Uva 6173 B-Casting
2013-04-28 23:12
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题目链接:https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=4184
求b进制下的a mod (b-1)的值。
(a*b^n)%(b-1) == a % (b-1),很好推导。
求b进制下的a mod (b-1)的值。
(a*b^n)%(b-1) == a % (b-1),很好推导。
#include <iostream> #include <stdio.h> #include <string.h> #include <stdlib.h> using namespace std; #define Maxn 10000005 int main() { #ifndef ONLINE_JUDGE freopen("in.txt","r",stdin); #endif int p; int m,b; int ans; int len; string str; scanf(" %d",&p); while(p--) { ans = 0; scanf(" %d %d",&m,&b); cin>>str; len = str.length(); for(int i=0;i<len;i++) { ans += (str[i] - '0')%(b-1); ans %= (b-1); } printf("%d %d\n",m,ans); } return 0; }
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