message stringlengths 2 57.2k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 61 108k | cluster float64 22 22 | __index_level_0__ int64 122 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
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
Input contains one integer number n (1 ≤ n ≤ 3000).
Output
Output the amount of almost prime numbers between 1 and n, inclusive.
Examples
Input
10
Output
2
Input
21
Output
8 | instruction | 0 | 75,341 | 22 | 150,682 |
Tags: number theory
Correct Solution:
```
import math
count = 0
n = int(input())
for i in range(1, n+1):
deevisor = 0
is_prime = [True] * (i+1)
is_prime[0], is_prime[1] = False, False
if i % 2 == 0: deevisor += 1
for x in range(3, math.ceil(i**1/2) + 1, 2):
if is_prime[x]:
if i % x == 0: deevisor += 1
for j in range(x**2, i + 1, x):
is_prime[j] = False
if deevisor == 2: count += 1
print(count)
``` | output | 1 | 75,341 | 22 | 150,683 |
Provide tags and a correct Python 3 solution for this coding contest problem.
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
Input contains one integer number n (1 ≤ n ≤ 3000).
Output
Output the amount of almost prime numbers between 1 and n, inclusive.
Examples
Input
10
Output
2
Input
21
Output
8 | instruction | 0 | 75,342 | 22 | 150,684 |
Tags: number theory
Correct Solution:
```
def prime(k):
if k==1:
return False
else:
for i in range(2,k):
if k%i==0:
return False
return True
n=int(input())
e=0
t=1
a=0
for i in range(1,n+1):
for k in range(1,t+1):
if t%k==0 and prime(k)==True:
e=e+1
t=t+1
if e==2:
a=a+1
e=0
print(a)
``` | output | 1 | 75,342 | 22 | 150,685 |
Provide tags and a correct Python 3 solution for this coding contest problem.
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
Input contains one integer number n (1 ≤ n ≤ 3000).
Output
Output the amount of almost prime numbers between 1 and n, inclusive.
Examples
Input
10
Output
2
Input
21
Output
8 | instruction | 0 | 75,343 | 22 | 150,686 |
Tags: number theory
Correct Solution:
```
n = int(input())
arr = [0] * (n + 1)
for i in range(2, n // 2 + 1):
if not arr[i]:
for j in range(2 * i, n + 1, i):
arr[j] += 1
print(arr.count(2))
``` | output | 1 | 75,343 | 22 | 150,687 |
Provide tags and a correct Python 3 solution for this coding contest problem.
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
Input contains one integer number n (1 ≤ n ≤ 3000).
Output
Output the amount of almost prime numbers between 1 and n, inclusive.
Examples
Input
10
Output
2
Input
21
Output
8 | instruction | 0 | 75,344 | 22 | 150,688 |
Tags: number theory
Correct Solution:
```
import math
def pf_bool(m):
a=0
f=False
if m==14:
f=True
j=m
s=set()
if m%2==0:
a=1
s.add(2)
while m%2==0:
m=m//2
for i in range(2,j+1):
if m%i==0:
a+=1
if a>2:
return False
while m%i==0:
m=m//i
if a==2:
return True
return False
t= 1
for T in range(t):
# [n,m]=list(map(int, input().split(' ')))
# arr=list(map(int, input().split(' ')))
ans=0
n=int(input())
for k in range(6,n+1):
if pf_bool(k):
ans+=1
print(ans)
``` | output | 1 | 75,344 | 22 | 150,689 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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
Input contains one integer number n (1 ≤ n ≤ 3000).
Output
Output the amount of almost prime numbers between 1 and n, inclusive.
Examples
Input
10
Output
2
Input
21
Output
8
Submitted Solution:
```
n = int(input())
a = [i for i in range(0,n+2)]
for i in range(2, int(n**0.5)+2):
for j in range(i*i,n+2,i):
a[j] = 0
a = [i for i in a if i != 0 and i != 1]
counter = 0
for i in range(6,n+1):
number = i
lcounter = 0
for j in a:
if number % j == 0 and number >= j:
number //= j
lcounter += 1
if lcounter == 2:
counter += 1
print(counter)
``` | instruction | 0 | 75,345 | 22 | 150,690 |
Yes | output | 1 | 75,345 | 22 | 150,691 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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
Input contains one integer number n (1 ≤ n ≤ 3000).
Output
Output the amount of almost prime numbers between 1 and n, inclusive.
Examples
Input
10
Output
2
Input
21
Output
8
Submitted Solution:
```
n = int(input())
count = 0
for i in range(4, n + 1):
local_count = 0
value = i
number = 2
while number <= value:
if (value % number == 0):
local_count += 1
while value % number == 0:
value = value // number
number += 1
if local_count == 2 and value == 1:
count += 1
break
print(count)
``` | instruction | 0 | 75,346 | 22 | 150,692 |
Yes | output | 1 | 75,346 | 22 | 150,693 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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
Input contains one integer number n (1 ≤ n ≤ 3000).
Output
Output the amount of almost prime numbers between 1 and n, inclusive.
Examples
Input
10
Output
2
Input
21
Output
8
Submitted Solution:
```
def check_prime(x):
for i in range(2,x):
if x%i ==0:
return False
return True
n = int(input())
count = 0
for i in range(6,n+1):
no_prime_count = 0
for j in range(2,i):
if(i%j ==0 and check_prime(j)):
no_prime_count+= 1
if no_prime_count == 2:
count +=1
print(count)
``` | instruction | 0 | 75,347 | 22 | 150,694 |
Yes | output | 1 | 75,347 | 22 | 150,695 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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
Input contains one integer number n (1 ≤ n ≤ 3000).
Output
Output the amount of almost prime numbers between 1 and n, inclusive.
Examples
Input
10
Output
2
Input
21
Output
8
Submitted Solution:
```
import math
def IsPrime(num):
if num == 2:
return True
if num % 2 == 0:
return False
limit = math.isqrt(num)
for i in range(3, limit + 1, 2):
if num % i == 0:
return False
return True
def HasExactlyTwoPrimeDivs(num):
counter = 0
for i in range(2, num):
if num % i == 0 and IsPrime(i):
counter += 1
return counter == 2
N, ans = int(input()), 0
for i in range(6, N + 1):
ans += HasExactlyTwoPrimeDivs(i)
print(ans)
``` | instruction | 0 | 75,348 | 22 | 150,696 |
Yes | output | 1 | 75,348 | 22 | 150,697 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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
Input contains one integer number n (1 ≤ n ≤ 3000).
Output
Output the amount of almost prime numbers between 1 and n, inclusive.
Examples
Input
10
Output
2
Input
21
Output
8
Submitted Solution:
```
n = int(input())
c = 0
for i in range(1,n+1):
cnt=0
if(n%i==0):
for j in range(2,i+1):
if(i%j!=0):
cnt=cnt+1
if(cnt==i):
c = c+1
print(c)
``` | instruction | 0 | 75,349 | 22 | 150,698 |
No | output | 1 | 75,349 | 22 | 150,699 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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
Input contains one integer number n (1 ≤ n ≤ 3000).
Output
Output the amount of almost prime numbers between 1 and n, inclusive.
Examples
Input
10
Output
2
Input
21
Output
8
Submitted Solution:
```
from bisect import bisect_right
from math import sqrt
sim = (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107,
109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229,
233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359,
367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491,
499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641,
643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787,
797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941,
947, 953, 967, 971, 977, 983, 991, 997, 1009, 1013, 1019, 1021, 1031, 1033, 1039, 1049, 1051, 1061, 1063, 1069,
1087, 1091, 1093, 1097, 1103, 1109, 1117, 1123, 1129, 1151, 1153, 1163, 1171, 1181, 1187, 1193, 1201, 1213, 1217,
1223, 1229, 1231, 1237, 1249, 1259, 1277, 1279, 1283, 1289, 1291, 1297, 1301, 1303, 1307, 1319, 1321, 1327, 1361,
1367, 1373, 1381, 1399, 1409, 1423, 1427, 1429, 1433, 1439, 1447, 1451, 1453, 1459, 1471, 1481, 1483, 1487, 1489,
1493, 1499)
n = int(input())
sim = sim[: bisect_right(sim, sqrt(n) + 3)]
ans = set()
for i in range(1, n + 1):
c = 0
for j in sim:
if i % j == 0:
c += 1
if c == 2:
ans.add(i)
print(len(ans))
``` | instruction | 0 | 75,350 | 22 | 150,700 |
No | output | 1 | 75,350 | 22 | 150,701 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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
Input contains one integer number n (1 ≤ n ≤ 3000).
Output
Output the amount of almost prime numbers between 1 and n, inclusive.
Examples
Input
10
Output
2
Input
21
Output
8
Submitted Solution:
```
import math
n=int(input())
s=0
t=1
for j in range(2,n+1):
for i in range(2,j):
if j%i==0:
s+=1
if s==0:
t+=1
s=0
r=2
e=2
while r**e<=n:
for i in range(r**e,n+1):
if e>=4 and e%4==0:
for j in range(3,i):
if j**e==i:
t=t+1
else:
for j in range(2,i):
if j**e==i:
t=t+1
e=e+1
print(n-t)
``` | instruction | 0 | 75,351 | 22 | 150,702 |
No | output | 1 | 75,351 | 22 | 150,703 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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
Input contains one integer number n (1 ≤ n ≤ 3000).
Output
Output the amount of almost prime numbers between 1 and n, inclusive.
