After the 40th ACM-ICPC, Diao Yang is thinking about finding a girlfriend because he feels very lonely when doing ACM all the time. But because of his philandering, he finally decided to find N girlfriends. To achieve his goal, he wanted to find one girlfriend every day for N days continue. That is to say, at the ith day, he will have i girlfriends exactly.
In order to make his N girlfriends happy, he decided to buy M candies everyday for N days continue. Every day all of his girlfriends can get candies, and he will give each of them the same amount of candies and the amount will be as much as possible. Then if there are some candies left, he will eat them by himself.
Now the problem is, Diao Yang want to know how many candies he can eat total by himself after N days continue.
The first line contains an integer T, indicating the total number of test cases. Each test case is a line with two integers N
15N" style="box-sizing: border-box; width: 9pt; height: 15.75pt;"> and M (
151鈮?/m:t>N&lolt;231" style="box-sizing: border-box; width: 63.75pt; height: 15.75pt;"> ,
150鈮?/m:t>M&lolt;231" style="box-sizing: border-box; width: 65.25pt; height: 15.75pt;"> ).
For each test case, output the answer in one line.
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define mod 1000000007
#define inf 999999999
#define esp 0.00000000001
//#pragma comment(linker, "/STACK:102400000,102400000")
int scan()
{
int res = 0 , ch ;
while( !( ( ch = getchar() ) >= ‘0‘ && ch <= ‘9‘ ) )
{
if( ch == EOF ) return 1 << 30 ;
}
res = ch - ‘0‘ ;
while( ( ch = getchar() ) >= ‘0‘ && ch <= ‘9‘ )
res = res * 10 + ( ch - ‘0‘ ) ;
return res ;
}
int main()
{
ll x,y,z,i,t;
scanf("%lld",&x);
while(x--)
{
scanf("%lld%lld",&y,&z);
ll ans=max(y-z,0LL)*z;
ll huan=min(y,z);
for(i=2;i<=huan;i++)
{
if(z%i!=0)
{
ll d=z/i;
ll maxx=min((z%i)/d+1,huan-i+1);
d=-d;
ans+=(z%i)*maxx+(maxx*(maxx-1)/2)*d;
}
i+=d-1;
}
printf("%lld\n",ans);
}
return 0;
}
