print(count)

Title: Books Time Limit: None seconds Memory Limit: None megabytes Problem Description: When Valera has got some free time, he goes to the library to read some books. Today he's got *t* free minutes to read. That's why Valera took *n* books in the library and for each book he estimated the time he is going to need to read it. Let's number the books by integers from 1 to *n*. Valera needs *a**i* minutes to read the *i*-th book. Valera decided to choose an arbitrary book with number *i* and read the books one by one, starting from this book. In other words, he will first read book number *i*, then book number *i*<=+<=1, then book number *i*<=+<=2 and so on. He continues the process until he either runs out of the free time or finishes reading the *n*-th book. Valera reads each book up to the end, that is, he doesn't start reading the book if he doesn't have enough free time to finish reading it. Print the maximum number of books Valera can read. Input Specification: The first line contains two integers *n* and *t* (1<=≤<=*n*<=≤<=105; 1<=≤<=*t*<=≤<=109) — the number of books and the number of free minutes Valera's got. The second line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104), where number *a**i* shows the number of minutes that the boy needs to read the *i*-th book. Output Specification: Print a single integer — the maximum number of books Valera can read. Demo Input: ['4 5\n3 1 2 1\n', '3 3\n2 2 3\n'] Demo Output: ['3\n', '1\n'] Note: none

```python import bisect n,m=map(int,input().split()) l=list(map(int,input().split())) l.sort() co=0 sm=[] for i in l: co+=i sm.append(co) # print(sm,l) mid=bisect.bisect(sm,m) if mid<n: if sm[mid]>m: mid-=1 timegone=sm[mid] count=mid+1 print(count) ```

-1

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" } ]

1,556,610,689

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

80

218

0

n = int(input()) sum = 0 for i in range(n): num = list(map(int, input().split())) for j in range(3): sum = sum + num[j] if sum == 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()) sum = 0 for i in range(n): num = list(map(int, input().split())) for j in range(3): sum = sum + num[j] if sum == 0: print("YES") else : print("NO") ```

0

228

A

Is your horseshoe on the other hoof?

PROGRAMMING

800

[ "implementation" ]

null

Valera the Horse is going to the party with friends. He has been following the fashion trends for a while, and he knows that it is very popular to wear all horseshoes of different color. Valera has got four horseshoes left from the last year, but maybe some of them have the same color. In this case he needs to go to the store and buy some few more horseshoes, not to lose face in front of his stylish comrades. Fortunately, the store sells horseshoes of all colors under the sun and Valera has enough money to buy any four of them. However, in order to save the money, he would like to spend as little money as possible, so you need to help Valera and determine what is the minimum number of horseshoes he needs to buy to wear four horseshoes of different colors to a party.

The first line contains four space-separated integers *s*1,<=*s*2,<=*s*3,<=*s*4 (1<=≤<=*s*1,<=*s*2,<=*s*3,<=*s*4<=≤<=109) — the colors of horseshoes Valera has. Consider all possible colors indexed with integers.

Print a single integer — the minimum number of horseshoes Valera needs to buy.

[ "1 7 3 3\n", "7 7 7 7\n" ]

[ "1\n", "3\n" ]

none

500

[ { "input": "1 7 3 3", "output": "1" }, { "input": "7 7 7 7", "output": "3" }, { "input": "81170865 673572653 756938629 995577259", "output": "0" }, { "input": "3491663 217797045 522540872 715355328", "output": "0" }, { "input": "251590420 586975278 916631563 586975278", "output": "1" }, { "input": "259504825 377489979 588153796 377489979", "output": "1" }, { "input": "652588203 931100304 931100304 652588203", "output": "2" }, { "input": "391958720 651507265 391958720 651507265", "output": "2" }, { "input": "90793237 90793237 90793237 90793237", "output": "3" }, { "input": "551651653 551651653 551651653 551651653", "output": "3" }, { "input": "156630260 609654355 668943582 973622757", "output": "0" }, { "input": "17061017 110313588 434481173 796661222", "output": "0" }, { "input": "24975422 256716298 337790533 690960249", "output": "0" }, { "input": "255635360 732742923 798648949 883146723", "output": "0" }, { "input": "133315691 265159773 734556507 265159773", "output": "1" }, { "input": "28442865 741657755 978106882 978106882", "output": "1" }, { "input": "131245479 174845575 497483467 131245479", "output": "1" }, { "input": "139159884 616215581 958341883 616215581", "output": "1" }, { "input": "147784432 947653080 947653080 947653080", "output": "2" }, { "input": "94055790 756126496 756126496 94055790", "output": "2" }, { "input": "240458500 511952208 240458500 511952208", "output": "2" }, { "input": "681828506 972810624 972810624 681828506", "output": "2" }, { "input": "454961014 454961014 454961014 454961014", "output": "3" }, { "input": "915819430 915819430 915819430 915819430", "output": "3" }, { "input": "671645142 671645142 671645142 671645142", "output": "3" }, { "input": "132503558 132503558 132503558 132503558", "output": "3" }, { "input": "5 5 999999 6", "output": "1" }, { "input": "1 1 2 5", "output": "1" }, { "input": "2 1 2 3", "output": "1" }, { "input": "1 1 3 5", "output": "1" }, { "input": "1 1 3 3", "output": "2" }, { "input": "2 2 2 1", "output": "2" }, { "input": "3 1 1 1", "output": "2" }, { "input": "1 2 2 2", "output": "2" } ]

1,688,671,411

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

0

62

0

l=list(map(int,input().split())) s=0 x=[l[0]] for i in range(1,len(l)): if l[i] in x: s += 1 else: x.append(l[i]) print(x) print(s)

Title: Is your horseshoe on the other hoof? Time Limit: None seconds Memory Limit: None megabytes Problem Description: Valera the Horse is going to the party with friends. He has been following the fashion trends for a while, and he knows that it is very popular to wear all horseshoes of different color. Valera has got four horseshoes left from the last year, but maybe some of them have the same color. In this case he needs to go to the store and buy some few more horseshoes, not to lose face in front of his stylish comrades. Fortunately, the store sells horseshoes of all colors under the sun and Valera has enough money to buy any four of them. However, in order to save the money, he would like to spend as little money as possible, so you need to help Valera and determine what is the minimum number of horseshoes he needs to buy to wear four horseshoes of different colors to a party. Input Specification: The first line contains four space-separated integers *s*1,<=*s*2,<=*s*3,<=*s*4 (1<=≤<=*s*1,<=*s*2,<=*s*3,<=*s*4<=≤<=109) — the colors of horseshoes Valera has. Consider all possible colors indexed with integers. Output Specification: Print a single integer — the minimum number of horseshoes Valera needs to buy. Demo Input: ['1 7 3 3\n', '7 7 7 7\n'] Demo Output: ['1\n', '3\n'] Note: none

```python l=list(map(int,input().split())) s=0 x=[l[0]] for i in range(1,len(l)): if l[i] in x: s += 1 else: x.append(l[i]) print(x) print(s) ```

0

839

C

Journey

PROGRAMMING

1,500

[ "dfs and similar", "dp", "graphs", "probabilities", "trees" ]

null

There are *n* cities and *n*<=-<=1 roads in the Seven Kingdoms, each road connects two cities and we can reach any city from any other by the roads. Theon and Yara Greyjoy are on a horse in the first city, they are starting traveling through the roads. But the weather is foggy, so they can’t see where the horse brings them. When the horse reaches a city (including the first one), it goes to one of the cities connected to the current city. But it is a strange horse, it only goes to cities in which they weren't before. In each such city, the horse goes with equal probabilities and it stops when there are no such cities. Let the length of each road be 1. The journey starts in the city 1. What is the expected length (expected value of length) of their journey? You can read about expected (average) value by the link [https://en.wikipedia.org/wiki/Expected_value](https://en.wikipedia.org/wiki/Expected_value).

The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100000) — number of cities. Then *n*<=-<=1 lines follow. The *i*-th line of these lines contains two integers *u**i* and *v**i* (1<=≤<=*u**i*,<=*v**i*<=≤<=*n*, *u**i*<=≠<=*v**i*) — the cities connected by the *i*-th road. It is guaranteed that one can reach any city from any other by the roads.

Print a number — the expected length of their journey. The journey starts in the city 1. Your answer will be considered correct if its absolute or relative error does not exceed 10<=-<=6. Namely: let's assume that your answer is *a*, and the answer of the jury is *b*. The checker program will consider your answer correct, if .

[ "4\n1 2\n1 3\n2 4\n", "5\n1 2\n1 3\n3 4\n2 5\n" ]

[ "1.500000000000000\n", "2.000000000000000\n" ]

In the first sample, their journey may end in cities 3 or 4 with equal probability. The distance to city 3 is 1 and to city 4 is 2, so the expected length is 1.5. In the second sample, their journey may end in city 4 or 5. The distance to the both cities is 2, so the expected length is 2.

1,500

[ { "input": "4\n1 2\n1 3\n2 4", "output": "1.500000000000000" }, { "input": "5\n1 2\n1 3\n3 4\n2 5", "output": "2.000000000000000" }, { "input": "70\n1 25\n57 1\n18 1\n65 1\n38 1\n1 41\n1 5\n1 69\n1 3\n31 1\n1 8\n1 9\n53 1\n70 1\n45 1\n1 24\n1 42\n1 30\n1 12\n1 37\n64 1\n1 28\n1 58\n1 22\n11 1\n1 4\n1 27\n1 16\n1 21\n54 1\n1 51\n1 43\n29 1\n56 1\n1 39\n32 1\n1 15\n1 17\n1 19\n1 40\n36 1\n48 1\n63 1\n1 7\n1 47\n1 13\n1 46\n60 1\n1 6\n23 1\n20 1\n1 52\n2 1\n26 1\n1 59\n1 66\n10 1\n1 62\n1 68\n1 55\n50 1\n33 1\n44 1\n1 34\n1 35\n1 61\n14 1\n67 1\n49 1", "output": "1.000000000000000" }, { "input": "10\n8 6\n9 10\n8 7\n1 4\n1 8\n9 5\n9 8\n2 5\n3 1", "output": "1.500000000000000" }, { "input": "1", "output": "0.000000000000000" } ]

1,610,319,800

2,000

PyPy 3

OK

TESTS

40

935

18,227,200

n=int(input()) graph=[] for i in range(n): graph.append([]) for _ in range(n-1): u,v=map(int,input().split()) graph[u-1].append(v-1) graph[v-1].append(u-1) ans=0 visited=[0]*n stack=[(0,1)] while stack: curr,prob=stack.pop() visited[curr]=1 ans+=prob r=[] for i in graph[curr]: if not visited[i]: r.append(i) if r: newprob=prob/len(r) for i in r: stack.append((i,newprob)) print(ans-1)

Title: Journey Time Limit: None seconds Memory Limit: None megabytes Problem Description: There are *n* cities and *n*<=-<=1 roads in the Seven Kingdoms, each road connects two cities and we can reach any city from any other by the roads. Theon and Yara Greyjoy are on a horse in the first city, they are starting traveling through the roads. But the weather is foggy, so they can’t see where the horse brings them. When the horse reaches a city (including the first one), it goes to one of the cities connected to the current city. But it is a strange horse, it only goes to cities in which they weren't before. In each such city, the horse goes with equal probabilities and it stops when there are no such cities. Let the length of each road be 1. The journey starts in the city 1. What is the expected length (expected value of length) of their journey? You can read about expected (average) value by the link [https://en.wikipedia.org/wiki/Expected_value](https://en.wikipedia.org/wiki/Expected_value). Input Specification: The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100000) — number of cities. Then *n*<=-<=1 lines follow. The *i*-th line of these lines contains two integers *u**i* and *v**i* (1<=≤<=*u**i*,<=*v**i*<=≤<=*n*, *u**i*<=≠<=*v**i*) — the cities connected by the *i*-th road. It is guaranteed that one can reach any city from any other by the roads. Output Specification: Print a number — the expected length of their journey. The journey starts in the city 1. Your answer will be considered correct if its absolute or relative error does not exceed 10<=-<=6. Namely: let's assume that your answer is *a*, and the answer of the jury is *b*. The checker program will consider your answer correct, if . Demo Input: ['4\n1 2\n1 3\n2 4\n', '5\n1 2\n1 3\n3 4\n2 5\n'] Demo Output: ['1.500000000000000\n', '2.000000000000000\n'] Note: In the first sample, their journey may end in cities 3 or 4 with equal probability. The distance to city 3 is 1 and to city 4 is 2, so the expected length is 1.5. In the second sample, their journey may end in city 4 or 5. The distance to the both cities is 2, so the expected length is 2.

```python n=int(input()) graph=[] for i in range(n): graph.append([]) for _ in range(n-1): u,v=map(int,input().split()) graph[u-1].append(v-1) graph[v-1].append(u-1) ans=0 visited=[0]*n stack=[(0,1)] while stack: curr,prob=stack.pop() visited[curr]=1 ans+=prob r=[] for i in graph[curr]: if not visited[i]: r.append(i) if r: newprob=prob/len(r) for i in r: stack.append((i,newprob)) print(ans-1) ```

3

492

B

Vanya and Lanterns

PROGRAMMING

1,200

[ "binary search", "implementation", "math", "sortings" ]

null

Vanya walks late at night along a straight street of length *l*, lit by *n* lanterns. Consider the coordinate system with the beginning of the street corresponding to the point 0, and its end corresponding to the point *l*. Then the *i*-th lantern is at the point *a**i*. The lantern lights all points of the street that are at the distance of at most *d* from it, where *d* is some positive number, common for all lanterns. Vanya wonders: what is the minimum light radius *d* should the lanterns have to light the whole street?

The first line contains two integers *n*, *l* (1<=≤<=*n*<=≤<=1000, 1<=≤<=*l*<=≤<=109) — the number of lanterns and the length of the street respectively. The next line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=*l*). Multiple lanterns can be located at the same point. The lanterns may be located at the ends of the street.

Print the minimum light radius *d*, needed to light the whole street. The answer will be considered correct if its absolute or relative error doesn't exceed 10<=-<=9.

[ "7 15\n15 5 3 7 9 14 0\n", "2 5\n2 5\n" ]

[ "2.5000000000\n", "2.0000000000\n" ]

Consider the second sample. At *d* = 2 the first lantern will light the segment [0, 4] of the street, and the second lantern will light segment [3, 5]. Thus, the whole street will be lit.

1,000

[ { "input": "7 15\n15 5 3 7 9 14 0", "output": "2.5000000000" }, { "input": "2 5\n2 5", "output": "2.0000000000" }, { "input": "46 615683844\n431749087 271781274 274974690 324606253 480870261 401650581 13285442 478090364 266585394 425024433 588791449 492057200 391293435 563090494 317950 173675329 473068378 356306865 311731938 192959832 321180686 141984626 578985584 512026637 175885185 590844074 47103801 212211134 330150 509886963 565955809 315640375 612907074 500474373 524310737 568681652 315339618 478782781 518873818 271322031 74600969 539099112 85129347 222068995 106014720 77282307", "output": "22258199.5000000000" }, { "input": "2 1000000000\n0 1000000000", "output": "500000000.0000000000" }, { "input": "2 555\n200 300", "output": "255.0000000000" }, { "input": "1 1\n1", "output": "1.0000000000" }, { "input": "1 1\n0", "output": "1.0000000000" }, { "input": "1 1000000000\n0", "output": "1000000000.0000000000" }, { "input": "1 1000000000\n1000000000", "output": "1000000000.0000000000" }, { "input": "1 999999999\n499999999", "output": "500000000.0000000000" }, { "input": "4 700\n0 250 475 700", "output": "125.0000000000" }, { "input": "4 700\n0 225 450 700", "output": "125.0000000000" }, { "input": "5 1000\n0 250 500 750 1000", "output": "125.0000000000" }, { "input": "5 2000\n1001 1001 1001 1001 1001", "output": "1001.0000000000" }, { "input": "1 999\n501", "output": "501.0000000000" }, { "input": "2 1000\n400 600", "output": "400.0000000000" }, { "input": "4 15\n1 3 9 11", "output": "4.0000000000" }, { "input": "1 5\n2", "output": "3.0000000000" } ]

1,693,933,652

2,147,483,647

Python 3

OK

TESTS

32

46

0

a,b=map(int,input().split()) l=list(map(int,input().split())) l=sorted(l) m=0 for i in range(a-1): d=l[i+1]-l[i] if d>m: m=d m=m/2 if l[0]!=0: if l[0]>m: m=l[0] if l[a-1]!=b: if b-l[a-1]>m: m=b-l[a-1] print(m)

Title: Vanya and Lanterns Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vanya walks late at night along a straight street of length *l*, lit by *n* lanterns. Consider the coordinate system with the beginning of the street corresponding to the point 0, and its end corresponding to the point *l*. Then the *i*-th lantern is at the point *a**i*. The lantern lights all points of the street that are at the distance of at most *d* from it, where *d* is some positive number, common for all lanterns. Vanya wonders: what is the minimum light radius *d* should the lanterns have to light the whole street? Input Specification: The first line contains two integers *n*, *l* (1<=≤<=*n*<=≤<=1000, 1<=≤<=*l*<=≤<=109) — the number of lanterns and the length of the street respectively. The next line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=*l*). Multiple lanterns can be located at the same point. The lanterns may be located at the ends of the street. Output Specification: Print the minimum light radius *d*, needed to light the whole street. The answer will be considered correct if its absolute or relative error doesn't exceed 10<=-<=9. Demo Input: ['7 15\n15 5 3 7 9 14 0\n', '2 5\n2 5\n'] Demo Output: ['2.5000000000\n', '2.0000000000\n'] Note: Consider the second sample. At *d* = 2 the first lantern will light the segment [0, 4] of the street, and the second lantern will light segment [3, 5]. Thus, the whole street will be lit.

```python a,b=map(int,input().split()) l=list(map(int,input().split())) l=sorted(l) m=0 for i in range(a-1): d=l[i+1]-l[i] if d>m: m=d m=m/2 if l[0]!=0: if l[0]>m: m=l[0] if l[a-1]!=b: if b-l[a-1]>m: m=b-l[a-1] print(m) ```

3

767

A

Snacktower

PROGRAMMING

1,100

[ "data structures", "implementation" ]

null

According to an old legeng, a long time ago Ankh-Morpork residents did something wrong to miss Fortune, and she cursed them. She said that at some time *n* snacks of distinct sizes will fall on the city, and the residents should build a Snacktower of them by placing snacks one on another. Of course, big snacks should be at the bottom of the tower, while small snacks should be at the top. Years passed, and once different snacks started to fall onto the city, and the residents began to build the Snacktower. However, they faced some troubles. Each day exactly one snack fell onto the city, but their order was strange. So, at some days the residents weren't able to put the new stack on the top of the Snacktower: they had to wait until all the bigger snacks fell. Of course, in order to not to anger miss Fortune again, the residents placed each snack on the top of the tower immediately as they could do it. Write a program that models the behavior of Ankh-Morpork residents.

The first line contains single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the total number of snacks. The second line contains *n* integers, the *i*-th of them equals the size of the snack which fell on the *i*-th day. Sizes are distinct integers from 1 to *n*.

Print *n* lines. On the *i*-th of them print the sizes of the snacks which the residents placed on the top of the Snacktower on the *i*-th day in the order they will do that. If no snack is placed on some day, leave the corresponding line empty.

[ "3\n3 1 2\n", "5\n4 5 1 2 3\n" ]

[ "3\n \n2 1", "5 4\n \n \n3 2 1\n" ]

In the example a snack of size 3 fell on the first day, and the residents immediately placed it. On the second day a snack of size 1 fell, and the residents weren't able to place it because they were missing the snack of size 2. On the third day a snack of size 2 fell, and the residents immediately placed it. Right after that they placed the snack of size 1 which had fallen before.

500

[ { "input": "3\n3 1 2", "output": "3 \n\n2 1 " }, { "input": "5\n4 5 1 2 3", "output": "5 4 \n\n\n3 2 1 " }, { "input": "1\n1", "output": "1 " }, { "input": "2\n1 2", "output": "2 1 " }, { "input": "10\n5 1 6 2 8 3 4 10 9 7", "output": "10 \n9 8 \n7 6 5 4 3 2 1 " }, { "input": "30\n16 10 4 29 5 28 12 21 11 30 18 6 14 3 17 22 20 15 9 1 27 19 24 26 13 25 2 23 8 7", "output": "30 29 28 \n\n\n\n\n\n\n\n\n\n\n27 \n\n\n26 \n\n25 24 \n\n23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 \n8 \n7 6 5 4 3 2 1 " }, { "input": "100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 4 80 72 39", "output": "100 \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\n99 98 \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\n97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 \n\n\n\n\n\n\n\n\n80 79 78 77 76 75 74 73 \n72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 \n39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 " }, { "input": "2\n2 1", "output": "2 \n1 " } ]

1,593,284,745

2,147,483,647

Python 3

OK

TESTS

25

452

14,028,800

import sys n = int(input()) s = [int(i) for i in input().split()] next = n a = (n + 1) * [bool] for p in s: a[p] = 1 while(a[next] == 1): print(next, end=" ") next -= 1 print("")

