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.

题意：

令F(i)表示满足x*y=i，且x和y都不包含平方数因子的合法(x,y)对数，求∑F(i)

思路：

很容易证明F(i)是积性函数，如果i中不存在3次方以上的质因子，F(i) = 2^(i中只出现一次的质因子个数)，否则F(i) = 0

那么就是个积性函数前缀和的问题了

很多项都为0，利用Min_25筛可以在1s内轻松解决 $10^{10}$ 范围内的所有询问

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<map>
#include<string>
#include<math.h>
#include<queue>
#include<stack>
#include<iostream>
using namespace std;
#define LL long long
#define mod 1000000007
#define er 500000004
int cnt, s[200005], id[200005], B, m, h[200005];
bool flag[200005];
LL n, w[200005], pri[200005];
void Primeset(int n)
{int i, j;for(i=2;i<=n;i++){if(flag[i]==0){pri[++cnt] = i;s[cnt] = s[cnt-1]+i;}for(j=1;j<=cnt&&i*pri[j]<=n;j++){flag[i*pri[j]] = 1;if(i%pri[j]==0)break;}}
}
LL F(LL x, int y)
{LL t1;int i, k, e, ans;if(x<=1 || pri[y]>x)return 0;if(x>B)k = n/x;if(x<=B)k = id[x];ans = 2*(h[k]-(y-1));for(i=y;i<=cnt&&(LL)pri[i]*pri[i]<=x;i++){t1 = pri[i];for(e=1;t1*pri[i]<=x;e++){if(e==1)ans = (ans+(F(x/t1, i+1)*2+1));else if(e==2)ans = (ans+F(x/t1, i+1));t1 *= pri[i];}}return ans;
}int main(void)
{LL i, j, k, last;int T;scanf("%d", &T);while(T--){m = cnt = 0;scanf("%lld", &n);B = sqrt(n), Primeset(B);for(i=1;i<=n;i=last+1){w[++m] = n/i;h[m] = w[m]-1;last = n/(n/i);if(n/i<=B)id[n/i] = m;}for(j=1;j<=cnt;j++){for(i=1;i<=m&&pri[j]*pri[j]<=w[i];i++){if(w[i]/pri[j]<=B)k = id[w[i]/pri[j]];elsek = n/(w[i]/pri[j]);h[i] = h[i]-(h[k]-(j-1));}}printf("%lld\n", (F(n, 1)+1));}return 0;
}

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2018ACM-ICPC南京赛区网络赛: J. Sum（积性函数前缀和）

J. Sum

Input

Output

Hint

题意：

思路：

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