HDU 1212 Big Number

Big number

Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 5984 Accepted Submission (s): 4182

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 had to ouput the result of A mod B.
 
Sample Input

2 312 7152455856554521 3250

 
Sample Output

251521

 
AuthorIGNATIUS.L 
SOURCE Hangzhou Electric ACM Province Race Training Team tryouts of the train of thought:for large numbers, you need to convert the large number (the original in the string) into the integer array, starting from the highest bit to the positive integer c, the final output of the remainder of C line! Code:

#include <stdio.h> #include <string.h> #define N 1005char a[n];int b[n];int main () {int len,n,i,j,k,t,s,c; while (scanf ("%s%d", A,&c)!=eof) {memset (b,0,sizeof (b)); Len=strlen (a);    for (i=0;i<len;i++)//Converts a bit character to a number and saves it in an array of shapes! (Note: a[0] is the highest bit saved! )        b[i]=a[i]-' 0 ';    for (i=0,t=0;i<len;i++)    {              t=t*10+b[i];          t=t%c;//from the highest bit to the C-bit to take the remainder!     }    printf ("%d\n", t);} return 0;}

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HDU 1212 Big number