Title: Snacktower Time Limit: None seconds Memory Limit: None megabytes Problem Description: According to an old legeng, a long time ago Ankh-Morpork residents did something wrong to miss Fortune, and she cursed them. She said that at some time *n* snacks of distinct sizes will fall on the city, and the residents should build a Snacktower of them by placing snacks one on another. Of course, big snacks should be at the bottom of the tower, while small snacks should be at the top. Years passed, and once different snacks started to fall onto the city, and the residents began to build the Snacktower. However, they faced some troubles. Each day exactly one snack fell onto the city, but their order was strange. So, at some days the residents weren't able to put the new stack on the top of the Snacktower: they had to wait until all the bigger snacks fell. Of course, in order to not to anger miss Fortune again, the residents placed each snack on the top of the tower immediately as they could do it. Write a program that models the behavior of Ankh-Morpork residents. Input Specification: The first line contains single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the total number of snacks. The second line contains *n* integers, the *i*-th of them equals the size of the snack which fell on the *i*-th day. Sizes are distinct integers from 1 to *n*. Output Specification: Print *n* lines. On the *i*-th of them print the sizes of the snacks which the residents placed on the top of the Snacktower on the *i*-th day in the order they will do that. If no snack is placed on some day, leave the corresponding line empty. Demo Input: ['3\n3 1 2\n', '5\n4 5 1 2 3\n'] Demo Output: ['3\n\xa0\n2 1', '5 4\n\xa0\n\xa0\n3 2 1\n'] Note: In the example a snack of size 3 fell on the first day, and the residents immediately placed it. On the second day a snack of size 1 fell, and the residents weren't able to place it because they were missing the snack of size 2. On the third day a snack of size 2 fell, and the residents immediately placed it. Right after that they placed the snack of size 1 which had fallen before.

```python import sys n = int(input()) s = [int(i) for i in input().split()] next = n a = (n + 1) * [bool] for p in s: a[p] = 1 while(a[next] == 1): print(next, end=" ") next -= 1 print("") ```

3

58

A

Chat room

PROGRAMMING

1,000

[ "greedy", "strings" ]

A. Chat room

1

256

Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.

The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.

If Vasya managed to say hello, print "YES", otherwise print "NO".

[ "ahhellllloou\n", "hlelo\n" ]

[ "YES\n", "NO\n" ]

none

500

[ { "input": "ahhellllloou", "output": "YES" }, { "input": "hlelo", "output": "NO" }, { "input": "helhcludoo", "output": "YES" }, { "input": "hehwelloho", "output": "YES" }, { "input": "pnnepelqomhhheollvlo", "output": "YES" }, { "input": "tymbzjyqhymedasloqbq", "output": "NO" }, { "input": "yehluhlkwo", "output": "NO" }, { "input": "hatlevhhalrohairnolsvocafgueelrqmlqlleello", "output": "YES" }, { "input": "hhhtehdbllnhwmbyhvelqqyoulretpbfokflhlhreeflxeftelziclrwllrpflflbdtotvlqgoaoqldlroovbfsq", "output": "YES" }, { "input": "rzlvihhghnelqtwlexmvdjjrliqllolhyewgozkuovaiezgcilelqapuoeglnwmnlftxxiigzczlouooi", "output": "YES" }, { "input": "pfhhwctyqdlkrwhebfqfelhyebwllhemtrmeblgrynmvyhioesqklclocxmlffuormljszllpoo", "output": "YES" }, { "input": "lqllcolohwflhfhlnaow", "output": "NO" }, { "input": "heheeellollvoo", "output": "YES" }, { "input": "hellooo", "output": "YES" }, { "input": "o", "output": "NO" }, { "input": "hhqhzeclohlehljlhtesllylrolmomvuhcxsobtsckogdv", "output": "YES" }, { "input": "yoegfuzhqsihygnhpnukluutocvvwuldiighpogsifealtgkfzqbwtmgghmythcxflebrkctlldlkzlagovwlstsghbouk", "output": "YES" }, { "input": "uatqtgbvrnywfacwursctpagasnhydvmlinrcnqrry", "output": "NO" }, { "input": "tndtbldbllnrwmbyhvqaqqyoudrstpbfokfoclnraefuxtftmgzicorwisrpfnfpbdtatvwqgyalqtdtrjqvbfsq", "output": "NO" }, { "input": "rzlvirhgemelnzdawzpaoqtxmqucnahvqnwldklrmjiiyageraijfivigvozgwngiulttxxgzczptusoi", "output": "YES" }, { "input": "kgyelmchocojsnaqdsyeqgnllytbqietpdlgknwwumqkxrexgdcnwoldicwzwofpmuesjuxzrasscvyuqwspm", "output": "YES" }, { "input": "pnyvrcotjvgynbeldnxieghfltmexttuxzyac", "output": "NO" }, { "input": "dtwhbqoumejligbenxvzhjlhosqojetcqsynlzyhfaevbdpekgbtjrbhlltbceobcok", "output": "YES" }, { "input": "crrfpfftjwhhikwzeedrlwzblckkteseofjuxjrktcjfsylmlsvogvrcxbxtffujqshslemnixoeezivksouefeqlhhokwbqjz", "output": "YES" }, { "input": "jhfbndhyzdvhbvhmhmefqllujdflwdpjbehedlsqfdsqlyelwjtyloxwsvasrbqosblzbowlqjmyeilcvotdlaouxhdpoeloaovb", "output": "YES" }, { "input": "hwlghueoemiqtjhhpashjsouyegdlvoyzeunlroypoprnhlyiwiuxrghekaylndhrhllllwhbebezoglydcvykllotrlaqtvmlla", "output": "YES" }, { "input": "wshiaunnqnqxodholbipwhhjmyeblhgpeleblklpzwhdunmpqkbuzloetmwwxmeltkrcomulxauzlwmlklldjodozxryghsnwgcz", "output": "YES" }, { "input": "shvksednttggehroewuiptvvxtrzgidravtnjwuqrlnnkxbplctzkckinpkgjopjfoxdbojtcvsuvablcbkrzajrlhgobkcxeqti", "output": "YES" }, { "input": "hyyhddqhxhekehkwfhlnlsihzefwchzerevcjtokefplholrbvxlltdlafjxrfhleglrvlolojoqaolagtbeyogxlbgfolllslli", "output": "YES" }, { "input": "iaagrdhhelxpdegueiulflquvlzidoprzkehthkldaytrjwhyhprwjxlltinxvuilxohqgjqcvkkdcuoliekcnlilwgqlnlzouoo", "output": "YES" }, { "input": "wfluaeseldgxyvxpwuhkptdmlflnlhktwxiabsvkolsquymrmhzczzoybvlilhmvsuunggvgxzgyyffk", "output": "NO" }, { "input": "loee", "output": "NO" }, { "input": "izdipwylefqmjbuoznfglgcfvedeouflkeehxbasaqmiooppfsqlhprospqgxvzrcpwlfdddngoqvpwvggsnvvxhmjoslwjudjlo", "output": "NO" }, { "input": "pplynflbjlseflaegzqpdxswgunhnlglqhycrbcwdfhclmcrwekkzsoosmfblljeenwfjyifcvxihhhbyhmhlkvtskeglaovjnsi", "output": "YES" }, { "input": "xqjqmenkodmlhzyzmmvofdngktygbbxbzpluzcohohmalkoeuwfikblltaaigv", "output": "NO" }, { "input": "hello", "output": "YES" }, { "input": "oohell", "output": "NO" }, { "input": "hell", "output": "NO" }, { "input": "eloellohoelo", "output": "NO" }, { "input": "helllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllo", "output": "YES" } ]

1,537,422,827

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

4

109

0

a=input().lower() b="" h=0 e=0 l=0 o=0 for x in a: if x =="h": if h==0: b=b+"h" h=h+1 elif x=="e": if e==0: b=b+"e" e=e+1 elif x=="l": if l!=2: b=b+"l" l=l+1 elif x=="o": if o==0: b=b+"o" o=o+1 if b.lower()=="hello": print("YES") else: print("NO")

Title: Chat room Time Limit: 1 seconds Memory Limit: 256 megabytes Problem Description: Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*. Input Specification: The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters. Output Specification: If Vasya managed to say hello, print "YES", otherwise print "NO". Demo Input: ['ahhellllloou\n', 'hlelo\n'] Demo Output: ['YES\n', 'NO\n'] Note: none

```python a=input().lower() b="" h=0 e=0 l=0 o=0 for x in a: if x =="h": if h==0: b=b+"h" h=h+1 elif x=="e": if e==0: b=b+"e" e=e+1 elif x=="l": if l!=2: b=b+"l" l=l+1 elif x=="o": if o==0: b=b+"o" o=o+1 if b.lower()=="hello": print("YES") else: print("NO") ```

0

289

B

Polo the Penguin and Matrix

PROGRAMMING

1,400

[ "brute force", "dp", "implementation", "sortings", "ternary search" ]

null

Little penguin Polo has an *n*<=×<=*m* matrix, consisting of integers. Let's index the matrix rows from 1 to *n* from top to bottom and let's index the columns from 1 to *m* from left to right. Let's represent the matrix element on the intersection of row *i* and column *j* as *a**ij*. In one move the penguin can add or subtract number *d* from some matrix element. Find the minimum number of moves needed to make all matrix elements equal. If the described plan is impossible to carry out, say so.

The first line contains three integers *n*, *m* and *d* (1<=≤<=*n*,<=*m*<=≤<=100,<=1<=≤<=*d*<=≤<=104) — the matrix sizes and the *d* parameter. Next *n* lines contain the matrix: the *j*-th integer in the *i*-th row is the matrix element *a**ij* (1<=≤<=*a**ij*<=≤<=104).

In a single line print a single integer — the minimum number of moves the penguin needs to make all matrix elements equal. If that is impossible, print "-1" (without the quotes).

[ "2 2 2\n2 4\n6 8\n", "1 2 7\n6 7\n" ]

[ "4\n", "-1\n" ]

none

1,000

[ { "input": "2 2 2\n2 4\n6 8", "output": "4" }, { "input": "1 2 7\n6 7", "output": "-1" }, { "input": "3 2 1\n5 7\n1 2\n5 100", "output": "104" }, { "input": "3 3 3\n5 8 5\n11 11 17\n14 5 2", "output": "12" }, { "input": "3 3 3\n5 8 5\n11 11 17\n14 5 3", "output": "-1" }, { "input": "2 2 4\n5 5\n5 5", "output": "0" }, { "input": "7 4 5\n7 7 7 12\n7 12 12 7\n7 7 7 7\n7 7 12 7\n7 7 12 12\n12 12 7 12\n7 7 7 7", "output": "9" }, { "input": "7 7 47\n91 91 91 91 91 91 91\n91 91 91 91 91 91 91\n91 91 91 91 91 91 91\n91 91 91 91 91 91 91\n91 91 91 91 91 91 91\n91 91 91 91 91 91 91\n91 91 91 91 91 91 91", "output": "0" }, { "input": "7 7 47\n47 47 47 47 47 47 47\n47 47 47 47 77 47 47\n47 47 47 47 47 47 47\n47 47 47 47 47 47 47\n47 47 47 47 47 47 47\n47 47 47 47 47 47 47\n47 47 47 47 47 127 47", "output": "-1" }, { "input": "7 5 47\n9583 1734 4601 5353 2110\n3802 5165 7985 6293 324\n7045 653 9160 7891 4930\n1781 3520 4178 2298 3943\n1405 2956 5447 5494 6528\n3097 1640 7750 4883 8032\n4225 8455 1875 4789 4366", "output": "1508" }, { "input": "1 1 1\n1", "output": "0" }, { "input": "1 1 8\n12", "output": "0" } ]

1,553,668,973

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

12

218

307,200

n,m,d = map(int,input().split()) a = [] check = 1 limit = 10001 mark = [0 for _ in range(limit)] ans = [] for _ in range(n): inn = list(map(int,input().split())) a = a+inn mem = {a[0]} total = a[0] mark[a[0]] = 1 for i in range(1,len(a)): mark[a[i]] +=1 mem.add(a[i]) if abs(a[0]-a[i])%d !=0: check = 0 break else: total +=a[i] meanSet = [] if check: mean = round(total/len(a)) if mark[mean]>0: meanSet.append(mean) need = 1 else: need = 2 l = mean r = mean run = need while(run): l-=1 if l >0 : if mark[l]>0: meanSet.append(l) run -=1 else: break run = need while (run): r += 1 if r < limit: if mark[r] > 0: meanSet.append(r) run -= 1 else: break for m in meanSet: temp = 0 for v in mem: mul = mark[v] temp += (abs(m-v)*mul)//d ans.append(temp) print(min(ans) if check else -1)

Title: Polo the Penguin and Matrix Time Limit: None seconds Memory Limit: None megabytes Problem Description: Little penguin Polo has an *n*<=×<=*m* matrix, consisting of integers. Let's index the matrix rows from 1 to *n* from top to bottom and let's index the columns from 1 to *m* from left to right. Let's represent the matrix element on the intersection of row *i* and column *j* as *a**ij*. In one move the penguin can add or subtract number *d* from some matrix element. Find the minimum number of moves needed to make all matrix elements equal. If the described plan is impossible to carry out, say so. Input Specification: The first line contains three integers *n*, *m* and *d* (1<=≤<=*n*,<=*m*<=≤<=100,<=1<=≤<=*d*<=≤<=104) — the matrix sizes and the *d* parameter. Next *n* lines contain the matrix: the *j*-th integer in the *i*-th row is the matrix element *a**ij* (1<=≤<=*a**ij*<=≤<=104). Output Specification: In a single line print a single integer — the minimum number of moves the penguin needs to make all matrix elements equal. If that is impossible, print "-1" (without the quotes). Demo Input: ['2 2 2\n2 4\n6 8\n', '1 2 7\n6 7\n'] Demo Output: ['4\n', '-1\n'] Note: none

```python n,m,d = map(int,input().split()) a = [] check = 1 limit = 10001 mark = [0 for _ in range(limit)] ans = [] for _ in range(n): inn = list(map(int,input().split())) a = a+inn mem = {a[0]} total = a[0] mark[a[0]] = 1 for i in range(1,len(a)): mark[a[i]] +=1 mem.add(a[i]) if abs(a[0]-a[i])%d !=0: check = 0 break else: total +=a[i] meanSet = [] if check: mean = round(total/len(a)) if mark[mean]>0: meanSet.append(mean) need = 1 else: need = 2 l = mean r = mean run = need while(run): l-=1 if l >0 : if mark[l]>0: meanSet.append(l) run -=1 else: break run = need while (run): r += 1 if r < limit: if mark[r] > 0: meanSet.append(r) run -= 1 else: break for m in meanSet: temp = 0 for v in mem: mul = mark[v] temp += (abs(m-v)*mul)//d ans.append(temp) print(min(ans) if check else -1) ```

0

761

A

Dasha and Stairs

PROGRAMMING

1,000

[ "brute force", "constructive algorithms", "implementation", "math" ]

null

On her way to programming school tiger Dasha faced her first test — a huge staircase! The steps were numbered from one to infinity. As we know, tigers are very fond of all striped things, it is possible that it has something to do with their color. So on some interval of her way she calculated two values — the number of steps with even and odd numbers. You need to check whether there is an interval of steps from the *l*-th to the *r*-th (1<=≤<=*l*<=≤<=*r*), for which values that Dasha has found are correct.

In the only line you are given two integers *a*, *b* (0<=≤<=*a*,<=*b*<=≤<=100) — the number of even and odd steps, accordingly.

In the only line print "YES", if the interval of steps described above exists, and "NO" otherwise.

[ "2 3\n", "3 1\n" ]

[ "YES\n", "NO\n" ]

In the first example one of suitable intervals is from 1 to 5. The interval contains two even steps — 2 and 4, and three odd: 1, 3 and 5.

500

[ { "input": "2 3", "output": "YES" }, { "input": "3 1", "output": "NO" }, { "input": "5 4", "output": "YES" }, { "input": "9 9", "output": "YES" }, { "input": "85 95", "output": "NO" }, { "input": "0 1", "output": "YES" }, { "input": "89 25", "output": "NO" }, { "input": "74 73", "output": "YES" }, { "input": "62 39", "output": "NO" }, { "input": "57 57", "output": "YES" }, { "input": "100 99", "output": "YES" }, { "input": "0 0", "output": "NO" }, { "input": "98 100", "output": "NO" }, { "input": "99 100", "output": "YES" }, { "input": "1 0", "output": "YES" }, { "input": "100 0", "output": "NO" }, { "input": "0 100", "output": "NO" }, { "input": "100 98", "output": "NO" }, { "input": "100 100", "output": "YES" }, { "input": "0 5", "output": "NO" }, { "input": "2 2", "output": "YES" } ]

1,536,896,556

2,147,483,647

Python 3

RUNTIME_ERROR

TESTS

0

92

0

a=int(input()) b=int(input()) if ((a!=0)or(b!=0))and(abs(a-b)<=1): print("YES") else: print("NO")

Title: Dasha and Stairs Time Limit: None seconds Memory Limit: None megabytes Problem Description: On her way to programming school tiger Dasha faced her first test — a huge staircase! The steps were numbered from one to infinity. As we know, tigers are very fond of all striped things, it is possible that it has something to do with their color. So on some interval of her way she calculated two values — the number of steps with even and odd numbers. You need to check whether there is an interval of steps from the *l*-th to the *r*-th (1<=≤<=*l*<=≤<=*r*), for which values that Dasha has found are correct. Input Specification: In the only line you are given two integers *a*, *b* (0<=≤<=*a*,<=*b*<=≤<=100) — the number of even and odd steps, accordingly. Output Specification: In the only line print "YES", if the interval of steps described above exists, and "NO" otherwise. Demo Input: ['2 3\n', '3 1\n'] Demo Output: ['YES\n', 'NO\n'] Note: In the first example one of suitable intervals is from 1 to 5. The interval contains two even steps — 2 and 4, and three odd: 1, 3 and 5.

```python a=int(input()) b=int(input()) if ((a!=0)or(b!=0))and(abs(a-b)<=1): print("YES") else: print("NO") ```

-1

698

A

Vacations

PROGRAMMING

1,400

[ "dp" ]

null

Vasya has *n* days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this *n* days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the *i*-th day there are four options: 1. on this day the gym is closed and the contest is not carried out; 1. on this day the gym is closed and the contest is carried out; 1. on this day the gym is open and the contest is not carried out; 1. on this day the gym is open and the contest is carried out. On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day). Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.

The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of days of Vasya's vacations. The second line contains the sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=3) separated by space, where: - *a**i* equals 0, if on the *i*-th day of vacations the gym is closed and the contest is not carried out; - *a**i* equals 1, if on the *i*-th day of vacations the gym is closed, but the contest is carried out; - *a**i* equals 2, if on the *i*-th day of vacations the gym is open and the contest is not carried out; - *a**i* equals 3, if on the *i*-th day of vacations the gym is open and the contest is carried out.

Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses: - to do sport on any two consecutive days, - to write the contest on any two consecutive days.

[ "4\n1 3 2 0\n", "7\n1 3 3 2 1 2 3\n", "2\n2 2\n" ]

[ "2\n", "0\n", "1\n" ]

In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days. In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day. In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.

500

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2 2 3 3 3 3 3 3 3 3 3 3 1 2 2 3 2 3 3 3 2 1 3 3 3 1 3", "output": "2" }, { "input": "35\n0 1 1 0 0 2 0 0 1 0 0 0 1 0 1 0 1 0 0 0 1 2 1 0 2 2 1 0 1 0 1 1 1 0 0", "output": "21" }, { "input": "35\n2 2 0 3 2 2 0 3 3 1 1 3 3 1 2 2 0 2 2 2 2 3 1 0 2 1 3 2 2 3 2 3 3 1 2", "output": "11" }, { "input": "35\n1 2 2 3 3 3 3 3 2 2 3 3 2 3 3 2 3 2 3 3 2 2 2 3 3 2 3 3 3 1 3 3 2 2 2", "output": "7" }, { "input": "40\n2 0 1 1 0 0 0 0 2 0 1 1 1 0 0 1 0 0 0 0 0 2 0 0 0 2 1 1 1 3 0 0 0 0 0 0 0 1 1 0", "output": "28" }, { "input": "40\n2 2 3 2 0 2 3 2 1 2 3 0 2 3 2 1 1 3 1 1 0 2 3 1 3 3 1 1 3 3 2 2 1 3 3 3 2 3 3 1", "output": "10" }, { "input": "40\n1 3 2 3 3 2 3 3 2 2 3 1 2 1 2 2 3 1 2 2 1 2 2 2 1 2 2 3 2 3 2 3 2 3 3 3 1 3 2 3", "output": "8" }, { "input": "45\n2 1 0 0 0 2 1 0 1 0 0 2 2 1 1 0 0 2 0 0 0 0 0 0 1 0 0 2 0 0 1 1 0 0 1 0 0 1 1 2 0 0 2 0 2", "output": "29" }, { "input": "45\n3 3 2 3 3 3 2 2 3 2 3 1 3 2 3 2 2 1 1 3 2 3 2 1 3 1 2 3 2 2 0 3 3 2 3 2 3 2 3 2 0 3 1 1 3", "output": "8" }, { "input": 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"60\n3 2 1 3 2 2 3 3 3 1 1 3 2 2 3 3 1 3 2 2 3 3 2 2 2 2 0 2 2 3 2 3 0 3 3 3 2 3 3 0 1 3 2 1 3 1 1 2 1 3 1 1 2 2 1 3 3 3 2 2", "output": "15" }, { "input": "60\n3 2 2 3 2 3 2 3 3 2 3 2 3 3 2 3 3 3 3 3 3 2 3 3 1 2 3 3 3 2 1 3 3 1 3 1 3 0 3 3 3 2 3 2 3 2 3 3 1 1 2 3 3 3 3 2 1 3 2 3", "output": "8" }, { "input": "65\n1 0 2 1 1 0 1 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 0 1 2 0 2 1 0 2 1 0 1 0 1 1 0 1 1 1 2 1 0 1 0 0 0 0 1 2 2 1 0 0 1 2 1 2 0 2 0 0 0 1 1", "output": "35" }, { "input": "65\n2 2 2 3 0 2 1 2 3 3 1 3 1 2 1 3 2 3 2 2 2 1 2 0 3 1 3 1 1 3 1 3 3 3 3 3 1 3 0 3 1 3 1 2 2 3 2 0 3 1 3 2 1 2 2 2 3 3 2 3 3 3 2 2 3", "output": "13" }, { "input": "65\n3 2 3 3 3 2 3 2 3 3 3 3 3 3 3 3 3 2 3 2 3 2 2 3 3 3 3 3 2 2 2 3 3 2 3 3 2 3 3 3 3 2 3 3 3 2 2 3 3 3 3 3 3 2 2 3 3 2 3 3 1 3 3 3 3", "output": "6" }, { "input": "70\n1 0 0 0 1 0 1 0 0 0 1 1 0 1 0 0 1 1 1 0 1 1 0 0 1 1 1 3 1 1 0 1 2 0 2 1 0 0 0 1 1 1 1 1 0 0 1 0 0 0 1 1 1 3 0 0 1 0 0 0 1 0 0 0 0 0 1 0 1 1", "output": "43" }, { "input": "70\n2 3 3 3 1 3 3 1 2 1 1 2 2 3 0 2 3 3 1 3 3 2 2 3 3 3 2 2 2 2 1 3 3 0 2 1 1 3 2 3 3 2 2 3 1 3 1 2 3 2 3 3 2 2 2 3 1 1 2 1 3 3 2 2 3 3 3 1 1 1", "output": "16" }, { "input": "70\n3 3 2 2 1 2 1 2 2 2 2 2 3 3 2 3 3 3 3 2 2 2 2 3 3 3 1 3 3 3 2 3 3 3 3 2 3 3 1 3 1 3 2 3 3 2 3 3 3 2 3 2 3 3 1 2 3 3 2 2 2 3 2 3 3 3 3 3 3 1", "output": "10" }, { "input": "75\n1 0 0 1 1 0 0 1 0 1 2 0 0 2 1 1 0 0 0 0 0 0 2 1 1 0 0 0 0 1 0 1 0 1 1 1 0 1 0 0 1 0 0 0 0 0 0 1 1 0 0 1 2 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 1 1 1 0 1 0", "output": "51" }, { "input": "75\n1 3 3 3 1 1 3 2 3 3 1 3 3 3 2 1 3 2 2 3 1 1 1 1 1 1 2 3 3 3 3 3 3 2 3 3 3 3 3 2 3 3 2 2 2 1 2 3 3 2 2 3 0 1 1 3 3 0 0 1 1 3 2 3 3 3 3 1 2 2 3 3 3 3 1", "output": "16" }, { "input": "75\n3 3 3 3 2 2 3 2 2 3 2 2 1 2 3 3 2 2 3 3 1 2 2 2 1 3 3 3 1 2 2 3 3 3 2 3 2 2 2 3 3 1 3 2 2 3 3 3 0 3 2 1 3 3 2 3 3 3 3 1 2 3 3 3 2 2 3 3 3 3 2 2 3 3 1", "output": "11" }, { "input": "80\n0 0 0 0 2 0 1 1 1 1 1 0 0 0 0 2 0 0 1 0 0 0 0 1 1 0 2 2 1 1 0 1 0 1 0 1 1 1 0 1 2 1 1 0 0 0 1 1 0 1 1 0 1 0 0 1 0 0 1 0 0 0 0 0 0 0 2 2 0 1 1 0 0 0 0 0 0 0 0 1", "output": "56" }, { "input": "80\n2 2 3 3 2 1 0 1 0 3 2 2 3 2 1 3 1 3 3 2 3 3 3 2 3 3 3 2 1 3 3 1 3 3 3 3 3 3 2 2 2 1 3 2 1 3 2 1 1 0 1 1 2 1 3 0 1 2 3 2 2 3 2 3 1 3 3 2 1 1 0 3 3 3 3 1 2 1 2 0", "output": "17" }, { "input": "80\n2 3 3 2 2 2 3 3 2 3 3 3 3 3 2 3 2 3 2 3 3 3 3 3 3 3 3 3 2 3 1 3 2 3 3 0 3 1 2 3 3 1 2 3 2 3 3 2 3 3 3 3 3 2 2 3 0 3 3 3 3 3 2 2 3 2 3 3 3 3 3 2 3 2 3 3 3 3 2 3", "output": "9" }, { "input": "85\n0 1 1 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 2 0 1 0 0 2 0 1 1 0 0 0 0 2 2 0 0 0 1 0 0 0 1 2 0 1 0 0 0 2 1 1 2 0 3 1 0 2 2 1 0 0 1 1 0 0 0 0 1 0 2 1 1 2 1 0 0 1 2 1 2 0 0 1 0 1 0", "output": "54" }, { "input": "85\n2 3 1 3 2 3 1 3 3 2 1 2 1 2 2 3 2 2 3 2 0 3 3 2 1 2 2 2 3 3 2 3 3 3 2 1 1 3 1 3 2 2 2 3 3 2 3 2 3 1 1 3 2 3 1 3 3 2 3 3 2 2 3 0 1 1 2 2 2 2 1 2 3 1 3 3 1 3 2 2 3 2 3 3 3", "output": "19" }, { "input": "85\n1 2 1 2 3 2 3 3 3 3 3 3 3 2 1 3 2 3 3 3 3 2 3 3 3 1 3 3 3 3 2 3 3 3 3 3 3 2 2 1 3 3 3 3 2 2 3 1 1 2 3 3 3 2 3 3 3 3 3 2 3 3 3 2 2 3 3 1 1 1 3 3 3 3 1 3 3 3 1 3 3 1 3 2 3", "output": "9" }, { "input": "90\n2 0 1 0 0 0 0 0 0 1 1 2 0 0 0 0 0 0 0 2 2 0 2 0 0 2 1 0 2 0 1 0 1 0 0 1 2 2 0 0 1 0 0 1 0 1 0 2 0 1 1 1 0 1 1 0 1 0 2 0 1 0 1 0 0 0 1 0 0 1 2 0 0 0 1 0 0 2 2 0 0 0 0 0 1 3 1 1 0 1", "output": "57" }, { "input": "90\n2 3 3 3 2 3 2 1 3 0 3 2 3 3 2 1 3 3 2 3 2 3 3 2 1 3 1 3 3 1 2 2 3 3 2 1 2 3 2 3 0 3 3 2 2 3 1 0 3 3 1 3 3 3 3 2 1 2 2 1 3 2 1 3 3 1 2 0 2 2 3 2 2 3 3 3 1 3 2 1 2 3 3 2 3 2 3 3 2 1", "output": "17" }, { "input": "90\n2 3 2 3 2 2 3 3 2 3 2 1 2 3 3 3 2 3 2 3 3 2 3 3 3 1 3 3 1 3 2 3 2 2 1 3 3 3 3 3 3 3 3 3 3 2 3 2 3 2 1 3 3 3 3 2 2 3 3 3 3 3 3 3 3 3 3 3 3 2 2 3 3 3 3 1 3 2 3 3 3 2 2 3 2 3 2 1 3 2", "output": "9" }, { "input": "95\n0 0 3 0 2 0 1 0 0 2 0 0 0 0 0 0 0 1 0 0 0 2 0 0 0 0 0 1 0 0 2 1 0 0 1 0 0 0 1 0 0 0 0 1 0 1 0 0 1 0 1 2 0 1 2 2 0 0 1 0 2 0 0 0 1 0 2 1 2 1 0 1 0 0 0 1 0 0 1 1 2 1 1 1 1 2 0 0 0 0 0 1 1 0 1", "output": "61" }, { "input": "95\n2 3 3 2 1 1 3 3 3 2 3 3 3 2 3 2 3 3 3 2 3 2 2 3 3 2 1 2 3 3 3 1 3 0 3 3 1 3 3 1 0 1 3 3 3 0 2 1 3 3 3 3 0 1 3 2 3 3 2 1 3 1 2 1 1 2 3 0 3 3 2 1 3 2 1 3 3 3 2 2 3 2 3 3 3 2 1 3 3 3 2 3 3 1 2", "output": "15" }, { "input": "95\n2 3 3 2 3 2 2 1 3 1 2 1 2 3 1 2 3 3 1 3 3 3 1 2 3 2 2 2 2 3 3 3 2 2 3 3 3 3 3 1 2 2 3 3 3 3 2 3 2 2 2 3 3 2 3 3 3 3 3 3 3 0 3 2 0 3 3 1 3 3 3 2 3 2 3 2 3 3 3 3 2 2 1 1 3 3 3 3 3 1 3 3 3 3 2", "output": "14" }, { "input": "100\n1 0 2 0 0 0 0 2 0 0 0 1 0 1 0 0 1 0 1 2 0 1 1 0 0 1 0 1 1 0 0 0 2 0 1 0 0 2 0 0 0 0 0 1 1 1 0 0 1 0 2 0 0 0 0 1 0 1 0 1 0 1 0 1 2 2 0 0 2 0 1 0 1 0 1 0 0 0 1 0 0 2 1 1 1 0 0 1 0 0 0 2 0 0 2 1 1 0 0 2", "output": "63" }, { "input": "100\n3 2 1 3 2 3 2 3 2 2 3 1 3 3 3 3 3 2 2 3 2 2 3 2 3 3 3 2 3 1 2 1 3 3 3 3 1 3 3 3 3 3 2 3 2 1 3 3 1 2 2 3 1 3 3 1 2 2 1 3 1 3 2 2 3 3 1 3 2 3 1 2 1 2 3 3 2 2 1 2 3 3 3 3 3 1 3 3 3 3 2 1 3 0 3 3 3 2 3 3", "output": "15" }, { "input": "100\n1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2", "output": "0" }, { "input": "100\n3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3", "output": "0" }, { "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": "50" }, { "input": "100\n2 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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2", "output": "50" }, { "input": "99\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", "output": "49" }, { "input": "100\n2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1", "output": "0" }, { "input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0", "output": "100" }, { "input": "2\n0 3", "output": "1" }, { "input": "2\n1 0", "output": "1" }, { "input": "2\n1 2", "output": "0" }, { "input": "2\n2 0", "output": "1" }, { "input": "2\n2 1", "output": "0" }, { "input": "2\n2 3", "output": "0" }, { "input": "2\n3 0", "output": "1" }, { "input": "2\n3 1", "output": "0" }, { "input": "100\n3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0", "output": "50" } ]

