Crawling in process... Crawling failed Time Limit:500MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Submit StatusDescription
The city park of IT City contains n east to west paths and n north to south paths. Each east to west path crosses each north to south path, so there are n2 intersections.
The city funded purchase of five benches. To make it seems that there are many benches it was decided to place them on as many paths as possible. Obviously this requirement is satisfied by the following scheme: each bench is placed on a cross of paths and each path contains not more than one bench.
Help the park administration count the number of ways to place the benches.
Input
The only line of the input contains one integer n (5 ≤ n ≤ 100) — the number of east to west paths and north to south paths.
Output
Output one integer — the number of ways to place the benches.
Sample Input
Input 5 Output 120 远看还以为是阶乘，--||然而n可以到一百，看了题才知道是组合数，东西向，南北向的路各n条，相互垂直的两条路一定会相交，形成了n*n个路口，现在5张长凳， 把凳子放到路口，并且每条路只能出现一张凳子，那摩我们就要从南北向，东西向的路中各挑出5条，然后，，，，，， #include<cstdio>#include<cstring>#include<iostream>using namespace std;__int64 C(__int64 m,__int64 n){__int64 ans=1;__int64 num=1;while(n--){ans*=(m-n);ans/=num;num++;}return ans;}int main(){__int64 n;while(cin>>n){cout<<120*C(n,5)*C(n,5)<<endl;}return 0;}