Casting

Casting

Time Limit: 3000ms, Special Time Limit:7500ms, Memory Limit:65536KB

Total submit users: 23, Accepted users: 23

Problem 12598 : No special judgement

Problem description

Casting around for problems leads us to combine modular arithmetic with different integer bases, particularly the problem of computing values modulo b - 1, where b is the base in which the value is represented. For example, 
7829₁₀ mod 9 = 8 
37777777777777773₈ mod 7 = 6 
123456₇ mod 6 = 3 
(Note that 37777777777777773₈ = 1125899906842619₁₀ and 123456₇ = 22875₁₀.) 
Your job is to write a program that reads integer values in various bases and computes the remainder after dividing these values by one less than the input base.

Input

The first line of input contains a single integer P, (1 ≤ P ≤ 1000), which is the number of data sets that follow. Each data set should be processed identically and independently. 
Each data set consists of a single line of input containing three space-separated values. The first is an integer which is the data set number. The second is an integer which is the number, B (2 ≤ B ≤ 10), denoting a numeric base. The third is an unsigned number, D, in base B representation. For this problem, the number of numeric characters in D will be limited to 10,000,000. 

Output

For each data set there is a single line of output. It contains the data set number followed by a single space which is then followed by the remainder resulting from dividing D by (B - 1 ). 

Sample Input

5
1 10 7829
2 7 123456
3 6 432504023545112
4 8 37777777777777773
5 2 10110100010101010101101110001010001010101010101010111

Sample Output

1 8
2 3
3 1
4 6
5 0

Problem Source

Greater New York Programming Contest 2012

 

#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
char a[10000000+10];
int main(void){
#ifndef ONLINE_JUDGE
  freopen("in", "r", stdin);
#endif
  int t; scanf("%d", &t);
  while (t--){
    int s, b; scanf("%d%d%s", &s, &b, a);
    int len = strlen(a), po = 1, temp = 0;
    for (int i = len-1; i >= 0; --i){
      temp += (a[i]-'0') * po; temp %= (b-1);
    }
    printf("%d %d\n", s, temp);
  }

  return 0;
}

求在某个进制的数字的余数，没卡。