1,597,119,184

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

6

108

6,963,200

n = int(input()) a = [int(i) for i in input().split()] dp = [{'rest': 0, 'contest': 0, 'gym': 0} for i in range(n)] for i in range(n): if a[i] == 1 or a[i] == 3: dp[i]['contest'] = (dp[i - 1]['gym'] | dp[i - 1]['rest']) if i == 0: dp[i]['contest'] = 1 if a[i] == 2 or a[i] == 3: dp[i]['gym'] = (dp[i - 1]['contest'] | dp[i - 1]['rest']) if i == 0: dp[i]['gym'] == 1 if (dp[i]['contest'] | dp[i]['gym']) == 0: dp[i]['rest'] = 1 count = 0 for i in range(n): count += dp[i]['rest'] print(count)

Title: Vacations Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vasya has *n* days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this *n* days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the *i*-th day there are four options: 1. on this day the gym is closed and the contest is not carried out; 1. on this day the gym is closed and the contest is carried out; 1. on this day the gym is open and the contest is not carried out; 1. on this day the gym is open and the contest is carried out. On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day). Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days. Input Specification: The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of days of Vasya's vacations. The second line contains the sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=3) separated by space, where: - *a**i* equals 0, if on the *i*-th day of vacations the gym is closed and the contest is not carried out; - *a**i* equals 1, if on the *i*-th day of vacations the gym is closed, but the contest is carried out; - *a**i* equals 2, if on the *i*-th day of vacations the gym is open and the contest is not carried out; - *a**i* equals 3, if on the *i*-th day of vacations the gym is open and the contest is carried out. Output Specification: Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses: - to do sport on any two consecutive days, - to write the contest on any two consecutive days. Demo Input: ['4\n1 3 2 0\n', '7\n1 3 3 2 1 2 3\n', '2\n2 2\n'] Demo Output: ['2\n', '0\n', '1\n'] Note: In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days. In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day. In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.

```python n = int(input()) a = [int(i) for i in input().split()] dp = [{'rest': 0, 'contest': 0, 'gym': 0} for i in range(n)] for i in range(n): if a[i] == 1 or a[i] == 3: dp[i]['contest'] = (dp[i - 1]['gym'] | dp[i - 1]['rest']) if i == 0: dp[i]['contest'] = 1 if a[i] == 2 or a[i] == 3: dp[i]['gym'] = (dp[i - 1]['contest'] | dp[i - 1]['rest']) if i == 0: dp[i]['gym'] == 1 if (dp[i]['contest'] | dp[i]['gym']) == 0: dp[i]['rest'] = 1 count = 0 for i in range(n): count += dp[i]['rest'] print(count) ```

0

760

B

Frodo and pillows

PROGRAMMING

1,500

[ "binary search", "greedy" ]

null

*n* hobbits are planning to spend the night at Frodo's house. Frodo has *n* beds standing in a row and *m* pillows (*n*<=≤<=*m*). Each hobbit needs a bed and at least one pillow to sleep, however, everyone wants as many pillows as possible. Of course, it's not always possible to share pillows equally, but any hobbit gets hurt if he has at least two pillows less than some of his neighbors have. Frodo will sleep on the *k*-th bed in the row. What is the maximum number of pillows he can have so that every hobbit has at least one pillow, every pillow is given to some hobbit and no one is hurt?

The only line contain three integers *n*, *m* and *k* (1<=≤<=*n*<=≤<=*m*<=≤<=109, 1<=≤<=*k*<=≤<=*n*) — the number of hobbits, the number of pillows and the number of Frodo's bed.

Print single integer — the maximum number of pillows Frodo can have so that no one is hurt.

[ "4 6 2\n", "3 10 3\n", "3 6 1\n" ]

[ "2\n", "4\n", "3\n" ]

In the first example Frodo can have at most two pillows. In this case, he can give two pillows to the hobbit on the first bed, and one pillow to each of the hobbits on the third and the fourth beds. In the second example Frodo can take at most four pillows, giving three pillows to each of the others. In the third example Frodo can take three pillows, giving two pillows to the hobbit in the middle and one pillow to the hobbit on the third bed.

1,000

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1,694,443,815

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

3

31

0

vals = input() vals = vals.split() vals = [int(num) for num in vals] n = vals[0] m = vals[1] k = vals[2] pillows_per_hobbit = m//n print(pillows_per_hobbit+1)

Title: Frodo and pillows Time Limit: None seconds Memory Limit: None megabytes Problem Description: *n* hobbits are planning to spend the night at Frodo's house. Frodo has *n* beds standing in a row and *m* pillows (*n*<=≤<=*m*). Each hobbit needs a bed and at least one pillow to sleep, however, everyone wants as many pillows as possible. Of course, it's not always possible to share pillows equally, but any hobbit gets hurt if he has at least two pillows less than some of his neighbors have. Frodo will sleep on the *k*-th bed in the row. What is the maximum number of pillows he can have so that every hobbit has at least one pillow, every pillow is given to some hobbit and no one is hurt? Input Specification: The only line contain three integers *n*, *m* and *k* (1<=≤<=*n*<=≤<=*m*<=≤<=109, 1<=≤<=*k*<=≤<=*n*) — the number of hobbits, the number of pillows and the number of Frodo's bed. Output Specification: Print single integer — the maximum number of pillows Frodo can have so that no one is hurt. Demo Input: ['4 6 2\n', '3 10 3\n', '3 6 1\n'] Demo Output: ['2\n', '4\n', '3\n'] Note: In the first example Frodo can have at most two pillows. In this case, he can give two pillows to the hobbit on the first bed, and one pillow to each of the hobbits on the third and the fourth beds. In the second example Frodo can take at most four pillows, giving three pillows to each of the others. In the third example Frodo can take three pillows, giving two pillows to the hobbit in the middle and one pillow to the hobbit on the third bed.

```python vals = input() vals = vals.split() vals = [int(num) for num in vals] n = vals[0] m = vals[1] k = vals[2] pillows_per_hobbit = m//n print(pillows_per_hobbit+1) ```

0

545

C

Woodcutters

PROGRAMMING

1,500

[ "dp", "greedy" ]

null

Little Susie listens to fairy tales before bed every day. Today's fairy tale was about wood cutters and the little girl immediately started imagining the choppers cutting wood. She imagined the situation that is described below. There are *n* trees located along the road at points with coordinates *x*1,<=*x*2,<=...,<=*x**n*. Each tree has its height *h**i*. Woodcutters can cut down a tree and fell it to the left or to the right. After that it occupies one of the segments [*x**i*<=-<=*h**i*,<=*x**i*] or [*x**i*;*x**i*<=+<=*h**i*]. The tree that is not cut down occupies a single point with coordinate *x**i*. Woodcutters can fell a tree if the segment to be occupied by the fallen tree doesn't contain any occupied point. The woodcutters want to process as many trees as possible, so Susie wonders, what is the maximum number of trees to fell.

The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of trees. Next *n* lines contain pairs of integers *x**i*,<=*h**i* (1<=≤<=*x**i*,<=*h**i*<=≤<=109) — the coordinate and the height of the *і*-th tree. The pairs are given in the order of ascending *x**i*. No two trees are located at the point with the same coordinate.

Print a single number — the maximum number of trees that you can cut down by the given rules.

[ "5\n1 2\n2 1\n5 10\n10 9\n19 1\n", "5\n1 2\n2 1\n5 10\n10 9\n20 1\n" ]

[ "3\n", "4\n" ]

In the first sample you can fell the trees like that: - fell the 1-st tree to the left — now it occupies segment [ - 1;1] - fell the 2-nd tree to the right — now it occupies segment [2;3] - leave the 3-rd tree — it occupies point 5 - leave the 4-th tree — it occupies point 10 - fell the 5-th tree to the right — now it occupies segment [19;20] In the second sample you can also fell 4-th tree to the right, after that it will occupy segment [10;19].

1,750

[ { "input": "5\n1 2\n2 1\n5 10\n10 9\n19 1", "output": "3" }, { "input": "5\n1 2\n2 1\n5 10\n10 9\n20 1", "output": "4" }, { "input": "4\n10 4\n15 1\n19 3\n20 1", "output": "4" }, { "input": "35\n1 7\n3 11\n6 12\n7 6\n8 5\n9 11\n15 3\n16 10\n22 2\n23 3\n25 7\n27 3\n34 5\n35 10\n37 3\n39 4\n40 5\n41 1\n44 1\n47 7\n48 11\n50 6\n52 5\n57 2\n58 7\n60 4\n62 1\n67 3\n68 12\n69 8\n70 1\n71 5\n72 5\n73 6\n74 4", "output": "10" }, { "input": "40\n1 1\n2 1\n3 1\n4 1\n5 1\n6 1\n7 1\n8 1\n9 1\n10 1\n11 1\n12 1\n13 1\n14 1\n15 1\n16 1\n17 1\n18 1\n19 1\n20 1\n21 1\n22 1\n23 1\n24 1\n25 1\n26 1\n27 1\n28 1\n29 1\n30 1\n31 1\n32 1\n33 1\n34 1\n35 1\n36 1\n37 1\n38 1\n39 1\n40 1", "output": "2" }, { "input": "67\n1 1\n3 8\n4 10\n7 8\n9 2\n10 1\n11 5\n12 8\n13 4\n16 6\n18 3\n19 3\n22 5\n24 6\n27 5\n28 3\n29 3\n30 5\n32 5\n33 10\n34 7\n35 8\n36 5\n41 3\n42 2\n43 5\n46 4\n48 4\n49 9\n52 4\n53 9\n55 1\n56 4\n59 7\n68 7\n69 4\n71 9\n72 10\n74 5\n76 4\n77 9\n80 7\n81 9\n82 5\n83 5\n84 9\n85 7\n86 9\n87 4\n88 7\n89 10\n90 3\n91 5\n92 10\n93 5\n94 8\n95 4\n96 2\n97 10\n98 1\n99 3\n100 1\n101 5\n102 4\n103 8\n104 8\n105 8", "output": "5" }, { "input": "1\n1000000000 1000000000", "output": "1" }, { "input": "10\n7 12\n10 2\n12 2\n15 1\n19 2\n20 1\n53 25\n63 10\n75 12\n87 1", "output": "9" }, { "input": "3\n1 1\n1000 1000\n1000000000 1000000000", "output": "3" }, { "input": "2\n1 999999999\n1000000000 1000000000", "output": "2" }, { "input": "10\n999999900 1000000000\n999999901 1000000000\n999999902 1000000000\n999999903 1000000000\n999999904 1000000000\n999999905 1000000000\n999999906 1000000000\n999999907 1000000000\n999999908 1000000000\n999999909 1000000000", "output": "2" }, { "input": "2\n100000000 1000000000\n1000000000 1000000000", "output": "2" } ]

1,697,717,570

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

2

46

0

n=int(input()) a=[[1,0]]+[[0]*2 for i in range(n-2)]+[[0,1]];x=set();c=[0,0];k=2 for i in range(n): b=list(map(int,input().split())) x.add(b[0]) if b[0]-b[1]>c[0] and c[0]+c[1]<b[0] and c[0]!=0 and i !=n-1: k+=2 elif b[0]-b[1]>c[0]and a[i-1]==[1,0]: a[i][0]=1;k+=1 elif c[0]+c[1]<b[0]and a[i][0]==0: a[i-1][1]=1;k+=1 c=b[:] print(k)

Title: Woodcutters Time Limit: None seconds Memory Limit: None megabytes Problem Description: Little Susie listens to fairy tales before bed every day. Today's fairy tale was about wood cutters and the little girl immediately started imagining the choppers cutting wood. She imagined the situation that is described below. There are *n* trees located along the road at points with coordinates *x*1,<=*x*2,<=...,<=*x**n*. Each tree has its height *h**i*. Woodcutters can cut down a tree and fell it to the left or to the right. After that it occupies one of the segments [*x**i*<=-<=*h**i*,<=*x**i*] or [*x**i*;*x**i*<=+<=*h**i*]. The tree that is not cut down occupies a single point with coordinate *x**i*. Woodcutters can fell a tree if the segment to be occupied by the fallen tree doesn't contain any occupied point. The woodcutters want to process as many trees as possible, so Susie wonders, what is the maximum number of trees to fell. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of trees. Next *n* lines contain pairs of integers *x**i*,<=*h**i* (1<=≤<=*x**i*,<=*h**i*<=≤<=109) — the coordinate and the height of the *і*-th tree. The pairs are given in the order of ascending *x**i*. No two trees are located at the point with the same coordinate. Output Specification: Print a single number — the maximum number of trees that you can cut down by the given rules. Demo Input: ['5\n1 2\n2 1\n5 10\n10 9\n19 1\n', '5\n1 2\n2 1\n5 10\n10 9\n20 1\n'] Demo Output: ['3\n', '4\n'] Note: In the first sample you can fell the trees like that: - fell the 1-st tree to the left — now it occupies segment [ - 1;1] - fell the 2-nd tree to the right — now it occupies segment [2;3] - leave the 3-rd tree — it occupies point 5 - leave the 4-th tree — it occupies point 10 - fell the 5-th tree to the right — now it occupies segment [19;20] In the second sample you can also fell 4-th tree to the right, after that it will occupy segment [10;19].

```python n=int(input()) a=[[1,0]]+[[0]*2 for i in range(n-2)]+[[0,1]];x=set();c=[0,0];k=2 for i in range(n): b=list(map(int,input().split())) x.add(b[0]) if b[0]-b[1]>c[0] and c[0]+c[1]<b[0] and c[0]!=0 and i !=n-1: k+=2 elif b[0]-b[1]>c[0]and a[i-1]==[1,0]: a[i][0]=1;k+=1 elif c[0]+c[1]<b[0]and a[i][0]==0: a[i-1][1]=1;k+=1 c=b[:] print(k) ```

0

910

A

The Way to Home

PROGRAMMING

800

[ "dfs and similar", "dp", "greedy", "implementation" ]

null

A frog lives on the axis *Ox* and needs to reach home which is in the point *n*. She starts from the point 1. The frog can jump to the right at a distance not more than *d*. So, after she jumped from the point *x* she can reach the point *x*<=+<=*a*, where *a* is an integer from 1 to *d*. For each point from 1 to *n* is known if there is a lily flower in it. The frog can jump only in points with a lilies. Guaranteed that there are lilies in the points 1 and *n*. Determine the minimal number of jumps that the frog needs to reach home which is in the point *n* from the point 1. Consider that initially the frog is in the point 1. If the frog can not reach home, print -1.

The first line contains two integers *n* and *d* (2<=≤<=*n*<=≤<=100, 1<=≤<=*d*<=≤<=*n*<=-<=1) — the point, which the frog wants to reach, and the maximal length of the frog jump. The second line contains a string *s* of length *n*, consisting of zeros and ones. If a character of the string *s* equals to zero, then in the corresponding point there is no lily flower. In the other case, in the corresponding point there is a lily flower. Guaranteed that the first and the last characters of the string *s* equal to one.

If the frog can not reach the home, print -1. In the other case, print the minimal number of jumps that the frog needs to reach the home which is in the point *n* from the point 1.

[ "8 4\n10010101\n", "4 2\n1001\n", "8 4\n11100101\n", "12 3\n101111100101\n" ]

[ "2\n", "-1\n", "3\n", "4\n" ]

In the first example the from can reach home in two jumps: the first jump from the point 1 to the point 4 (the length of the jump is three), and the second jump from the point 4 to the point 8 (the length of the jump is four). In the second example the frog can not reach home, because to make it she need to jump on a distance three, but the maximum length of her jump equals to two.

500

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13\n1000000001101001110000010000011001000000000000001010000000100001001010000000000000000100010000000001", "output": "-1" }, { "input": "100 18\n1000000000000000110000000000000000010000000001000001000001000000000100000000000010000000000000000001", "output": "-1" }, { "input": "100 32\n1000000000000000000000000001000000000000000000000101000000000000000000000000000000000001000000000001", "output": "-1" }, { "input": "100 79\n1000000001000000000101000000000000000000000000000000000000000000000000000000000000000000000000000001", "output": "2" }, { "input": "100 41\n1000000000000000000000000000000000010000000000000000000000000000000000000000100000000000000000000001", "output": "3" }, { "input": "100 82\n1000000000000000000100000000000000000000000000000000000000000000000000000000000000000000000000000001", "output": "2" }, { "input": "100 96\n1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001", "output": "-1" }, { "input": "43 30\n1001000001111111010100100100110101011101101", "output": "2" }, { "input": "7 1\n1111111", "output": "6" }, { "input": "9 3\n101000001", "output": "-1" }, { "input": "10 3\n1100000001", "output": "-1" }, { "input": "8 2\n10000101", "output": "-1" }, { "input": "2 1\n11", "output": "1" } ]

1,618,255,309

2,147,483,647

Python 3

OK

TESTS

63

77

0

# https://codeforces.com/problemset/problem/910/A n, d = list(map(int, input().split())) arr = input() i, count = 0, 0 while i < n - 1 and count < n: if arr[i] == '1': i += d count += 1 else: i -= 1 # we move back since we have found a '0' print(-1 if count >= n else count)

Title: The Way to Home Time Limit: None seconds Memory Limit: None megabytes Problem Description: A frog lives on the axis *Ox* and needs to reach home which is in the point *n*. She starts from the point 1. The frog can jump to the right at a distance not more than *d*. So, after she jumped from the point *x* she can reach the point *x*<=+<=*a*, where *a* is an integer from 1 to *d*. For each point from 1 to *n* is known if there is a lily flower in it. The frog can jump only in points with a lilies. Guaranteed that there are lilies in the points 1 and *n*. Determine the minimal number of jumps that the frog needs to reach home which is in the point *n* from the point 1. Consider that initially the frog is in the point 1. If the frog can not reach home, print -1. Input Specification: The first line contains two integers *n* and *d* (2<=≤<=*n*<=≤<=100, 1<=≤<=*d*<=≤<=*n*<=-<=1) — the point, which the frog wants to reach, and the maximal length of the frog jump. The second line contains a string *s* of length *n*, consisting of zeros and ones. If a character of the string *s* equals to zero, then in the corresponding point there is no lily flower. In the other case, in the corresponding point there is a lily flower. Guaranteed that the first and the last characters of the string *s* equal to one. Output Specification: If the frog can not reach the home, print -1. In the other case, print the minimal number of jumps that the frog needs to reach the home which is in the point *n* from the point 1. Demo Input: ['8 4\n10010101\n', '4 2\n1001\n', '8 4\n11100101\n', '12 3\n101111100101\n'] Demo Output: ['2\n', '-1\n', '3\n', '4\n'] Note: In the first example the from can reach home in two jumps: the first jump from the point 1 to the point 4 (the length of the jump is three), and the second jump from the point 4 to the point 8 (the length of the jump is four). In the second example the frog can not reach home, because to make it she need to jump on a distance three, but the maximum length of her jump equals to two.

```python # https://codeforces.com/problemset/problem/910/A n, d = list(map(int, input().split())) arr = input() i, count = 0, 0 while i < n - 1 and count < n: if arr[i] == '1': i += d count += 1 else: i -= 1 # we move back since we have found a '0' print(-1 if count >= n else count) ```

3

523

A

Rotate, Flip and Zoom

PROGRAMMING

1,200

[ "*special", "implementation" ]

null

Polycarp is writing the prototype of a graphic editor. He has already made up his mind that the basic image transformations in his editor will be: rotate the image 90 degrees clockwise, flip the image horizontally (symmetry relative to the vertical line, that is, the right part of the image moves to the left, and vice versa) and zooming on the image. He is sure that that there is a large number of transformations that can be expressed through these three. He has recently stopped implementing all three transformations for monochrome images. To test this feature, he asked you to write a code that will consecutively perform three actions with a monochrome image: first it will rotate the image 90 degrees clockwise, then it will flip the image horizontally and finally, it will zoom in twice on the image (that is, it will double all the linear sizes). Implement this feature to help Polycarp test his editor.

