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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1212
Big Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7083 Accepted Submission(s): 4884
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
Author
Ignatius.L
Source
题目大意：给你一个长度不超过1000的大数A，还有一个不超过100000的B，快速求A % B。
解题思路：第一种：按照正常除法的方法一位一位除下去即可。
第二种：
举个例子，1314 % 7= 5
由秦九韶公式：
1314= （（1*10+3）*10+1）*10+4
所以有
1314 % 7= （ （ （1 * 10 % 7 +3 ）*10 % 7 +1）*10 % 7 +4 ）%7
详见代码。
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
char a[100000+10];
int main()
{
int b;
while (~scanf("%s%d",a,&b))
{
int len=strlen(a);
int pre=0,ans;
for (int i=0; i<len; i++)
{
ans=pre*10+a[i]-'0';
pre=ans%b;
}
printf ("%d\n",pre);
}
return 0;
}
hdu 1212 Big Number(大数取模)
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原文地址：http://blog.csdn.net/qiqi_skystar/article/details/51863651
