codeforces 616E. Sum of Remainders 数学

题目链接

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 ;
}