The first line contains two integers, *w* and *h* (1<=≤<=*w*,<=*h*<=≤<=100) — the width and height of an image in pixels. The picture is given in *h* lines, each line contains *w* characters — each character encodes the color of the corresponding pixel of the image. The line consists only of characters "." and "*", as the image is monochrome.

Print 2*w* lines, each containing 2*h* characters — the result of consecutive implementing of the three transformations, described above.

[ "3 2\n.*.\n.*.\n", "9 20\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" ]

none

500

[ { "input": "3 2\n.*.\n.*.", "output": "....\n....\n****\n****\n....\n...." }, { "input": "9 20\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......*..." }, { "input": "1 100\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.\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........................................................................................................................................................................................................" }, { "input": "1 100\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*\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********************************************************************************************************************************************************************************************************" }, { "input": "1 100\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.\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..**......**......******............**......**..**....****..**....**....****..........**..**..**........******........**....********..**********..********************....**************....**....****.." }, { "input": "100 1\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**\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**\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..." }, { "input": "100 1\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**\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**\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..." }, { "input": "1 1\n.", "output": "..\n.." }, { "input": "1 1\n*", "output": "**\n**" }, { "input": "2 2\n.*\n*.", "output": "..**\n..**\n**..\n**.." }, { "input": "1 2\n*\n.", "output": "**..\n**.." }, { "input": "2 1\n*.", "output": "**\n**\n..\n.." } ]

1,427,878,066

2,147,483,647

Python 3

OK

TESTS

24

77

0

w,h = input().split() w,h = int(w),int(h) S = [] for i in range(0,h): ts = input() S += [ts] for i in range(0,2*w): idxi = i // 2 tans = "" for j in range(0,2*h): idxj = j // 2 tans += S[idxj][idxi] print(tans)

Title: Rotate, Flip and Zoom Time Limit: None seconds Memory Limit: None megabytes Problem Description: Polycarp is writing the prototype of a graphic editor. He has already made up his mind that the basic image transformations in his editor will be: rotate the image 90 degrees clockwise, flip the image horizontally (symmetry relative to the vertical line, that is, the right part of the image moves to the left, and vice versa) and zooming on the image. He is sure that that there is a large number of transformations that can be expressed through these three. He has recently stopped implementing all three transformations for monochrome images. To test this feature, he asked you to write a code that will consecutively perform three actions with a monochrome image: first it will rotate the image 90 degrees clockwise, then it will flip the image horizontally and finally, it will zoom in twice on the image (that is, it will double all the linear sizes). Implement this feature to help Polycarp test his editor. Input Specification: The first line contains two integers, *w* and *h* (1<=≤<=*w*,<=*h*<=≤<=100) — the width and height of an image in pixels. The picture is given in *h* lines, each line contains *w* characters — each character encodes the color of the corresponding pixel of the image. The line consists only of characters "." and "*", as the image is monochrome. Output Specification: Print 2*w* lines, each containing 2*h* characters — the result of consecutive implementing of the three transformations, described above. Demo Input: ['3 2\n.*.\n.*.\n', '9 20\n**.......\n****.....\n******...\n*******..\n..******.\n....****.\n......***\n*.....***\n*********\n*********\n*********\n*********\n....**...\n...****..\n..******.\n.********\n****..***\n***...***\n**.....**\n*.......*\n'] Demo 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'] Note: none

```python w,h = input().split() w,h = int(w),int(h) S = [] for i in range(0,h): ts = input() S += [ts] for i in range(0,2*w): idxi = i // 2 tans = "" for j in range(0,2*h): idxj = j // 2 tans += S[idxj][idxi] print(tans) ```

3

0

none

0

[ "none" ]

null

You are given two lists of non-zero digits. Let's call an integer pretty if its (base 10) representation has at least one digit from the first list and at least one digit from the second list. What is the smallest positive pretty integer?

The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=9) — the lengths of the first and the second lists, respectively. The second line contains *n* distinct digits *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=9) — the elements of the first list. The third line contains *m* distinct digits *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=9) — the elements of the second list.

Print the smallest pretty integer.

[ "2 3\n4 2\n5 7 6\n", "8 8\n1 2 3 4 5 6 7 8\n8 7 6 5 4 3 2 1\n" ]

[ "25\n", "1\n" ]

In the first example 25, 46, 24567 are pretty, as well as many other integers. The smallest among them is 25. 42 and 24 are not pretty because they don't have digits from the second list. In the second example all integers that have at least one digit different from 9 are pretty. It's obvious that the smallest among them is 1, because it's the smallest positive integer.

0

[ { "input": "2 3\n4 2\n5 7 6", "output": "25" }, { "input": "8 8\n1 2 3 4 5 6 7 8\n8 7 6 5 4 3 2 1", "output": "1" }, { "input": "1 1\n9\n1", "output": "19" }, { "input": "9 1\n5 4 2 3 6 1 7 9 8\n9", "output": "9" }, { "input": "5 3\n7 2 5 8 6\n3 1 9", "output": "12" }, { "input": "4 5\n5 2 6 4\n8 9 1 3 7", "output": "12" }, { "input": "5 9\n4 2 1 6 7\n2 3 4 5 6 7 8 9 1", "output": "1" }, { "input": "9 9\n5 4 3 2 1 6 7 8 9\n3 2 1 5 4 7 8 9 6", "output": "1" }, { "input": "9 5\n2 3 4 5 6 7 8 9 1\n4 2 1 6 7", "output": "1" }, { "input": "9 9\n1 2 3 4 5 6 7 8 9\n1 2 3 4 5 6 7 8 9", "output": "1" }, { "input": "9 9\n1 2 3 4 5 6 7 8 9\n9 8 7 6 5 4 3 2 1", "output": "1" }, { "input": "9 9\n9 8 7 6 5 4 3 2 1\n1 2 3 4 5 6 7 8 9", "output": "1" }, { "input": "9 9\n9 8 7 6 5 4 3 2 1\n9 8 7 6 5 4 3 2 1", "output": "1" }, { "input": "1 1\n8\n9", "output": "89" }, { "input": "1 1\n9\n8", "output": "89" }, { "input": "1 1\n1\n2", "output": "12" }, { "input": "1 1\n2\n1", "output": "12" }, { "input": "1 1\n9\n9", "output": "9" }, { "input": "1 1\n1\n1", "output": "1" }, { "input": "4 5\n3 2 4 5\n1 6 5 9 8", "output": "5" }, { "input": "3 2\n4 5 6\n1 5", "output": "5" }, { "input": "5 4\n1 3 5 6 7\n2 4 3 9", "output": "3" }, { "input": "5 5\n1 3 5 7 9\n2 4 6 8 9", "output": "9" }, { "input": "2 2\n1 8\n2 8", "output": "8" }, { "input": "5 5\n5 6 7 8 9\n1 2 3 4 5", "output": "5" }, { "input": "5 5\n1 2 3 4 5\n1 2 3 4 5", "output": "1" }, { "input": "5 5\n1 2 3 4 5\n2 3 4 5 6", "output": "2" }, { "input": "2 2\n1 5\n2 5", "output": "5" }, { "input": "4 4\n1 3 5 8\n2 4 6 8", "output": "8" }, { "input": "3 3\n1 5 3\n2 5 7", "output": "5" }, { "input": "3 3\n3 6 8\n2 6 9", "output": "6" }, { "input": "2 2\n1 4\n2 4", "output": "4" }, { "input": "5 3\n3 4 5 6 7\n1 5 9", "output": "5" }, { "input": "4 4\n1 2 3 4\n2 5 6 7", "output": "2" }, { "input": "5 5\n1 2 3 4 5\n9 2 1 7 5", "output": "1" }, { "input": "2 2\n1 3\n2 3", "output": "3" }, { "input": "3 3\n3 2 1\n3 2 1", "output": "1" }, { "input": "3 3\n1 3 5\n2 3 6", "output": "3" }, { "input": "3 3\n5 6 7\n5 6 7", "output": "5" }, { "input": "1 2\n5\n2 5", "output": "5" }, { "input": "3 3\n2 4 9\n7 8 9", "output": "9" }, { "input": "3 3\n1 2 4\n3 4 5", "output": "4" }, { "input": "3 2\n1 4 9\n2 4", "output": "4" }, { "input": "3 3\n3 5 6\n1 5 9", "output": "5" }, { "input": "3 2\n1 2 4\n3 4", "output": "4" }, { "input": "2 4\n8 9\n1 2 3 9", "output": "9" }, { "input": "1 2\n9\n8 9", "output": "9" }, { "input": "3 2\n1 2 4\n4 2", "output": "2" }, { "input": "2 3\n4 5\n1 3 5", "output": "5" }, { "input": "3 2\n1 2 3\n2 3", "output": "2" }, { "input": "4 3\n1 3 5 9\n2 8 9", "output": "9" }, { "input": "2 2\n1 9\n9 2", "output": "9" } ]

1,508,055,320

620

Python 3

WRONG_ANSWER

PRETESTS

3

46

0

#CF440 yl1 n,m=list(map(int,input().split())) a=min(map(int,input().split())) b=min(map(int,input().split())) print(''.join(list(map(str,sorted(set([a,b]))))))

Title: none Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given two lists of non-zero digits. Let's call an integer pretty if its (base 10) representation has at least one digit from the first list and at least one digit from the second list. What is the smallest positive pretty integer? Input Specification: The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=9) — the lengths of the first and the second lists, respectively. The second line contains *n* distinct digits *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=9) — the elements of the first list. The third line contains *m* distinct digits *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=9) — the elements of the second list. Output Specification: Print the smallest pretty integer. Demo Input: ['2 3\n4 2\n5 7 6\n', '8 8\n1 2 3 4 5 6 7 8\n8 7 6 5 4 3 2 1\n'] Demo Output: ['25\n', '1\n'] Note: In the first example 25, 46, 24567 are pretty, as well as many other integers. The smallest among them is 25. 42 and 24 are not pretty because they don't have digits from the second list. In the second example all integers that have at least one digit different from 9 are pretty. It's obvious that the smallest among them is 1, because it's the smallest positive integer.

```python #CF440 yl1 n,m=list(map(int,input().split())) a=min(map(int,input().split())) b=min(map(int,input().split())) print(''.join(list(map(str,sorted(set([a,b])))))) ```

0

26

A

Almost Prime

PROGRAMMING

900

[ "number theory" ]

A. Almost Prime

2

256

A number is called almost prime if it has exactly two distinct prime divisors. For example, numbers 6, 18, 24 are almost prime, while 4, 8, 9, 42 are not. Find the amount of almost prime numbers which are between 1 and *n*, inclusive.

Input contains one integer number *n* (1<=≤<=*n*<=≤<=3000).

Output the amount of almost prime numbers between 1 and *n*, inclusive.

[ "10\n", "21\n" ]

[ "2\n", "8\n" ]

none

500

[ { "input": "10", "output": "2" }, { "input": "21", "output": "8" }, { "input": "1", "output": "0" }, { "input": "2", "output": "0" }, { "input": "4", "output": "0" }, { "input": "3", "output": "0" }, { "input": "8", "output": "1" }, { "input": "19", "output": "6" }, { "input": "40", "output": "19" }, { "input": "77", "output": "41" }, { "input": "222", "output": "125" }, { "input": "987", "output": "501" }, { "input": "1000", "output": "508" }, { "input": "2000", "output": "958" }, { "input": "3000", "output": "1375" }, { "input": "2999", "output": "1375" }, { "input": "2998", "output": "1375" }, { "input": "2997", "output": "1374" }, { "input": "1429", "output": "706" }, { "input": "1673", "output": "808" }, { "input": "1500", "output": "732" }, { "input": "500", "output": "266" }, { "input": "856", "output": "439" } ]

1,610,142,458

2,147,483,647

Python 3

OK

TESTS

23

124

0

import math n = int(input()) if n<6: print(0) exit() maximum_count = 0 for i in range(6,n+1): count = 0 for j in range(2,int(math.sqrt(i))+1): if i%j==0: count+=1 while i%j==0: i = i//j if i!=1: count+=1 # print(count) if count==2: maximum_count+=1 print(maximum_count)

Title: Almost Prime Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: A number is called almost prime if it has exactly two distinct prime divisors. For example, numbers 6, 18, 24 are almost prime, while 4, 8, 9, 42 are not. Find the amount of almost prime numbers which are between 1 and *n*, inclusive. Input Specification: Input contains one integer number *n* (1<=≤<=*n*<=≤<=3000). Output Specification: Output the amount of almost prime numbers between 1 and *n*, inclusive. Demo Input: ['10\n', '21\n'] Demo Output: ['2\n', '8\n'] Note: none

```python import math n = int(input()) if n<6: print(0) exit() maximum_count = 0 for i in range(6,n+1): count = 0 for j in range(2,int(math.sqrt(i))+1): if i%j==0: count+=1 while i%j==0: i = i//j if i!=1: count+=1 # print(count) if count==2: maximum_count+=1 print(maximum_count) ```

3.969

378

A

Playing with Dice

PROGRAMMING

800

[ "brute force" ]

null

Two players are playing a game. First each of them writes an integer from 1 to 6, and then a dice is thrown. The player whose written number got closer to the number on the dice wins. If both payers have the same difference, it's a draw. The first player wrote number *a*, the second player wrote number *b*. How many ways to throw a dice are there, at which the first player wins, or there is a draw, or the second player wins?

The single line contains two integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=6) — the numbers written on the paper by the first and second player, correspondingly.

Print three integers: the number of ways to throw the dice at which the first player wins, the game ends with a draw or the second player wins, correspondingly.

[ "2 5\n", "2 4\n" ]

[ "3 0 3\n", "2 1 3\n" ]

The dice is a standard cube-shaped six-sided object with each side containing a number from 1 to 6, and where all numbers on all sides are distinct. You can assume that number *a* is closer to number *x* than number *b*, if |*a* - *x*| < |*b* - *x*|.

500

[ { "input": "2 5", "output": "3 0 3" }, { "input": "2 4", "output": "2 1 3" }, { "input": "5 3", "output": "2 1 3" }, { "input": "1 6", "output": "3 0 3" }, { "input": "5 1", "output": "3 1 2" }, { "input": "6 3", "output": "2 0 4" }, { "input": "2 3", "output": "2 0 4" }, { "input": "5 6", "output": "5 0 1" }, { "input": "4 4", "output": "0 6 0" }, { "input": "1 1", "output": "0 6 0" }, { "input": "6 4", "output": "1 1 4" }, { "input": "1 4", "output": "2 0 4" }, { "input": "5 5", "output": "0 6 0" }, { "input": "4 5", "output": "4 0 2" }, { "input": "4 3", "output": "3 0 3" }, { "input": "1 5", "output": "2 1 3" }, { "input": "6 5", "output": "1 0 5" }, { "input": "2 2", "output": "0 6 0" }, { "input": "1 3", "output": "1 1 4" }, { "input": "3 6", "output": "4 0 2" }, { "input": "3 1", "output": "4 1 1" }, { "input": "3 2", "output": "4 0 2" }, { "input": "3 5", "output": "3 1 2" }, { "input": "3 3", "output": "0 6 0" }, { "input": "6 2", "output": "2 1 3" }, { "input": "4 1", "output": "4 0 2" }, { "input": "5 2", "output": "3 0 3" }, { "input": "4 2", "output": "3 1 2" }, { "input": "2 1", "output": "5 0 1" }, { "input": "6 1", "output": "3 0 3" }, { "input": "4 6", "output": "4 1 1" }, { "input": "2 6", "output": "3 1 2" }, { "input": "3 4", "output": "3 0 3" }, { "input": "1 2", "output": "1 0 5" }, { "input": "6 6", "output": "0 6 0" }, { "input": "5 4", "output": "2 0 4" }, { "input": "3 3", "output": "0 6 0" }, { "input": "1 1", "output": "0 6 0" } ]

1,680,780,521

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

1

15

0

a,b = input().split() a=int(a) b=int(b) x=[1,2,3,4,5,6] a1=0 b1=0 e1=0 if(1<=a<=6): if(1<=b<=6): for i in range(0,6): if(abs(x[i]-a)>abs(x[i]-b)): a1=a1+1 elif(abs(x[i]-b)>abs(x[i]-a)): b1=b1+1 elif(abs(x[i]-b)==abs(x[i]-a)): e1=e1+1 print(a1,e1,b1)

Title: Playing with Dice Time Limit: None seconds Memory Limit: None megabytes Problem Description: Two players are playing a game. First each of them writes an integer from 1 to 6, and then a dice is thrown. The player whose written number got closer to the number on the dice wins. If both payers have the same difference, it's a draw. The first player wrote number *a*, the second player wrote number *b*. How many ways to throw a dice are there, at which the first player wins, or there is a draw, or the second player wins? Input Specification: The single line contains two integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=6) — the numbers written on the paper by the first and second player, correspondingly. Output Specification: Print three integers: the number of ways to throw the dice at which the first player wins, the game ends with a draw or the second player wins, correspondingly. Demo Input: ['2 5\n', '2 4\n'] Demo Output: ['3 0 3\n', '2 1 3\n'] Note: The dice is a standard cube-shaped six-sided object with each side containing a number from 1 to 6, and where all numbers on all sides are distinct. You can assume that number *a* is closer to number *x* than number *b*, if |*a* - *x*| < |*b* - *x*|.

```python a,b = input().split() a=int(a) b=int(b) x=[1,2,3,4,5,6] a1=0 b1=0 e1=0 if(1<=a<=6): if(1<=b<=6): for i in range(0,6): if(abs(x[i]-a)>abs(x[i]-b)): a1=a1+1 elif(abs(x[i]-b)>abs(x[i]-a)): b1=b1+1 elif(abs(x[i]-b)==abs(x[i]-a)): e1=e1+1 print(a1,e1,b1) ```

0

28

C

Bath Queue

PROGRAMMING

2,200

[ "combinatorics", "dp", "probabilities" ]

C. Bath Queue

2

256

There are *n* students living in the campus. Every morning all students wake up at the same time and go to wash. There are *m* rooms with wash basins. The *i*-th of these rooms contains *a**i* wash basins. Every student independently select one the rooms with equal probability and goes to it. After all students selected their rooms, students in each room divide into queues by the number of wash basins so that the size of the largest queue is the least possible. Calculate the expected value of the size of the largest queue among all rooms.

The first line contains two positive integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=50) — the amount of students and the amount of rooms. The second line contains *m* integers *a*1,<=*a*2,<=... ,<=*a**m* (1<=≤<=*a**i*<=≤<=50). *a**i* means the amount of wash basins in the *i*-th room.

Output single number: the expected value of the size of the largest queue. Your answer must have an absolute or relative error less than 10<=-<=9.

[ "1 1\n2\n", "2 2\n1 1\n", "2 3\n1 1 1\n", "7 5\n1 1 2 3 1\n" ]

[ "1.00000000000000000000\n", "1.50000000000000000000\n", "1.33333333333333350000\n", "2.50216960000000070000\n" ]

none

1,500

[ { "input": "1 1\n2", "output": "1.00000000000000000000" }, { "input": "2 2\n1 1", "output": "1.50000000000000000000" }, { "input": "2 3\n1 1 1", "output": "1.33333333333333350000" }, { "input": "7 5\n1 1 2 3 1", "output": "2.50216960000000070000" }, { "input": "10 4\n8 4 7 6", "output": "1.08210754394531210000" }, { "input": "5 5\n5 5 5 5 5", "output": "1.00000000000000020000" }, { "input": "7 4\n1 2 3 4", "output": "2.11712646484374910000" }, { "input": "50 50\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", "output": "3.80545467981579130000" }, { "input": "30 30\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", "output": "3.49298980907245000000" }, { "input": "20 20\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "3.23123684379753670000" }, { "input": "1 50\n47 24 49 50 4 21 42 22 34 48 45 15 31 18 12 10 4 45 45 42 49 13 12 9 7 5 30 18 22 50 15 16 25 18 5 41 3 26 19 18 22 5 8 10 16 50 43 44 6 43", "output": "1.00000000000000000000" }, { "input": "1 50\n46 45 44 49 48 48 47 42 48 47 47 48 39 47 48 49 50 48 50 46 48 46 50 47 45 50 41 49 39 44 46 47 43 47 42 47 49 40 49 50 50 50 48 50 48 47 49 46 46 42", "output": "1.00000000000000000000" }, { "input": "1 50\n9 1 1 4 1 9 7 4 3 10 1 7 4 7 2 5 13 2 3 3 2 1 2 1 1 7 7 5 2 6 1 8 2 6 2 15 2 3 1 2 4 8 6 2 6 11 1 2 1 1", "output": "1.00000000000000000000" }, { "input": "50 1\n27", "output": "2.00000000000000000000" }, { "input": "50 1\n48", "output": "2.00000000000000000000" }, { "input": "50 1\n4", "output": "13.00000000000000000000" }, { "input": "20 35\n48 40 49 37 36 44 48 42 37 42 18 44 47 47 41 45 49 47 47 50 16 24 42 24 36 37 45 48 36 43 44 25 34 30 42", "output": "0.99999999999999978000" }, { "input": "50 50\n3 12 1 3 6 2 5 14 2 4 4 1 6 9 4 2 3 19 7 6 4 1 7 4 1 3 6 3 2 4 4 1 6 1 3 1 1 4 1 6 1 2 2 4 12 12 1 5 5 2", "output": "2.83614403586073620000" }, { "input": "50 50\n21 35 15 42 44 1 50 4 26 21 43 41 50 33 47 3 21 14 33 34 43 44 16 41 35 27 3 4 7 14 15 35 27 36 46 13 3 48 32 20 15 33 38 36 39 22 45 7 16 50", "output": "1.40898003277183290000" }, { "input": "50 50\n50 48 38 45 50 48 38 48 49 49 50 50 49 50 38 48 45 40 43 49 48 43 50 50 50 42 47 50 49 34 48 47 48 50 50 50 49 47 48 49 48 48 50 39 45 47 48 47 45 50", "output": "0.99999999999999156000" }, { "input": "50 50\n50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 49 49 49 50 49 50 50 50 50 48 50 49 50 50 50 50 48 50 50 50 49 50 50 50 50 50 50 50 50 50 50 49 50 50", "output": "0.99999999999999156000" }, { "input": "50 50\n2 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 2 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 1 2 1 1", "output": "3.71403384155135140000" }, { "input": "44 5\n40 48 45 43 17", "output": "1.00121533621041100000" }, { "input": "3 49\n4 2 9 21 22 25 6 9 15 10 37 3 8 6 14 1 3 3 18 1 9 11 8 5 20 21 10 25 35 16 14 18 2 5 12 6 9 8 3 6 19 18 1 13 12 33 4 2 16", "output": "1.00374843815077060000" }, { "input": "19 17\n50 46 38 48 41 41 40 45 47 50 49 33 46 44 46 48 36", "output": "0.99999999999999967000" }, { "input": "12 34\n47 50 49 45 48 50 49 45 50 48 43 49 50 47 49 49 50 50 45 43 45 44 50 47 50 49 47 49 49 42 50 50 50 49", "output": "1.00000000000000000000" }, { "input": "35 40\n12 1 4 8 1 9 1 11 1 1 8 8 16 1 6 5 3 1 6 4 6 2 4 6 2 1 1 16 2 2 3 1 1 2 2 3 8 12 1 4", "output": "2.65978492228475400000" }, { "input": "10 6\n1 1 1 1 1 1", "output": "3.44474669607021380000" }, { "input": "50 50\n50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50", "output": "0.99999999999999156000" }, { "input": "50 50\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", "output": "3.80545467981579130000" }, { "input": "50 50\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", "output": "1.44158938050050490000" }, { "input": "1 1\n50", "output": "1.00000000000000000000" } ]

1,676,700,619

2,147,483,647

Python 3

OK

TESTS

32

1,778

9,932,800

import functools,math,itertools,time u=functools.lru_cache(maxsize=None) n,m=map(int,input().split()) s=[*map(int,input().split())] t=u(lambda x:1 if x<2 else x*t(x-1)) c=u(lambda r,n:t(n)/t(r)/t(n-r)) p=u(lambda n,k:n**k) w=u(lambda n,k:math.ceil(n/k)) r=u(lambda k,n:max(k,n)) h=u(lambda i,j,l:c(l,i)*p(j-1,i-l)/p(j,i)) for i,j in itertools.product(range(1,n+1),range(1,m+1)): for l in range(i+1): h(i,j,l) @u def d(i,j,k): if j: q=0 for l in range(i + 1): q+=h(i,j,l)*d(i-l,j-1,r(k,w(l,s[j-1]))) return q return k print(d(n,m,0))