Examples
Input
10
Output
2
Input
21
Output
8
Submitted Solution:
```
k = [6, 10, 12, 14, 15, 18, 20, 21, 22, 24, 26, 28, 33, 35, 36, 39, 40, 44, 45, 48, 50, 52, 54, 55, 56, 63, 65, 72, 75, 77, 80, 88, 91, 96, 98, 99, 100, 102, 104, 108, 112, 114, 117, 135, 138, 143, 144, 147, 160, 162, 170, 174, 175, 176, 186, 189, 190, 192, 196, 200, 204, 208, 216, 222, 224, 225, 228, 230, 238, 242, 245, 246, 250, 255, 258, 266, 275, 276, 282, 285, 288, 290, 297, 306, 310, 318, 320, 322, 324, 325, 338, 340, 342, 345, 348, 351, 352, 354, 357, 363, 366, 370, 372, 374, 375, 380, 384, 392, 399, 400, 402, 405, 406, 408, 410, 414, 416, 418, 426, 430, 432, 434, 435, 438, 441, 442, 444, 448, 456, 460, 465, 470, 474, 476, 483, 484, 486, 492, 494, 498, 500, 506, 507, 516, 518, 522, 530, 532, 534, 539, 552, 555, 558, 561, 564, 567, 574, 576, 580, 582, 590, 595, 598, 602, 605, 606, 609, 610, 612, 615, 618, 620, 627, 636, 637, 638, 640, 642, 644, 645, 648, 651, 654, 658, 663, 665, 666, 670, 675, 676, 678, 680, 682, 684, 686, 696, 704, 705, 708, 710, 730, 732, 738, 740, 741, 742, 744, 748, 754, 759, 760, 762, 765, 768, 774, 777, 784, 786, 790, 795, 800, 804, 805, 806, 812, 814, 816, 820, 822, 826, 828, 830, 832, 834, 836, 845, 846, 847, 850, 852, 854, 855, 860, 861, 864, 868, 875, 876, 884, 885, 888, 890, 891, 894, 896, 897, 902, 903, 906, 912, 915, 918, 920, 935, 938, 940, 942, 946, 948, 950, 952, 954, 957, 962, 968, 970, 972, 978, 984, 987, 988, 994, 996, 1000, 1002, 1005, 1010, 1012, 1015, 1022, 1023, 1026, 1029, 1030, 1032, 1034, 1035, 1036, 1038, 1044, 1045, 1053, 1060, 1062, 1064, 1065, 1066, 1068, 1070, 1071, 1074, 1085, 1086, 1089, 1090, 1095, 1098, 1104, 1105, 1106, 1113, 1116, 1118, 1125, 1128, 1130, 1131, 1146, 1148, 1150, 1152, 1158, 1160, 1162, 1164, 1166, 1180, 1182, 1183, 1185, 1194, 1196, 1197, 1204, 1206, 1209, 1212, 1215, 1220, 1221, 1222, 1224, 1225, 1235, 1236, 1239, 1240, 1242, 1245, 1246, 1250, 1265, 1266, 1270, 1272, 1275, 1276, 1278, 1280, 1281, 1284, 1288, 1295, 1296, 1298, 1305, 1308, 1309, 1310, 1314, 1316, 1323, 1332, 1335, 1338, 1340, 1342, 1352, 1353, 1356, 1358, 1360, 1362, 1364, 1368, 1370, 1372, 1374, 1375, 1378, 1390, 1392, 1395, 1398, 1407, 1408, 1414, 1416, 1419, 1420, 1422, 1425, 1434, 1435, 1442, 1443, 1446, 1449, 1450, 1455, 1458, 1460, 1463, 1464, 1474, 1476, 1480, 1484, 1488, 1490, 1491, 1494, 1495, 1496, 1498, 1505, 1506, 1508, 1510, 1515, 1520, 1521, 1524, 1526, 1533, 1534, 1536, 1542, 1545, 1547, 1548, 1550, 1551, 1562, 1566, 1568, 1570, 1572, 1573, 1578, 1580, 1582, 1586, 1595, 1599, 1600, 1602, 1605, 1606, 1608, 1612, 1614, 1624, 1625, 1626, 1628, 1630, 1632, 1635, 1640, 1644, 1645, 1652, 1656, 1659, 1660, 1662, 1664, 1665, 1666, 1668, 1670, 1672, 1674, 1677, 1683, 1686, 1692, 1695, 1698, 1700, 1701, 1704, 1705, 1708, 1715, 1720, 1725, 1728, 1729, 1730, 1734, 1736, 1738, 1742, 1743, 1746, 1749, 1752, 1758, 1768, 1771, 1776, 1778, 1780, 1788, 1790, 1792, 1804, 1810, 1812, 1818, 1824, 1826, 1827, 1833, 1834, 1836, 1840, 1842, 1845, 1846, 1850, 1854, 1855, 1859, 1862, 1866, 1869, 1875, 1876, 1878, 1880, 1881, 1884, 1885, 1892, 1896, 1898, 1900, 1902, 1904, 1905, 1908, 1910, 1918, 1924, 1926, 1930, 1935, 1936, 1938, 1940, 1944, 1946, 1947, 1953, 1956, 1958, 1962, 1965, 1968, 1970, 1976, 1986, 1988, 1989, 1990, 1992, 1998, 2000, 2004, 2013, 2015, 2020, 2022, 2024, 2025, 2034, 2035, 2037, 2044, 2050, 2052, 2054, 2055, 2060, 2064, 2065, 2067, 2068, 2072, 2076, 2082, 2085, 2086, 2088, 2093, 2094, 2110, 2114, 2115, 2118, 2120, 2121, 2124, 2128, 2132, 2134, 2135, 2136, 2140, 2148, 2150, 2154, 2158, 2163, 2166, 2172, 2175, 2180, 2196, 2198, 2202, 2208, 2211, 2212, 2214, 2222, 2223, 2230, 2232, 2233, 2235, 2236, 2238, 2247, 2254, 2255, 2256, 2260, 2265, 2266, 2270, 2274, 2277, 2282, 2286, 2289, 2290, 2292, 2295, 2296, 2298, 2300, 2301, 2304, 2314, 2316, 2320, 2322, 2324, 2325, 2328, 2330, 2331, 2332, 2334, 2338, 2343, 2345, 2346, 2350, 2354, 2355, 2358, 2360, 2364, 2365, 2373, 2379, 2382, 2385, 2387, 2388, 2390, 2392, 2398, 2405, 2406, 2408, 2409, 2410, 2412, 2422, 2424, 2431, 2440, 2444, 2445, 2448, 2454, 2466, 2472, 2480, 2484, 2485, 2486, 2492, 2499, 2500, 2502, 2505, 2506, 2510, 2514, 2522, 2526, 2532, 2534, 2538, 2540, 2544, 2552, 2555, 2556, 2560, 2565, 2568, 2570, 2576, 2583, 2585, 2586, 2592, 2595, 2596, 2598, 2607, 2613, 2616, 2620, 2622, 2626, 2628, 2630, 2632, 2634, 2639, 2650, 2655, 2658, 2662, 2664, 2665, 2667, 2673, 2674, 2676, 2678, 2680, 2682, 2684, 2685, 2690, 2691, 2694, 2702, 2704, 2709, 2710, 2712, 2715, 2716, 2717, 2718, 2720, 2724, 2728, 2736, 2739, 2740, 2742, 2744, 2745, 2748, 2751, 2754, 2756, 2758, 2765, 2766, 2769, 2770, 2775, 2778, 2780, 2782, 2784, 2786, 2793, 2794, 2795, 2796, 2802, 2810, 2816, 2821, 2826, 2828, 2830, 2832, 2834, 2840, 2842, 2844, 2847, 2849, 2862, 2865, 2868, 2871, 2874, 2877, 2882, 2884, 2890, 2892, 2895, 2900, 2905, 2915, 2916, 2919, 2920, 2922, 2928, 2930, 2934, 2937, 2938, 2946, 2948, 2950, 2952, 2954, 2955, 2958, 2960, 2961, 2968, 2975, 2976, 2980, 2985, 2988, 2992, 2994, 2996]
i = int(input())
x = 0
for num in k:
if num > i:
break
else:
x += 1
print(x)
``` | instruction | 0 | 75,352 | 22 | 150,704 |
No | output | 1 | 75,352 | 22 | 150,705 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It can be shown that any positive integer x can be uniquely represented as x = 1 + 2 + 4 + ... + 2k - 1 + r, where k and r are integers, k ≥ 0, 0 < r ≤ 2k. Let's call that representation prairie partition of x.
For example, the prairie partitions of 12, 17, 7 and 1 are:
12 = 1 + 2 + 4 + 5,
17 = 1 + 2 + 4 + 8 + 2,
7 = 1 + 2 + 4,
1 = 1.
Alice took a sequence of positive integers (possibly with repeating elements), replaced every element with the sequence of summands in its prairie partition, arranged the resulting numbers in non-decreasing order and gave them to Borys. Now Borys wonders how many elements Alice's original sequence could contain. Find all possible options!
Input
The first line contains a single integer n (1 ≤ n ≤ 105) — the number of numbers given from Alice to Borys.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1012; a1 ≤ a2 ≤ ... ≤ an) — the numbers given from Alice to Borys.
Output
Output, in increasing order, all possible values of m such that there exists a sequence of positive integers of length m such that if you replace every element with the summands in its prairie partition and arrange the resulting numbers in non-decreasing order, you will get the sequence given in the input.
If there are no such values of m, output a single integer -1.
Examples
Input
8
1 1 2 2 3 4 5 8
Output
2
Input
6
1 1 1 2 2 2
Output
2 3
Input
5
1 2 4 4 4
Output
-1
Note
In the first example, Alice could get the input sequence from [6, 20] as the original sequence.
In the second example, Alice's original sequence could be either [4, 5] or [3, 3, 3]. | instruction | 0 | 75,553 | 22 | 151,106 |
Tags: binary search, constructive algorithms, greedy, math
Correct Solution:
```
from collections import Counter
from math import log2
powers = [2 ** p for p in range(64)]
def can():
near = Counter()
for num in set(nums):
p = int(log2(num))
near[p] += nums[num]
seq_cp = seqs - Counter()
for num in set(near):
cnt = near[num]
while cnt != 0 and num < 64:
dif = min(cnt, seq_cp[powers[num]])
cnt -= dif
seq_cp[powers[num]] -= dif
num += 1
if cnt > 0:
return False
return True
n = int(input())
seq = list(map(int, input().split()))
cnt = Counter(seq)
nums = Counter(seq)
seqs = Counter()
cur_cnt = cnt[1]
cur_pow = 1
while cur_cnt != 0:
nums[cur_pow] -= cur_cnt
cur_pow *= 2
if cur_cnt > cnt[cur_pow]:
seqs[cur_pow // 2] = cur_cnt - cnt[cur_pow]
cur_cnt = cnt[cur_pow]
for num in set(nums):
addition = nums[num]
nums[num] = 0
nums[2 ** int(log2(num))] += addition
# remove elements with zero count
nums -= Counter()
seqs -= Counter()
cur_len = sum(seqs[num] for num in set(seqs))
res = []
cur_pow = 1
while can():
res.append(cur_len)
while seqs[cur_pow] == 0 and cur_pow <= 2 ** 63:
cur_pow *= 2
if cur_pow > 2 ** 63:
break
other_pow = 1
while other_pow <= cur_pow:
nums[other_pow] += 1
other_pow *= 2
seqs[cur_pow] -= 1
cur_len -= 1
print(-1 if len(res) == 0 else ' '.join(map(str, reversed(res))))
``` | output | 1 | 75,553 | 22 | 151,107 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob begin their day with a quick game. They first choose a starting number X0 ≥ 3 and try to reach one million by the process described below.
Alice goes first and then they take alternating turns. In the i-th turn, the player whose turn it is selects a prime number smaller than the current number, and announces the smallest multiple of this prime number that is not smaller than the current number.
Formally, he or she selects a prime p < Xi - 1 and then finds the minimum Xi ≥ Xi - 1 such that p divides Xi. Note that if the selected prime p already divides Xi - 1, then the number does not change.
Eve has witnessed the state of the game after two turns. Given X2, help her determine what is the smallest possible starting number X0. Note that the players don't necessarily play optimally. You should consider all possible game evolutions.
Input
The input contains a single integer X2 (4 ≤ X2 ≤ 106). It is guaranteed that the integer X2 is composite, that is, is not prime.
Output
Output a single integer — the minimum possible X0.
Examples
Input
14
Output
6
Input
20
Output
15
Input
8192
Output
8191
Note
In the first test, the smallest possible starting number is X0 = 6. One possible course of the game is as follows:
* Alice picks prime 5 and announces X1 = 10
* Bob picks prime 7 and announces X2 = 14.
In the second case, let X0 = 15.
* Alice picks prime 2 and announces X1 = 16
* Bob picks prime 5 and announces X2 = 20. | instruction | 0 | 75,601 | 22 | 151,202 |
Tags: math, number theory
Correct Solution:
```
N = int(input())
P = [0 for _ in range(1000010)]
# store the largest prime factor of a number --> if number is prime, mark its multiples
for i in range(2, N):
if P[i] == 0:
for j in range(2*i, N+1, i):
P[j] = i
ans = 1000010
for i in range(N-P[N]+1, N+1):
ans = min(ans, i-P[i]+1)
print(ans)
``` | output | 1 | 75,601 | 22 | 151,203 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's assume that
* v(n) is the largest prime number, that does not exceed n;
* u(n) is the smallest prime number strictly greater than n.
Find <image>.
Input
The first line contains integer t (1 ≤ t ≤ 500) — the number of testscases.
Each of the following t lines of the input contains integer n (2 ≤ n ≤ 109).
Output
Print t lines: the i-th of them must contain the answer to the i-th test as an irreducible fraction "p/q", where p, q are integers, q > 0.
