kestory的博客

辗转相除法求最大公约数gcd再利用 gcd(a, b)*lcm(a, b) = a * b求得最小公倍数lcm#include <iostream>#define D(x) cout<<#x<<"="<<x<<endl;using namespace std;int gcd(int a, int b) { ...

参考图：https://blog.youkuaiyun.com/jopus/article/details/19091643#include <iostream>#define D(x) cout<<#x<<"="<<x<<endl;using namespace std;int main() { int a;// = 3; cin...

#include <iostream>#include <algorithm>#include <vector>#define D(x) cout<<#x<<"="<<x<<endl;using namespace std;int main() { int line, x; ve

#include <iostream>#include <string>#define D(x) cout<<#x<<"="<<x<<endl;using namespace std;int main() { string s1, s2; int result = 0; cin >> s1 &gt...

待：搞懂里面的数论知识&动归的方法数论：互质数的最大不能组合数#include <iostream>using namespace std;int main(int argc, char *argv[]) { int a, b; cin >> a >> b; cout << a*b - a - b;...

#include <iostream>using namespace std;#define D(x) cout<<#x<<"="<<x<<endl;int main(int argc, char *argv[]) { int n; cin >> n; bool flag; for (int i0 = 1; ...

// 相遇后是否向相反方向可以看做两只蚂蚁擦肩而过/*--> lr --> rr cold<-- ll <-- rl*/#include <iostream>#include <cmath>using namespace std;int main(int argc, char *ar...