BZOJ1257: [CQOI2007]余数之和

设数列 Ai=X mod i 

分析可得   余数是一些等差数列头尾相接组合起来的  然后就是二分找这些等差数列喽

#include<cstdio>#define SUM(L,R) (((L+R)*(R-L+1))>>1)using namespace std;#define LL long longLL n,k,ans;int main(){scanf("%lld%lld",&n,&k);ans=n*k;LL s=1,L,R,M;for(;s<=n&&s<=k;s=L+1){L=s;R=n;while(L<R){M=(L+R+1)>>1;if(k/s>k/M)R=M-1;else L=M;}ans-=SUM(s,R)*(k/s);}printf("%lld\n",ans);return 0;}