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$\displaystyle \sum_{i=1}^n(k-i\lfloor\frac{k}{i}\rfloor)=\displaystyle \sum_{i=1}^nk-\sum_{i=1}^ni\lfloor\frac{k}{i}\rfloor$

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-;

const int mod = 1e9+;

const int maxn = 2e6+;

const int maxm = 2e6+;

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 = , r = ; l <= min(k, n); l = r+){

        r = min(n, k/(k/l));

        ans -= (((l+r)*(k/l))*(r-l+))>>;

    }

    printf("%lld",ans);

    return ;

}