uva 1323 - Vivian's Problem(梅森素数)
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题目链接:uva 1323 - Vivian's Problem
题目大意:给定N个数,然后为每个数添加一个幂ei,最后N项垒乘的结果为M,要是得M的所有因子的和可以写成2x,求x的最大值,如果没有条件满足,输出NO
解题思路:若一个数可以写成若干个不同的梅森素数的乘积,那么这个数的所以因子和可以写成2x.
232−1的范围内只有8个梅森素数,所以可以用状压处理。
梅森素数即为2^i-1形式的素数
/********************** * 梅森素数,(2^k) - 1 * * 一个数若能差分成若干个不同的梅森素数的积,那么该数的所有因子和可以写成2^n形式。 ***********************/#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long ll;const int maxp = 8;const int f[maxp+5] = {2, 3, 5, 7, 13, 17, 19, 31};int N, ans, s[105];ll mpri[maxp+5];void insert (ll u) { s[N] = 0; for (int i = 0; i < maxp; i++) { if (u % mpri[i] == 0) { s[N] |= (1<<i); u /= mpri[i]; if (u % mpri[i] == 0) return; } } if (u == 1) N++;}int solve (int S) { int ret = 0; for (int i = 0; i < maxp; i++) if (S&(1<<i)) ret += f[i]; return ret;}void dfs (int d, int S) { ans = max(ans, solve(S)); for (int i = d; i < N; i++) if ((S&s[i]) == 0) dfs(i+1, S|s[i]);}int main () { for (int i = 0; i < maxp; i++) mpri[i] = (1LL<<f[i]) - 1; ll n, a; while (scanf("%lld", &n) == 1) { N = 0; for (int i = 0; i < n; i++) { scanf("%lld", &a); insert(a); } if (N == 0) printf("NO\n"); else { ans = 0; dfs(0, 0); printf("%d\n", ans); } } return 0;}
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