BZOJ1257 [CQOI2007]余数之和 (数论分块)

题意：

给定n, k,求$displaystyle sum_{i=1}^nk;mod;i$

n,k<=1e9

思路：

先转化为$displaystyle sum_{i=1}^n(k-ilfloorfrac{k}{i} floor)=displaystyle sum_{i=1}^nk-sum_{i=1}^nilfloorfrac{k}{i} floor$

而k/i在一定范围内是不变的，所以分块求等差数列就可以了

代码：

/**************************************************************
    Problem: 1257
    User: wrjlinkkkkkk
    Language: C++
    Result: Accepted
    Time:60 ms
    Memory:1288 kb
****************************************************************/
 
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<functional>
 
#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
#define lowbit(x) ((x)&(-x)) 
 
using namespace std;
 
typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;
 
const db eps = 1e-6;
const int mod = 1e9+7;
const int maxn = 2e6+100;
const int maxm = 2e6+100;
const int inf = 0x3f3f3f3f;
const db pi = acos(-1.0);
 
 
int main() {
    ll n, k;
    scanf("%lld %lld", &n, &k);
    ll ans = n*k;
    for(ll l = 1, r = 0; l <= min(k, n); l = r+1){
        r = min(n, k/(k/l));
        ans -= (((l+r)*(k/l))*(r-l+1))>>1;
    }
    printf("%lld",ans);
    return 0;
}