摘要：
1 #include 2 #include 3 #include 4 using namespace std; 5 int gcd(int a,int b) 6 { 7 return b==0?a:gcd(b,a%b); 8 } 9 int main() 10 { 11 int b,a; 12 while(~scanf("%d%d",&a,&b))... 阅读全文
摘要：
Ignatius's puzzle Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9783 Accepted Submission(s): 68 阅读全文