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Description
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
5
120
题意：n条南北向的路，n条东西向的路，相交形成n*n的方格，交点处是路口，现在有5个凳子要往路口方，要求1、每条路上只能放一个凳子问总共有多少种方法
#include<stdio.h>
#include<string.h>
#include<cstdio>
#include<string>
#include<math.h>
#include<algorithm>
#define LL long long
#define PI atan(1.0)*4
#define DD double
#define MAX 30000
#define mod 100
#define dian 1.000000011
#define INF 0x3f3f3f
using namespace std;
LL C(LL n,LL m)
{
LL ans=1;
int num=1;
while(m--)
{
ans*=(n-m);
ans/=num;
num++;
}
return ans;
}
int main()
{
LL n,m,j,i,t;
LL sum;
while(scanf("%lld",&n)!=EOF)
{
if(n<=5)
{
sum=1;
for(i=1;i<=n;i++)
sum*=i;
printf("%lld\n",sum);
continue;
}
sum=C(n,5)*n*(n-1)*(n-2)*(n-3)*(n-4);
printf("%lld\n",sum);
}
return 0;
}
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原文地址：http://www.cnblogs.com/tonghao/p/5267051.html
