UVA 11859 - Division Game
题意:给定一个矩阵。每次能选一行中几个数字,把他们变成他们的因子,最后不能变的人输。问能否先手必胜
思路:转变成因子等价于删去一些素数。这样问题转化为了Nim游戏
代码:
#include#include const int N = 10005;int t, n, m, num, cnt[N], vis[N], prime[N], pn = 0;int main() { for (int i = 2; i < N; i++) { if (vis[i]) continue; prime[pn++] = i; for (int j = i; j < N; j += i) { vis[j] = 1; } } for (int i = 2; i < N; i++) { int num = i; for (int j = 0; j < pn && prime[j] <= i; j++) { while (num % prime[j] == 0) { cnt[i]++; num /= prime[j]; } } } int cas = 0; scanf("%d", &t); while (t--) { int ans = 0; scanf("%d%d", &n, &m); for (int i = 0; i < n; i++) { int sum = 0; for (int j = 0; j < m; j++) { scanf("%d", &num); sum += cnt[num]; } ans ^= sum; } printf("Case #%d: %s\n", ++cas, ans == 0?"NO":"YES"); } return 0;}
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