UVa1363 Joseph's Problem

把整个序列进行拆分成[k,k/2),[k/2, k/3), [k/3,k/4)...k[k/a, k/b)的形式，对于k/i（整除）相同的项，k%i成等差数列。

 /*by SilverN*/
 #include<algorithm>
 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<cmath>
 using namespace std;
 long long n,m,k;
 int main(){
     while(scanf("%lld%lld",&n,&k)!=EOF){
         int i,j;
         long long ans=;
         for(i=;i<=n;i=m+){
             m=k/i;
             if(!m)m=n;
             else m=min(k/m,n);
             ans+=(k%i+k%m)*(m-i+)/;
         }
         printf("%lld\n",ans);
     }
     return ;
 }