（分块除法，题目数学式子的提取，%的除法转化和%的取余数的应用）

F. Fair Distribution
time limit per test1 second
memory limit per test512 megabytes
inputstandard input
outputstandard output
There are n robots and m energy bars in the Dream Kingdom. DreamGrid, the king, is trying to make a fair distribution of the energy bars. A fair distribution exists if and only if the number of the energy bars is a multiple of the number of robots.

The only tool DreamGrid has is a powerful laser gun. Every time he turns on the laser gun, he can do exactly one of the two things:

Create a new energy bar.
Destroy a robot.
To avoid the extinction of robots, it's forbidden to destroy all the n robots. It takes one dollar to turn on the laser gun once. You are asked to find the minimum cost of making a fair distribution.

Input
There are multiple test cases. The first line of the input contains an integer T (1≤T≤1000), indicating the number of test cases. For each test case:

The only line contains two integers n and m (1≤n,m≤108), indicating the initial number of robots and energy bars.

Output
For each test case output one line containing an integer, indicating the minimum cost to get a fair distribution.

Example
inputCopy
3
3 12
10 6
8 20
outputCopy
0
4
2
Note
For the third sample, the best way is to destroy a robot and create an energy bar. After that, we have 7 robots and 21 energy bars, which leads to a fair distribution.

#include <bits/stdc++.h>
using namespace std;
#define ri register int 
#define M 10005

template <class G> void read(G &x)
{
    x=0;int f=0;char ch=getchar();
    while(ch<'0'||ch>'9'){f|=ch=='-';ch=getchar();}
    while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
    x=f?-x:x;
    return ;
}

int  T,n,m;
int main(){
    read(T);
    while(T--)
    {
        read(n);read(m);
        if(n>=m)
        {
            printf("%d\n",n-m);
            continue;
        }
        if(m%n==0)
        {
            printf("0\n");
            continue;
        }
        int l=1;int ans=1e8;
        int aa=1e8;
        while(l<=n)
        {
            int  r=((m-1)/((m-1)/l));
            int tmp=l*((m-1)/l);
            ans=min(ans,tmp);
            l=r+1;
        }
        ans=n-m+ans;

        printf("%d\n",ans);
        
    }
   
    
    
} 

思路： 更具题目 题目数学式子的提取，%的除法转化和%的取余数的应用

主题在求余数时，有整除可以特判的情况，就利用 这个m-1%x+1 来做就行了 X- 这个东西，不会影响结果的。

%可以用除法来代替，同时有除法就有分块除法来节省时间。