KK's Point
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 357 Accepted Submission(s): 124
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.
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
2
3
4
Sample Output
3
5
Source
题意：n个不同的点 两点画一条线 最多两条线交于一点 问最多有多少个交点 包括边界点
题解
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 C（n，4）
最后我们再把边界上的N个点加上
bc 结束了交了几遍wa 爆long long
改用 usigned long long
#include<iostream>
#include<cstring>
#include<cstdio>
#define LL unsigned long long
using namespace std;
LL t;
LL n;
int main()
{
while(scanf("%I64d",&t)!=EOF)
{
for(LL i=1; i<=t; i++)
{
scanf("%I64d",&n);
printf("%I64u
",n+n*(n-1)/2*(n-2)/3*(n-3)/4);
}
}
return 0;
}