Examples
Input
2
2
3
Output
1/6
7/30 | instruction | 0 | 76,206 | 22 | 152,412 |
Tags: math, number theory
Correct Solution:
```
T = int( input() )
MAX = 33000
bePrime = [0] * MAX;
primNum = []
for j in range(2, MAX):
if bePrime[j] == 0:
primNum.append( j )
i = j
while i < MAX:
bePrime[i] = 1
i = i + j
def isPrime( a ):
for j in primNum:
if j >= a:
return True
if a % j == 0:
return False
return True
def gcd( a, b ):
if b == 0:
return a
return gcd( b, a % b );
while T > 0:
num = 0;
n = int( input() )
m = n
while isPrime(m) == False:
m -= 1
while isPrime(n + 1) == False:
n += 1
num += 1
a = n - 1
b = 2 * ( n+1 )
a = a * (n+1) * m - num * b
b = b * (n+1) * m
g = gcd( a, b)
a //= g
b //= g
print( '{0}/{1}'.format( a, b ) )
T -= 1;
``` | output | 1 | 76,206 | 22 | 152,413 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's assume that
* v(n) is the largest prime number, that does not exceed n;
* u(n) is the smallest prime number strictly greater than n.
Find <image>.
Input
The first line contains integer t (1 ≤ t ≤ 500) — the number of testscases.
Each of the following t lines of the input contains integer n (2 ≤ n ≤ 109).
Output
Print t lines: the i-th of them must contain the answer to the i-th test as an irreducible fraction "p/q", where p, q are integers, q > 0.
Examples
Input
2
2
3
Output
1/6
7/30 | instruction | 0 | 76,207 | 22 | 152,414 |
Tags: math, number theory
Correct Solution:
```
def isPrime(n):
i = 2
while i * i <= n:
if n % i == 0:
return False
i += 1
return True
def prevPrime(n):
while not isPrime(n):
n -= 1
return n
def nextPrime(n):
n += 1
while not isPrime(n):
n += 1
return n
def gcd(a, b):
while(a):
b %= a
a, b = b, a
return b;
def solve():
n = int(input())
prev = prevPrime(n);
next = nextPrime(n);
num1 = prev - 2;
den1 = prev * 2;
g = gcd(num1, den1);
num1 //= g;
den1 //= g;
num2 = n - prev + 1;
den2 = prev * next;
g = gcd(num2, den2);
num2 //= g;
den2 //= g;
g = gcd(num1 * den2 + num2 * den1, den1 * den2);
x = (num1 * den2 + num2 * den1) // g;
y = den1 * den2 // g;
print('{}/{}'.format(x, y))
t = int(input())
while t:
t -= 1
solve()
``` | output | 1 | 76,207 | 22 | 152,415 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's assume that
* v(n) is the largest prime number, that does not exceed n;
* u(n) is the smallest prime number strictly greater than n.
Find <image>.
Input
The first line contains integer t (1 ≤ t ≤ 500) — the number of testscases.
Each of the following t lines of the input contains integer n (2 ≤ n ≤ 109).
Output
Print t lines: the i-th of them must contain the answer to the i-th test as an irreducible fraction "p/q", where p, q are integers, q > 0.
Examples
Input
2
2
3
Output
1/6
7/30 | instruction | 0 | 76,208 | 22 | 152,416 |
Tags: math, number theory
Correct Solution:
```
T = int( input() )
#for every prime x
#(b-a)/ab
#1/a-1/b
MAX = 33000
bePrime = [0] * MAX;
primNum = []
for j in range(2, MAX):
if bePrime[j] == 0:
primNum.append( j )
i = j
while i < MAX:
bePrime[i] = 1
i = i + j
def isPrime( a ):
for j in primNum:
if j >= a:
return True
if a % j == 0:
return False
return True
def gcd( a, b ):
if b == 0:
return a
return gcd( b, a % b );
while T > 0:
num = 0;
n = int( input() )
m = n
while isPrime(m) == False:
m -= 1
while isPrime(n + 1) == False:
n += 1
num += 1
a = n - 1
b = 2 * ( n+1 )
a = a * (n+1) * m - num * b
b = b * (n+1) * m
g = gcd( a, b)
a //= g
b //= g
print( '{0}/{1}'.format( a, b ) )
T -= 1;
``` | output | 1 | 76,208 | 22 | 152,417 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's assume that
* v(n) is the largest prime number, that does not exceed n;
* u(n) is the smallest prime number strictly greater than n.
Find <image>.
Input
The first line contains integer t (1 ≤ t ≤ 500) — the number of testscases.
Each of the following t lines of the input contains integer n (2 ≤ n ≤ 109).
Output
Print t lines: the i-th of them must contain the answer to the i-th test as an irreducible fraction "p/q", where p, q are integers, q > 0.
Examples
Input
2
2
3
Output
1/6
7/30 | instruction | 0 | 76,209 | 22 | 152,418 |
Tags: math, number theory
Correct Solution:
```
from math import gcd
def prime(n):
for i in range(2, int(n ** 0.5) + 1):
if n % i == 0:
return False
return True
for i in range(int(input())):
n = int(input())
x = n
y = n + 1
while not prime(x):
x -= 1
while not prime(y):
y += 1
p = 2 * n - 2 * x + 2 + y * x - 2 * y
q = 2 * y * x
d = gcd(p, q)
p //= d
q //= d
print(str(p) + '/' + str(q))
``` | output | 1 | 76,209 | 22 | 152,419 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's assume that
* v(n) is the largest prime number, that does not exceed n;
* u(n) is the smallest prime number strictly greater than n.
Find <image>.
Input
The first line contains integer t (1 ≤ t ≤ 500) — the number of testscases.
Each of the following t lines of the input contains integer n (2 ≤ n ≤ 109).
Output
Print t lines: the i-th of them must contain the answer to the i-th test as an irreducible fraction "p/q", where p, q are integers, q > 0.
Examples
Input
2
2
3
Output
1/6
7/30 | instruction | 0 | 76,210 | 22 | 152,420 |
Tags: math, number theory
Correct Solution:
```
def prime(n):
m = int(n ** 0.5) + 1
t = [1] * (n + 1)
for i in range(3, m):
if t[i]: t[i * i :: 2 * i] = [0] * ((n - i * i) // (2 * i) + 1)
return [2] + [i for i in range(3, n + 1, 2) if t[i]]
def gcd(a, b):
c = a % b
return gcd(b, c) if c else b
p = prime(31650)
def g(n):
m = int(n ** 0.5)
for j in p:
if n % j == 0: return True
if j > m: return False
def f(n):
a, b = n, n + 1
while g(a): a -= 1
while g(b): b += 1
p, q = (b - 2) * a + 2 * (n - b + 1), 2 * a * b
d = gcd(p, q)
print(str(p // d) + '/' + str(q // d))
for i in range(int(input())): f(int(input()))
``` | output | 1 | 76,210 | 22 | 152,421 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's assume that
* v(n) is the largest prime number, that does not exceed n;
* u(n) is the smallest prime number strictly greater than n.
Find <image>.
Input
The first line contains integer t (1 ≤ t ≤ 500) — the number of testscases.
Each of the following t lines of the input contains integer n (2 ≤ n ≤ 109).
Output
Print t lines: the i-th of them must contain the answer to the i-th test as an irreducible fraction "p/q", where p, q are integers, q > 0.
Examples
Input
2
2
3
Output
1/6
7/30
Submitted Solution:
```
from math import*
t = int(input())
arr = [0]*t
def snt(n):
for i in range(2,int(sqrt(n))+1):
if(n%i==0):
return 0
return 1
def ucln(a,b):
while(a > 0 and b > 0):
if(a > b):
a=a%b
else:
b=b%a
if(b > 0):
return b
return a
max = 0
tu = 0
mau = 1
for i in range(t):
arr[i] = int(input())
if(arr[i] > max):
max = arr[i]
for i in range(2,max+1):
tu1 = 1
v1 = i
while(snt(v1) == 0):
v1-=1
u1 = i + 1
while(snt(u1) == 0):
u1+=1
mau1 = v1*u1
tu = tu*mau1+tu1*mau
mau = mau1*mau
uc = ucln(tu,mau)
tu = int(tu/uc)
mau = int(mau/uc)
for j in range(t):
if(i == arr[j]):
print(str(tu),'/',str(mau))
``` | instruction | 0 | 76,211 | 22 | 152,422 |
No | output | 1 | 76,211 | 22 | 152,423 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's assume that
* v(n) is the largest prime number, that does not exceed n;
* u(n) is the smallest prime number strictly greater than n.
Find <image>.
Input
The first line contains integer t (1 ≤ t ≤ 500) — the number of testscases.
Each of the following t lines of the input contains integer n (2 ≤ n ≤ 109).
Output
Print t lines: the i-th of them must contain the answer to the i-th test as an irreducible fraction "p/q", where p, q are integers, q > 0.
Examples
Input
2
2
3
Output
1/6
7/30
Submitted Solution:
```
from fractions import Fraction
import math
from bisect import bisect_left
primes = [2, 3, 5]
def getbiggerandsmallerprimemultiplied(number):
pos = bisect_left(primes, number)
if primes[pos]<=number:
return primes[pos + 1]*primes[pos]
else:
return primes[pos - 1]*primes[pos]
until=2
def addprimesuntil(number):
global until
for n in range(primes[-1],number,2):
if primes[until] < int(math.sqrt(n)) + 1:
until+=1
if all(n % i for i in primes[:until]):
primes.append(n)
for t in range(int(input())):
fraction = Fraction(0)
actualcalculation=int(input())
if actualcalculation==1000000000:
print('999999941999999673/1999999887999999118')
continue
for n in range(2,actualcalculation+1):
if n >= primes[-1]:
addprimesuntil(primes[-1]*2)
fraction += Fraction(1,getbiggerandsmallerprimemultiplied(n))
print(fraction)
``` | instruction | 0 | 76,212 | 22 | 152,424 |
No | output | 1 | 76,212 | 22 | 152,425 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's assume that
* v(n) is the largest prime number, that does not exceed n;
* u(n) is the smallest prime number strictly greater than n.
Find <image>.
Input
The first line contains integer t (1 ≤ t ≤ 500) — the number of testscases.
Each of the following t lines of the input contains integer n (2 ≤ n ≤ 109).
Output
Print t lines: the i-th of them must contain the answer to the i-th test as an irreducible fraction "p/q", where p, q are integers, q > 0.
Examples
Input
2
2
3
Output
1/6
7/30
Submitted Solution:
```
from math import gcd
def check(n):
i = 2
while i * i <= n:
if n % 2 == 0:
return False
i += 1
return True
for _ in range(int(input())):
n = int(input())
x = n + 1
while check(x) is False:
x += 1
l = n
while check(l) is False:
l -= 1
a, b = (x - 2) * l - 2 * (x - n - 1), 2 * x * l
lol = gcd(a, b)
a //= lol
b //= lol
print(str(a) + '/' + str(b))
``` | instruction | 0 | 76,213 | 22 | 152,426 |
No | output | 1 | 76,213 | 22 | 152,427 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's assume that
* v(n) is the largest prime number, that does not exceed n;
* u(n) is the smallest prime number strictly greater than n.
Find <image>.
Input
The first line contains integer t (1 ≤ t ≤ 500) — the number of testscases.
Each of the following t lines of the input contains integer n (2 ≤ n ≤ 109).
Output
Print t lines: the i-th of them must contain the answer to the i-th test as an irreducible fraction "p/q", where p, q are integers, q > 0.
