美味果冻

链接：https://ac.nowcoder.com/acm/problem/50418
来源：牛客网

题目描述
\sum_{i=1}^{{n}{}\sum_{j=1}}{i}{i*\left[ \frac{i}{j}\right]^{j}}∑
i=1
n
​
∑
j=1
i
​
i∗[
j
i
​
]
j
由于n越大jelly越美味，这里n<=3000000,只需求这个式子对1e9+7取模的值。
输入描述:
第一行输入一个整数 n。 1<=n<=3000000。
输出描述:
输出一个整数表示答案。
示例1
输入
复制
3
输出
复制
22

更换一下枚举顺序更加容易得到

我这里枚举的是j和那个结果然后求出i。

/*                         _
                        _ooOoo_
                       o8888888o
                       88" . "88
                       (| -_- |)
                  .'  \\|     |//  `.
                 /  \\|||  :  |||//  \
                /  _||||| -:- |||||_  \
                |   | \\\  -  /'| |   |
                | \_|  `\`---'//  |_/ |
                \  .-\__ `-. -'__/-.  /
              ___`. .'  /--.--\  `. .'___
           ."" '<  `.___\_<|>_/___.' _> \"".
          | | :  `- \`. ;`. _/; .'/ /  .' ; |
          \  \ `-.   \_\_`. _.'_/_/  -' _.' /
===========`-.`___`-.__\ \___  /__.-'_.'_.-'================
 
                  Please give me AC.
*/

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <set>
#include <string>
#include <queue>
#include <map>
#include <stack>
#include <map>
#include <unordered_map>
#include <vector>
#include <cmath>
//#include <ext/rope>
#include <bits/stdc++.h> 

using namespace std;
//using namespace __gnu_cxx;

#define gt(x) x = read()
#define int long long
#define ios ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define endl "\n"
//#define x first
//#define y second

int dx[4] = {0, 1, 0, -1};
int dy[4] = {1, 0, -1, 0}; 

typedef __int128 INT;
typedef pair<double, int> PDI;
typedef pair<int, int> PII;
typedef unsigned long long ULL;

inline int read(){
    int x = 0, f = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9'){
        if (ch == '-')
            f = -1;
        ch = getchar();
    }
    while(ch >= '0' && ch <= '9'){
        x = (x<<1) + (x<<3) + (ch^48);
        ch = getchar();
    }
    return x * f;
}

inline void print(INT x) {
  if (x < 0) { putchar('-'); x = -x; }
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}

const int N = 3e6 + 10;
const int M = 3 * N;
const int mod = 1e9 + 7;
const int PP = 131;
const int inf = 0x3f3f3f3f;
const int INF = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-10;
const double PI = acos(-1);

int primes[N], cnt;
bool st[N];

void get_primes(int n)
{
    for (int i = 2; i <= n; i ++ )
    {
        if (!st[i]) primes[cnt ++ ] = i;
        for (int j = 0; primes[j] <= n / i; j ++ )
        {
            st[primes[j] * i] = true;
            if (i % primes[j] == 0) break;
        }
    }
}


int qmi(int a, int b){
	int res = 1;
	while(b){
		if (b & 1)    res = res % mod * a % mod;
		a = a % mod * a % mod;
		b >>= 1;
	}
	return res % mod;
}

signed main(){ 
	int ans = 0;
	int n;
	gt(n);
	
	for (int i = 1; i <= n; i ++){
		ans = ans % mod + i * (i - (i / 2)) % mod;
		ans %= mod;
	}
	
	//get_primes(n);
	for (int i = 1; i <= n; i ++){
		for (int j = 2; j * i <= n; j ++){
			int temp = i * j;
			//while(temp / i == j && temp <= n)    ans = ans % mod + temp * qmi(j, i) % mod, temp ++;
			int temp2 = min((j + 1) * i - 1, n);
			for (int k = temp; k <= temp2; k ++)   ans = ans % mod + k * qmi(j, i) % mod;
		}
	}
	
	ans %= mod;
	print(ans);
	
	return 0;
}

标签：美味,const,int,果冻,ch,primes,include,mod
来源： https://blog.csdn.net/qq_45772483/article/details/113794462