• hdu 5621


    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(2N105) 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(1T10) , which indicates the number of test cases.

    For each test case, there are one lines,includes a integer N(2N105) ,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;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/hsd-/p/5184969.html
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