Title: Bath Queue Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: There are *n* students living in the campus. Every morning all students wake up at the same time and go to wash. There are *m* rooms with wash basins. The *i*-th of these rooms contains *a**i* wash basins. Every student independently select one the rooms with equal probability and goes to it. After all students selected their rooms, students in each room divide into queues by the number of wash basins so that the size of the largest queue is the least possible. Calculate the expected value of the size of the largest queue among all rooms. Input Specification: The first line contains two positive integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=50) — the amount of students and the amount of rooms. The second line contains *m* integers *a*1,<=*a*2,<=... ,<=*a**m* (1<=≤<=*a**i*<=≤<=50). *a**i* means the amount of wash basins in the *i*-th room. Output Specification: Output single number: the expected value of the size of the largest queue. Your answer must have an absolute or relative error less than 10<=-<=9. Demo Input: ['1 1\n2\n', '2 2\n1 1\n', '2 3\n1 1 1\n', '7 5\n1 1 2 3 1\n'] Demo Output: ['1.00000000000000000000\n', '1.50000000000000000000\n', '1.33333333333333350000\n', '2.50216960000000070000\n'] Note: none

```python import functools,math,itertools,time u=functools.lru_cache(maxsize=None) n,m=map(int,input().split()) s=[*map(int,input().split())] t=u(lambda x:1 if x<2 else x*t(x-1)) c=u(lambda r,n:t(n)/t(r)/t(n-r)) p=u(lambda n,k:n**k) w=u(lambda n,k:math.ceil(n/k)) r=u(lambda k,n:max(k,n)) h=u(lambda i,j,l:c(l,i)*p(j-1,i-l)/p(j,i)) for i,j in itertools.product(range(1,n+1),range(1,m+1)): for l in range(i+1): h(i,j,l) @u def d(i,j,k): if j: q=0 for l in range(i + 1): q+=h(i,j,l)*d(i-l,j-1,r(k,w(l,s[j-1]))) return q return k print(d(n,m,0)) ```

3.536999

169

A

Chores

PROGRAMMING

800

[ "sortings" ]

null

Petya and Vasya are brothers. Today is a special day for them as their parents left them home alone and commissioned them to do *n* chores. Each chore is characterized by a single parameter — its complexity. The complexity of the *i*-th chore equals *h**i*. As Petya is older, he wants to take the chores with complexity larger than some value *x* (*h**i*<=><=*x*) to leave to Vasya the chores with complexity less than or equal to *x* (*h**i*<=≤<=*x*). The brothers have already decided that Petya will do exactly *a* chores and Vasya will do exactly *b* chores (*a*<=+<=*b*<==<=*n*). In how many ways can they choose an integer *x* so that Petya got exactly *a* chores and Vasya got exactly *b* chores?

The first input line contains three integers *n*,<=*a* and *b* (2<=≤<=*n*<=≤<=2000; *a*,<=*b*<=≥<=1; *a*<=+<=*b*<==<=*n*) — the total number of chores, the number of Petya's chores and the number of Vasya's chores. The next line contains a sequence of integers *h*1,<=*h*2,<=...,<=*h**n* (1<=≤<=*h**i*<=≤<=109), *h**i* is the complexity of the *i*-th chore. The numbers in the given sequence are not necessarily different. All numbers on the lines are separated by single spaces.

Print the required number of ways to choose an integer value of *x*. If there are no such ways, print 0.

[ "5 2 3\n6 2 3 100 1\n", "7 3 4\n1 1 9 1 1 1 1\n" ]

[ "3\n", "0\n" ]

In the first sample the possible values of *x* are 3, 4 or 5. In the second sample it is impossible to find such *x*, that Petya got 3 chores and Vasya got 4.

500

[ { "input": "5 2 3\n6 2 3 100 1", "output": "3" }, { "input": "7 3 4\n1 1 9 1 1 1 1", "output": "0" }, { "input": "2 1 1\n10 2", "output": "8" }, { "input": "2 1 1\n7 7", "output": "0" }, { "input": "2 1 1\n1 1000000000", "output": "999999999" }, { "input": "3 1 2\n6 5 5", "output": "1" }, { "input": "3 2 1\n10 10 8", "output": "2" }, { "input": "8 3 5\n42 55 61 72 83 10 22 33", "output": "6" }, { "input": "10 5 5\n1 2 3 4 5 999999999 999999998 999999997 999999996 999999995", "output": "999999990" }, { "input": "4 1 3\n10 8 7 3", "output": "2" }, { "input": "4 2 2\n402 10 10 402", "output": "392" }, { "input": "4 1 3\n10 402 402 10", "output": "0" }, { "input": "4 3 1\n100 100 200 200", "output": "0" }, { "input": "150 10 140\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 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "0" }, { "input": "102 101 1\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 1 1", "output": "0" } ]

1,547,265,926

2,147,483,647

Python 3

OK

TESTS

29

109

307,200

total_chores, p_chores, v_chores = map(int, input().split()) chores = list(map(int, input().split())) chores.sort() v = chores[:v_chores] p = chores[v_chores:] answer = max(0, p[0] - v[-1]) print(answer)

Title: Chores Time Limit: None seconds Memory Limit: None megabytes Problem Description: Petya and Vasya are brothers. Today is a special day for them as their parents left them home alone and commissioned them to do *n* chores. Each chore is characterized by a single parameter — its complexity. The complexity of the *i*-th chore equals *h**i*. As Petya is older, he wants to take the chores with complexity larger than some value *x* (*h**i*<=><=*x*) to leave to Vasya the chores with complexity less than or equal to *x* (*h**i*<=≤<=*x*). The brothers have already decided that Petya will do exactly *a* chores and Vasya will do exactly *b* chores (*a*<=+<=*b*<==<=*n*). In how many ways can they choose an integer *x* so that Petya got exactly *a* chores and Vasya got exactly *b* chores? Input Specification: The first input line contains three integers *n*,<=*a* and *b* (2<=≤<=*n*<=≤<=2000; *a*,<=*b*<=≥<=1; *a*<=+<=*b*<==<=*n*) — the total number of chores, the number of Petya's chores and the number of Vasya's chores. The next line contains a sequence of integers *h*1,<=*h*2,<=...,<=*h**n* (1<=≤<=*h**i*<=≤<=109), *h**i* is the complexity of the *i*-th chore. The numbers in the given sequence are not necessarily different. All numbers on the lines are separated by single spaces. Output Specification: Print the required number of ways to choose an integer value of *x*. If there are no such ways, print 0. Demo Input: ['5 2 3\n6 2 3 100 1\n', '7 3 4\n1 1 9 1 1 1 1\n'] Demo Output: ['3\n', '0\n'] Note: In the first sample the possible values of *x* are 3, 4 or 5. In the second sample it is impossible to find such *x*, that Petya got 3 chores and Vasya got 4.

```python total_chores, p_chores, v_chores = map(int, input().split()) chores = list(map(int, input().split())) chores.sort() v = chores[:v_chores] p = chores[v_chores:] answer = max(0, p[0] - v[-1]) print(answer) ```

3

797

A

k-Factorization

PROGRAMMING

1,100

[ "implementation", "math", "number theory" ]

null

Given a positive integer *n*, find *k* integers (not necessary distinct) such that all these integers are strictly greater than 1, and their product is equal to *n*.

The first line contains two integers *n* and *k* (2<=≤<=*n*<=≤<=100000, 1<=≤<=*k*<=≤<=20).

If it's impossible to find the representation of *n* as a product of *k* numbers, print -1. Otherwise, print *k* integers in any order. Their product must be equal to *n*. If there are multiple answers, print any of them.

[ "100000 2\n", "100000 20\n", "1024 5\n" ]

[ "2 50000 \n", "-1\n", "2 64 2 2 2 \n" ]

none

0

[ { "input": "100000 2", "output": "2 50000 " }, { "input": "100000 20", "output": "-1" }, { "input": "1024 5", "output": "2 64 2 2 2 " }, { "input": "100000 10", "output": "2 2 2 2 2 5 5 5 5 5 " }, { "input": "99999 3", "output": "3 813 41 " }, { "input": "99999 4", "output": "3 3 41 271 " }, { "input": "99999 5", "output": "-1" }, { "input": "1024 10", "output": "2 2 2 2 2 2 2 2 2 2 " }, { "input": "1024 11", "output": "-1" }, { "input": "2048 11", "output": "2 2 2 2 2 2 2 2 2 2 2 " }, { "input": "2 1", "output": "2 " }, { "input": "2 2", "output": "-1" }, { "input": "2 3", "output": "-1" }, { "input": "2 4", "output": "-1" }, { "input": "2 5", "output": "-1" }, { "input": "2 1", "output": "2 " }, { "input": "3 1", "output": "3 " }, { "input": "3 2", "output": "-1" }, { "input": "349 2", "output": "-1" }, { "input": "8 1", "output": "8 " }, { "input": "66049 2", "output": "257 257 " }, { "input": "6557 2", "output": "83 79 " }, { "input": "9 2", "output": "3 3 " }, { "input": "4 2", "output": "2 2 " }, { "input": "2 2", "output": "-1" }, { "input": "4 4", "output": "-1" }, { "input": "12 1", "output": "12 " }, { "input": "17 1", "output": "17 " }, { "input": "8 2", "output": "2 4 " }, { "input": "14 2", "output": "7 2 " }, { "input": "99991 1", "output": "99991 " }, { "input": "30 2", "output": "3 10 " }, { "input": "97 1", "output": "97 " }, { "input": "92 2", "output": "2 46 " }, { "input": "4 1", "output": "4 " }, { "input": "4 3", "output": "-1" }, { "input": "30 4", "output": "-1" }, { "input": "2 6", "output": "-1" }, { "input": "3 1", "output": "3 " }, { "input": "3 2", "output": "-1" }, { "input": "3 3", "output": "-1" }, { "input": "3 4", "output": "-1" }, { "input": "3 5", "output": "-1" }, { "input": "3 6", "output": "-1" }, { "input": "4 1", "output": "4 " }, { "input": "4 2", "output": "2 2 " }, { "input": "4 3", "output": "-1" }, { "input": "4 4", "output": "-1" }, { "input": "4 5", "output": "-1" }, { "input": "4 6", "output": "-1" }, { "input": "5 1", "output": "5 " }, { "input": "5 2", "output": "-1" }, { "input": "5 3", "output": "-1" }, { "input": "5 4", "output": "-1" }, { "input": "5 5", "output": "-1" }, { "input": "5 6", "output": "-1" }, { "input": "6 1", "output": "6 " }, { "input": "6 2", "output": "3 2 " }, { "input": "6 3", "output": "-1" }, { "input": "6 4", "output": "-1" }, { "input": "6 5", "output": "-1" }, { "input": "6 6", "output": "-1" }, { "input": "7 1", "output": "7 " }, { "input": "7 2", "output": "-1" }, { "input": "7 3", "output": "-1" }, { "input": "7 4", "output": "-1" }, { "input": "7 5", "output": "-1" }, { "input": "7 6", "output": "-1" }, { "input": "8 1", "output": "8 " }, { "input": "8 2", "output": "2 4 " }, { "input": "8 3", "output": "2 2 2 " }, { "input": "8 4", "output": "-1" }, { "input": "8 5", "output": "-1" }, { "input": "8 6", "output": "-1" }, { "input": "9 1", "output": "9 " }, { "input": "9 2", "output": "3 3 " }, { "input": "9 3", "output": "-1" }, { "input": "9 4", "output": "-1" }, { "input": "9 5", "output": "-1" }, { "input": "9 6", "output": "-1" }, { "input": "10 1", "output": "10 " }, { "input": "10 2", "output": "5 2 " }, { "input": "10 3", "output": "-1" }, { "input": "10 4", "output": "-1" }, { "input": "10 5", "output": "-1" }, { "input": "10 6", "output": "-1" }, { "input": "11 1", "output": "11 " }, { "input": "11 2", "output": "-1" }, { "input": "11 3", "output": "-1" }, { "input": "11 4", "output": "-1" }, { "input": "11 5", "output": "-1" }, { "input": "11 6", "output": "-1" }, { "input": "12 1", "output": "12 " }, { "input": "12 2", "output": "2 6 " }, { "input": "12 3", "output": "2 2 3 " }, { "input": "12 4", "output": "-1" }, { "input": "12 5", "output": "-1" }, { "input": "12 6", "output": "-1" }, { "input": "13 1", "output": "13 " }, { "input": "13 2", "output": "-1" }, { "input": "13 3", "output": "-1" }, { "input": "13 4", "output": "-1" }, { "input": "13 5", "output": "-1" }, { "input": "13 6", "output": "-1" }, { "input": "14 1", "output": "14 " }, { "input": "14 2", "output": "7 2 " }, { "input": "14 3", "output": "-1" }, { "input": "14 4", "output": "-1" }, { "input": "14 5", "output": "-1" }, { "input": "14 6", "output": "-1" }, { "input": "15 1", "output": "15 " }, { "input": "15 2", "output": "5 3 " }, { "input": "15 3", "output": "-1" }, { "input": "15 4", "output": "-1" }, { "input": "15 5", "output": "-1" }, { "input": "15 6", "output": "-1" }, { "input": "16 1", "output": "16 " }, { "input": "16 2", "output": "2 8 " }, { "input": "16 3", "output": "2 4 2 " }, { "input": "16 4", "output": "2 2 2 2 " }, { "input": "16 5", "output": "-1" }, { "input": "16 6", "output": "-1" }, { "input": "17 1", "output": "17 " }, { "input": "17 2", "output": "-1" }, { "input": "17 3", "output": "-1" }, { "input": "17 4", "output": "-1" }, { "input": "17 5", "output": "-1" }, { "input": "17 6", "output": "-1" }, { "input": "18 1", "output": "18 " }, { "input": "18 2", "output": "3 6 " }, { "input": "18 3", "output": "3 2 3 " }, { "input": "18 4", "output": "-1" }, { "input": "18 5", "output": "-1" }, { "input": "18 6", "output": "-1" }, { "input": "19 1", "output": "19 " }, { "input": "19 2", "output": "-1" }, { "input": "19 3", "output": "-1" }, { "input": "19 4", "output": "-1" }, { "input": "19 5", "output": "-1" }, { "input": "19 6", "output": "-1" }, { "input": "20 1", "output": "20 " }, { "input": "20 2", "output": "2 10 " }, { "input": "20 3", "output": "2 2 5 " }, { "input": "20 4", "output": "-1" }, { "input": "20 5", "output": "-1" }, { "input": "20 6", "output": "-1" }, { "input": "94249 1", "output": "94249 " }, { "input": "94249 2", "output": "307 307 " }, { "input": "94249 3", "output": "-1" }, { "input": "94249 4", "output": "-1" }, { "input": "94249 5", "output": "-1" }, { "input": "95477 1", "output": "95477 " }, { "input": "95477 2", "output": "311 307 " }, { "input": "95477 3", "output": "-1" }, { "input": "95477 4", "output": "-1" }, { "input": "95477 5", "output": "-1" }, { "input": "35557 1", "output": "35557 " }, { "input": "35557 2", "output": "31 1147 " }, { "input": "35557 3", "output": "31 31 37 " }, { "input": "35557 4", "output": "-1" }, { "input": "35557 5", "output": "-1" }, { "input": "42439 1", "output": "42439 " }, { "input": "42439 2", "output": "37 1147 " }, { "input": "42439 3", "output": "37 31 37 " }, { "input": "42439 4", "output": "-1" }, { "input": "42439 5", "output": "-1" } ]

1,666,377,438

2,147,483,647

Python 3

RUNTIME_ERROR

TESTS

1

30

0

n,k = map(int, input().split(' ')) #Taking the input of two integers factorlst = [] #Creating a list to store the number of factors we need of n for i in range(k-1): #Using for loop to get the desired number of factors for j in range(2,n): #Using nested for loop to get the factors of n if n%j == 0: n = n/j #Dividing n by the factor we get i.e. j to get the remaining factors through new n factorlst.append(j) #Appending the factors in our list break #We are breaking the if loop to keep only one factor we got in the list if len(factorlst) == k-1: #Checking whether we have correct number i.e. k-1 of factors as we have n which is one of the factors print(*factorlst,int(n)) #'*' allows us to print elements of a list seperately, along with the elements we print n else: print('-1') #If the length is not the one we desired for then we print -1

Title: k-Factorization Time Limit: None seconds Memory Limit: None megabytes Problem Description: Given a positive integer *n*, find *k* integers (not necessary distinct) such that all these integers are strictly greater than 1, and their product is equal to *n*. Input Specification: The first line contains two integers *n* and *k* (2<=≤<=*n*<=≤<=100000, 1<=≤<=*k*<=≤<=20). Output Specification: If it's impossible to find the representation of *n* as a product of *k* numbers, print -1. Otherwise, print *k* integers in any order. Their product must be equal to *n*. If there are multiple answers, print any of them. Demo Input: ['100000 2\n', '100000 20\n', '1024 5\n'] Demo Output: ['2 50000 \n', '-1\n', '2 64 2 2 2 \n'] Note: none

```python n,k = map(int, input().split(' ')) #Taking the input of two integers factorlst = [] #Creating a list to store the number of factors we need of n for i in range(k-1): #Using for loop to get the desired number of factors for j in range(2,n): #Using nested for loop to get the factors of n if n%j == 0: n = n/j #Dividing n by the factor we get i.e. j to get the remaining factors through new n factorlst.append(j) #Appending the factors in our list break #We are breaking the if loop to keep only one factor we got in the list if len(factorlst) == k-1: #Checking whether we have correct number i.e. k-1 of factors as we have n which is one of the factors print(*factorlst,int(n)) #'*' allows us to print elements of a list seperately, along with the elements we print n else: print('-1') #If the length is not the one we desired for then we print -1 ```

-1

789

A

Anastasia and pebbles

PROGRAMMING

1,100

[ "implementation", "math" ]

null

Anastasia loves going for a walk in Central Uzhlyandian Park. But she became uninterested in simple walking, so she began to collect Uzhlyandian pebbles. At first, she decided to collect all the pebbles she could find in the park. She has only two pockets. She can put at most *k* pebbles in each pocket at the same time. There are *n* different pebble types in the park, and there are *w**i* pebbles of the *i*-th type. Anastasia is very responsible, so she never mixes pebbles of different types in same pocket. However, she can put different kinds of pebbles in different pockets at the same time. Unfortunately, she can't spend all her time collecting pebbles, so she can collect pebbles from the park only once a day. Help her to find the minimum number of days needed to collect all the pebbles of Uzhlyandian Central Park, taking into consideration that Anastasia can't place pebbles of different types in same pocket.

The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105, 1<=≤<=*k*<=≤<=109) — the number of different pebble types and number of pebbles Anastasia can place in one pocket. The second line contains *n* integers *w*1,<=*w*2,<=...,<=*w**n* (1<=≤<=*w**i*<=≤<=104) — number of pebbles of each type.

The only line of output contains one integer — the minimum number of days Anastasia needs to collect all the pebbles.

[ "3 2\n2 3 4\n", "5 4\n3 1 8 9 7\n" ]

[ "3\n", "5\n" ]

In the first sample case, Anastasia can collect all pebbles of the first type on the first day, of second type — on the second day, and of third type — on the third day. Optimal sequence of actions in the second sample case: - In the first day Anastasia collects 8 pebbles of the third type. - In the second day she collects 8 pebbles of the fourth type. - In the third day she collects 3 pebbles of the first type and 1 pebble of the fourth type. - In the fourth day she collects 7 pebbles of the fifth type. - In the fifth day she collects 1 pebble of the second type.

500

[ { "input": "3 2\n2 3 4", "output": "3" }, { "input": "5 4\n3 1 8 9 7", "output": "5" }, { "input": "1 22\n1", "output": "1" }, { "input": "3 57\n78 165 54", "output": "3" }, { "input": "5 72\n74 10 146 189 184", "output": "6" }, { "input": "9 13\n132 87 200 62 168 51 185 192 118", "output": "48" }, { "input": "1 1\n10000", "output": "5000" }, { "input": "10 1\n1 1 1 1 1 1 1 1 1 1", "output": "5" }, { "input": "2 2\n2 2", "output": "1" } ]

1,595,571,510

2,147,483,647

Python 3

OK

TESTS

31

280

14,438,400

n,k = map(int,input().split()) a = list(map(int,input().split())) a.sort() ans = 0 prev = 0 for i in range(n-1): ans += max((a[i]-prev*k-1)//k + 1,0) prev = max((a[i]-prev*k-1)//k + 1,0) ans += max((a[-1]-prev*k-1)//(2*k) + 1,0) print (ans)

Title: Anastasia and pebbles Time Limit: None seconds Memory Limit: None megabytes Problem Description: Anastasia loves going for a walk in Central Uzhlyandian Park. But she became uninterested in simple walking, so she began to collect Uzhlyandian pebbles. At first, she decided to collect all the pebbles she could find in the park. She has only two pockets. She can put at most *k* pebbles in each pocket at the same time. There are *n* different pebble types in the park, and there are *w**i* pebbles of the *i*-th type. Anastasia is very responsible, so she never mixes pebbles of different types in same pocket. However, she can put different kinds of pebbles in different pockets at the same time. Unfortunately, she can't spend all her time collecting pebbles, so she can collect pebbles from the park only once a day. Help her to find the minimum number of days needed to collect all the pebbles of Uzhlyandian Central Park, taking into consideration that Anastasia can't place pebbles of different types in same pocket. Input Specification: The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105, 1<=≤<=*k*<=≤<=109) — the number of different pebble types and number of pebbles Anastasia can place in one pocket. The second line contains *n* integers *w*1,<=*w*2,<=...,<=*w**n* (1<=≤<=*w**i*<=≤<=104) — number of pebbles of each type. Output Specification: The only line of output contains one integer — the minimum number of days Anastasia needs to collect all the pebbles. Demo Input: ['3 2\n2 3 4\n', '5 4\n3 1 8 9 7\n'] Demo Output: ['3\n', '5\n'] Note: In the first sample case, Anastasia can collect all pebbles of the first type on the first day, of second type — on the second day, and of third type — on the third day. Optimal sequence of actions in the second sample case: - In the first day Anastasia collects 8 pebbles of the third type. - In the second day she collects 8 pebbles of the fourth type. - In the third day she collects 3 pebbles of the first type and 1 pebble of the fourth type. - In the fourth day she collects 7 pebbles of the fifth type. - In the fifth day she collects 1 pebble of the second type.

```python n,k = map(int,input().split()) a = list(map(int,input().split())) a.sort() ans = 0 prev = 0 for i in range(n-1): ans += max((a[i]-prev*k-1)//k + 1,0) prev = max((a[i]-prev*k-1)//k + 1,0) ans += max((a[-1]-prev*k-1)//(2*k) + 1,0) print (ans) ```

3

15

A

Cottage Village

PROGRAMMING

1,200

[ "implementation", "sortings" ]

A. Cottage Village

2

64

A new cottage village called «Flatville» is being built in Flatland. By now they have already built in «Flatville» *n* square houses with the centres on the *Оx*-axis. The houses' sides are parallel to the coordinate axes. It's known that no two houses overlap, but they can touch each other. The architect bureau, where Peter works, was commissioned to build a new house in «Flatville». The customer wants his future house to be on the *Оx*-axis, to be square in shape, have a side *t*, and touch at least one of the already built houses. For sure, its sides should be parallel to the coordinate axes, its centre should be on the *Ox*-axis and it shouldn't overlap any of the houses in the village. Peter was given a list of all the houses in «Flatville». Would you help him find the amount of possible positions of the new house?

The first line of the input data contains numbers *n* and *t* (1<=≤<=*n*,<=*t*<=≤<=1000). Then there follow *n* lines, each of them contains two space-separated integer numbers: *x**i* *a**i*, where *x**i* — *x*-coordinate of the centre of the *i*-th house, and *a**i* — length of its side (<=-<=1000<=≤<=*x**i*<=≤<=1000, 1<=≤<=*a**i*<=≤<=1000).

Output the amount of possible positions of the new house.

[ "2 2\n0 4\n6 2\n", "2 2\n0 4\n5 2\n", "2 3\n0 4\n5 2\n" ]

[ "4\n", "3\n", "2\n" ]

It is possible for the *x*-coordinate of the new house to have non-integer value.

