标签:bottom 转化 ace miss src inpu mem code accept
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2469 Accepted Submission(s): 1022
代码如下:
#include <map> #include <stack> #include <queue> #include <cmath> #include <vector> #include <cstdio> #include <cstring> #include <cstdlib> #include <iostream> #include <algorithm> using namespace std; typedef long long ll; ll MOD; ll mi(ll a,ll b) { ll ans=1; a=a%MOD; while(b>0) { if(b&1)ans=(ans*a)%MOD; b=(b>>1); a=(a*a)%MOD; } return ans; } int main() { ll t,x,m,k,c,r,Case=0; scanf("%lld",&t); while(t--) { Case++; scanf("%lld%lld%lld%lld",&x,&m,&k,&c); MOD=9*k; r=(mi(10,m)-1)/9*x%k; printf("Case #%lld:\n",Case); if(r==c)puts("Yes"); else puts("No"); } return 0; }
i
标签:bottom 转化 ace miss src inpu mem code accept
原文地址:http://www.cnblogs.com/a249189046/p/7628931.html