【WOJ-1420-可见点】采用数论知识求互质对

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题目来源： http://acm.whu.edu.cn/learn/problem/detail?problem_id=1420

代码：

#include<stdio.h>#include<vector>using namespace std;vector<int> frac;int m, n;long long int ans=0;void bfs(int step, int fraci, int tol,int num){if (step >= frac.size()){if (tol & 1)ans -= (long long) m / fraci;else{ans += (long long )m / fraci;}return;}bfs(step + 1, fraci*frac[step], tol + 1,num);bfs(step + 1, fraci, tol,num);}void prime(int num){frac.clear();int cur = num;for (int i = 2; i*i <= num; i++){if (cur%i == 0&&i!=num){frac.push_back(i);while (cur%i == 0)cur /= i;}}bfs(0, 1, 0,num);}int main(){int k;scanf_s("%d", &k);for (int h = 0; h < k; h++){ans = 0;scanf_s("%d%d", &m, &n);for (int i = 1; i <= n; i++){prime(i);if ((i <= m)&&(i!=1)) ans -= 1;}if(h!=k-1) printf("%d\n",ans);else  printf("%d", ans);}}

void prime(int num) 该函数求num的质因数

bfs() 求num与1...m互质对的个数。

把(1,1)看作互质对，把(2,2)不看作。

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