How to use congruence theorem to find remainder "Hdoj 1212 big number" for remainder ""

Source: Internet
Author: User

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 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 Input2 312 7 152455856554521 3250

Sample Output2 5 1521 This topic needs to use the same remainder theorem: The common congruence theorem has the following three kinds: (1) ((a mod m) * (B mod m)) mod m = = (a*b) mod m (2) ((a mod m) + (B mod m)) m   od m = = (a+b) mod m (3) ((A mod m)-(B mod m) + m) mod m = = (a) mod m in subtraction, since A mod n may be less than B mod n, you need to add N to the result. Let us first understand how to use the congruence theorem to find the remainder:
/* For example 10000 to M redundancy: * 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* The code is: * Assuming that 10000 is the length of the string is Len *//** such as 123 to M for redundancy * 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;}

For detailed modulo operations please refer to:

This problem has been used in (1) (2) Two:
#include <stdio.h> #include <string.h> #define MAX 1100int 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;}

How to use congruence theorem to find remainder "Hdoj 1212 big number" for remainder ""

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