Ayoub's function

思维,就是反过来想,题解太强了  

#include <bits/stdc++.h>
using namespace std;
int main()
{
    long long t;
    cin>>t;
    while(t--)
    {
        long long n,m;
        scanf("%lld%lld",&n,&m);
        if(m==0)
        {
            cout<<0<<endl;
            continue;
        }
        if(n==m)
        { 
            cout<<n*(n+1)/2<<endl;
            continue;
        }
        long long ans=(n+1)*n/2;
        long long noz=n-m;
        if(noz-1<=m)
        {
            ans-=noz;
            cout<<ans<<endl;
            continue;
        }
        long long num =noz/(m+1);//每部分0的个数 
        long long res_group=noz%(m+1);
        ans=ans-(m+1-res_group)*num*(num+1)/2-res_group*(num+1)*(num+2)/2;
        printf("%lld\n",ans);
    }
    return 0;
}

 

posted @ 2020-02-27 22:33  SunCY  阅读(...)  评论(...编辑  收藏