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| Time Limit: 1000MS | Memory Limit: 32768KB | 64bit IO Format: %lld & %llu |
Description
Given two integers, a and b, you should check whether a is divisible by b or not. We know that an integer a is divisible by an integer b if and only if there exists an integer c such that a = b * c.
Input
Input starts with an integer T (≤ 525), denoting the number of test cases.
Each case starts with a line containing two integers a (-10200 ≤ a ≤ 10200) and b (|b| > 0, b fits into a 32 bit signed integer). Numbers will not contain leading zeroes.
Output
For each case, print the case number first. Then print ‘divisible‘ if a is divisible by b. Otherwise print ‘not divisible‘.
Sample Input
6
101 101
0 67
-101 101
7678123668327637674887634 101
11010000000000000000 256
-202202202202000202202202 -101
Sample Output
Case 1: divisible
Case 2: divisible
Case 3: divisible
Case 4: not divisible
Case 5: divisible
Case 6: divisible
Source
b fits into a 32 bit signed integer ？ 为毛long long.
#include <cstdio> #include <cstring> int main() { int t, Q=1; scanf("%d", &t); while(t--) { char num[205]; long long div; scanf("%s%lld", num, &div); int lent=strlen(num); long long sum=0; for(int i=0; i<lent; i++) { if(num[i]==‘-‘) continue; sum=sum*10+num[i]-‘0‘; sum=sum%div; } if(sum==0) printf("Case %d: divisible\n", Q++); else printf("Case %d: not divisible\n", Q++); } }
LightOJ - 1214 Large Division(同余)
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原文地址：http://www.cnblogs.com/fengshun/p/5333267.html
