https://cn.vjudge.net/problem/POJ-2992
这个题数据非常严格,极容易TLE。。。需要预处理出阶乘对应的质数的幂次然后查询。对于阶乘中某个质数的指数有结论
然后枚举质数,把分子分母加加减减就行了
#include#include #include using namespace std; const int N=505; typedef long long LL; LL prime[100]={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431,1000000}; LL cal[500][500]; void getcal() { LL t,res; for (int i=1;i<=431;i++) { for (int j=0;prime[j]<=431;j++) { res=0; t=i; while(t) { t/=prime[j]; res+=t; } cal[i][prime[j]]=res; } } } int main() { getcal(); int a,b; LL res; while(~scanf("%d%d",&a,&b)) { res=1; for (int j=0;prime[j]<=a;j++) { res*=(cal[a][prime[j]]-cal[b][prime[j]]-cal[a-b][prime[j]]+1); } printf("%lld\n",res); } return 0; }