hdoj5621KK's Point

KK's Point

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 503    Accepted Submission(s): 171

Problem Description

Our lovely KK has a difficult mathematical problem:He points N(2≤N≤105) points on a circle,there are all different.Now he's going to connect the N points with each other(There are no three lines in the circle to hand over a point.).KK wants to know how many points are there in the picture(Including the dots of boundary).

 

Input

The first line of the input file contains an integer T(1≤T≤10), which indicates the number of test cases.

For each test case, there are one lines,includes a integer N(2≤N≤105),indicating the number of dots of the polygon.

 

Output

For each test case, there are one lines,includes a integer,indicating the number of the dots.

 

Sample Input

234

 

Sample Output

35

 

我们先撇开边界上的点不管，那么所有的点都是有两条线所构成的

手算得出N=4的时候，能形成一个点

那么，我们只要知道N个点可以构成几个四边形即可

即求CN4{{C}^{4}_{N}}C​N​4​​

最后我们再把边界上的N个点加上，最后的结果是CN4−N{{C}^{4}_{N}}-NC​N​4​​ +N

#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>#include<list>#include<queue>#include<vector>using namespace std;const int maxn=100010;long long C[maxn][5];void init(){      for(int i=0;i<maxn;++i){          C[i][0]=1;      }      for(int i=1;i<maxn;++i){          for(int j=1;j<=4;++j){              C[i][j]=C[i-1][j]+C[i-1][j-1];          }      }  }  int main() {int t;long long n;init();scanf("%d",&t);while(t--){scanf("%lld",&n);if(n<=3){printf("%lld\n",n);}else {printf("%lld\n",C[n][4]+n);}}return 0;}