【HDU】 1787 GCD Again

GCD Again

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题目链接

• GCD Again

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题目大意

    给定一个n，现在要求m的个数,满足gcd(n,m)＞1 且(m＜n)

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题解

    这一题很明显可以用欧拉函数，还可以用容斥原理，鉴于好久没写过容斥原理了，又把容斥原理写了一遍…

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代码

    欧拉函数

#include <iostream>#include <cstring>#include <cstdio>#include <cmath>using namespace std;int n;int eular(int x){    int t=x,ans=x;    for (int i=2;i<=sqrt(t);i++) if (t%i==0)    {        ans=(ans/i)*(i-1);        while(t%i==0) t/=i;    }    if (t>1) ans=(ans/t)*(t-1);    return ans;}int main(){    while(scanf("%d",&n),n!=0)    {        printf("%d\n",n-1-eular(n));    }    return 0;}

    容斥原理

#include <iostream>#include <cstring>#include <cstdio>#include <vector>#include <cmath>using namespace std;int n;vector<int> V;int solve(int n){    int cnt=1,ans=0,t;    for (int i=1;i<(1<<V.size());i++)    {        cnt=1; t=0;        for (int j=0;j<V.size();j++) if (i&(1<<j))        {            cnt*=V[j];            t++;        }        cnt=(n-1)/cnt;        if (t%2) ans+=cnt;        else ans-=cnt;    }    return ans;}int main(){    while(scanf("%d",&n),n!=0)    {        V.clear();        int t=n;        for (int i=2;i<=sqrt(n);i++) if (n%i==0)        {            V.push_back(i);            while (n%i==0) n/=i;        }        if (n>1) V.push_back(n);        printf("%d\n",solve(t));    }    return 0;}