Examples
Input
2
2
3
Output
1/6
7/30
Submitted Solution:
```
import math
def prime(n):
root = int(math.sqrt(n))
i=2
while(i<=root):
if(n%i==0):
return 0
i+=1
return 1
def diviseur(num):
root = int(math.sqrt(num))
i = 2
div = []
while(i < root):
if(num % i == 0):
div.append(i)
num /= i
while(num % i == 0):
num /= i
i += 1
return div
def pgcd(a, b):
while b != 0:
r = a % b
a, b = b, r
return a
def reduc(a,b):
pg = pgcd(a,b)
return a/pg,b/pg
nb = int(input())
for i in range(nb):
l = []
a = int(input())
l ,m= a,a+1
"""for x in range(2,a+1):
vn = x
for j in range(x,1,-1):
if(prime(j)==1):
vn = j
break
x+=1
un = x
while(1):
if(prime(x)==1):
un = x
break
x+=1
l.append(un*vn)"""
while(prime(l)==0):
l-=1
while(prime(m)==0):
m+=1
aa = l * m-2*m+2*(a-l+1)
bb = 2*l*m
g = pgcd(aa,bb)
print(str(int(aa/g))+"/"+str(int((bb/g))))
``` | instruction | 0 | 76,214 | 22 | 152,428 |
No | output | 1 | 76,214 | 22 | 152,429 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array of n elements, you must make it a co-prime array in as few moves as possible.
In each move you can insert any positive integral number you want not greater than 109 in any place in the array.
An array is co-prime if any two adjacent numbers of it are co-prime.
In the number theory, two integers a and b are said to be co-prime if the only positive integer that divides both of them is 1.
Input
The first line contains integer n (1 ≤ n ≤ 1000) — the number of elements in the given array.
The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.
Output
Print integer k on the first line — the least number of elements needed to add to the array a to make it co-prime.
The second line should contain n + k integers aj — the elements of the array a after adding k elements to it. Note that the new array should be co-prime, so any two adjacent values should be co-prime. Also the new array should be got from the original array a by adding k elements to it.
If there are multiple answers you can print any one of them.
Example
Input
3
2 7 28
Output
1
2 7 9 28 | instruction | 0 | 76,318 | 22 | 152,636 |
Tags: greedy, implementation, math, number theory
Correct Solution:
```
import math
num_de_elementos = int(input())
elementos = list(map(int,input().split())) + [1]
aux = []
cont = 0
for i in range(num_de_elementos):
aux.append(elementos[i])
if(math.gcd(elementos[i], elementos[i+1]) != 1):
aux.append(1)
cont += 1
co_primos = ''
for i in aux:
co_primos += str(i) + ' '
print(cont)
print(co_primos)
``` | output | 1 | 76,318 | 22 | 152,637 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array of n elements, you must make it a co-prime array in as few moves as possible.
In each move you can insert any positive integral number you want not greater than 109 in any place in the array.
An array is co-prime if any two adjacent numbers of it are co-prime.
In the number theory, two integers a and b are said to be co-prime if the only positive integer that divides both of them is 1.
Input
The first line contains integer n (1 ≤ n ≤ 1000) — the number of elements in the given array.
The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.
Output
Print integer k on the first line — the least number of elements needed to add to the array a to make it co-prime.
The second line should contain n + k integers aj — the elements of the array a after adding k elements to it. Note that the new array should be co-prime, so any two adjacent values should be co-prime. Also the new array should be got from the original array a by adding k elements to it.
If there are multiple answers you can print any one of them.
Example
Input
3
2 7 28
Output
1
2 7 9 28 | instruction | 0 | 76,319 | 22 | 152,638 |
Tags: greedy, implementation, math, number theory
Correct Solution:
```
from fractions import gcd
n = int(input())
l = list(map(int, input().split()))
ans = 0
new = [l[0]]
for i in range(1, n):
if gcd(l[i], l[i - 1]) != 1:
ans += 1
new.append(1)
new.append(l[i])
print(ans)
print(*new)
``` | output | 1 | 76,319 | 22 | 152,639 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array of n elements, you must make it a co-prime array in as few moves as possible.
In each move you can insert any positive integral number you want not greater than 109 in any place in the array.
An array is co-prime if any two adjacent numbers of it are co-prime.
In the number theory, two integers a and b are said to be co-prime if the only positive integer that divides both of them is 1.
Input
The first line contains integer n (1 ≤ n ≤ 1000) — the number of elements in the given array.
The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.
Output
Print integer k on the first line — the least number of elements needed to add to the array a to make it co-prime.
The second line should contain n + k integers aj — the elements of the array a after adding k elements to it. Note that the new array should be co-prime, so any two adjacent values should be co-prime. Also the new array should be got from the original array a by adding k elements to it.
If there are multiple answers you can print any one of them.
Example
Input
3
2 7 28
Output
1
2 7 9 28 | instruction | 0 | 76,320 | 22 | 152,640 |
Tags: greedy, implementation, math, number theory
Correct Solution:
```
from fractions import gcd
def is_co_prime(x, y):
return gcd(x, y) != 1
def solve():
N = int(input())
L = list(map(int, input().split()))
ans = [str(L[0])]
x = 0
for i in range(1, N):
if is_co_prime(L[i], L[i - 1]):
x += 1
ans.append('1')
ans.append(str(L[i]))
print(x)
print(' '.join(ans))
if __name__ == '__main__':
solve()
``` | output | 1 | 76,320 | 22 | 152,641 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array of n elements, you must make it a co-prime array in as few moves as possible.
In each move you can insert any positive integral number you want not greater than 109 in any place in the array.
An array is co-prime if any two adjacent numbers of it are co-prime.
In the number theory, two integers a and b are said to be co-prime if the only positive integer that divides both of them is 1.
Input
The first line contains integer n (1 ≤ n ≤ 1000) — the number of elements in the given array.
The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.
Output
Print integer k on the first line — the least number of elements needed to add to the array a to make it co-prime.
The second line should contain n + k integers aj — the elements of the array a after adding k elements to it. Note that the new array should be co-prime, so any two adjacent values should be co-prime. Also the new array should be got from the original array a by adding k elements to it.
If there are multiple answers you can print any one of them.
Example
Input
3
2 7 28
Output
1
2 7 9 28 | instruction | 0 | 76,321 | 22 | 152,642 |
Tags: greedy, implementation, math, number theory
Correct Solution:
```
def mdc(a, b):
if b == 0:
return a
else:
return mdc(b, a % b)
input()
lista = list(map(int, input().split()))
tamanhoIni = len(lista)
i = 0
while i < len(lista) - 1:
mdcPar = 1
if lista[i] > lista[i + 1]:
mdcPar = mdc(lista[i], lista[i + 1])
else:
mdcPar = mdc(lista[i + 1], lista[i])
if mdcPar == 1:
i += 1
else:
lista.insert(i + 1, 1)
i += 2
print(len(lista) - tamanhoIni)
print(*lista)
``` | output | 1 | 76,321 | 22 | 152,643 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array of n elements, you must make it a co-prime array in as few moves as possible.
In each move you can insert any positive integral number you want not greater than 109 in any place in the array.
An array is co-prime if any two adjacent numbers of it are co-prime.
In the number theory, two integers a and b are said to be co-prime if the only positive integer that divides both of them is 1.
Input
The first line contains integer n (1 ≤ n ≤ 1000) — the number of elements in the given array.
The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.
Output
Print integer k on the first line — the least number of elements needed to add to the array a to make it co-prime.
The second line should contain n + k integers aj — the elements of the array a after adding k elements to it. Note that the new array should be co-prime, so any two adjacent values should be co-prime. Also the new array should be got from the original array a by adding k elements to it.
If there are multiple answers you can print any one of them.
Example
Input
3
2 7 28
Output
1
2 7 9 28 | instruction | 0 | 76,322 | 22 | 152,644 |
Tags: greedy, implementation, math, number theory
Correct Solution:
```
def gcd(a, b):
if b == 0: return a
else: return gcd(b, a % b)
if __name__ == "__main__":
n = int(input())
A = list(map(int, input().split()))
ans, cnt = [A[0]], 0
for i in range(1, n):
if gcd(A[i], A[i - 1]) != 1:
ans.append(1)
cnt += 1
ans.append(A[i])
print(cnt)
print(' '.join(str(i) for i in ans))
``` | output | 1 | 76,322 | 22 | 152,645 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array of n elements, you must make it a co-prime array in as few moves as possible.
In each move you can insert any positive integral number you want not greater than 109 in any place in the array.
An array is co-prime if any two adjacent numbers of it are co-prime.
In the number theory, two integers a and b are said to be co-prime if the only positive integer that divides both of them is 1.
Input
The first line contains integer n (1 ≤ n ≤ 1000) — the number of elements in the given array.
The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.
Output
Print integer k on the first line — the least number of elements needed to add to the array a to make it co-prime.
The second line should contain n + k integers aj — the elements of the array a after adding k elements to it. Note that the new array should be co-prime, so any two adjacent values should be co-prime. Also the new array should be got from the original array a by adding k elements to it.
If there are multiple answers you can print any one of them.
Example
Input
3
2 7 28
Output
1
2 7 9 28 | instruction | 0 | 76,323 | 22 | 152,646 |
Tags: greedy, implementation, math, number theory
Correct Solution:
```
# all number 1~1e9
def gcd(a,b):
while b != 0:
t = b
b = a % b
a = t
return a
def coprime(a,b):
if gcd(a,b) == 1:
return True
else:
return False
def isPrime(n):
if (n <= 1): return False
if (n <= 3): return True
# This is checked so that we can skip
# middle five numbers in below loop
if (n % 2 == 0 or n % 3 == 0): return False
i = 5
while (i * i <= n):
if (n % i == 0 or n % (i + 2) == 0):
return False
i = i + 6
return True
# find a number coprime with both inputs
def find_coprime(a,b,primelist):
for p in primelist:
if (coprime(p,a) and coprime(p,b)):
return p
n = len(primelist)
p = primelist[n-1]+2 # last prime is odd
while True:
if not isPrime(p):
p += 2
primelist.append(p)
if (coprime(p, a) and coprime(p, b)):
return p
p += 2
return p
def coprime_array(a,b,primelist):
n = len(a)
k = 0
if n == 1:
b.append(a[0])
return k
for i in range(n):
if i == n - 1:
b.append(a[i])
break
b.append(a[i])
if not coprime(a[i], a[i + 1]):
b.append(find_coprime(a[i], a[i + 1], primelist))
k += 1
return k
n = int(input())
a = [int(x) for x in input().split()]
primelist = [2,3,5,7]
b = []
k = coprime_array(a,b,primelist)
print(k)
print(*b, sep=' ')
``` | output | 1 | 76,323 | 22 | 152,647 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array of n elements, you must make it a co-prime array in as few moves as possible.
In each move you can insert any positive integral number you want not greater than 109 in any place in the array.
An array is co-prime if any two adjacent numbers of it are co-prime.
In the number theory, two integers a and b are said to be co-prime if the only positive integer that divides both of them is 1.
Input
The first line contains integer n (1 ≤ n ≤ 1000) — the number of elements in the given array.
The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.
Output
Print integer k on the first line — the least number of elements needed to add to the array a to make it co-prime.
The second line should contain n + k integers aj — the elements of the array a after adding k elements to it. Note that the new array should be co-prime, so any two adjacent values should be co-prime. Also the new array should be got from the original array a by adding k elements to it.
If there are multiple answers you can print any one of them.
Example
Input
3
2 7 28
Output
1
2 7 9 28 | instruction | 0 | 76,324 | 22 | 152,648 |
Tags: greedy, implementation, math, number theory
Correct Solution:
```
import math
n = input()
numbers = list(map(int, input().split()))
i = 0
out = []
last_insert = 0
while i < len(numbers) - 1:
if math.gcd(numbers[i], numbers[i+1]) == 1:
if str(numbers[i]) != last_insert:
out.append(str(numbers[i]))
last_insert = str(numbers[i+ 1])
out.append(last_insert)
else:
if str(numbers[i]) != last_insert:
out.append(str(numbers[i]))
out.append('1')
last_insert = str(numbers[i+ 1])
out.append(last_insert)
i += 1
added = len(out) - len(numbers)
if out == []:
out = [str(i) for i in numbers]
added = 0
print(added)
print(' '.join(out))
``` | output | 1 | 76,324 | 22 | 152,649 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array of n elements, you must make it a co-prime array in as few moves as possible.
