如何运用同余定理求余数【hdoj 1212 Big Number【大数求余数】】

Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5930    Accepted Submission(s): 4146

Problem Description

As we know, Big Number is always troublesome. But it's really important in our ACM. And today, your task is to write a program to calculate A mod B.

To make the problem easier, I promise that B will be smaller than 100000.

Is it too hard? No, I work it out in 10 minutes, and my program contains less than 25 lines.

 

 

Input

The input contains several test cases. Each test case consists of two positive integers A and B. The length of A will not exceed 1000, and B will be smaller than 100000. Process to the end of file.

 

 

Output

For each test case, you have to ouput the result of A mod B.

 

 

Sample Input

2 3

12 7

152455856554521 3250

 

 

Sample Output

2

5

1521

 

此题需要用到同余定理：

   常用的同余定理有以下三种：

   （1）（（A mod m）*（B mod m））mod m ==（A*B）mod m

   （2）（（A mod m）+（B mod m））mod m ==（A+B）mod m

   （3）（（A mod m）-（B mod m）+ m）mod m ==(A-B)mod m

    在减法中，由于a mod n 可能小于b mod n，需要在结果上加上n.

   我们先了解下如何运用同余定理求余数：

   对大数求余数：

 

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/*例如10000对m求余： * 10000%m * ==(10%m*1000%m)%m * ==(10%m*(10%m*100%m)%m)%m  * ==(10%m*(10%m*(10%m*10%m)%m)%m)%m * 用代码表示就是: * 假设10000是字符串长度是len */   /* * 如123对m求余 * 123%m * ==((12%m*10%m)%m+3%m)%m * ==(((10%m+2%m)%m*10%m)%m+3%m)%m * ==((((1%m*10%m)%m+2%m)%m*10%m)%m+3%m)%m */ gets(str); int ans=0; for(i=0;i<len;i++) {     ans=ans*10+str[i];     ans=ans%m; }

 对幂求余数：

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/* * 对幂取模如对37的4次方取模 * （37*37*37*37）%m *  ==(37%m*(37*37*37)%m)%m *  ==(37%m*(37%m*(37*37)%m)%m)%m *  ==(37%m*(37%m*(37%m*37%m)%m)%m)%m */ //求n^m%1000 s=n; for(i=1;i<m;i++) {     s=s*n;     s=s%1000;   }

 

详细的模运算请参考：http://blog.csdn.net/chocolate_22/article/details/6458029

此题用到了（1）（2）两个：

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#include<stdio.h> #include<string.h> #define MAX 1100 int main() {     int n,m,j,i,s,t;     char p[MAX];     while(scanf("%s",p)!=EOF)     {         scanf("%d",&n);         int l=strlen(p);         s=0;         for(i=0;i<l;i++)         {             s=s*10+p[i]-'0';             s=s%n;         }         printf("%d\n",s);     }     return 0; }