hdu 4485 B-Casting （数学）

B-Casting

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 370    Accepted Submission(s): 187

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 o 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

41 10 78292 7 1234563 6 4325040235451124 8 37777777777777773 

 

Sample Output

1 82 33 14 6

 

对于一个b进制的数,每一位分别是an,...a5,a4,a3,a2,a1,a0;

则变成十进制就是((b)^0*a0)+((b^1)*a1)+......((b^n)*an);

对于其中的一项 求(b^k *ak )%(b-1)

用t=b-1

则变成((t+1)^k*ak)%t

而(t+1)^k%t   展开后显然只剩1了,其他都是t的倍数

所以(b^k *ak )%(b-1)=ak%(b-1)

.......

#include<stdio.h>#include<string.h>#include<iostream>#include<string>#include<algorithm>using namespace std;int p,a,b;char s[10000000+1];int main(){    scanf("%d",&p);    while(p--)    {        scanf("%d%d%s",&a,&b,s);        int j=0;        int ans=0;        int l=strlen(s)-1;        for(int i=l;i>=0;i--)        {            ans+=(s[i]-'0')%(b-1);            ans%=b-1;        }        printf("%d %d\n",a,ans);    }    return 0;}