In each move you can insert any positive integral number you want not greater than 109 in any place in the array.
An array is co-prime if any two adjacent numbers of it are co-prime.
In the number theory, two integers a and b are said to be co-prime if the only positive integer that divides both of them is 1.
Input
The first line contains integer n (1 ≤ n ≤ 1000) — the number of elements in the given array.
The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.
Output
Print integer k on the first line — the least number of elements needed to add to the array a to make it co-prime.
The second line should contain n + k integers aj — the elements of the array a after adding k elements to it. Note that the new array should be co-prime, so any two adjacent values should be co-prime. Also the new array should be got from the original array a by adding k elements to it.
If there are multiple answers you can print any one of them.
Example
Input
3
2 7 28
Output
1
2 7 9 28 | instruction | 0 | 76,325 | 22 | 152,650 |
Tags: greedy, implementation, math, number theory
Correct Solution:
```
import math
n=int(input())
a=list(map(int,input().split()))
b=[]
for i in range(0,n-1):
b+=[a[i]]
if math.gcd(a[i],a[i+1])-1:
b+=[1]
b+=[a[-1]]
print(len(b)-n)
for i in b:
print(i,end=' ')
# Made By Mostafa_Khaled
``` | output | 1 | 76,325 | 22 | 152,651 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array of n elements, you must make it a co-prime array in as few moves as possible.
In each move you can insert any positive integral number you want not greater than 109 in any place in the array.
An array is co-prime if any two adjacent numbers of it are co-prime.
In the number theory, two integers a and b are said to be co-prime if the only positive integer that divides both of them is 1.
Input
The first line contains integer n (1 ≤ n ≤ 1000) — the number of elements in the given array.
The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.
Output
Print integer k on the first line — the least number of elements needed to add to the array a to make it co-prime.
The second line should contain n + k integers aj — the elements of the array a after adding k elements to it. Note that the new array should be co-prime, so any two adjacent values should be co-prime. Also the new array should be got from the original array a by adding k elements to it.
If there are multiple answers you can print any one of them.
Example
Input
3
2 7 28
Output
1
2 7 9 28
Submitted Solution:
```
# @oj: codeforces
# @id: hitwanyang
# @email: 296866643@qq.com
# @date: 2020-11-14 21:23
# @url:https://codeforc.es/contest/660/problem/A
import sys,os
from io import BytesIO, IOBase
import collections,itertools,bisect,heapq,math,string
from decimal import *
# region fastio
BUFSIZE = 8192
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------------------
## 注意嵌套括号!!!!!!
## 先有思路,再写代码,别着急!!!
## 先有朴素解法,不要有思维定式,试着换思路解决
## 精度 print("%.10f" % ans)
## sqrt:int(math.sqrt(n))+1
## 字符串拼接不要用+操作，会超时
## 二进制转换:bin(1)[2:].rjust(32,'0')
## array copy:cur=array[::]
## oeis:example 1, 3, _, 1260, _, _, _, _, _, 12164510040883200
def main():
n=int(input())
a=list(map(int,input().split()))
ans=[a[0]]
for i in range(1,n):
if math.gcd(a[i],a[i-1])!=1:
ans.append(1)
ans.append(a[i])
print (len(ans)-len(a))
print (" ".join([str(x) for x in ans]))
if __name__ == "__main__":
main()
``` | instruction | 0 | 76,326 | 22 | 152,652 |
Yes | output | 1 | 76,326 | 22 | 152,653 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array of n elements, you must make it a co-prime array in as few moves as possible.
In each move you can insert any positive integral number you want not greater than 109 in any place in the array.
An array is co-prime if any two adjacent numbers of it are co-prime.
In the number theory, two integers a and b are said to be co-prime if the only positive integer that divides both of them is 1.
Input
The first line contains integer n (1 ≤ n ≤ 1000) — the number of elements in the given array.
The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.
Output
Print integer k on the first line — the least number of elements needed to add to the array a to make it co-prime.
The second line should contain n + k integers aj — the elements of the array a after adding k elements to it. Note that the new array should be co-prime, so any two adjacent values should be co-prime. Also the new array should be got from the original array a by adding k elements to it.
If there are multiple answers you can print any one of them.
Example
Input
3
2 7 28
Output
1
2 7 9 28
Submitted Solution:
```
import math
n=int(input())
list1=list(map(int,input().split()))
list2=list1.copy()
c=0
for i in range(n-1):
x=list1[i]
y=list1[i+1]
jj=1
if(math.gcd(x,y)>1):
# for j in range(1,min(min(x,y),max(x,y)//2)+1):
# if(x%j==0 and y%j==0):
# if(j>jj):
# jj=j
# break
# if(jj>1):
list2.insert(i+1+c,1)
c+=1
print(c)
print(*list2)
``` | instruction | 0 | 76,327 | 22 | 152,654 |
Yes | output | 1 | 76,327 | 22 | 152,655 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array of n elements, you must make it a co-prime array in as few moves as possible.
In each move you can insert any positive integral number you want not greater than 109 in any place in the array.
An array is co-prime if any two adjacent numbers of it are co-prime.
In the number theory, two integers a and b are said to be co-prime if the only positive integer that divides both of them is 1.
Input
The first line contains integer n (1 ≤ n ≤ 1000) — the number of elements in the given array.
The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.
Output
Print integer k on the first line — the least number of elements needed to add to the array a to make it co-prime.
The second line should contain n + k integers aj — the elements of the array a after adding k elements to it. Note that the new array should be co-prime, so any two adjacent values should be co-prime. Also the new array should be got from the original array a by adding k elements to it.
If there are multiple answers you can print any one of them.
Example
Input
3
2 7 28
Output
1
2 7 9 28
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
def getGcd(a, b):
while True:
if a < b:
num = b % a
b = num
else:
num = a % b
a = num
if a == 0:
return b
elif b == 0:
return a
i, count = 0, 0
while True:
if i + 1 > len(a) - 1:
break
num, prox = a[i], a[i + 1]
gcd = getGcd(num, prox)
if gcd != 1:
a.insert(i + 1, 1)
i += 2
count += 1
else:
i += 1
print(count)
print(*a)
``` | instruction | 0 | 76,328 | 22 | 152,656 |
Yes | output | 1 | 76,328 | 22 | 152,657 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array of n elements, you must make it a co-prime array in as few moves as possible.
In each move you can insert any positive integral number you want not greater than 109 in any place in the array.
An array is co-prime if any two adjacent numbers of it are co-prime.
In the number theory, two integers a and b are said to be co-prime if the only positive integer that divides both of them is 1.
Input
The first line contains integer n (1 ≤ n ≤ 1000) — the number of elements in the given array.
The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.
Output
Print integer k on the first line — the least number of elements needed to add to the array a to make it co-prime.
The second line should contain n + k integers aj — the elements of the array a after adding k elements to it. Note that the new array should be co-prime, so any two adjacent values should be co-prime. Also the new array should be got from the original array a by adding k elements to it.
If there are multiple answers you can print any one of them.
Example
Input
3
2 7 28
Output
1
2 7 9 28
Submitted Solution:
```
import math
n=int(input())
l=list(map(int,input().split()))
a=[l[0]]
cnt=0
for i in range(1,n):
if math.gcd(l[i-1],l[i])==1:
a.append(l[i])
else:
a.append(1)
cnt+=1
a.append(l[i])
print(cnt)
print(*a)
``` | instruction | 0 | 76,329 | 22 | 152,658 |
Yes | output | 1 | 76,329 | 22 | 152,659 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array of n elements, you must make it a co-prime array in as few moves as possible.
In each move you can insert any positive integral number you want not greater than 109 in any place in the array.
An array is co-prime if any two adjacent numbers of it are co-prime.
In the number theory, two integers a and b are said to be co-prime if the only positive integer that divides both of them is 1.
Input
The first line contains integer n (1 ≤ n ≤ 1000) — the number of elements in the given array.
The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.
Output
Print integer k on the first line — the least number of elements needed to add to the array a to make it co-prime.
The second line should contain n + k integers aj — the elements of the array a after adding k elements to it. Note that the new array should be co-prime, so any two adjacent values should be co-prime. Also the new array should be got from the original array a by adding k elements to it.
If there are multiple answers you can print any one of them.
Example
Input
3
2 7 28
Output
1
2 7 9 28
Submitted Solution:
```
from math import gcd
n = int(input())
a = [int(x) for x in input().split()]
count = 0
b = []
for i in range(n):
if i != 0 and gcd(a[i], a[i - 1]) != 1:
b.append(a[i - 1] + 1)
count += 1
b.append(a[i])
print(count)
response = ""
for num in b:
response += str(num) + " "
print(response)
``` | instruction | 0 | 76,330 | 22 | 152,660 |
No | output | 1 | 76,330 | 22 | 152,661 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array of n elements, you must make it a co-prime array in as few moves as possible.
In each move you can insert any positive integral number you want not greater than 109 in any place in the array.
An array is co-prime if any two adjacent numbers of it are co-prime.
In the number theory, two integers a and b are said to be co-prime if the only positive integer that divides both of them is 1.
Input
The first line contains integer n (1 ≤ n ≤ 1000) — the number of elements in the given array.
The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.
Output
Print integer k on the first line — the least number of elements needed to add to the array a to make it co-prime.
The second line should contain n + k integers aj — the elements of the array a after adding k elements to it. Note that the new array should be co-prime, so any two adjacent values should be co-prime. Also the new array should be got from the original array a by adding k elements to it.
If there are multiple answers you can print any one of them.
Example
Input
3
2 7 28
Output
1
2 7 9 28
Submitted Solution:
```
def gcd(a,b):
if b == 0:
return a
else:
return gcd(b, a%b)
n = int(input())
a = list(map(int, input().split()))
b = []
b.append(a[0])
s = 0
for i in range(1, len(a)):
if gcd(a[i], a[i-1]) != 1:
b.append(29)
s+=1
b.append(a[i])
print(s)
print(*b)
``` | instruction | 0 | 76,331 | 22 | 152,662 |
No | output | 1 | 76,331 | 22 | 152,663 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array of n elements, you must make it a co-prime array in as few moves as possible.
In each move you can insert any positive integral number you want not greater than 109 in any place in the array.
An array is co-prime if any two adjacent numbers of it are co-prime.
In the number theory, two integers a and b are said to be co-prime if the only positive integer that divides both of them is 1.
Input
The first line contains integer n (1 ≤ n ≤ 1000) — the number of elements in the given array.
The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.
Output
Print integer k on the first line — the least number of elements needed to add to the array a to make it co-prime.
The second line should contain n + k integers aj — the elements of the array a after adding k elements to it. Note that the new array should be co-prime, so any two adjacent values should be co-prime. Also the new array should be got from the original array a by adding k elements to it.
If there are multiple answers you can print any one of them.
Example
Input
3
2 7 28
Output
1
2 7 9 28
Submitted Solution:
```
from sys import stdin
from math import gcd
n = int(input())
l = list(map(int, stdin.readline().split()))
t = []
for i in range(n-1):
t.append(l[i])
if gcd(l[i], l[i+1])!=1:
a = min(l[i], l[i+1])
b = max(l[i], l[i+1])
# print(a, b)
while a<b:
a+=1
if gcd(a, b)==1:
t.append(a)
break
t.append(l[n-1])
print(abs(n-len(t)))
print(' '.join(map(str, t)))
``` | instruction | 0 | 76,332 | 22 | 152,664 |
No | output | 1 | 76,332 | 22 | 152,665 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array of n elements, you must make it a co-prime array in as few moves as possible.
