Lightoj 1214-large Division (large number of surplus)

1214-large Division

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Time Limit: 1 second (s)

Memory Limit: MB

Given-Integers, a and b, you should checkwhethera is divisible by b or not. We knowA integer ais divisible by an integer b if and only if there exists an integerCs Uch that a = b * C.

Input

Input starts with an integer T (≤525), denoting the number of test cases.

Starts with a line containing-integers a ( -10200≤a≤10200) and B (|b| >0, B fits into a 32 Bit signed integer). Numbers would not contain leadingzeroes.

Output

For each case, print the case number first. Then print ' divisible 'ifa was divisible by b. Otherwise print ' not divisible '.

Sample Input	Output for Sample Input
6 101 101 0 67 -101 101 7678123668327637674887634 101 11010000000000000000 256 -202202202202000202202202-101	Case 1:divisible Case 2:divisible Case 3:divisible Case 4:not Divisible Case 5:divisible Case 6:divisible

Double type can not take redundancy;

#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #define LL Long Longusing namespace Std;char s[400];int main () {    LL n,m,i,x,y;    LL CLA;    bool BJ;    scanf ("%lld", &CLA);    for (int gr=1; gr<=cla; gr++)    {        bj=true;        scanf ("%s%lld", s,&m);        printf ("Case%LLD:", GR);        x=0;        For (I=0;i<strlen (s); i++)//A large number of the type of an integer, handling the words enumerated by a large number of each character for the remainder        {            if (s[i]== '-')            continue;            x= (x*10+s[i]-' 0 ')%m;        }        if (!x)//Whether the final result is 0            printf ("divisible\n");        else            printf ("Not divisible\n");    }    return 0;}

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Lightoj 1214-large Division (large number of surplus)