題面在這裡

把答案的形式寫出來就是這樣的：

示例程式：

#include<cstdio>
#include<cmath>
#define LL long long

LL n,nn,m,ans,a[3162300];
int main(){
    scanf("%lld%lld",&m,&n);nn=sqrt(n);
    ans=n*(m);
    for (int k=1;k<=nn;k++){
        a[k]=n/k;
        LL l=n/(k+1
 
)+1,r=n/k;
        if (l>m) continue;
        if (r>m) r=m;
        ans-=(k*((l+r)*(r-l+1)/2));
    }
    for (int i=1;i<nn;i++){
        LL k=a[i],l=n/(k+1)+1,r=n/k;
        if (l>m) continue;
        if (r>m) r=m;
        ans-=(k*((l+r)*(r-l+1)/2));
    }
    if (n/nn!=nn){
        LL k=a[nn],l=n/(k+1
 
)+1,r=n/k;
        ans-=(k*((l+r)*(r-l+1)/2));
    }
    printf("%lld",ans);
    return 0;
}