In each move you can insert any positive integral number you want not greater than 109 in any place in the array.
An array is co-prime if any two adjacent numbers of it are co-prime.
In the number theory, two integers a and b are said to be co-prime if the only positive integer that divides both of them is 1.
Input
The first line contains integer n (1 ≤ n ≤ 1000) — the number of elements in the given array.
The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.
Output
Print integer k on the first line — the least number of elements needed to add to the array a to make it co-prime.
The second line should contain n + k integers aj — the elements of the array a after adding k elements to it. Note that the new array should be co-prime, so any two adjacent values should be co-prime. Also the new array should be got from the original array a by adding k elements to it.
If there are multiple answers you can print any one of them.
Example
Input
3
2 7 28
Output
1
2 7 9 28
Submitted Solution:
```
from math import gcd
n = int(input())
coprime_array = list(map(int, input().split()))
pairs = []
for i in range(1,n):
if gcd(coprime_array[i],coprime_array[i-1]) > 1:
pairs.append([max(coprime_array[i-1], coprime_array[i]) + 1, i-1])
print(len(pairs))
answer = ''
for i in range(n):
answer += str(coprime_array[i]) + ' '
if len(pairs) and i == pairs[0][1]:
answer += str(pairs[0][0]) + ' '
del pairs[0]
print(answer)
``` | instruction | 0 | 76,333 | 22 | 152,666 |
No | output | 1 | 76,333 | 22 | 152,667 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Author has gone out of the stories about Vasiliy, so here is just a formal task description.
You are given q queries and a multiset A, initially containing only integer 0. There are three types of queries:
1. "+ x" — add integer x to multiset A.
2. "- x" — erase one occurrence of integer x from multiset A. It's guaranteed that at least one x is present in the multiset A before this query.
3. "? x" — you are given integer x and need to compute the value <image>, i.e. the maximum value of bitwise exclusive OR (also know as XOR) of integer x and some integer y from the multiset A.
Multiset is a set, where equal elements are allowed.
Input
The first line of the input contains a single integer q (1 ≤ q ≤ 200 000) — the number of queries Vasiliy has to perform.
Each of the following q lines of the input contains one of three characters '+', '-' or '?' and an integer xi (1 ≤ xi ≤ 109). It's guaranteed that there is at least one query of the third type.
Note, that the integer 0 will always be present in the set A.
Output
For each query of the type '?' print one integer — the maximum value of bitwise exclusive OR (XOR) of integer xi and some integer from the multiset A.
Example
Input
10
+ 8
+ 9
+ 11
+ 6
+ 1
? 3
- 8
? 3
? 8
? 11
Output
11
10
14
13
Note
After first five operations multiset A contains integers 0, 8, 9, 11, 6 and 1.
The answer for the sixth query is integer <image> — maximum among integers <image>, <image>, <image>, <image> and <image>. | instruction | 0 | 76,334 | 22 | 152,668 |
Tags: binary search, bitmasks, data structures, trees
Correct Solution:
```
# 10
# + 8
# + 9
# + 11
# + 6
# + 1
# ? 3
# - 8
# ? 3
# ? 8
# ? 11
MAX_BIT = 30
class Node:
def __init__(self):
self.left = None
self.right = None
self.leftCnt = 0
self.rightCnt = 0
def AddRight(self):
if self.right == None:
self.right = Node()
self.rightCnt = 1
else:
self.rightCnt += 1
def AddLeft(self):
if self.left == None:
self.left = Node()
self.leftCnt = 1
else:
self.leftCnt += 1
def RemRight(self):
self.rightCnt -= 1
def RemLeft(self):
self.leftCnt -= 1
def Left(self):
return self.left != None and self.leftCnt > 0
def Right(self):
return self.right != None and self.rightCnt > 0
def insert(u, num, dig=MAX_BIT):
if dig < 0:
return
bit = (num>>dig)&1
if bit > 0: #insert to right
u.AddRight()
insert(u.right, num, dig-1)
else:
u.AddLeft()
insert(u.left, num, dig-1)
def remove(u, num, dig=MAX_BIT):
if dig < 0:
return
bit = (num>>dig)&1
if bit > 0: #remove right
u.RemRight()
remove(u.right, num, dig-1)
else:
u.RemLeft()
remove(u.left, num, dig-1)
def cal(u, num, dig=MAX_BIT):
if dig < 0 or u == None:
return 0
bit = (num>>dig)&1
if bit > 0: #try to go to left first
if u.Left(): #if valid
return (1<<dig) + cal(u.left, num, dig-1)
elif u.Right():
return cal(u.right, num, dig-1)
else: #try to go to right first
if u.Right():
return (1<<dig) + cal(u.right, num, dig-1)
elif u.Left():
return cal(u.left, num, dig-1)
return 0
def main():
root = Node()
insert(root, 0)
n = int(input())
for i in range(n):
tmp = input().split()
num = int(tmp[1])
if tmp[0] == "+":
insert(root, num)
elif tmp[0] == "-":
remove(root, num)
else:
print(cal(root, num))
main()
``` | output | 1 | 76,334 | 22 | 152,669 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Author has gone out of the stories about Vasiliy, so here is just a formal task description.
You are given q queries and a multiset A, initially containing only integer 0. There are three types of queries:
1. "+ x" — add integer x to multiset A.
2. "- x" — erase one occurrence of integer x from multiset A. It's guaranteed that at least one x is present in the multiset A before this query.
3. "? x" — you are given integer x and need to compute the value <image>, i.e. the maximum value of bitwise exclusive OR (also know as XOR) of integer x and some integer y from the multiset A.
Multiset is a set, where equal elements are allowed.
Input
The first line of the input contains a single integer q (1 ≤ q ≤ 200 000) — the number of queries Vasiliy has to perform.
Each of the following q lines of the input contains one of three characters '+', '-' or '?' and an integer xi (1 ≤ xi ≤ 109). It's guaranteed that there is at least one query of the third type.
Note, that the integer 0 will always be present in the set A.
Output
For each query of the type '?' print one integer — the maximum value of bitwise exclusive OR (XOR) of integer xi and some integer from the multiset A.
Example
Input
10
+ 8
+ 9
+ 11
+ 6
+ 1
? 3
- 8
? 3
? 8
? 11
Output
11
10
14
13
Note
After first five operations multiset A contains integers 0, 8, 9, 11, 6 and 1.
The answer for the sixth query is integer <image> — maximum among integers <image>, <image>, <image>, <image> and <image>. | instruction | 0 | 76,336 | 22 | 152,672 |
Tags: binary search, bitmasks, data structures, trees
Correct Solution:
```
import collections
max_bits = 30
#root = collections.Counter()
#vals = collections.defaultdict(int)
class BNode:
def __init__(self, ct=0, zero=None, one=None):
self.ct = ct
self.zero = None
self.one = None
def __str__(self):
return ' '.join([str(self.ct), str(self.zero), str(self.one)])
root = BNode()
def bits(x):
bit = 2**(max_bits-1)
for i in range(max_bits):
if x & bit:
yield 1
else:
yield 0
bit >>= 1
def add(x, root):
root.ct += 1
for b in bits(x):
if b:
if not root.one:
root.one = BNode()
root = root.one
else:
if not root.zero:
root.zero = BNode()
root = root.zero
root.ct += 1
def sub(x, root):
root.ct -= 1
for b in bits(x):
if b:
if root.one.ct == 1:
root.one = None
break
root = root.one
else:
if root.zero.ct == 1:
root.zero = None
break
root = root.zero
root.ct -= 1
def question(x, root):
y = 0
for b in bits(x):
if b:
if root.zero and root.zero.ct > 0:
root = root.zero
y = y*2
else:
root = root.one
y = y*2 + 1
else:
if root.one and root.one.ct > 0:
root = root.one
y = y*2 + 1
else:
root = root.zero
y = y*2
return x ^ y
add(0,root)
q = int(input())
output = []
for i in range(q):
qtype, x = input().split()
x = int(x)
if qtype == '+':
add(x, root)
if qtype == '-':
sub(x, root)
if qtype == '?':
output.append(str(question(x,root)))
print("\n".join(output))
``` | output | 1 | 76,336 | 22 | 152,673 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Author has gone out of the stories about Vasiliy, so here is just a formal task description.
You are given q queries and a multiset A, initially containing only integer 0. There are three types of queries:
1. "+ x" — add integer x to multiset A.
2. "- x" — erase one occurrence of integer x from multiset A. It's guaranteed that at least one x is present in the multiset A before this query.
3. "? x" — you are given integer x and need to compute the value <image>, i.e. the maximum value of bitwise exclusive OR (also know as XOR) of integer x and some integer y from the multiset A.
Multiset is a set, where equal elements are allowed.
Input
The first line of the input contains a single integer q (1 ≤ q ≤ 200 000) — the number of queries Vasiliy has to perform.
Each of the following q lines of the input contains one of three characters '+', '-' or '?' and an integer xi (1 ≤ xi ≤ 109). It's guaranteed that there is at least one query of the third type.
Note, that the integer 0 will always be present in the set A.
Output
For each query of the type '?' print one integer — the maximum value of bitwise exclusive OR (XOR) of integer xi and some integer from the multiset A.
Example
Input
10
+ 8
+ 9
+ 11
+ 6
+ 1
? 3
- 8
? 3
? 8
? 11
Output
11
10
14
13
Note
After first five operations multiset A contains integers 0, 8, 9, 11, 6 and 1.
The answer for the sixth query is integer <image> — maximum among integers <image>, <image>, <image>, <image> and <image>. | instruction | 0 | 76,337 | 22 | 152,674 |
Tags: binary search, bitmasks, data structures, trees
Correct Solution:
```
from collections import defaultdict
class Trie():
def __init__(self,value):
self.value = value
self.left = None
self.right = None
self.terminal = False
self.par = None
self.child = 0
def insert(n,head):
cur_node = head
for i in range(31,-1,-1):
z = (n>>i)&1
if z == 0:
if cur_node.left == None:
cur_node.left = Trie(z)
cur_node.left.par = cur_node
cur_node.child+=1
cur_node = cur_node.left
else:
if cur_node.right == None:
cur_node.right = Trie(z)
cur_node.right.par = cur_node
cur_node.child+=1
cur_node = cur_node.right
cur_node.terminal = True
def maxi(n,head):
cur_node = head
cur_xor = 0
for i in range(31,-1,-1):
z = (n>>i)&1
if cur_node == None:
break
if z == 0:
if cur_node.right!=None:
cur_node = cur_node.right
cur_xor+=pow(2,i)
else:
cur_node = cur_node.left
if z == 1:
if cur_node.left!=None:
cur_node = cur_node.left
cur_xor+=pow(2,i)
else:
cur_node = cur_node.right
return cur_xor
def delete(head,n):
cur_node = head
for i in range(31,-1,-1):
z = (n>>i)&1
if z == 0:
if cur_node.left == None:
cur_node.left = Trie(z)
cur_node = cur_node.left
else:
if cur_node.right == None:
cur_node.right = Trie(z)
cur_node = cur_node.right
# print(cur_node.value)
z = cur_node.value
temp = cur_node.par
if cur_node.left!=None:
cur_node.left = None
if cur_node.right!=None:
cur_node.right = None
if z == 0:
if temp.left != None:
if temp.left.value == 0:
temp.left = None
elif z == 1:
if temp.right != None:
if temp.right.value == 1:
temp.right = None
del(cur_node)
cur_node = temp
while cur_node.par!=None:
if cur_node.child-1 == 0:
if cur_node.terminal == False:
z = cur_node.value
temp = cur_node.par
if cur_node.left!=None:
cur_node.left = None
if cur_node.right!=None:
cur_node.right = None
if z == 0:
if temp.left != None:
if temp.left.value == 0:
temp.left = None
elif z == 1:
if temp.right != None:
if temp.right.value == 1:
temp.right = None
del(cur_node)
cur_node = temp
if cur_node == head:
cur_node.child-=1
break
else:
break
else:
cur_node.child-=1
break
hash = defaultdict(int)
head = Trie('y')
head.par = None
insert(0,head)
ans = []
q = int(input())
for i in range(q):
a,b = map(str,input().split())
b = int(b)
if a == '+':
hash[b]+=1
if hash[b] == 1:
insert(b,head)
if a == '-':
hash[b]-=1
if hash[b] == 0:
delete(head,b)
if a == '?':
ans.append(maxi(b,head))
print(*ans)
``` | output | 1 | 76,337 | 22 | 152,675 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Author has gone out of the stories about Vasiliy, so here is just a formal task description.
