思路

简单的模拟，答案就是\(min\{(\lfloor\frac{d\times n}{u+d}\rfloor+1)\times(u+d)-d\times n\}\)

代码

#include <cstdio>

#include <algorithm>

#include <cstring>

#define int long long

using namespace std;

int n,m,ans=0x3f3f3f3f3f3f3f3fLL;

signed main(){

    while(scanf("%lld %lld",&n,&m)==2){

        ans=0x3f3f3f3f3f3f3f3fLL;

        for(int i=1;i<=m;i++){

            int u,d;

            scanf("%lld %lld",&u,&d);

            int t=(d*n)/(u+d);

            t++;

            ans=min(ans,(u+d)*t-d*n);

        }

        printf("%lld\n",ans);

    }

    return 0;

}