CF GYM 100548 Color(2014ACM西安现场赛Problem F)
2016-10-28 22:30
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题意
有n个连续的花需要图色, 共m种颜色,相邻的花的颜色不能相同,求刚好用k种颜色涂满这些花的方案数题解
先不管m。设g[i] 为用 <=i 种颜色涂n朵花的方案数
f[i] 为刚好i种颜色涂n朵花的方案数
所以有
为简单的容斥
即用(<= k 种颜色的方案数)-k * (<=k-1的方案数)
但是此时会剪多,所以加上C(k,2)* (<=k-2的方案数),以此类推
之前写了一发一直TLE, 后发现要预处理1~k的模拟元,经验问题。。
哦,最后的结果要乘以C(m,k)
代码
#include <iostream> #include <cstdio> #include <cstring> #include <string> #include <algorithm> #define mod 1000000007 #define p 1000000007 using namespace std; typedef long long LL; const int MAX_K = 1e6+10; LL quick_mod(LL a, LL b) { LL ans = 1; a %= p; while(b) { if(b & 1) { ans = ans * a % p; b--; } b >>= 1; a = a * a % p; } return ans; } LL C(LL n, LL m) { if(m > n) return 0; LL ans = 1; for(LL i=1; i<=m; i++) { LL a = (n + i - m) % p; LL b = i % p; ans = ans * (a * quick_mod(b, p-2) % p) % p; } return ans; } LL Lucas(LL n, LL m) { if(m == 0) return 1; return C(n % p, m % p) * Lucas(n / p, m / p) % p; } LL g[MAX_K]; LL inv[MAX_K]; void getinv() { for (LL i = 1; i < MAX_K-1; i++) inv[i] = quick_mod(i,mod-2); } int main() { int t; scanf("%d",&t); getinv(); for(int tt = 1;tt<=t;tt++) { LL n,m,k; scanf("%lld%lld%lld",&n,&m,&k); LL ans1 = Lucas(m,k); LL ans = 0; LL tmp = 1; for(int i=1;i<=k;i++) { tmp = ((tmp * (k-i+1))%mod * inv[i] )%mod;// 为C(k,i) g[i] =( i * quick_mod(i-1, n-1) ) % mod; if((k-i)%2 == 0) { ans = (ans + (tmp * g[i]) % mod) % mod; } else ans = (ans - tmp * g[i] % mod + mod) % mod; } ans = ans*ans1%mod; printf("Case #%d: %lld\n",tt,ans); } return 0; }
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