hdu 多校联赛 Rikka with Graph

Rikka with Graph

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

Problem Description

As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them:

For an undirected graph G with n nodes and m edges, we can define the distance between (i,j) (\text{dist}(i,j)) as the length of the shortest path between i and j. The length of a path is equal to the number of the edges on it. Specially, if there are no path between i and j, we make \text{dist}(i,j) equal to n.

Then, we can define the weight of the graph G (w_G) as \sum_{i=1}^n \sum_{j=1}^n \text{dist}(i,j).

Now, Yuta has n nodes, and he wants to choose no more than m pairs of nodes (i,j)(i \neq j) and then link edges between each pair. In this way, he can get an undirected graph G with n nodes and no more than m edges.

Yuta wants to know the minimal value of w_G.

It is too difficult for Rikka. Can you help her?  

In the sample, Yuta can choose (1,2),(1,4),(2,4),(2,3),(3,4).

 

Input

The first line contains a number t(1 \leq t \leq 10), the number of the testcases. 

For each testcase, the first line contains two numbers n,m(1 \leq n \leq 10^6,1 \leq m \leq 10^{12}).

 

Output

For each testcase, print a single line with a single number -- the answer.

 

Sample Input

14 5

 

Sample Output

14

 

这道题做的时候掉到坑里去了 忘了环状和链状的路径都比放射状的长度长 加上后面一个变量没用long long类型 就这么傻傻的错了十几遍 最近智商下降有点严重(┬＿┬)、、

ac代码：

#include<bits/stdc++.h>using namespace std;int main(){    int t;    scanf("%d",&t);    while(t--)    {        long long n,m;        long long ans=0;        scanf("%lld%lld",&n,&m);        if(m>=n)        {            if(n*(n-1)/2<=m)                {ans=n*(n-1);                }            else if(n*(n-1)/2>m)                {ans=n*(n-1)+(n*(n-1)/2-m)*2;                }        }        else if(m==n-1)        {            ans=(n-2)*(n-1)*2+2*(n-1);        }        else if(m<n-1)        {            ans+=(m-1)*m*2+2*m;            long long g=n-(m+1);            ans+=g*(m+1)*n*2;            ans+=g*(g-1)*n;        }        cout<<ans<<endl;    }    return 0;}