You are given q queries and a multiset A, initially containing only integer 0. There are three types of queries:
1. "+ x" — add integer x to multiset A.
2. "- x" — erase one occurrence of integer x from multiset A. It's guaranteed that at least one x is present in the multiset A before this query.
3. "? x" — you are given integer x and need to compute the value <image>, i.e. the maximum value of bitwise exclusive OR (also know as XOR) of integer x and some integer y from the multiset A.
Multiset is a set, where equal elements are allowed.
Input
The first line of the input contains a single integer q (1 ≤ q ≤ 200 000) — the number of queries Vasiliy has to perform.
Each of the following q lines of the input contains one of three characters '+', '-' or '?' and an integer xi (1 ≤ xi ≤ 109). It's guaranteed that there is at least one query of the third type.
Note, that the integer 0 will always be present in the set A.
Output
For each query of the type '?' print one integer — the maximum value of bitwise exclusive OR (XOR) of integer xi and some integer from the multiset A.
Example
Input
10
+ 8
+ 9
+ 11
+ 6
+ 1
? 3
- 8
? 3
? 8
? 11
Output
11
10
14
13
Note
After first five operations multiset A contains integers 0, 8, 9, 11, 6 and 1.
The answer for the sixth query is integer <image> — maximum among integers <image>, <image>, <image>, <image> and <image>. | instruction | 0 | 76,338 | 22 | 152,676 |
Tags: binary search, bitmasks, data structures, trees
Correct Solution:
```
class D: u, v, n = None, None, 0
p = [1 << 30 - j for j in range(31)]
def sub(d, y):
for j in p:
d = d.u if y & j else d.v
d.n -= 1
def add(d, y):
for j in p:
if not d.u: d.u, d.v = D(), D()
d = d.u if y & j else d.v
d.n += 1
def get(d, y):
s = 0
for j in p:
if y & j:
if d.v.n:
d = d.v
s += j
else: d = d.u
else:
if d.u.n:
d = d.u
s += j
else: d = d.v
print(s)
d = D()
add(d, 0)
for i in range(int(input())):
k, x = input().split()
y = int(x)
if k == '-': sub(d, y)
elif k == '+': add(d, y)
else: get(d, y)
``` | output | 1 | 76,338 | 22 | 152,677 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Author has gone out of the stories about Vasiliy, so here is just a formal task description.
You are given q queries and a multiset A, initially containing only integer 0. There are three types of queries:
1. "+ x" — add integer x to multiset A.
2. "- x" — erase one occurrence of integer x from multiset A. It's guaranteed that at least one x is present in the multiset A before this query.
3. "? x" — you are given integer x and need to compute the value <image>, i.e. the maximum value of bitwise exclusive OR (also know as XOR) of integer x and some integer y from the multiset A.
Multiset is a set, where equal elements are allowed.
Input
The first line of the input contains a single integer q (1 ≤ q ≤ 200 000) — the number of queries Vasiliy has to perform.
Each of the following q lines of the input contains one of three characters '+', '-' or '?' and an integer xi (1 ≤ xi ≤ 109). It's guaranteed that there is at least one query of the third type.
Note, that the integer 0 will always be present in the set A.
Output
For each query of the type '?' print one integer — the maximum value of bitwise exclusive OR (XOR) of integer xi and some integer from the multiset A.
Example
Input
10
+ 8
+ 9
+ 11
+ 6
+ 1
? 3
- 8
? 3
? 8
? 11
Output
11
10
14
13
Note
After first five operations multiset A contains integers 0, 8, 9, 11, 6 and 1.
The answer for the sixth query is integer <image> — maximum among integers <image>, <image>, <image>, <image> and <image>. | instruction | 0 | 76,339 | 22 | 152,678 |
Tags: binary search, bitmasks, data structures, trees
Correct Solution:
```
import sys
input = sys.stdin.readline
from collections import defaultdict
class Node:
def __init__(self):
self.cnt = 0
self.next = [None] * 2
root = Node()
def add(x):
root.cnt += 1
cur = root
for k in range(32, -1, -1):
bit = (x>>k) & 1
if not cur.next[bit]:
cur.next[bit] = Node()
cur = cur.next[bit]
cur.cnt += 1
def remove(x):
root.cnt -= 1
cur = root
for k in range(32, -1, -1):
bit = (x>>k) & 1
cur = cur.next[bit]
cur.cnt -= 1
def query(x):
cur = root
ret = 0
for k in range(32, -1, -1):
bit = (x>>k) & 1
if cur.next[bit^1] and cur.next[bit^1].cnt > 0:
ret += (1<<k)
cur = cur.next[bit^1]
else:
cur = cur.next[bit]
return ret
add(0)
q = int(input())
for _ in range(q):
comm, x = input().split()
x = int(x)
if comm == '+':
add(x)
elif comm == '-':
remove(x)
else:
print(query(x))
``` | output | 1 | 76,339 | 22 | 152,679 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Author has gone out of the stories about Vasiliy, so here is just a formal task description.
You are given q queries and a multiset A, initially containing only integer 0. There are three types of queries:
1. "+ x" — add integer x to multiset A.
2. "- x" — erase one occurrence of integer x from multiset A. It's guaranteed that at least one x is present in the multiset A before this query.
3. "? x" — you are given integer x and need to compute the value <image>, i.e. the maximum value of bitwise exclusive OR (also know as XOR) of integer x and some integer y from the multiset A.
Multiset is a set, where equal elements are allowed.
Input
The first line of the input contains a single integer q (1 ≤ q ≤ 200 000) — the number of queries Vasiliy has to perform.
Each of the following q lines of the input contains one of three characters '+', '-' or '?' and an integer xi (1 ≤ xi ≤ 109). It's guaranteed that there is at least one query of the third type.
Note, that the integer 0 will always be present in the set A.
Output
For each query of the type '?' print one integer — the maximum value of bitwise exclusive OR (XOR) of integer xi and some integer from the multiset A.
Example
Input
10
+ 8
+ 9
+ 11
+ 6
+ 1
? 3
- 8
? 3
? 8
? 11
Output
11
10
14
13
Note
After first five operations multiset A contains integers 0, 8, 9, 11, 6 and 1.
The answer for the sixth query is integer <image> — maximum among integers <image>, <image>, <image>, <image> and <image>. | instruction | 0 | 76,340 | 22 | 152,680 |
Tags: binary search, bitmasks, data structures, trees
Correct Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO,IOBase
class Node:
def __init__(self):
self.zero = None
self.one = None
self.onpath = 0
class Trie:
def __init__(self, node):
self.root = node
def add(self, num):
curr = self.root
for bit in num:
if bit == '1':
if not curr.one:
curr.one = Node()
curr = curr.one
else:
if not curr.zero:
curr.zero = Node()
curr = curr.zero
curr.onpath += 1
def findmaxxor(self, num):
curr,val = self.root,0
for i in num:
if not curr:
val *= 2
continue
z = '0' if i == '1' else '1'
if z == '1':
if curr.one:
curr = curr.one
val = val*2+1
else:
curr = curr.zero
val *= 2
else:
if curr.zero:
curr = curr.zero
val = val*2+1
else:
curr = curr.one
val *= 2
return val
def __delitem__(self, num):
curr = self.root
for bit in num:
if bit == '1':
if curr.one.onpath == 1:
curr.one = None
break
curr = curr.one
else:
if curr.zero.onpath == 1:
curr.zero = None
break
curr = curr.zero
curr.onpath -= 1
def main():
z = Trie(Node())
z.add('0' * 30)
for _ in range(int(input())):
r = input().split()
x = bin(int(r[1]))[2:].zfill(30)
if r[0] == '+':
z.add(x)
elif r[0] == '-':
del z[x]
else:
print(z.findmaxxor(x))
# Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self,file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd,max(os.fstat(self._fd).st_size,BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0,2),self.buffer.write(b),self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd,max(os.fstat(self._fd).st_size,BUFSIZE))
self.newlines = b.count(b"\n")+(not b)
ptr = self.buffer.tell()
self.buffer.seek(0,2),self.buffer.write(b),self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd,self.buffer.getvalue())
self.buffer.truncate(0),self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self,file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s:self.buffer.write(s.encode("ascii"))
self.read = lambda:self.buffer.read().decode("ascii")
self.readline = lambda:self.buffer.readline().decode("ascii")
sys.stdin,sys.stdout = IOWrapper(sys.stdin),IOWrapper(sys.stdout)
input = lambda:sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
``` | output | 1 | 76,340 | 22 | 152,681 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Author has gone out of the stories about Vasiliy, so here is just a formal task description.
You are given q queries and a multiset A, initially containing only integer 0. There are three types of queries:
1. "+ x" — add integer x to multiset A.
2. "- x" — erase one occurrence of integer x from multiset A. It's guaranteed that at least one x is present in the multiset A before this query.
3. "? x" — you are given integer x and need to compute the value <image>, i.e. the maximum value of bitwise exclusive OR (also know as XOR) of integer x and some integer y from the multiset A.
Multiset is a set, where equal elements are allowed.
Input
The first line of the input contains a single integer q (1 ≤ q ≤ 200 000) — the number of queries Vasiliy has to perform.
Each of the following q lines of the input contains one of three characters '+', '-' or '?' and an integer xi (1 ≤ xi ≤ 109). It's guaranteed that there is at least one query of the third type.
Note, that the integer 0 will always be present in the set A.
Output
For each query of the type '?' print one integer — the maximum value of bitwise exclusive OR (XOR) of integer xi and some integer from the multiset A.
Example
Input
10
+ 8
+ 9
+ 11
+ 6
+ 1
? 3
- 8
? 3
? 8
? 11
Output
11
10
14
13
Note
After first five operations multiset A contains integers 0, 8, 9, 11, 6 and 1.
