求 n % 1 + n %2 + n % 3 .....n % n 。結果取餘1e9 + 7

n <= 1e12

顯然是用到了整除分塊。

這裡的for迴圈可以當作模板

    for (ll l = 1, r; l <= n; l = r + 1) {
        if (n / l != 0) r = min(n / (n / i), n);
        else r = n;
        // do something
    }

當然，此題還要注意取模

#pragma warning(disable:4996)

#include<iostream>
#include
 
<algorithm>
#include<bitset>
#include<tuple>
#include<unordered_map>
#include<fstream>
#include<iomanip>
#include<string>
#include<cmath>
#include<cstring>
#include<vector>
#include<map>
#include<set>
#include<list>
#include
 
<queue>
#include<stack>
#include<sstream>
#include<cstdio>
#include<ctime>
#include<cstdlib>
#define INF 0x3f3f3f3f
#define inf 0x7FFFFFFF
#define moD 1000000003
#define pii pair<int,string>
#define eps 1e-8
#define equals(a,b) (fabs(a-b)<eps)
#define bug puts("bug")
#define
 
 re  register
#define fi first
#define se second
const int maxn = 1e6 + 5;
const double Inf = 10000.0;
const double PI = acos(-1.0);
typedef  long long ll;
typedef unsigned long long ull;
using namespace std;

const ll MOD = 1e9 + 7;
ll inv = 5e8 + 4;

int main() {
    ll n;
    scanf("%lld", &n);
    ll res = (n % MOD) * (n % MOD) % MOD;
    ll ans = 0;
    for (ll l = 1, r; l <= n; l = r + 1) {
        if (n / l != 0) r = min(n / (n / l), n);
        else r = n;
        ans += ((n / l) % MOD) * ((((l + r) % MOD) * ((r - l + 1) % MOD) % MOD) * inv % MOD);
        ans %= MOD;
    }
    printf("%lld", ((res - ans + MOD) % MOD + MOD) % MOD);
}