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=1n​f(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=1n​f(i).

Hint

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

2
5
8

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 = 0; 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+10;
int vis[N],f[N],sum[N];
int cnt;
void init()
{
    memset(vis,1,sizeof(vis));
    for(int i=2;i*i<N;i++)
        for(int j=i*i;j<N;j+=i*i)
        vis[j]=0;
    for(int i=1;i<N;i++)
    {
        if(vis[i])
        {
            sum[i]=sum[i-1]+1;
            f[cnt++]=i;
        }
        else
            sum[i]=sum[i-1];
    }
}
int main()
{
    int n,t;
    cnt=1;
    init();
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        ll ans=0;
        for(int i=1;i<=cnt&&f[i]<=n;i++)
        {
            int res=(int)(n/f[i]);
            ans+=sum[res];
        }
        printf("%lld\n",ans);
    }
    return 0;
}