“电脑运行速度很快！24的阶乘也不过就620448401733239439360000,yyy你快写个程序来算一算”
我想你可能需要神威·太湖之光
因为一共就24张卡片，可以状压
设f[S]，表示使用集合(S)的卡片的方案数，dis[S]表示使用集合(S)的卡片所能到达的距离
转移就很简单了，但这道题比较卡常，所以要用(lowbit)来快速找出所有的'(1)'
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define LL long long
#define mod 1000000007
#define lb(__a) (__a & (-__a))
using namespace std;
LL read() {
LL k = 0, f = 1; char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9')
k = k * 10 + c - 48, c = getchar();
return k * f;
}
int f[1 << 25], dis[1 << 25];
LL b[3];
int main() {
int n = read();
for(int i = 1; i <= n; ++i) dis[1 << i-1] = read();
int m = read();
for(int i = 1; i <= m; ++i) b[i] = read();
f[0] = 1; int cnt = (1 << n) - 1;
for(int i = 1; i <= cnt; ++i) {
int k = lb(i);
dis[i] = dis[k] + dis[i^k];
if(dis[i] == b[1] || dis[i] == b[2]) continue;
for(int j = i; j; j ^= k) {
k = lb(j); f[i] = f[i] + f[i^k];
while(f[i] >= mod) f[i] -= mod;
}
}
printf("%d
", f[cnt]);
return 0;
}
双倍经验：CF327E