1214 - Large Division
PDF (English) | Statistics | Forum |
Time Limit: 1 second(s) | Memory Limit: 32 MB |
Given two integers, a and b, you should checkwhether
a is divisible by b or not. We know that an integer
ais divisible by an integer b if and only if there exists an integer
csuch 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 leadingzeroes.
Output
For each case, print the case number first. Then print ‘divisible‘if
a is 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类型不可以取余;
#include<iostream> #include<cstdio> #include<cstring> #include<cmath> #define LL long long using 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++)//一个大数一个整数的类型,处理的话枚举大数的每个字符一位一位的进行取余 { if(s[i]=='-') continue; x=(x*10+s[i]-'0')%m; } if(!x)//最终的结果是否为0 printf("divisible\n"); else printf("not divisible\n"); } return 0; }
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