0

[ { "input": "2 2\n0 4\n6 2", "output": "4" }, { "input": "2 2\n0 4\n5 2", "output": "3" }, { "input": "2 3\n0 4\n5 2", "output": "2" }, { "input": "1 1\n1 1", "output": "2" }, { "input": "1 2\n2 1", "output": "2" }, { "input": "2 1\n2 1\n1 1", "output": "2" }, { "input": "2 2\n0 4\n7 4", "output": "4" }, { "input": "4 1\n-12 1\n-14 1\n4 1\n-11 1", "output": "5" }, { "input": "6 15\n19 1\n2 3\n6 2\n-21 2\n-15 2\n23 1", "output": "2" }, { "input": "10 21\n-61 6\n55 2\n-97 1\n37 1\n-39 1\n26 2\n21 1\n64 3\n-68 1\n-28 6", "output": "6" }, { "input": "26 51\n783 54\n-850 6\n-997 59\n573 31\n-125 20\n472 52\n101 5\n-561 4\n625 35\n911 14\n-47 33\n677 55\n-410 54\n13 53\n173 31\n968 30\n-497 7\n832 42\n271 59\n-638 52\n-301 51\n378 36\n-813 7\n-206 22\n-737 37\n-911 9", "output": "35" }, { "input": "14 101\n121 88\n-452 91\n635 28\n-162 59\n-872 26\n-996 8\n468 86\n742 63\n892 89\n-249 107\n300 51\n-753 17\n-620 31\n-13 34", "output": "16" }, { "input": "3 501\n827 327\n-85 480\n-999 343", "output": "6" }, { "input": "2 999\n-999 471\n530 588", "output": "4" }, { "input": "22 54\n600 43\n806 19\n-269 43\n-384 78\n222 34\n392 10\n318 30\n488 73\n-756 49\n-662 22\n-568 50\n-486 16\n-470 2\n96 66\n864 16\n934 15\n697 43\n-154 30\n775 5\n-876 71\n-33 78\n-991 31", "output": "30" }, { "input": "17 109\n52 7\n216 24\n-553 101\n543 39\n391 92\n-904 67\n95 34\n132 14\n730 103\n952 118\n-389 41\n-324 36\n-74 2\n-147 99\n-740 33\n233 1\n-995 3", "output": "16" }, { "input": "4 512\n-997 354\n-568 216\n-234 221\n603 403", "output": "4" }, { "input": "3 966\n988 5\n15 2\n-992 79", "output": "6" }, { "input": "2 1000\n-995 201\n206 194", "output": "4" }, { "input": "50 21\n-178 1\n49 1\n-98 1\n-220 1\n152 1\n-160 3\n17 2\n77 1\n-24 1\n214 2\n-154 2\n-141 1\n79 1\n206 1\n8 1\n-208 1\n36 1\n231 3\n-2 2\n-130 2\n-14 2\n34 1\n-187 2\n14 1\n-83 2\n-241 1\n149 2\n73 1\n-233 3\n-45 1\n197 1\n145 2\n-127 2\n-229 4\n-85 1\n-66 1\n-76 2\n104 1\n175 1\n70 1\n131 3\n-108 1\n-5 4\n140 1\n33 1\n248 3\n-36 3\n134 1\n-183 1\n56 2", "output": "9" }, { "input": "50 1\n37 1\n-38 1\n7 1\n47 1\n-4 1\n24 1\n-32 1\n-23 1\n-3 1\n-19 1\n5 1\n-50 1\n11 1\n-11 1\n49 1\n-39 1\n0 1\n43 1\n-10 1\n6 1\n19 1\n1 1\n27 1\n29 1\n-47 1\n-40 1\n-46 1\n-26 1\n-42 1\n-37 1\n13 1\n-29 1\n-30 1\n3 1\n44 1\n10 1\n4 1\n-14 1\n-2 1\n34 1\n18 1\n-33 1\n-44 1\n9 1\n-36 1\n-7 1\n25 1\n22 1\n-20 1\n-41 1", "output": "43" }, { "input": "50 1\n-967 7\n696 7\n-366 4\n557 1\n978 2\n800 4\n-161 2\n-773 2\n-248 2\n134 3\n869 6\n-932 2\n-262 14\n191 3\n669 2\n72 5\n0 1\n757 8\n859 2\n-131 8\n-169 3\n543 10\n-120 2\n-87 8\n-936 6\n-620 3\n-281 11\n684 3\n886 10\n497 4\n380 4\n833 1\n-727 6\n470 11\n584 9\n66 6\n-609 12\n-661 4\n-57 8\n628 7\n635 4\n-924 3\n-982 4\n-201 7\n-9 8\n-560 9\n712 7\n-330 8\n-191 1\n-892 7", "output": "96" }, { "input": "1 1000\n0 1000", "output": "2" } ]

1,432,708,665

2,147,483,647

Python 3

OK

TESTS

35

124

0

__author__ = 'Darren' def solve(): n, t = map(int, input().split()) houses = [list(map(int, input().split())) for _i in range(n)] houses.sort() count = 1 for i in range(n-1): space = houses[i+1][0] - houses[i+1][1] / 2 - (houses[i][0] + houses[i][1] / 2) if space == t: count += 1 elif space > t: count += 2 else: pass count += 1 print(count) if __name__ == '__main__': solve()

Title: Cottage Village Time Limit: 2 seconds Memory Limit: 64 megabytes Problem Description: A new cottage village called «Flatville» is being built in Flatland. By now they have already built in «Flatville» *n* square houses with the centres on the *Оx*-axis. The houses' sides are parallel to the coordinate axes. It's known that no two houses overlap, but they can touch each other. The architect bureau, where Peter works, was commissioned to build a new house in «Flatville». The customer wants his future house to be on the *Оx*-axis, to be square in shape, have a side *t*, and touch at least one of the already built houses. For sure, its sides should be parallel to the coordinate axes, its centre should be on the *Ox*-axis and it shouldn't overlap any of the houses in the village. Peter was given a list of all the houses in «Flatville». Would you help him find the amount of possible positions of the new house? Input Specification: The first line of the input data contains numbers *n* and *t* (1<=≤<=*n*,<=*t*<=≤<=1000). Then there follow *n* lines, each of them contains two space-separated integer numbers: *x**i* *a**i*, where *x**i* — *x*-coordinate of the centre of the *i*-th house, and *a**i* — length of its side (<=-<=1000<=≤<=*x**i*<=≤<=1000, 1<=≤<=*a**i*<=≤<=1000). Output Specification: Output the amount of possible positions of the new house. Demo Input: ['2 2\n0 4\n6 2\n', '2 2\n0 4\n5 2\n', '2 3\n0 4\n5 2\n'] Demo Output: ['4\n', '3\n', '2\n'] Note: It is possible for the *x*-coordinate of the new house to have non-integer value.

```python __author__ = 'Darren' def solve(): n, t = map(int, input().split()) houses = [list(map(int, input().split())) for _i in range(n)] houses.sort() count = 1 for i in range(n-1): space = houses[i+1][0] - houses[i+1][1] / 2 - (houses[i][0] + houses[i][1] / 2) if space == t: count += 1 elif space > t: count += 2 else: pass count += 1 print(count) if __name__ == '__main__': solve() ```

3.969

581

A

Vasya the Hipster

PROGRAMMING

800

[ "implementation", "math" ]

null

One day Vasya the Hipster decided to count how many socks he had. It turned out that he had *a* red socks and *b* blue socks. According to the latest fashion, hipsters should wear the socks of different colors: a red one on the left foot, a blue one on the right foot. Every day Vasya puts on new socks in the morning and throws them away before going to bed as he doesn't want to wash them. Vasya wonders, what is the maximum number of days when he can dress fashionable and wear different socks, and after that, for how many days he can then wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got. Can you help him?

The single line of the input contains two positive integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=100) — the number of red and blue socks that Vasya's got.

Print two space-separated integers — the maximum number of days when Vasya can wear different socks and the number of days when he can wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got. Keep in mind that at the end of the day Vasya throws away the socks that he's been wearing on that day.

[ "3 1\n", "2 3\n", "7 3\n" ]

[ "1 1\n", "2 0\n", "3 2\n" ]

In the first sample Vasya can first put on one pair of different socks, after that he has two red socks left to wear on the second day.

500

[ { "input": "3 1", "output": "1 1" }, { "input": "2 3", "output": "2 0" }, { "input": "7 3", "output": "3 2" }, { "input": "100 100", "output": "100 0" }, { "input": "4 10", "output": "4 3" }, { "input": "6 10", "output": "6 2" }, { "input": "6 11", "output": "6 2" }, { "input": "10 40", "output": "10 15" }, { "input": "11 56", "output": "11 22" }, { "input": "34 30", "output": "30 2" }, { "input": "33 33", "output": "33 0" }, { "input": "100 45", "output": "45 27" }, { "input": "100 23", "output": "23 38" }, { "input": "45 12", "output": "12 16" }, { "input": "1 1", "output": "1 0" }, { "input": "1 100", "output": "1 49" }, { "input": "100 1", "output": "1 49" }, { "input": "68 59", "output": "59 4" }, { "input": "45 99", "output": "45 27" }, { "input": "99 100", "output": "99 0" }, { "input": "100 98", "output": "98 1" }, { "input": "59 12", "output": "12 23" }, { "input": "86 4", "output": "4 41" }, { "input": "68 21", "output": "21 23" }, { "input": "100 11", "output": "11 44" }, { "input": "100 10", "output": "10 45" }, { "input": "15 45", "output": "15 15" }, { "input": "11 32", "output": "11 10" }, { "input": "34 96", "output": "34 31" }, { "input": "89 89", "output": "89 0" } ]

1,684,267,309

2,147,483,647

PyPy 3

OK

TESTS

30

93

0

a,b = map(int,input().split()) c = min(a,b) a = a-c b = b-c d = max(int(a/2),int(b/2)) print(c,d)

Title: Vasya the Hipster Time Limit: None seconds Memory Limit: None megabytes Problem Description: One day Vasya the Hipster decided to count how many socks he had. It turned out that he had *a* red socks and *b* blue socks. According to the latest fashion, hipsters should wear the socks of different colors: a red one on the left foot, a blue one on the right foot. Every day Vasya puts on new socks in the morning and throws them away before going to bed as he doesn't want to wash them. Vasya wonders, what is the maximum number of days when he can dress fashionable and wear different socks, and after that, for how many days he can then wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got. Can you help him? Input Specification: The single line of the input contains two positive integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=100) — the number of red and blue socks that Vasya's got. Output Specification: Print two space-separated integers — the maximum number of days when Vasya can wear different socks and the number of days when he can wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got. Keep in mind that at the end of the day Vasya throws away the socks that he's been wearing on that day. Demo Input: ['3 1\n', '2 3\n', '7 3\n'] Demo Output: ['1 1\n', '2 0\n', '3 2\n'] Note: In the first sample Vasya can first put on one pair of different socks, after that he has two red socks left to wear on the second day.

```python a,b = map(int,input().split()) c = min(a,b) a = a-c b = b-c d = max(int(a/2),int(b/2)) print(c,d) ```

3

41

A

Translation

PROGRAMMING

800

[ "implementation", "strings" ]

A. Translation

2

256

The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.

The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.

If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.

[ "code\nedoc\n", "abb\naba\n", "code\ncode\n" ]

[ "YES\n", "NO\n", "NO\n" ]

none

500

[ { "input": "code\nedoc", "output": "YES" }, { "input": "abb\naba", "output": "NO" }, { "input": "code\ncode", "output": "NO" }, { "input": "abacaba\nabacaba", "output": "YES" }, { "input": "q\nq", "output": "YES" }, { "input": "asrgdfngfnmfgnhweratgjkk\nasrgdfngfnmfgnhweratgjkk", "output": "NO" }, { "input": "z\na", "output": "NO" }, { "input": "asd\ndsa", "output": "YES" }, { "input": "abcdef\nfecdba", "output": "NO" }, { "input": "ywjjbirapvskozubvxoemscfwl\ngnduubaogtfaiowjizlvjcu", "output": "NO" }, { "input": "mfrmqxtzvgaeuleubcmcxcfqyruwzenguhgrmkuhdgnhgtgkdszwqyd\nmfxufheiperjnhyczclkmzyhcxntdfskzkzdwzzujdinf", "output": "NO" }, { "input": "bnbnemvybqizywlnghlykniaxxxlkhftppbdeqpesrtgkcpoeqowjwhrylpsziiwcldodcoonpimudvrxejjo\ntiynnekmlalogyvrgptbinkoqdwzuiyjlrldxhzjmmp", "output": "NO" }, { "input": "pwlpubwyhzqvcitemnhvvwkmwcaawjvdiwtoxyhbhbxerlypelevasmelpfqwjk\nstruuzebbcenziscuoecywugxncdwzyfozhljjyizpqcgkyonyetarcpwkqhuugsqjuixsxptmbnlfupdcfigacdhhrzb", "output": "NO" }, { "input": "gdvqjoyxnkypfvdxssgrihnwxkeojmnpdeobpecytkbdwujqfjtxsqspxvxpqioyfagzjxupqqzpgnpnpxcuipweunqch\nkkqkiwwasbhezqcfeceyngcyuogrkhqecwsyerdniqiocjehrpkljiljophqhyaiefjpavoom", "output": "NO" }, { "input": "umeszdawsvgkjhlqwzents\nhxqhdungbylhnikwviuh", "output": "NO" }, { "input": "juotpscvyfmgntshcealgbsrwwksgrwnrrbyaqqsxdlzhkbugdyx\nibqvffmfktyipgiopznsqtrtxiijntdbgyy", "output": "NO" }, { "input": "zbwueheveouatecaglziqmudxemhrsozmaujrwlqmppzoumxhamwugedikvkblvmxwuofmpafdprbcftew\nulczwrqhctbtbxrhhodwbcxwimncnexosksujlisgclllxokrsbnozthajnnlilyffmsyko", "output": "NO" }, { "input": "nkgwuugukzcv\nqktnpxedwxpxkrxdvgmfgoxkdfpbzvwsduyiybynbkouonhvmzakeiruhfmvrktghadbfkmwxduoqv", "output": "NO" }, { "input": "incenvizhqpcenhjhehvjvgbsnfixbatrrjstxjzhlmdmxijztphxbrldlqwdfimweepkggzcxsrwelodpnryntepioqpvk\ndhjbjjftlvnxibkklxquwmzhjfvnmwpapdrslioxisbyhhfymyiaqhlgecpxamqnocizwxniubrmpyubvpenoukhcobkdojlybxd", "output": "NO" }, { "input": "w\nw", "output": "YES" }, { "input": "vz\nzv", "output": "YES" }, { "input": "ry\nyr", "output": "YES" }, { "input": "xou\nuox", "output": "YES" }, { "input": "axg\ngax", "output": "NO" }, { "input": "zdsl\nlsdz", "output": "YES" }, { "input": "kudl\nldku", "output": "NO" }, { "input": "zzlzwnqlcl\nlclqnwzlzz", "output": "YES" }, { "input": "vzzgicnzqooejpjzads\nsdazjpjeooqzncigzzv", "output": "YES" }, { "input": "raqhmvmzuwaykjpyxsykr\nxkysrypjkyawuzmvmhqar", "output": "NO" }, { "input": "ngedczubzdcqbxksnxuavdjaqtmdwncjnoaicvmodcqvhfezew\nwezefhvqcdomvciaonjcnwdmtqajdvauxnskxbqcdzbuzcdegn", "output": "YES" }, { "input": "muooqttvrrljcxbroizkymuidvfmhhsjtumksdkcbwwpfqdyvxtrlymofendqvznzlmim\nmimlznzvqdnefomylrtxvydqfpwwbckdskmutjshhmfvdiumykziorbxcjlrrvttqooum", "output": "YES" }, { "input": "vxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaivg\ngviayyikkitmuomcpiakhbxszgbnhvwyzkftwoagzixaearxpjacrnvpvbuzenvovehkmmxvblqyxvctroddksdsgebcmlluqpxv", "output": "YES" }, { "input": "mnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfdc\ncdfmkdgrdptkpewbsqvszipgxvgvuiuzbkkwuowbafkikgvnqdkxnayzdjygvezmtsgywnupocdntipiyiorblqkrzjpzatxahnm", "output": "NO" }, { "input": "dgxmzbqofstzcdgthbaewbwocowvhqpinehpjatnnbrijcolvsatbblsrxabzrpszoiecpwhfjmwuhqrapvtcgvikuxtzbftydkw\nwkdytfbztxukivgctvparqhuwmjfhwpceiozsprzbaxrslbbqasvlocjirbnntajphenipthvwocowbweabhtgdcztsfoqbzmxgd", "output": "NO" }, { "input": "gxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwgeh\nhegwxvocotmzstqfbmpjvijgkcyodlxyjawrpkczpmdspsuhoiruavnnnuwvtwohglkdxjetshkboalvzqbgjgthoteceixioxg", "output": "YES" }, { "input": "sihxuwvmaambplxvjfoskinghzicyfqebjtkysotattkahssumfcgrkheotdxwjckpvapbkaepqrxseyfrwtyaycmrzsrsngkh\nhkgnsrszrmcyaytwrfyesxrqpeakbpavpkcjwxdtoehkrgcfmusshakttatosyktjbeqfycizhgniksofjvxlpbmaamvwuxhis", "output": "YES" }, { "input": "ycnahksbughnonldzrhkysujmylcgcfuludjvjiahtkyzqvkopzqcnwhltbzfugzojqkjjlggmvnultascmygelkiktmfieok\nkoeifmtkiklegkmcsatlunvmggkjjlqjozgufzbtlhwncqzpokvqzykthaijvjdulufcgclymjusyyhrzdlnonhgubskhancy", "output": "NO" }, { "input": "wbqasaehtkfojruzyhrlgwmtyiovmzyfifslvlemhqheyaelzwnthrenjsbmntwaoryzwfbxmscmypvxlfmzpnkkjlvwvmtz\nztmvwvljkknpzmflxvpymcsmxbfwzyroawtnmbsjnerhtnwzleayehqhmelvlsfifyzmvoiytmwglrhyzurjofktheasaqbw", "output": "YES" }, { "input": "imippqurprbhfugngtgifelytadegwrgaefnfhbjjnmzikvjaccotqzemufqieqldgnbmviisgkynzeldlhqxuqphjfmyij\njiymfjhpquxqhldleznykgsiivmbngdlqeiqfumezqtoccajvkizmnjjbhfnfeagrwgedatylefigtgngufhbrpruqppimi", "output": "YES" }, { "input": "bikydffiuisckpvzqlteqfhegsagimodb\nbdomigasgehfqetlqzvpkcsiuiffdykib", "output": "YES" } ]

1,690,379,247

2,147,483,647

Python 3

RUNTIME_ERROR

TESTS

0

30

0

n = str(input('Hãy nhập n: ')) m = str(input('Hãy nhập m: ')) o = sorted(n, reverse=True) o = ''.join([str(c) for c in o]) if m == o: print('Yes') else: print('No')

Title: Translation Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly. Input Specification: The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols. Output Specification: If the word *t* is a word *s*, written reversely, print YES, otherwise print NO. Demo Input: ['code\nedoc\n', 'abb\naba\n', 'code\ncode\n'] Demo Output: ['YES\n', 'NO\n', 'NO\n'] Note: none

```python n = str(input('Hãy nhập n: ')) m = str(input('Hãy nhập m: ')) o = sorted(n, reverse=True) o = ''.join([str(c) for c in o]) if m == o: print('Yes') else: print('No') ```

-1

43

A

Football

PROGRAMMING

1,000

[ "strings" ]

A. Football

2

256

One day Vasya decided to have a look at the results of Berland 1910 Football Championship’s finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie.

The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the description. Then follow *n* lines — for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams.

Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner.

[ "1\nABC\n", "5\nA\nABA\nABA\nA\nA\n" ]

[ "ABC\n", "A\n" ]

none

500

[ { "input": "1\nABC", "output": "ABC" }, { "input": "5\nA\nABA\nABA\nA\nA", "output": "A" }, { "input": "2\nXTSJEP\nXTSJEP", "output": "XTSJEP" }, { "input": "3\nXZYDJAEDZ\nXZYDJAEDZ\nXZYDJAEDZ", "output": "XZYDJAEDZ" }, { "input": "3\nQCCYXL\nQCCYXL\nAXGLFQDD", "output": "QCCYXL" }, { "input": "3\nAZID\nEERWBC\nEERWBC", "output": "EERWBC" }, { "input": "3\nHNCGYL\nHNCGYL\nHNCGYL", "output": "HNCGYL" }, { "input": "4\nZZWZTG\nZZWZTG\nZZWZTG\nZZWZTG", "output": "ZZWZTG" }, { "input": "4\nA\nA\nKUDLJMXCSE\nA", "output": "A" }, { "input": "5\nPHBTW\nPHBTW\nPHBTW\nPHBTW\nPHBTW", "output": "PHBTW" }, { "input": "5\nPKUZYTFYWN\nPKUZYTFYWN\nSTC\nPKUZYTFYWN\nPKUZYTFYWN", "output": "PKUZYTFYWN" }, { "input": "5\nHH\nHH\nNTQWPA\nNTQWPA\nHH", "output": "HH" }, { "input": "10\nW\nW\nW\nW\nW\nD\nW\nD\nD\nW", "output": "W" }, { "input": "19\nXBCP\nTGACNIH\nXBCP\nXBCP\nXBCP\nXBCP\nXBCP\nTGACNIH\nXBCP\nXBCP\nXBCP\nXBCP\nXBCP\nTGACNIH\nXBCP\nXBCP\nTGACNIH\nTGACNIH\nXBCP", "output": "XBCP" }, { "input": "33\nOWQWCKLLF\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS", "output": "PYPAS" }, { "input": "51\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC", "output": "NC" }, { "input": "89\nH\nVOCI\nVOCI\nH\nVOCI\nH\nH\nVOCI\nVOCI\nVOCI\nH\nH\nH\nVOCI\nVOCI\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nH\nVOCI\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nH\nVOCI\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nH\nVOCI\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nH\nH\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI", "output": "VOCI" }, { "input": "100\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nM\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA", "output": "HA" }, { "input": "100\nG\nG\nS\nS\nG\nG\nS\nS\nG\nS\nS\nS\nG\nS\nG\nG\nS\nG\nS\nS\nG\nS\nS\nS\nS\nS\nG\nS\nG\nS\nS\nG\nG\nG\nS\nS\nS\nS\nG\nS\nS\nG\nG\nG\nG\nG\nS\nG\nG\nS\nS\nS\nS\nS\nG\nG\nS\nG\nG\nG\nG\nG\nS\nS\nG\nS\nS\nS\nS\nG\nS\nS\nG\nS\nG\nG\nG\nG\nG\nG\nG\nG\nG\nG\nG\nS\nS\nG\nS\nS\nS\nS\nG\nG\nG\nS\nG\nG\nG\nS", "output": "G" }, { "input": "100\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nOBH\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL", "output": "WL" } ]

1,592,475,689

2,147,483,647

Python 3

OK

TESTS

34

216

0

n = int(input()) ans = [] for i in range(n): tmp = input() ans.append(tmp) teams = set(ans) if len(teams) == 1: print(ans[0]) else: t1 = teams.pop() t2 = teams.pop() if ans.count(t1) > ans.count(t2): print(t1) else: print(t2)

Title: Football Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: One day Vasya decided to have a look at the results of Berland 1910 Football Championship’s finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie. Input Specification: The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the description. Then follow *n* lines — for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams. Output Specification: Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner. Demo Input: ['1\nABC\n', '5\nA\nABA\nABA\nA\nA\n'] Demo Output: ['ABC\n', 'A\n'] Note: none

```python n = int(input()) ans = [] for i in range(n): tmp = input() ans.append(tmp) teams = set(ans) if len(teams) == 1: print(ans[0]) else: t1 = teams.pop() t2 = teams.pop() if ans.count(t1) > ans.count(t2): print(t1) else: print(t2) ```

3.946

344

A

Magnets

PROGRAMMING

800

[ "implementation" ]

null

Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other. Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed.

The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100000) — the number of magnets. Then *n* lines follow. The *i*-th line (1<=≤<=*i*<=≤<=*n*) contains either characters "01", if Mike put the *i*-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position.

On the single line of the output print the number of groups of magnets.

[ "6\n10\n10\n10\n01\n10\n10\n", "4\n01\n01\n10\n10\n" ]

[ "3\n", "2\n" ]

The first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets. The second testcase has two groups, each consisting of two magnets.

