ACM-ICPC 2018 南京赛区网络预赛 J.sum

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

这题显示用sum【i】表示第i及i前面的非平方因子的数量，f【cnt】用来存储非平方因子i，

主要是这部分：

for(int i = ; i < cnt && f[i] <= n; i ++)

{

   int res = (int)(n/f[i]);

   ans += sum[res];

}

题目要求∑a*b=i （i= 1:n）其实就可以看成有多少个a*b <= n （a b由非平分因子构成）

假设n = 100 非平分因子有2,3,5,6,7.......

当a=7时，计算有多少个b 100/7 = 14 所以b最大才14 只要加上前14个非平分因子的个数就行了

贴上代码：

#include <iostream>

#include <cstdio>

#include <algorithm>

#include <cstring>

using namespace std;

#define ll long long

const int N=2e7+;

int vis[N],f[N],sum[N];

int cnt;

void init()

{

    memset(vis,,sizeof(vis));

    for(int i=;i*i<N;i++)

        for(int j=i*i;j<N;j+=i*i)

        vis[j]=;

    for(int i=;i<N;i++)

    {

        if(vis[i])

        {

            sum[i]=sum[i-]+;

            f[cnt++]=i;

        }

        else

            sum[i]=sum[i-];

    }

}

int main()

{

    int n,t;

    cnt=;

    init();

    scanf("%d",&t);

    while(t--)

    {

        scanf("%d",&n);

        ll ans=;

        for(int i=;i<=cnt&&f[i]<=n;i++)

        {

            int res=(int)(n/f[i]);

            ans+=sum[res];

        }

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

    }

    return ;

}

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