luogu2261余数求和题解--整除分块

题目链接

https://www.luogu.org/problemnew/show/P2261

分析

显然(k) (mod) (i=k-lfloor {k/i} floor) ( imes) (i),于是我们只需要求(N * k-sum_{i=1}^N {lfloor {k/i} floor imes i})

这里就需要数论分块,也称作整除分块的知识

结论:

(forall{i} in [x,lfloor {k/{lfloor {k/x} floor }} floor]),(lfloor k/i floor)的值都相等

证明

先咕了....

于是这道题再套个等差数列求和就完了...

代码

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cctype>
#define ll long long 
#define ri register int 
using std::min;
using std::max;
ll n,k,ans=0,g;
int main(){
	scanf("%lld %lld",&n,&k);
	ans=n*k;
	for(ri i=1;i<=n;i=g+1){
		g= k/i ? min(k/(k/i),n) : n;//如果i大于k的话直接一步把后面的算完 
		ans -= (i+g)*(g-i+1)/2 * (k/i);	
		//     等差数列求和      数论分块 
	}
	printf("%lld
",ans);
	return 0;
}