摘要:
题目:109A - Lucky Sum of Digits思路:扩展欧几里得#include #include #include #include #include #include using namespace std;long long exgcd(long long a,long long b,long long &x,long long &y){ if(b==0) { x=1; y=0; return a; } else { long long ans=exgcd(b,a%b,x,y); ... 阅读全文