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题意是给定了一个叫“jamcoin”的定义,让你生成足够数量满足条件的jamcoin。
jamcoin其实就可以理解成一个二进制整数,题目要求的要么长度为16位,要么为32位,一头一尾两个位必须是1,然后就是这个数字串在各种进制下表示的数都不能是质数。
我的做法很简单,因为大致口算了一下,满足条件的jamcoin应该挺多的,随便找50个或500个就好了。就从最小的串开始检查,找到一个就输出一个,直到找到要求的个数为止。
检查的方法也很简单,就是把这个串表示的数用素数表去除,能找到整数这个数的,就记下来,否则就认为这个串不满足条件。为了加快速度,我的素数表很小,只有1000以内的素数。
当然,因为当N为32时,串表示的数会超过long long,所以没必要先表数表示成整数再去除,而是直接用快速幂取余去做即可。比如100011,作为二进制串,(2^5+2^1+2^0) % 5 = 0
/* * Author : ben */ #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <ctime> #include <iostream> #include <algorithm> #include <queue> #include <set> #include <map> #include <stack> #include <string> #include <vector> #include <deque> #include <list> #include <functional> #include <numeric> #include <cctype> #include <bitset> using namespace std; typedef long long LL; int modular_exp(int a, int b, int c) { LL res, temp; res = 1 % c, temp = a % c; while (b) { if (b & 1) { res = res * temp % c; } temp = temp * temp % c; b >>= 1; } return (int) res; } int N, J; int arr[11]; const int ptn = 168; const int pt[ptn] = {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}; bool isprime(LL d, int k) { for (int i = 0; i < ptn; i++) { int p = pt[i]; int mod = 0; LL tmp = d; int b = 0; while (tmp > 0) { if (tmp % 2 == 1) { mod += modular_exp(k, b, p); mod %= p; } b++; tmp /= 2; } if (mod == 0) { arr[k] = p; return false; } } return true; } bool judge(LL d) { for (int k = 2; k <= 10; k++) { if (isprime(d, k)) { return false; } } return true; } void work() { printf("Case #1:\n"); LL s = (1LL << (N - 1)) + 1; LL e = 1LL << N; int pn = 0; for (; s < e && pn < J; s += 2) { if (judge(s)) { if (N == 16) { bitset<16> bs(s); cout << bs; } else { bitset<32> bs(s); cout << bs; } for (int i = 2; i <= 10; i++) { printf(" %d", arr[i]); } putchar(‘\n‘); pn++; } } } int main1() { if (judge(35)) { cout << "true!" << endl; } else { cout << "false" << endl; } return 0; } int main() { int T; scanf("%d", &T); for (int t = 1; t <= T; t++) { scanf("%d %d", &N, &J); work(); } return 0; }
GCJ Qualification Round 2016 C题
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原文地址:http://www.cnblogs.com/moonbay/p/5372410.html