传送门
文章目录
- 题意:
- 思路:
题意:
n,k≤1e9n,k\le1e9n,k≤1e9
思路:
考虑转换式子,∑i=1nkmodi=∑i=1n(k−⌊ki⌋∗i)=n∗k−∑i=1n⌊ki⌋∗i\sum_{i=1}^{n}k\bmod i=\sum_{i=1}^n(k-\left \lfloor \frac{k}{i} \right \rfloor*i)=n*k-\sum_{i=1}^n\left \lfloor \frac{k}{i} \right \rfloor *i∑i=1nkmodi=∑i=1n(k−⌊ik⌋∗i)=n∗k−∑i=1n⌊ik⌋∗i,对于∑i=1n⌊ki⌋∗i\sum_{i=1}^n\left \lfloor \frac{k}{i} \right \rfloor *i∑i=1n⌊ik⌋∗i我们可以整除分块来解决,实际上他是一段等差数列,直接套求和公式即可。
// Problem: P2261 [CQOI2007]余数求和
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P2261
// Memory Limit: 125 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#include<random>
#include<cassert>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid ((tr[u].l+tr[u].r)>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=1000010,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;LL n,k;int main()
{
// ios::sync_with_stdio(false);
// cin.tie(0);cin>>n>>k;LL ans=n*k;for(LL l=1,r;l<=n;l=r+1) {if(k/l!=0) r=min(n,k/(k/l));else r=n;LL a=l*(k/l),d=k/l;ans-=(r-l+1)*(a+a+d*(r-l))/2;}cout<<ans<<endl;return 0;
}
/**/