题解 [51nod1225]余数之和

题解 [51nod1225]余数之和

题面

解析

首先可以发现,(a)%(b)(=a-b*lfloor a/b floor).

而对于一段连续的(b)来说(lfloor a/b floor)是一样的.

并且这一段(b)是等差数列.

因此整除分块搞一搞就行了.

数据范围真的恶心(爆longlong)

code:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#define int __int128
#define fre(x) freopen(x".in","r",stdin),freopen(x".out","w",stdout)
using namespace std;

inline int read(){
	int sum=0,f=1;char ch=getchar();
	while(ch>'9'||ch<'0'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){sum=sum*10+ch-'0';ch=getchar();}
	return f*sum;
}

const int Mod=1000000007;
int n,ans;

inline int fpow(int a,int b){
	int ret=1;
	for(;b;a=a*a%Mod,b>>=1) if(b&1) ret=ret*a%Mod;
	return ret;
}

signed main(){
	n=read();int inv=fpow(2,Mod-2);
	for(int l=1,r;l<=n;l=r+1){
		r=n/(n/l);
		ans=(ans+(n/l)*inv%Mod*(l+r)%Mod*(r-l+1)%Mod)%Mod;
	}
	ans=((n%Mod)*(n%Mod)-ans)%Mod;
	long long ret=(ans+Mod)%Mod;
	printf("%lld
",ret);
	return 0;
}