The answer for the sixth query is integer <image> — maximum among integers <image>, <image>, <image>, <image> and <image>. | instruction | 0 | 76,341 | 22 | 152,682 |
Tags: binary search, bitmasks, data structures, trees
Correct Solution:
```
#!/usr/bin/env pypy3
import sys
class BitTreeNode(object):
__slots__ = ["left", "right", "count"]
def __init__(self, cnt):
self.left = None
self.right = None
self.count = cnt
class BitTree(object):
def __init__(self):
self.root = BitTreeNode(0)
self.add(reversed(get_bits(0)))
def add(self, bits):
node = self.root
node.count += 1
for bit in bits:
if bit:
if not node.right:
node.right = BitTreeNode(1)
else:
node.right.count += 1
node = node.right
else:
if not node.left:
node.left = BitTreeNode(1)
else:
node.left.count += 1
node = node.left
def remove(self, bits):
node = self.root
node.count -= 1
for bit in bits:
if bit == 0:
node.left.count -= 1
if node.left.count == 0:
node.left = None
return
node = node.left
else:
node.right.count -= 1
if node.right.count == 0:
node.right = None
return
node = node.right
def examine_xor(self, bits):
best = 0
node = self.root
for i, bit in reversed(list(enumerate(bits))):
assert node.count > 0
if bit == 0:
if node.right is not None:
best += 1 << i
node = node.right
else:
node = node.left
else:
if node.left is not None:
best += 1 << i
node = node.left
else:
node = node.right
return best
@classmethod
def print_tree(cls, node, indent):
print(" " * indent + "+", node.count)
if node.right:
cls.print_tree(node.right, indent + 2)
if node.left:
cls.print_tree(node.left, indent + 2)
def read_query():
qType, number = next(sys.stdin).split()
return qType, int(number)
def get_bits(number):
ans = []
for _ in range(32):
ans.append(number % 2)
number //= 2
return ans
def solve_and_print(q):
tree = BitTree()
for _ in range(q):
qType, number = read_query()
#print()
#print(qType, number)
if qType == "+":
tree.add(reversed(get_bits(number)))
#tree.print_tree(tree.root, 0)
elif qType == "-":
tree.remove(reversed(get_bits(number)))
#tree.print_tree(tree.root, 0)
elif qType == "?":
print(tree.examine_xor(get_bits(number)))
if __name__ == "__main__":
q = int(next(sys.stdin))
solve_and_print(q)
``` | output | 1 | 76,341 | 22 | 152,683 |
Provide a correct Python 3 solution for this coding contest problem.
Find the number of sequences of length K consisting of positive integers such that the product of any two adjacent elements is at most N, modulo 10^9+7.
Constraints
* 1\leq N\leq 10^9
* ~~1~~ 2\leq K\leq 100 (fixed at 21:33 JST)
* N and K are integers.
Input
Input is given from Standard Input in the following format:
N K
Output
Print the number of sequences, modulo 10^9+7.
Examples
Input
3 2
Output
5
Input
10 3
Output
147
Input
314159265 35
Output
457397712 | instruction | 0 | 76,509 | 22 | 153,018 |
"Correct Solution:
```
import sys
input = sys.stdin.readline
sys.setrecursionlimit(1000000)
from collections import deque, Counter, defaultdict
def getN():
return int(input())
def getList():
return list(map(int, input().split()))
import math
import copy
import bisect
MOD = 10**9 + 7
from logging import getLogger, StreamHandler, DEBUG, WARNING
logger = getLogger(__name__)
handler = StreamHandler()
handler.setLevel(DEBUG)
logger.setLevel(DEBUG)
# handler.setLevel(WARNING)
# logger.setLevel(WARNING)
logger.addHandler(handler)
def main():
n,k = getList()
# =================約数列挙=================
divisors = []
tmp_div = n + 1
for i in range(1, int(math.sqrt(n)) + 3):
if tmp_div * i > n:
other = n // i
if i > other:
break
elif i == other:
divisors.append(i)
break
else:
divisors.append(i)
divisors.append(other)
divisors.sort()
n_div = len(divisors)
# =================約数列挙=================
# =================dp[0]の作成=============
diff = [divisors[0]]
for i, j in zip(divisors, divisors[1:]):
diff.append(j - i)
dp = copy.copy(diff)
# print(dp)
# =================dp[0]の作成=============
for iteration in range(k-1):
dp_copy = []
tmp = sum(dp) % MOD
dp_copy.append((tmp * diff[0]) % MOD)
for cid, dp_content in enumerate(range(n_div - 1)):
tmp -= dp[n_div - dp_content - 1]
dp_copy.append((tmp * diff[cid+1]) % MOD)
dp = copy.copy(dp_copy)
# print(dp)
print(sum(dp) % MOD)
# print(acc)
# print(n_div)
if __name__ == "__main__":
main()
``` | output | 1 | 76,509 | 22 | 153,019 |
Provide tags and a correct Python 3 solution for this coding contest problem.
While doing some spring cleaning, Daniel found an old calculator that he loves so much. However, it seems like it is broken. When he tries to compute 1 + 3 using the calculator, he gets 2 instead of 4. But when he tries computing 1 + 4, he gets the correct answer, 5. Puzzled by this mystery, he opened up his calculator and found the answer to the riddle: the full adders became half adders!
So, when he tries to compute the sum a + b using the calculator, he instead gets the xorsum a ⊕ b (read the definition by the link: <https://en.wikipedia.org/wiki/Exclusive_or>).
As he saw earlier, the calculator sometimes gives the correct answer. And so, he wonders, given integers l and r, how many pairs of integers (a, b) satisfy the following conditions: $$$a + b = a ⊕ b l ≤ a ≤ r l ≤ b ≤ r$$$
However, Daniel the Barman is going to the bar and will return in two hours. He tells you to solve the problem before he returns, or else you will have to enjoy being blocked.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of testcases.
Then, t lines follow, each containing two space-separated integers l and r (0 ≤ l ≤ r ≤ 10^9).
Output
Print t integers, the i-th integer should be the answer to the i-th testcase.
Example
Input
3
1 4
323 323
1 1000000
Output
8
0
3439863766
Note
a ⊕ b denotes the bitwise XOR of a and b.
For the first testcase, the pairs are: (1, 2), (1, 4), (2, 1), (2, 4), (3, 4), (4, 1), (4, 2), and (4, 3). | instruction | 0 | 76,832 | 22 | 153,664 |
Tags: bitmasks, brute force, combinatorics, dp
Correct Solution:
```
def get_bin (a):
nums=[]
for i in range (32):
if ((1<<i)&a):
nums.append(1)
else:
nums.append(0)
while(len(nums)>0 and nums[-1]==0):
nums.pop()
return nums
dp={}
def get_num (a, b):
global dp
if ((a,b) in dp):
return dp[(a,b)]
if (a < 0 or b < 0):
return 0
if (a == 0 and b == 0):
return 1
a_bin = get_bin(a)
b_bin = get_bin(b)
if(b>a):
a_bin,b_bin=b_bin,a_bin
a,b=b,a
if (len(a_bin)>len(b_bin)):
big_bit = 1 << (len(a_bin) - 1)
to_ret=((get_num(big_bit-1,b) + get_num(a-big_bit, b)))
dp[(a,b)]=to_ret
return to_ret
if(sum(a_bin)==len(a_bin) and sum(b_bin)==len(b_bin)):
to_ret = pow(3, len(a_bin))
dp[(a,b)]=to_ret
return to_ret
big_bit = 1 << (len(a_bin) - 1)
to_ret=(get_num(big_bit-1, b-big_bit) + get_num(a, big_bit-1))
dp[(a,b)]=to_ret
return to_ret
tc = int(input(""))
for i in range (int(tc)):
nums = input("").split(' ')
l = int(nums[0])
r = int(nums[1])
ans = get_num(r, r) - 2 * get_num(r, l - 1) + get_num(l - 1, l - 1)
print(ans)
``` | output | 1 | 76,832 | 22 | 153,665 |
Provide tags and a correct Python 3 solution for this coding contest problem.
While doing some spring cleaning, Daniel found an old calculator that he loves so much. However, it seems like it is broken. When he tries to compute 1 + 3 using the calculator, he gets 2 instead of 4. But when he tries computing 1 + 4, he gets the correct answer, 5. Puzzled by this mystery, he opened up his calculator and found the answer to the riddle: the full adders became half adders!
So, when he tries to compute the sum a + b using the calculator, he instead gets the xorsum a ⊕ b (read the definition by the link: <https://en.wikipedia.org/wiki/Exclusive_or>).
As he saw earlier, the calculator sometimes gives the correct answer. And so, he wonders, given integers l and r, how many pairs of integers (a, b) satisfy the following conditions: $$$a + b = a ⊕ b l ≤ a ≤ r l ≤ b ≤ r$$$
However, Daniel the Barman is going to the bar and will return in two hours. He tells you to solve the problem before he returns, or else you will have to enjoy being blocked.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of testcases.
Then, t lines follow, each containing two space-separated integers l and r (0 ≤ l ≤ r ≤ 10^9).
Output
Print t integers, the i-th integer should be the answer to the i-th testcase.
Example
Input
3
1 4
323 323
1 1000000
Output
8
0
3439863766
Note
a ⊕ b denotes the bitwise XOR of a and b.
For the first testcase, the pairs are: (1, 2), (1, 4), (2, 1), (2, 4), (3, 4), (4, 1), (4, 2), and (4, 3). | instruction | 0 | 76,833 | 22 | 153,666 |
Tags: bitmasks, brute force, combinatorics, dp
Correct Solution:
```
def solve(L, R):
res = 0
for i in range(32):
for j in range(32):
l = (L >> i) << i
r = (R >> j) << j
#print(l, r)
if l>>i&1==0 or r>>j&1==0:
continue
l -= 1<<i
r -= 1<<j
if l & r:
continue
lr = l ^ r
ma = max(i, j)
mi = min(i, j)
mask = (1<<ma)-1
p = bin(lr&mask).count("1")
ip = ma - mi - p
res += 3**mi * 2**ip
#print(l, r, mi, ip, 3**mi * 2**ip)
return res
T = int(input())
for _ in range(T):
l, r = map(int, input().split())
print(solve(r+1, r+1) + solve(l, l) - solve(l, r+1) * 2)
``` | output | 1 | 76,833 | 22 | 153,667 |
Provide tags and a correct Python 3 solution for this coding contest problem.
While doing some spring cleaning, Daniel found an old calculator that he loves so much. However, it seems like it is broken. When he tries to compute 1 + 3 using the calculator, he gets 2 instead of 4. But when he tries computing 1 + 4, he gets the correct answer, 5. Puzzled by this mystery, he opened up his calculator and found the answer to the riddle: the full adders became half adders!
So, when he tries to compute the sum a + b using the calculator, he instead gets the xorsum a ⊕ b (read the definition by the link: <https://en.wikipedia.org/wiki/Exclusive_or>).
As he saw earlier, the calculator sometimes gives the correct answer. And so, he wonders, given integers l and r, how many pairs of integers (a, b) satisfy the following conditions: $$$a + b = a ⊕ b l ≤ a ≤ r l ≤ b ≤ r$$$
However, Daniel the Barman is going to the bar and will return in two hours. He tells you to solve the problem before he returns, or else you will have to enjoy being blocked.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of testcases.
Then, t lines follow, each containing two space-separated integers l and r (0 ≤ l ≤ r ≤ 10^9).
Output
Print t integers, the i-th integer should be the answer to the i-th testcase.
Example
Input
3
1 4
323 323
1 1000000
Output
8
0
3439863766
Note
a ⊕ b denotes the bitwise XOR of a and b.
For the first testcase, the pairs are: (1, 2), (1, 4), (2, 1), (2, 4), (3, 4), (4, 1), (4, 2), and (4, 3). | instruction | 0 | 76,834 | 22 | 153,668 |
Tags: bitmasks, brute force, combinatorics, dp
Correct Solution:
```
def g( a , b ):
cur = 1
res = 0
ze = 0
while cur <= b:
if b & cur:
b ^= cur
if a & b == 0:
res += ( 1 << ze )
if a & cur == 0:
ze = ze + 1
cur <<= 1
return res
def f( a , b ):
res = 0
if a == b:
return 0
if a == 0:
return 2 * b - 1 + f( 1 , b )
if a & 1:
res = res + 2 * ( g( a , b ) - g( a , a ) )
a = a + 1
if b & 1:
res = res + 2 * ( g( b - 1 , b ) - g( b - 1 , a ) )
return 3 * f( a >> 1 , b >> 1 ) + res
t = int(input())
while t > 0:
t = t - 1
l , r = map(int , input().split())
print( f( l , r + 1 ) )
``` | output | 1 | 76,834 | 22 | 153,669 |
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