ACM-ICPC 2018 南京赛区网络预赛 J.Sum [ 类打表 ]

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=22∗312=22∗3 is not, because 2222
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∗6=6∗1=2∗3=3∗2,n=ab,n=ba6=1∗6=6∗1=2∗3=3∗2,n=ab,n=ba, are considered different if a≠ba≠b. f(n) is the number of decomposition ways that n=ab such that a and b are square-free integers. The problem is calculating∑ni=1f(i)∑i=1nf(i)

Input
The first line contains an integer T(T≤20), denoting the number of test cases.
For each test case, there first line has a integer n(n≤2⋅107)n(n≤2⋅107)

Output
For each test case, print the answer ∑ni=1f(i)∑i=1nf(i)

Hint
∑8i=1f(i)=f(1)+⋯+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.

题意：找出一个数的所有乘数组合，使这些乘数组合中的任一因数都不存在完全平方因数。求从1到n的的所有数的乘数组合之和。

思路：
打个表找到每一个数的最小的质因数 ，然后
if ( i % (mprim[i] * mprim[i] * mprim[i]) == 0 ) ans[i] = 0;
else if ( i % (mprim[i] * mprim[i]) == 0 ) ans[i] = ans[i / (mprim[i]*mprim[i])];
else ans[i] = ans[i / mprim[i]] * 2;

AC code：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long ll;
typedef pair<int,int>pii;

const int maxn = 2e7+50;

int p[maxn] ,mprim[maxn] ,tot ;
bool vis[maxn];
ll ans[maxn];

void init() {
    tot = 0;
    memset(vis ,true ,sizeof(vis) );
    for (int i = 2;i < maxn;i++) {
        if ( vis[i] ) p[tot++] = i,mprim[i] = i;
        for (int j = 0; j < tot && i * p[j]< maxn ;j++) {
            vis[i * p[j]] = false;
            mprim[i * p[j]] = p[j];
            if ( i % p[j] == 0 ) break;
        }
    }
    ans[1] = 1;
    for (int i = 2;i < maxn ;i++) {
        if ( i % (mprim[i] * mprim[i] * mprim[i]) == 0 ) ans[i] = 0;
        else if ( i % (mprim[i] * mprim[i]) == 0 ) ans[i] = ans[i / (mprim[i]*mprim[i])];
        else ans[i] = ans[i / mprim[i]] * 2;
    }
}

int main() {
    init();
    int t ,n ; cin>>t;
    for (int i = 1;i <= t;i++) {
        scanf("%d",&n);
        ll res = 0;
        for (int i = 1;i<=n;i++) res += ans[i];
        printf("%lld
",res);
    }
    return 0;
}