500

[ { "input": "6\n10\n10\n10\n01\n10\n10", "output": "3" }, { "input": "4\n01\n01\n10\n10", "output": "2" }, { "input": "1\n10", "output": "1" }, { "input": "2\n01\n10", "output": "2" }, { "input": "2\n10\n10", "output": "1" }, { "input": "3\n10\n01\n10", "output": "3" }, { "input": "1\n01", "output": "1" }, { "input": "2\n01\n01", "output": "1" }, { "input": "2\n10\n01", "output": "2" }, { "input": "3\n01\n01\n01", "output": "1" }, { "input": "3\n10\n10\n01", "output": "2" }, { "input": "3\n01\n10\n10", "output": "2" }, { "input": "115\n10\n10\n10\n10\n01\n01\n10\n10\n10\n01\n01\n10\n01\n01\n10\n10\n10\n01\n10\n01\n10\n10\n01\n01\n10\n10\n10\n10\n01\n10\n01\n01\n10\n10\n10\n10\n01\n10\n10\n10\n01\n10\n01\n10\n10\n10\n10\n01\n01\n01\n10\n10\n01\n01\n01\n10\n10\n01\n10\n01\n01\n01\n01\n10\n10\n01\n10\n01\n01\n01\n01\n01\n10\n01\n10\n10\n01\n01\n01\n10\n01\n01\n10\n10\n01\n01\n01\n01\n01\n10\n01\n10\n01\n10\n01\n01\n01\n10\n01\n10\n10\n01\n10\n10\n01\n01\n01\n10\n10\n10\n10\n10\n10\n10\n10", "output": "55" } ]

1,696,951,960

2,147,483,647

PyPy 3-64

TIME_LIMIT_EXCEEDED

TESTS

13

1,000

11,059,200

a = int(input()) d = [] for _ in range(a): b = str(input()) d += b[0] ans = 1 c = d[0] for k in d: if (c!=k): ans += 1 c = k print (ans)

Title: Magnets Time Limit: None seconds Memory Limit: None megabytes Problem Description: Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other. Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. Input Specification: The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100000) — the number of magnets. Then *n* lines follow. The *i*-th line (1<=≤<=*i*<=≤<=*n*) contains either characters "01", if Mike put the *i*-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. Output Specification: On the single line of the output print the number of groups of magnets. Demo Input: ['6\n10\n10\n10\n01\n10\n10\n', '4\n01\n01\n10\n10\n'] Demo Output: ['3\n', '2\n'] Note: The first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets. The second testcase has two groups, each consisting of two magnets.

```python a = int(input()) d = [] for _ in range(a): b = str(input()) d += b[0] ans = 1 c = d[0] for k in d: if (c!=k): ans += 1 c = k print (ans) ```

0

862

E

Mahmoud and Ehab and the function

PROGRAMMING

2,100

[ "binary search", "data structures", "sortings" ]

null

Dr. Evil is interested in math and functions, so he gave Mahmoud and Ehab array *a* of length *n* and array *b* of length *m*. He introduced a function *f*(*j*) which is defined for integers *j*, which satisfy 0<=≤<=*j*<=≤<=*m*<=-<=*n*. Suppose, *c**i*<==<=*a**i*<=-<=*b**i*<=+<=*j*. Then *f*(*j*)<==<=|*c*1<=-<=*c*2<=+<=*c*3<=-<=*c*4... *c**n*|. More formally, . Dr. Evil wants Mahmoud and Ehab to calculate the minimum value of this function over all valid *j*. They found it a bit easy, so Dr. Evil made their task harder. He will give them *q* update queries. During each update they should add an integer *x**i* to all elements in *a* in range [*l**i*;*r**i*] i.e. they should add *x**i* to *a**l**i*,<=*a**l**i*<=+<=1,<=... ,<=*a**r**i* and then they should calculate the minimum value of *f*(*j*) for all valid *j*. Please help Mahmoud and Ehab.

The first line contains three integers *n*,<=*m* and *q* (1<=≤<=*n*<=≤<=*m*<=≤<=105, 1<=≤<=*q*<=≤<=105) — number of elements in *a*, number of elements in *b* and number of queries, respectively. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n*. (<=-<=109<=≤<=*a**i*<=≤<=109) — elements of *a*. The third line contains *m* integers *b*1,<=*b*2,<=...,<=*b**m*. (<=-<=109<=≤<=*b**i*<=≤<=109) — elements of *b*. Then *q* lines follow describing the queries. Each of them contains three integers *l**i* *r**i* *x**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*, <=-<=109<=≤<=*x*<=≤<=109) — range to be updated and added value.

The first line should contain the minimum value of the function *f* before any update. Then output *q* lines, the *i*-th of them should contain the minimum value of the function *f* after performing the *i*-th update .

[ "5 6 3\n1 2 3 4 5\n1 2 3 4 5 6\n1 1 10\n1 1 -9\n1 5 -1\n" ]

[ "0\n9\n0\n0\n" ]

For the first example before any updates it's optimal to choose *j* = 0, *f*(0) = |(1 - 1) - (2 - 2) + (3 - 3) - (4 - 4) + (5 - 5)| = |0| = 0. After the first update *a* becomes {11, 2, 3, 4, 5} and it's optimal to choose *j* = 1, *f*(1) = |(11 - 2) - (2 - 3) + (3 - 4) - (4 - 5) + (5 - 6) = |9| = 9. After the second update *a* becomes {2, 2, 3, 4, 5} and it's optimal to choose *j* = 1, *f*(1) = |(2 - 2) - (2 - 3) + (3 - 4) - (4 - 5) + (5 - 6)| = |0| = 0. After the third update *a* becomes {1, 1, 2, 3, 4} and it's optimal to choose *j* = 0, *f*(0) = |(1 - 1) - (1 - 2) + (2 - 3) - (3 - 4) + (4 - 5)| = |0| = 0.

2,000

[ { "input": "5 6 3\n1 2 3 4 5\n1 2 3 4 5 6\n1 1 10\n1 1 -9\n1 5 -1", "output": "0\n9\n0\n0" }, { "input": "1 1 1\n937982044\n179683049\n1 1 821220804", "output": "758298995\n1579519799" } ]

1,513,604,528

2,147,483,647

Python 3

WRONG_ANSWER

TESTS

0

46

5,529,600

from bisect import * f = lambda: list(map(int, input().split())) n, m, q = f() k = m - n + 1 a = f() s = sum(a[0:n:2]) - sum(a[1:n:2]) b = [0] + f() for i in range(2, m + 1, 2): b[i] = -b[i] for i in range(m): b[i + 1] += b[i] u = [b[j] - b[j + n] for j in range(1, k, 2)] v = [b[j + n] - b[j] for j in range(0, k, 2)] d = sorted(u + v) def g(s): j = bisect_right(d, s) print(min(abs(s - d[j % k]), abs(s - d[j - 1]))) g(s) for i in range(q): l, r, x = f() s += x * (r % 2 + l % 2 - 1) if not r-l&1: if not l&1:x=-x s+=x g(s)

Title: Mahmoud and Ehab and the function Time Limit: None seconds Memory Limit: None megabytes Problem Description: Dr. Evil is interested in math and functions, so he gave Mahmoud and Ehab array *a* of length *n* and array *b* of length *m*. He introduced a function *f*(*j*) which is defined for integers *j*, which satisfy 0<=≤<=*j*<=≤<=*m*<=-<=*n*. Suppose, *c**i*<==<=*a**i*<=-<=*b**i*<=+<=*j*. Then *f*(*j*)<==<=|*c*1<=-<=*c*2<=+<=*c*3<=-<=*c*4... *c**n*|. More formally, . Dr. Evil wants Mahmoud and Ehab to calculate the minimum value of this function over all valid *j*. They found it a bit easy, so Dr. Evil made their task harder. He will give them *q* update queries. During each update they should add an integer *x**i* to all elements in *a* in range [*l**i*;*r**i*] i.e. they should add *x**i* to *a**l**i*,<=*a**l**i*<=+<=1,<=... ,<=*a**r**i* and then they should calculate the minimum value of *f*(*j*) for all valid *j*. Please help Mahmoud and Ehab. Input Specification: The first line contains three integers *n*,<=*m* and *q* (1<=≤<=*n*<=≤<=*m*<=≤<=105, 1<=≤<=*q*<=≤<=105) — number of elements in *a*, number of elements in *b* and number of queries, respectively. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n*. (<=-<=109<=≤<=*a**i*<=≤<=109) — elements of *a*. The third line contains *m* integers *b*1,<=*b*2,<=...,<=*b**m*. (<=-<=109<=≤<=*b**i*<=≤<=109) — elements of *b*. Then *q* lines follow describing the queries. Each of them contains three integers *l**i* *r**i* *x**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*, <=-<=109<=≤<=*x*<=≤<=109) — range to be updated and added value. Output Specification: The first line should contain the minimum value of the function *f* before any update. Then output *q* lines, the *i*-th of them should contain the minimum value of the function *f* after performing the *i*-th update . Demo Input: ['5 6 3\n1 2 3 4 5\n1 2 3 4 5 6\n1 1 10\n1 1 -9\n1 5 -1\n'] Demo Output: ['0\n9\n0\n0\n'] Note: For the first example before any updates it's optimal to choose *j* = 0, *f*(0) = |(1 - 1) - (2 - 2) + (3 - 3) - (4 - 4) + (5 - 5)| = |0| = 0. After the first update *a* becomes {11, 2, 3, 4, 5} and it's optimal to choose *j* = 1, *f*(1) = |(11 - 2) - (2 - 3) + (3 - 4) - (4 - 5) + (5 - 6) = |9| = 9. After the second update *a* becomes {2, 2, 3, 4, 5} and it's optimal to choose *j* = 1, *f*(1) = |(2 - 2) - (2 - 3) + (3 - 4) - (4 - 5) + (5 - 6)| = |0| = 0. After the third update *a* becomes {1, 1, 2, 3, 4} and it's optimal to choose *j* = 0, *f*(0) = |(1 - 1) - (1 - 2) + (2 - 3) - (3 - 4) + (4 - 5)| = |0| = 0.

```python from bisect import * f = lambda: list(map(int, input().split())) n, m, q = f() k = m - n + 1 a = f() s = sum(a[0:n:2]) - sum(a[1:n:2]) b = [0] + f() for i in range(2, m + 1, 2): b[i] = -b[i] for i in range(m): b[i + 1] += b[i] u = [b[j] - b[j + n] for j in range(1, k, 2)] v = [b[j + n] - b[j] for j in range(0, k, 2)] d = sorted(u + v) def g(s): j = bisect_right(d, s) print(min(abs(s - d[j % k]), abs(s - d[j - 1]))) g(s) for i in range(q): l, r, x = f() s += x * (r % 2 + l % 2 - 1) if not r-l&1: if not l&1:x=-x s+=x g(s) ```

0

156

D

Clues

PROGRAMMING

2,500

[ "combinatorics", "graphs" ]

null

As Sherlock Holmes was investigating another crime, he found a certain number of clues. Also, he has already found direct links between some of those clues. The direct links between the clues are mutual. That is, the direct link between clues *A* and *B* and the direct link between clues *B* and *A* is the same thing. No more than one direct link can exist between two clues. Of course Sherlock is able to find direct links between all clues. But it will take too much time and the criminals can use this extra time to hide. To solve the crime, Sherlock needs each clue to be linked to all other clues (maybe not directly, via some other clues). Clues *A* and *B* are considered linked either if there is a direct link between them or if there is a direct link between *A* and some other clue *C* which is linked to *B*. Sherlock Holmes counted the minimum number of additional direct links that he needs to find to solve the crime. As it turns out, it equals *T*. Please count the number of different ways to find exactly *T* direct links between the clues so that the crime is solved in the end. Two ways to find direct links are considered different if there exist two clues which have a direct link in one way and do not have a direct link in the other way. As the number of different ways can turn out rather big, print it modulo *k*.

The first line contains three space-separated integers *n*,<=*m*,<=*k* (1<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105, 1<=≤<=*k*<=≤<=109) — the number of clues, the number of direct clue links that Holmes has already found and the divisor for the modulo operation. Each of next *m* lines contains two integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=*n*,<=*a*<=≠<=*b*), that represent a direct link between clues. It is guaranteed that any two clues are linked by no more than one direct link. Note that the direct links between the clues are mutual.

Print the single number — the answer to the problem modulo *k*.

[ "2 0 1000000000\n", "3 0 100\n", "4 1 1000000000\n1 4\n" ]

[ "1\n", "3\n", "8\n" ]

The first sample only has two clues and Sherlock hasn't found any direct link between them yet. The only way to solve the crime is to find the link. The second sample has three clues and Sherlock hasn't found any direct links between them. He has to find two of three possible direct links between clues to solve the crime — there are 3 ways to do it. The third sample has four clues and the detective has already found one direct link between the first and the fourth clue. There are 8 ways to find two remaining clues to solve the crime.

2,500

[ { "input": "2 0 1000000000", "output": "1" }, { "input": "3 0 100", "output": "3" }, { "input": "4 1 1000000000\n1 4", "output": "8" }, { "input": "6 4 100000\n1 4\n4 6\n6 1\n2 5", "output": "36" }, { "input": "10 0 123456789", "output": "100000000" }, { "input": "10 5 1000000000\n1 2\n4 3\n5 6\n8 7\n10 9", "output": "32000" }, { "input": "8 4 17\n1 2\n2 3\n3 4\n4 1", "output": "8" }, { "input": "9 6 342597160\n1 2\n3 4\n4 5\n6 7\n7 8\n8 9", "output": "216" }, { "input": "1 0 1", "output": "0" }, { "input": "15 10 1\n1 2\n4 5\n6 3\n11 8\n8 5\n5 9\n9 1\n11 12\n12 1\n2 8", "output": "0" }, { "input": "8 8 999999937\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n8 7", "output": "1" }, { "input": "100000 0 1000000000", "output": "0" }, { "input": "100000 0 1", "output": "0" }, { "input": "9 11 498920381\n2 8\n5 4\n1 8\n8 3\n4 9\n3 6\n8 9\n1 7\n5 1\n5 6\n9 6", "output": "1" }, { "input": "2 0 753780649", "output": "1" }, { "input": "1 0 997185958", "output": "1" }, { "input": "10 36 279447540\n10 7\n10 8\n1 8\n1 5\n4 5\n9 5\n3 9\n7 3\n10 4\n8 9\n2 10\n6 2\n4 8\n10 3\n1 4\n10 1\n10 6\n8 3\n3 6\n9 7\n10 5\n6 9\n3 1\n8 6\n4 9\n5 3\n9 10\n7 2\n2 4\n7 4\n5 6\n5 8\n7 5\n5 2\n6 7\n1 9", "output": "1" }, { "input": "7 7 302838679\n5 3\n4 1\n5 4\n6 5\n1 6\n3 2\n6 4", "output": "6" }, { "input": "6 1 310732484\n4 2", "output": "432" }, { "input": "10 7 587143295\n1 10\n7 1\n1 8\n7 10\n8 10\n4 8\n6 8", "output": "6000" }, { "input": "3 3 975373207\n1 2\n1 3\n3 2", "output": "1" }, { "input": "9 33 321578376\n9 5\n6 3\n8 4\n4 1\n3 5\n2 6\n8 2\n7 6\n7 9\n8 6\n4 5\n1 6\n1 2\n5 6\n9 4\n7 8\n3 9\n9 6\n4 7\n7 2\n1 8\n4 6\n8 3\n3 7\n8 9\n5 7\n3 4\n7 1\n9 2\n5 1\n2 5\n9 1\n3 2", "output": "1" }, { "input": "5 10 93196990\n1 5\n1 4\n4 2\n1 3\n3 4\n1 2\n5 2\n5 4\n5 3\n2 3", "output": "1" }, { "input": "1 0 773734495", "output": "1" }, { "input": "2 1 719418546\n1 2", "output": "1" }, { "input": "3 2 21502109\n3 2\n1 2", "output": "1" }, { "input": "9 35 480175322\n6 3\n8 6\n7 5\n7 9\n3 4\n2 8\n5 3\n4 5\n4 6\n7 1\n7 6\n2 5\n8 3\n6 9\n8 4\n8 5\n6 1\n8 1\n3 2\n5 1\n8 9\n3 1\n8 7\n5 6\n5 9\n4 9\n7 4\n2 7\n3 9\n2 4\n7 3\n9 1\n2 9\n1 4\n1 2", "output": "1" }, { "input": "5 7 729985252\n2 3\n3 1\n2 5\n1 5\n1 2\n1 4\n4 3", "output": "1" }, { "input": "2 1 819865995\n2 1", "output": "1" }, { "input": "10 0 766953983", "output": "100000000" }, { "input": "2 1 855341703\n2 1", "output": "1" }, { "input": "10 30 407595309\n3 6\n6 10\n6 7\n7 10\n7 8\n3 10\n3 4\n1 4\n9 10\n8 4\n3 7\n5 1\n2 4\n6 2\n8 9\n10 5\n7 5\n10 4\n5 8\n8 2\n10 2\n1 6\n4 7\n2 3\n5 6\n8 10\n3 5\n1 8\n9 7\n1 9", "output": "1" }, { "input": "1 0 21080115", "output": "1" }, { "input": "57 28 776442742\n31 10\n25 28\n51 45\n14 40\n21 52\n53 51\n52 53\n4 6\n51 35\n53 15\n17 16\n40 44\n37 51\n33 43\n55 40\n42 16\n30 8\n19 45\n7 27\n31 8\n49 8\n43 44\n45 3\n16 22\n32 36\n52 36\n5 26\n2 23", "output": "135540294" }, { "input": "30 70 288262020\n27 18\n5 19\n23 17\n16 17\n29 17\n1 22\n23 5\n10 13\n22 26\n14 3\n8 3\n29 9\n9 1\n3 9\n16 4\n9 22\n10 22\n20 1\n3 7\n23 19\n26 8\n24 1\n5 7\n28 29\n20 11\n16 12\n6 9\n24 29\n30 4\n5 26\n18 21\n5 21\n30 6\n12 13\n16 23\n28 14\n30 1\n7 27\n7 19\n27 17\n5 30\n30 27\n28 30\n12 28\n27 9\n30 26\n20 18\n21 16\n8 30\n4 26\n13 22\n2 14\n12 30\n4 2\n6 12\n29 25\n19 29\n14 15\n3 23\n10 28\n7 1\n21 10\n4 12\n1 14\n7 21\n21 8\n17 26\n7 6\n26 29\n9 8", "output": "1" }, { "input": "99 12 832839308\n66 23\n36 5\n16 57\n70 62\n94 96\n63 33\n99 23\n63 10\n6 85\n73 23\n69 46\n72 95", "output": "71450536" }, { "input": "30 18 918975816\n30 18\n23 1\n21 14\n14 8\n18 9\n23 29\n3 23\n29 19\n18 4\n27 19\n30 2\n9 10\n9 28\n16 15\n10 6\n18 12\n23 9\n19 14", "output": "782410104" }, { "input": "83 33 367711297\n14 74\n26 22\n55 19\n8 70\n6 42\n53 49\n54 56\n52 17\n62 44\n78 61\n76 4\n78 30\n51 2\n31 42\n33 67\n45 41\n64 62\n15 25\n33 35\n37 20\n38 65\n65 83\n61 14\n20 67\n62 47\n7 34\n78 41\n38 83\n26 69\n54 58\n11 62\n30 55\n15 74", "output": "131377693" }, { "input": "24 68 862907549\n6 9\n16 22\n11 23\n12 17\n18 2\n15 5\n5 22\n16 4\n21 9\n7 11\n19 16\n9 13\n21 20\n5 24\n7 12\n17 1\n24 21\n23 7\n16 17\n16 18\n10 13\n18 7\n8 21\n13 5\n10 18\n4 11\n21 6\n15 13\n2 1\n20 16\n11 16\n22 19\n2 4\n21 1\n6 18\n24 12\n21 19\n6 14\n22 24\n11 20\n2 19\n1 11\n24 18\n14 8\n10 24\n5 3\n11 3\n17 4\n4 20\n2 10\n12 11\n24 7\n23 16\n2 3\n19 24\n22 1\n22 4\n4 6\n3 4\n11 13\n6 5\n18 23\n4 23\n22 13\n20 5\n2 5\n2 11\n9 5", "output": "1" }, { "input": "86 23 608266393\n62 78\n44 84\n42 37\n20 24\n40 36\n41 76\n14 38\n80 72\n39 52\n31 58\n71 17\n81 6\n32 65\n11 69\n43 86\n85 59\n28 77\n78 64\n15 19\n36 39\n53 49\n48 75\n33 85", "output": "235915236" }, { "input": "47 51 283106191\n18 14\n30 26\n24 2\n18 41\n35 31\n16 24\n29 39\n6 12\n17 21\n7 19\n36 16\n27 39\n28 34\n22 35\n28 43\n40 5\n2 26\n18 16\n27 13\n21 6\n19 5\n35 30\n13 31\n7 10\n25 7\n44 42\n45 1\n35 47\n11 28\n47 46\n18 15\n27 16\n24 41\n10 8\n25 41\n4 40\n5 11\n24 6\n10 17\n41 38\n47 28\n8 29\n25 24\n35 37\n44 17\n24 47\n8 32\n33 11\n26 28\n23 9\n5 9", "output": "189974" }, { "input": "67 2 818380264\n4 52\n15 44", "output": "517849052" }, { "input": "10 45 220178113\n9 1\n8 1\n5 8\n1 5\n7 8\n6 7\n7 9\n6 2\n3 2\n1 4\n8 3\n8 9\n3 6\n4 5\n5 3\n10 4\n3 9\n9 6\n5 9\n2 9\n10 7\n1 10\n9 4\n3 10\n2 5\n7 1\n6 10\n6 5\n8 6\n8 4\n8 10\n1 6\n4 2\n9 10\n2 10\n7 3\n6 4\n7 5\n1 2\n4 3\n10 5\n4 7\n3 1\n7 2\n8 2", "output": "1" }, { "input": "588 32 634894588\n535 26\n562 406\n70 368\n357 513\n108 361\n515 5\n159 56\n522 81\n169 229\n312 252\n492 43\n476 405\n524 555\n537 169\n142 149\n586 112\n7 159\n76 370\n295 376\n33 455\n278 225\n377 88\n526 308\n517 303\n300 576\n230 493\n588 525\n177 312\n356 215\n515 34\n196 236\n323 9", "output": "478655040" }, { "input": "3 2 11\n1 2\n2 3", "output": "1" }, { "input": "2 1 1000\n1 2", "output": "1" }, { "input": "1 0 10000", "output": "1" }, { "input": "5 4 100000\n1 2\n2 3\n3 4\n4 5", "output": "1" }, { "input": "2 1 100000\n1 2", "output": "1" }, { "input": "1 0 10", "output": "1" }, { "input": "3 3 100\n1 2\n2 3\n3 1", "output": "1" }, { "input": "2 1 100\n1 2", "output": "1" }, { "input": "3 2 42\n1 2\n2 3", "output": "1" } ]

1,605,149,105

2,147,483,647

PyPy 3

RUNTIME_ERROR

TESTS

16

1,964

13,721,600

def dfs(node): vis[node] = True acc[cc]+=1 for i in adj[node]: if not vis[i]: dfs(i) if __name__ == '__main__': _input = input().split() n = int(_input[0]) m = int(_input[1]) k = int(_input[2]) adj = [] vis = [] acc = [] cc = 0 for i in range(n): vis.append(False) adj.append([]) acc.append(0) for i in range(m): _in2 = input().split() v = int(_in2[0]) - 1 w = int(_in2[1]) - 1 adj[v].append(w) adj[w].append(v) for i in range(n): if not vis[i]: dfs(i) cc+=1 if cc == 1: print(1 % k) exit() ans = 1 for i in range(cc - 2): ans = ans * n ans = ans % k for i in range(cc): ans = ans * acc[i] ans = ans % k print(ans)

