南京网络赛J-Sum【数论】

A square-free integer is an integer which is indivisible by any square number except 11. For example, 6 = 2 \cdot 36=2⋅3 is square-free, but 12 = 2^2 \cdot 312=22⋅3 is not, because 2^222 is a square number. Some integers could be decomposed into product of two square-free integers, there may be more than one decomposition ways. For example, 6 = 1\cdot 6=6 \cdot 1=2\cdot 3=3\cdot 2, n=ab6=1⋅6=6⋅1=2⋅3=3⋅2,n=ab and n=ban=ba are considered different if a \not = ba̸=b. f(n)f(n) is the number of decomposition ways that n=abn=ab such that aa and bb are square-free integers. The problem is calculating \sum_{i = 1}^nf(i)∑i=1nf(i).

Input

The first line contains an integer T(T\le 20)T(T≤20), denoting the number of test cases.

For each test case, there first line has a integer n(n \le 2\cdot 10^7)n(n≤2⋅107).

Output

For each test case, print the answer \sum_{i = 1}^n f(i)∑i=1nf(i).

Hint

\sum_{i = 1}^8 f(i)=f(1)+ \cdots +f(8)∑i=18f(i)=f(1)+⋯+f(8)

=1+2+2+1+2+4+2+0=14=1+2+2+1+2+4+2+0=14.

样例输入复制

样例输出复制

8

14

题目来源

ACM-ICPC 2018 南京赛区网络预赛

用素数筛类似的方法先把f的表打出来

一个数可以分解成 n = p1^a1 * p2^a2 * p3^a3.......其中p1,p2,p3...都是素数

如果a1,a2,a3...中有一个大于2，那么f（n）=0因为不管怎样组合都会使一个因子是平方数

如果a1,a2,a3...中有m个是1，剩下的都是2 那么只有那些是2的就必须分开，剩下的每一个数都有2中可能所以f(n)=2^m

打表 prime[p] = p prime[i*p]=1

对于比i小的所有素数 i是作为他们的倍数进行处理的

如果i%prime[j]==0 而且i%(prime[j]*prime[j])==0那么i*prime[j]肯定指数有一个大于2了 f[i*prime[j]] = 0

如果i%prime[j]==0 那么i*prime[j]要少掉一个数可以选择所以f[i*prime[j]] = f[i] / 2

如果上面一个都不是那么i*prime[j]之后就多了一个数可以选择所以f[i*prime[j]] = f[i] * 2

不加if(i%prime[j]==0)break;会T，还没想懂为什么....



#include<iostream>

#include<stdio.h>

#include<string.h>

#include<algorithm>

#include<stack>

#include<queue>

#include<map>

#include<vector>

#include<set>

//#include<bits/stdc++.h>

#define inf 0x3f3f3f3f

using namespace std;

typedef long long LL;

int t, n, cnt;

const int maxn = 2e7 + 5;

int f[maxn], prime[maxn];

void table()

{

    memset(f, 0, sizeof(f));

    memset(prime, 0, sizeof(prime));

    cnt = 0;

    f[1] = 1;

    for(int i = 2; i < maxn; i++){

        if(!prime[i]){

            prime[cnt++] = i;

            f[i] = 2;

        }

        for(int j = 0; j < cnt && prime[j] * i < maxn; j++){

            prime[prime[j] * i] = 1;

            if(i % prime[j] == 0){

                if(i % (prime[j] * prime[j]) == 0){

                    f[i * prime[j]] = 0;

                }

                else{

                    f[i * prime[j]] = f[i] / 2;

                }

            }

            else{

                f[i * prime[j]] = f[i] * 2;

            }

            if(i % prime[j] == 0)break;

        }

    }

}

int main()

{

    cin>>t;

    table();

    while(t--){

        scanf("%d", &n);

        LL ans = 0;

        for(int i = 1; i <= n; i++){

            ans += f[i];

        }

        printf("%lld\n", ans);

    }

	return 0;

}

巴特西

南京网络赛J-Sum【数论】

Input

Output

Hint

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