codeforces 616E. Sum of Remainders 数学

给两个数n, m. 求n%1+n%2+.......+n%m的值。

首先， n%i = n-n/i*i, 那么原式转化为n*m-sigma(i:1 to m)(n/i*i)。

然后我们可以发现 1/4 = 2/4 = 3/4 = 0， 4/4 = 5/4 = 6/4 = 7/4 = 1. 所以可以将这些结果分成很多块，按块算结果。

注意计算过程中时刻避免爆longlong。

#include <iostream>

#include <vector>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <map>

#include <set>

#include <string>

#include <queue>

#include <stack>

#include <bitset>

using namespace std;

#define pb(x) push_back(x)

#define ll long long

#define mk(x, y) make_pair(x, y)

#define lson l, m, rt<<1

#define mem(a) memset(a, 0, sizeof(a))

#define rson m+1, r, rt<<1|1

#define mem1(a) memset(a, -1, sizeof(a))

#define mem2(a) memset(a, 0x3f, sizeof(a))

#define rep(i, n, a) for(int i = a; i<n; i++)

#define fi first

#define se second

typedef pair<int, int> pll;

const double PI = acos(-1.0);

const double eps = 1e-;

const ll mod = 1e9+;

const int inf = ;

const int dir[][] = { {-, }, {, }, {, -}, {, } };

vector <pair<ll, ll> > v;

ll getsum(ll a, ll b) {

    if((b-a)%==) {

        return ((b-a)/%mod)*((a+b+)%mod)%mod;

    }

    return ((a+b+)/%mod)*((b-a)%mod)%mod;

}

int main()

{

    ll n, m;

    cin>>n>>m;

    for(ll i = ; i*i<=n; i++) {

        ll tmp = n/i;

        v.pb(mk(i, tmp));

        if(tmp != i) {

            v.pb(mk(tmp, i));

        }

    }

    v.pb(mk(, ));

    sort(v.begin(), v.end());

    ll ans = (n%mod)*(m%mod)%mod;

    int i;

    for(i = ; i<v.size()&&m>=v[i].fi; i++) {

        ans = (ans- getsum(v[i-].fi, v[i].fi)%mod*(v[i].se%mod)%mod+mod)%mod;

    }

    if(m<n)

        ans = (ans - getsum(v[i-].fi, m)%mod*(v[i].se%mod)%mod+mod)%mod;

    cout<<ans<<endl;

    return ;

}