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长度为l的用k种字符组成的字符串有k^l中 其中m个字符要不相同 那就是k^l*C(l, m)*(k-1)^m 有重复 要除以2 但是你mod n了 不能直接除 n不一定是素数 所以不能乘以逆元
所以我都mod 2倍的n 最后的结果再除以2 特判l = 1 和 m = 0的情况
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
typedef long long LL;
int vis[100010];
int prime[100010], c;void sieve(int n)
{int m = sqrt(n+0.5);memset(vis, 0, sizeof(vis));vis[0] = vis[1] = 1;for(int i = 2; i <= m; i++)if(!vis[i])for(int j = i*i; j <= n; j += i)vis[j] = 1;
}int get_primes(int n)
{sieve(n);int c = 0;for(int i = 2; i <= n; i++)if(!vis[i])prime[c++] = i;return c;
}
LL pow(LL a, LL b, LL n)
{LL ans = 1;while(b){if(b&1){ans *= a;ans %= n;}b >>= 1;a *= a;a %= n;}return ans;
}
LL work(LL x, LL y)
{LL ans = 0;while(x){ans += x/y;x /= y;}return ans;
}
LL cm(LL n, LL m, LL p)
{LL ans = 1;for(int i = 0; prime[i] <= n && i < c; i++){LL x = work(n, prime[i]);LL y = work(n-m, prime[i]);LL z = work(m, prime[i]);x -= y+z;ans *= pow(prime[i], x, p);ans %= p;}return ans;
}
LL cal(LL n, LL k, LL l, LL m)
{LL ans = 1;ans = ans * pow(k, l, n) % n;ans = ans * pow(k-1, m, n) % n;ans = ans * cm(l, m, n) % n;return ans;
}
int main()
{c = get_primes(100000);int T;int cas = 1;scanf("%d", &T);while(T--){LL n, k, l, m;scanf("%lld %lld %lld %lld", &n, &k, &l, &m);if(m == 0){printf("Case %d: %lld\n", cas++, pow(k, l, n)+1);}else if(k == 1)printf("Case %d: 1\n", cas++);elseprintf("Case %d: %lld\n", cas++, cal(2*n, k, l, m)/2+1);}return 0;
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