HDU 1212 Big number (large numbers modulo)

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=1212

Big numberTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others) Total Submission (s): 7083 Accepted Submission (s): 4884

Problem Descriptionas 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.

The problem easier, I promise that B'll be smaller than 100000.

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

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

Outputfor Each test case, you have to ouput the result of A mod B.

Sample Input

2 312 7152455856554521 3250

Sample Output

251521

Authorignatius.l
SOURCE Hangzhou Electric ACM Provincial Training Team Qualifying matchThe main topic: give you a length of not more than 1000 of the large number A, there is a not more than 100000 B, quickly seek a% B.
Problem-solving ideas: the first kind: according to the normal division of the method of a person except down.
the second type:
As an example, 1314 7 = 5
By Qin Jiushao formula:
1314= ((1*10+3) *10+1) *10+4
So there are
1314% 7 = (((1 * 10 7 +3) *10% 7 + 1) *10% 7 +4)%7

See the code.

#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 (large numbers modulo)