Title: Clues Time Limit: None seconds Memory Limit: None megabytes Problem Description: As Sherlock Holmes was investigating another crime, he found a certain number of clues. Also, he has already found direct links between some of those clues. The direct links between the clues are mutual. That is, the direct link between clues *A* and *B* and the direct link between clues *B* and *A* is the same thing. No more than one direct link can exist between two clues. Of course Sherlock is able to find direct links between all clues. But it will take too much time and the criminals can use this extra time to hide. To solve the crime, Sherlock needs each clue to be linked to all other clues (maybe not directly, via some other clues). Clues *A* and *B* are considered linked either if there is a direct link between them or if there is a direct link between *A* and some other clue *C* which is linked to *B*. Sherlock Holmes counted the minimum number of additional direct links that he needs to find to solve the crime. As it turns out, it equals *T*. Please count the number of different ways to find exactly *T* direct links between the clues so that the crime is solved in the end. Two ways to find direct links are considered different if there exist two clues which have a direct link in one way and do not have a direct link in the other way. As the number of different ways can turn out rather big, print it modulo *k*. Input Specification: The first line contains three space-separated integers *n*,<=*m*,<=*k* (1<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105, 1<=≤<=*k*<=≤<=109) — the number of clues, the number of direct clue links that Holmes has already found and the divisor for the modulo operation. Each of next *m* lines contains two integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=*n*,<=*a*<=≠<=*b*), that represent a direct link between clues. It is guaranteed that any two clues are linked by no more than one direct link. Note that the direct links between the clues are mutual. Output Specification: Print the single number — the answer to the problem modulo *k*. Demo Input: ['2 0 1000000000\n', '3 0 100\n', '4 1 1000000000\n1 4\n'] Demo Output: ['1\n', '3\n', '8\n'] Note: The first sample only has two clues and Sherlock hasn't found any direct link between them yet. The only way to solve the crime is to find the link. The second sample has three clues and Sherlock hasn't found any direct links between them. He has to find two of three possible direct links between clues to solve the crime — there are 3 ways to do it. The third sample has four clues and the detective has already found one direct link between the first and the fourth clue. There are 8 ways to find two remaining clues to solve the crime.

```python def dfs(node): vis[node] = True acc[cc]+=1 for i in adj[node]: if not vis[i]: dfs(i) if __name__ == '__main__': _input = input().split() n = int(_input[0]) m = int(_input[1]) k = int(_input[2]) adj = [] vis = [] acc = [] cc = 0 for i in range(n): vis.append(False) adj.append([]) acc.append(0) for i in range(m): _in2 = input().split() v = int(_in2[0]) - 1 w = int(_in2[1]) - 1 adj[v].append(w) adj[w].append(v) for i in range(n): if not vis[i]: dfs(i) cc+=1 if cc == 1: print(1 % k) exit() ans = 1 for i in range(cc - 2): ans = ans * n ans = ans % k for i in range(cc): ans = ans * acc[i] ans = ans % k print(ans) ```

-1

352

A

Jeff and Digits

PROGRAMMING

1,000

[ "brute force", "implementation", "math" ]

null

Jeff's got *n* cards, each card contains either digit 0, or digit 5. Jeff can choose several cards and put them in a line so that he gets some number. What is the largest possible number divisible by 90 Jeff can make from the cards he's got? Jeff must make the number without leading zero. At that, we assume that number 0 doesn't contain any leading zeroes. Jeff doesn't have to use all the cards.

The first line contains integer *n* (1<=≤<=*n*<=≤<=103). The next line contains *n* integers *a*1, *a*2, ..., *a**n* (*a**i*<==<=0 or *a**i*<==<=5). Number *a**i* represents the digit that is written on the *i*-th card.

In a single line print the answer to the problem — the maximum number, divisible by 90. If you can't make any divisible by 90 number from the cards, print -1.

[ "4\n5 0 5 0\n", "11\n5 5 5 5 5 5 5 5 0 5 5\n" ]

[ "0\n", "5555555550\n" ]

In the first test you can make only one number that is a multiple of 90 — 0. In the second test you can make number 5555555550, it is a multiple of 90.

500

[ { "input": "4\n5 0 5 0", "output": "0" }, { "input": "11\n5 5 5 5 5 5 5 5 0 5 5", "output": "5555555550" }, { "input": "7\n5 5 5 5 5 5 5", "output": "-1" }, { "input": "1\n5", "output": "-1" }, { "input": "1\n0", "output": "0" }, { "input": "11\n5 0 5 5 5 0 0 5 5 5 5", "output": "0" }, { "input": "23\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 0 0 0 0 0", "output": "55555555555555555500000" }, { "input": "9\n5 5 5 5 5 5 5 5 5", "output": "-1" }, { "input": "24\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 0 0 0 0 0", "output": "55555555555555555500000" }, { "input": "10\n0 0 0 0 0 0 0 0 0 0", "output": "0" }, { "input": "10\n5 5 5 5 5 0 0 5 0 5", "output": "0" }, { "input": "3\n5 5 0", "output": "0" }, { "input": "5\n5 5 0 5 5", "output": "0" }, { "input": "14\n0 5 5 0 0 0 0 0 0 5 5 5 5 5", "output": "0" }, { "input": "3\n5 5 5", "output": "-1" }, { "input": "3\n0 5 5", "output": "0" }, { "input": "13\n0 0 5 0 5 0 5 5 0 0 0 0 0", "output": "0" }, { "input": "9\n5 5 0 5 5 5 5 5 5", "output": "0" }, { "input": "8\n0 0 0 0 0 0 0 0", "output": "0" }, { "input": "101\n5 0 0 0 0 0 0 0 5 0 0 0 0 5 0 0 5 0 0 0 0 0 5 0 0 0 0 0 0 0 0 5 0 0 5 0 0 0 0 0 0 0 5 0 0 5 0 0 0 5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 5 0 0 0 0 0 0 0 0 0 0 5 0 0 0 0 5 0 0 0 0 0 0 0 0 0 5 0 0 5 0 0 0 0 5 0 0", "output": "5555555550000000000000000000000000000000000000000000000000000000000000000000000000000000000000" }, { "input": "214\n5 0 5 0 5 0 0 0 5 5 0 5 0 5 5 0 5 0 0 0 0 5 5 0 0 5 5 0 0 0 0 5 5 5 5 0 5 0 0 0 0 0 0 5 0 0 0 5 0 0 5 0 0 5 5 0 0 5 5 0 0 0 0 0 5 0 5 0 5 5 0 5 0 0 5 5 5 0 5 0 5 0 5 5 0 5 0 0 0 5 5 0 5 0 5 5 5 5 5 0 0 0 0 0 0 5 0 5 5 0 5 0 5 0 5 5 0 0 0 0 5 0 5 0 5 0 0 5 0 0 5 5 5 5 5 0 0 5 0 0 5 0 0 5 0 0 5 0 0 5 0 5 0 0 0 5 0 0 5 5 5 0 0 5 5 5 0 0 5 5 0 0 0 5 0 0 5 5 5 5 5 5 0 5 0 0 5 5 5 5 0 5 5 0 0 0 5 5 5 5 0 0 0 0 5 0 0 5 0 0 5 5 0 0", "output": "5555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555550000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000" }, { "input": "80\n0 0 0 0 5 0 5 5 5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 5 0 0 5 0 0 0 0 0 0 0 0 0 5 5 0 5 0 0 0 0 0 0 5 0 0 0 0 0 0 0 5 0 0 0 0 5 0 5 5 0 0 0", "output": "555555555000000000000000000000000000000000000000000000000000000000000000000" }, { "input": "2\n0 0", "output": "0" }, { "input": "3\n5 0 0", "output": "0" }, { "input": "4\n5 5 5 5", "output": "-1" }, { "input": "2\n0 5", "output": "0" }, { "input": "14\n5 5 5 5 5 5 5 5 5 5 5 5 5 0", "output": "5555555550" }, { "input": "18\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5", "output": "-1" }, { "input": "10\n5 5 5 5 5 5 5 5 5 0", "output": "5555555550" }, { "input": "10\n5 5 5 5 5 5 5 5 5 5", "output": "-1" }, { "input": "20\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5", "output": "-1" } ]

1,638,034,263

2,147,483,647

Python 3

OK

TESTS

42

92

0

n=int(input()) l=input() a=l.count("0") if a<1:print("-1") else:print(int("5"*((9)*(l.count("5")//9))+"0"*a))

Title: Jeff and Digits Time Limit: None seconds Memory Limit: None megabytes Problem Description: Jeff's got *n* cards, each card contains either digit 0, or digit 5. Jeff can choose several cards and put them in a line so that he gets some number. What is the largest possible number divisible by 90 Jeff can make from the cards he's got? Jeff must make the number without leading zero. At that, we assume that number 0 doesn't contain any leading zeroes. Jeff doesn't have to use all the cards. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=103). The next line contains *n* integers *a*1, *a*2, ..., *a**n* (*a**i*<==<=0 or *a**i*<==<=5). Number *a**i* represents the digit that is written on the *i*-th card. Output Specification: In a single line print the answer to the problem — the maximum number, divisible by 90. If you can't make any divisible by 90 number from the cards, print -1. Demo Input: ['4\n5 0 5 0\n', '11\n5 5 5 5 5 5 5 5 0 5 5\n'] Demo Output: ['0\n', '5555555550\n'] Note: In the first test you can make only one number that is a multiple of 90 — 0. In the second test you can make number 5555555550, it is a multiple of 90.

```python n=int(input()) l=input() a=l.count("0") if a<1:print("-1") else:print(int("5"*((9)*(l.count("5")//9))+"0"*a)) ```

3

33

A

What is for dinner?

PROGRAMMING

1,200

[ "greedy", "implementation" ]

A. What is for dinner?

2

256

In one little known, but very beautiful country called Waterland, lives a lovely shark Valerie. Like all the sharks, she has several rows of teeth, and feeds on crucians. One of Valerie's distinguishing features is that while eating one crucian she uses only one row of her teeth, the rest of the teeth are "relaxing". For a long time our heroine had been searching the sea for crucians, but a great misfortune happened. Her teeth started to ache, and she had to see the local dentist, lobster Ashot. As a professional, Ashot quickly relieved Valerie from her toothache. Moreover, he managed to determine the cause of Valerie's developing caries (for what he was later nicknamed Cap). It turned that Valerie eats too many crucians. To help Valerie avoid further reoccurrence of toothache, Ashot found for each Valerie's tooth its residual viability. Residual viability of a tooth is a value equal to the amount of crucians that Valerie can eat with this tooth. Every time Valerie eats a crucian, viability of all the teeth used for it will decrease by one. When the viability of at least one tooth becomes negative, the shark will have to see the dentist again. Unhappy, Valerie came back home, where a portion of crucians was waiting for her. For sure, the shark couldn't say no to her favourite meal, but she had no desire to go back to the dentist. That's why she decided to eat the maximum amount of crucians from the portion but so that the viability of no tooth becomes negative. As Valerie is not good at mathematics, she asked you to help her to find out the total amount of crucians that she can consume for dinner. We should remind you that while eating one crucian Valerie uses exactly one row of teeth and the viability of each tooth from this row decreases by one.

The first line contains three integers *n*, *m*, *k* (1<=≤<=*m*<=≤<=*n*<=≤<=1000,<=0<=≤<=*k*<=≤<=106) — total amount of Valerie's teeth, amount of tooth rows and amount of crucians in Valerie's portion for dinner. Then follow *n* lines, each containing two integers: *r* (1<=≤<=*r*<=≤<=*m*) — index of the row, where belongs the corresponding tooth, and *c* (0<=≤<=*c*<=≤<=106) — its residual viability. It's guaranteed that each tooth row has positive amount of teeth.

In the first line output the maximum amount of crucians that Valerie can consume for dinner.

[ "4 3 18\n2 3\n1 2\n3 6\n2 3\n", "2 2 13\n1 13\n2 12\n" ]

[ "11\n", "13\n" ]

none

500

[ { "input": "4 3 18\n2 3\n1 2\n3 6\n2 3", "output": "11" }, { "input": "2 2 13\n1 13\n2 12", "output": "13" }, { "input": "5 4 8\n4 6\n4 5\n1 3\n2 0\n3 3", "output": "8" }, { "input": "1 1 0\n1 3", "output": "0" }, { "input": "7 1 30\n1 8\n1 15\n1 5\n1 17\n1 9\n1 16\n1 16", "output": "5" }, { "input": "4 2 8\n1 9\n1 10\n1 4\n2 6", "output": "8" }, { "input": "10 4 14\n2 6\n1 5\n2 8\n2 6\n2 5\n4 1\n4 0\n2 4\n3 4\n1 0", "output": "8" }, { "input": "54 22 1009\n15 7\n17 7\n11 9\n5 11\n12 9\n13 8\n13 12\n22 11\n20 9\n20 7\n16 11\n19 12\n3 12\n15 9\n1 12\n2 10\n16 10\n16 10\n14 10\n9 11\n9 9\n14 8\n10 10\n16 12\n1 8\n3 8\n21 11\n18 12\n2 6\n9 11\n11 7\n15 9\n20 11\n6 8\n8 8\n19 11\n15 7\n9 9\n22 7\n12 9\n17 9\n17 11\n12 7\n15 9\n5 11\n11 6\n18 8\n7 10\n4 10\n13 12\n4 8\n20 6\n15 12\n9 7", "output": "180" }, { "input": "1 1 1000000\n1 1000000", "output": "1000000" }, { "input": "4 3 181818\n3 1299\n1 1694\n3 1164\n2 1278", "output": "4136" }, { "input": "50 50 0\n33 0\n26 0\n30 0\n41 0\n38 0\n36 0\n16 0\n7 0\n23 0\n5 0\n39 0\n29 0\n40 0\n47 0\n24 0\n43 0\n14 0\n10 0\n20 0\n3 0\n6 0\n34 0\n13 0\n18 0\n21 0\n9 0\n8 0\n22 0\n25 0\n4 0\n11 0\n17 0\n32 0\n44 0\n49 0\n50 0\n27 0\n28 0\n45 0\n12 0\n35 0\n15 0\n42 0\n31 0\n2 0\n1 0\n48 0\n19 0\n37 0\n46 0", "output": "0" }, { "input": "19 12 199\n7 1\n8 6\n6 14\n1 7\n4 1\n6 6\n3 4\n1 5\n9 2\n5 3\n11 3\n9 4\n1 12\n4 7\n7 3\n12 14\n2 1\n10 8\n6 12", "output": "54" } ]

1,679,538,123

2,147,483,647

PyPy 3-64

RUNTIME_ERROR

TESTS

0

122

2,867,200

low = 0 high=1000000 for i in range(25): mid=(low+high)//2 print(mid,flush=True) o=input() if o=='<': high=mid else: low=mid print(low)

Title: What is for dinner? Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: In one little known, but very beautiful country called Waterland, lives a lovely shark Valerie. Like all the sharks, she has several rows of teeth, and feeds on crucians. One of Valerie's distinguishing features is that while eating one crucian she uses only one row of her teeth, the rest of the teeth are "relaxing". For a long time our heroine had been searching the sea for crucians, but a great misfortune happened. Her teeth started to ache, and she had to see the local dentist, lobster Ashot. As a professional, Ashot quickly relieved Valerie from her toothache. Moreover, he managed to determine the cause of Valerie's developing caries (for what he was later nicknamed Cap). It turned that Valerie eats too many crucians. To help Valerie avoid further reoccurrence of toothache, Ashot found for each Valerie's tooth its residual viability. Residual viability of a tooth is a value equal to the amount of crucians that Valerie can eat with this tooth. Every time Valerie eats a crucian, viability of all the teeth used for it will decrease by one. When the viability of at least one tooth becomes negative, the shark will have to see the dentist again. Unhappy, Valerie came back home, where a portion of crucians was waiting for her. For sure, the shark couldn't say no to her favourite meal, but she had no desire to go back to the dentist. That's why she decided to eat the maximum amount of crucians from the portion but so that the viability of no tooth becomes negative. As Valerie is not good at mathematics, she asked you to help her to find out the total amount of crucians that she can consume for dinner. We should remind you that while eating one crucian Valerie uses exactly one row of teeth and the viability of each tooth from this row decreases by one. Input Specification: The first line contains three integers *n*, *m*, *k* (1<=≤<=*m*<=≤<=*n*<=≤<=1000,<=0<=≤<=*k*<=≤<=106) — total amount of Valerie's teeth, amount of tooth rows and amount of crucians in Valerie's portion for dinner. Then follow *n* lines, each containing two integers: *r* (1<=≤<=*r*<=≤<=*m*) — index of the row, where belongs the corresponding tooth, and *c* (0<=≤<=*c*<=≤<=106) — its residual viability. It's guaranteed that each tooth row has positive amount of teeth. Output Specification: In the first line output the maximum amount of crucians that Valerie can consume for dinner. Demo Input: ['4 3 18\n2 3\n1 2\n3 6\n2 3\n', '2 2 13\n1 13\n2 12\n'] Demo Output: ['11\n', '13\n'] Note: none

```python low = 0 high=1000000 for i in range(25): mid=(low+high)//2 print(mid,flush=True) o=input() if o=='<': high=mid else: low=mid print(low) ```

-1

257

D

Sum

PROGRAMMING

1,900

[ "greedy", "math" ]

null

Vasya has found a piece of paper with an array written on it. The array consists of *n* integers *a*1,<=*a*2,<=...,<=*a**n*. Vasya noticed that the following condition holds for the array *a**i*<=≤<=*a**i*<=+<=1<=≤<=2·*a**i* for any positive integer *i* (*i*<=<<=*n*). Vasya wants to add either a "+" or a "-" before each number of array. Thus, Vasya will get an expression consisting of *n* summands. The value of the resulting expression is the sum of all its elements. The task is to add signs "+" and "-" before each number so that the value of expression *s* meets the limits 0<=≤<=*s*<=≤<=*a*1. Print a sequence of signs "+" and "-", satisfying the given limits. It is guaranteed that the solution for the problem exists.

The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the size of the array. The second line contains space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) — the original array. It is guaranteed that the condition *a**i*<=≤<=*a**i*<=+<=1<=≤<=2·*a**i* fulfills for any positive integer *i* (*i*<=<<=*n*).

In a single line print the sequence of *n* characters "+" and "-", where the *i*-th character is the sign that is placed in front of number *a**i*. The value of the resulting expression *s* must fit into the limits 0<=≤<=*s*<=≤<=*a*1. If there are multiple solutions, you are allowed to print any of them.

[ "4\n1 2 3 5\n", "3\n3 3 5\n" ]

[ "+++-", "++-" ]

none

2,000

[ { "input": "4\n1 2 3 5", "output": "+++-" }, { "input": "3\n3 3 5", "output": "++-" }, { "input": "4\n2 4 5 6", "output": "-++-" }, { "input": "6\n3 5 10 11 12 20", "output": "++-++-" }, { "input": "10\n10 14 17 22 43 72 74 84 88 93", "output": "++---++--+" }, { "input": "11\n3 6 7 11 13 16 26 52 63 97 97", "output": "++--+-++--+" }, { "input": "12\n3 3 4 7 14 26 51 65 72 72 85 92", "output": "+-+--++-+--+" }, { "input": "40\n3 3 3 6 10 10 18 19 34 66 107 150 191 286 346 661 1061 1620 2123 3679 5030 8736 10539 19659 38608 47853 53095 71391 135905 255214 384015 694921 1357571 1364832 2046644 2595866 2918203 3547173 4880025 6274651", "output": "+-++-+-+-+-++-++-+-++--++--++--+-+-+-++-" }, { "input": "41\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0", "output": "-----------------------------------------" }, { "input": "42\n2 2 2 3 6 8 14 22 37 70 128 232 330 472 473 784 1481 2008 3076 4031 7504 8070 8167 11954 17832 24889 27113 41190 48727 92327 148544 186992 247329 370301 547840 621571 868209 1158781 1725242 3027208 4788036 5166155", "output": "-++-+-++--+-+-+-+-+-+-+-+--++-+-++--+--++-" }, { "input": "43\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0", "output": "-------------------------------------------" }, { "input": "44\n4 6 8 14 28 36 43 76 78 151 184 217 228 245 469 686 932 1279 2100 2373 4006 4368 8173 10054 18409 28333 32174 53029 90283 161047 293191 479853 875055 1206876 1423386 1878171 2601579 3319570 4571631 4999760 6742654 12515994 22557290 29338426", "output": "+-+-+-+--++-+-+-++--+-+--+--+++--++--+-+-++-" }, { "input": "45\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0", "output": "---------------------------------------------" }, { "input": "46\n3 6 6 8 16 19 23 46 53 90 114 131 199 361 366 523 579 1081 1457 2843 4112 4766 7187 8511 15905 22537 39546 70064 125921 214041 324358 392931 547572 954380 1012122 1057632 1150405 1393895 1915284 1969248 2541748 4451203 8201302 10912223 17210988 24485089", "output": "-+++-+--+-++-+--++-+-+-++-++-+--++--++--++-++-" }, { "input": "47\n3 3 5 6 9 13 13 14 22 33 50 76 83 100 168 303 604 1074 1417 2667 3077 4821 5129 7355 11671 22342 24237 34014 66395 73366 105385 205561 387155 756780 965476 1424160 1624526 2701046 4747339 5448855 6467013 9133423 11001389 18298303 23824100 41393164 58364321", "output": "-++--+-+-+-+-+-+-++-+-++-+--+----++-+-+--++-++-" }, { "input": "48\n4 7 12 16 23 43 61 112 134 141 243 267 484 890 1427 1558 1653 2263 2889 3313 3730 5991 10176 18243 18685 36555 40006 62099 70557 106602 122641 125854 213236 309698 379653 713328 999577 1021356 2007207 2886237 4994645 5812125 11576387 14215887 26060277 35989707 36964781 57933366", "output": "++-++-+-++---+-+-+-+--+--+-+-+-+-+-+--+-++-+-++-" }, { "input": "1\n1000000000", "output": "+" }, { "input": "2\n5 8", "output": "-+" }, { "input": "3\n1000000000 1000000000 1000000000", "output": "++-" } ]

1,587,630,982

2,147,483,647

PyPy 3

TIME_LIMIT_EXCEEDED

TESTS

7

2,000

10,956,800

n=int(input()) a=list(map(int,input().split(' '))) summ=[0]*n summ[0]=1 s=0 for i in range(n): if(s>=a[n-i-1]): s-=a[n-i-1] summ[i]=-1 # print(summ) else: summ[i]=1 s=a[n-i-1]-s if(i==0): continue # print('before:',summ,i) for j in range(i): summ[j]*=(-1) # print('after:',summ) # print(s) summ.reverse() for x in summ: if(x==1): print('+',end='') else: print('-',end='')

Title: Sum Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vasya has found a piece of paper with an array written on it. The array consists of *n* integers *a*1,<=*a*2,<=...,<=*a**n*. Vasya noticed that the following condition holds for the array *a**i*<=≤<=*a**i*<=+<=1<=≤<=2·*a**i* for any positive integer *i* (*i*<=<<=*n*). Vasya wants to add either a "+" or a "-" before each number of array. Thus, Vasya will get an expression consisting of *n* summands. The value of the resulting expression is the sum of all its elements. The task is to add signs "+" and "-" before each number so that the value of expression *s* meets the limits 0<=≤<=*s*<=≤<=*a*1. Print a sequence of signs "+" and "-", satisfying the given limits. It is guaranteed that the solution for the problem exists. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the size of the array. The second line contains space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) — the original array. It is guaranteed that the condition *a**i*<=≤<=*a**i*<=+<=1<=≤<=2·*a**i* fulfills for any positive integer *i* (*i*<=<<=*n*). Output Specification: In a single line print the sequence of *n* characters "+" and "-", where the *i*-th character is the sign that is placed in front of number *a**i*. The value of the resulting expression *s* must fit into the limits 0<=≤<=*s*<=≤<=*a*1. If there are multiple solutions, you are allowed to print any of them. Demo Input: ['4\n1 2 3 5\n', '3\n3 3 5\n'] Demo Output: ['+++-', '++-'] Note: none

```python n=int(input()) a=list(map(int,input().split(' '))) summ=[0]*n summ[0]=1 s=0 for i in range(n): if(s>=a[n-i-1]): s-=a[n-i-1] summ[i]=-1 # print(summ) else: summ[i]=1 s=a[n-i-1]-s if(i==0): continue # print('before:',summ,i) for j in range(i): summ[j]*=(-1) # print('after:',summ) # print(s) summ.reverse() for x in summ: if(x==1): print('+',end='') else: print('-',end='') ```

0

994

A

Fingerprints

PROGRAMMING

800

[ "implementation" ]

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" } ]

1,654,001,403

2,147,483,647

Python 3

RUNTIME_ERROR

TESTS

0

15

0

an = lambda: (int, input().split()) n, a, b = an(), an(), an() print(' '.join(map(str, (k for k in k if k in b))))

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 an = lambda: (int, input().split()) n, a, b = an(), an(), an() print(' '.join(map(str, (k for k in k if k in b)))) ```

-1