Light Oj 1214 大数整除
Large Division
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu
Submit
Status
Practice
LightOJ 1214
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
<span style="color:#6600cc;">/******************************************* author : Grant Yuan time : 2014.8.7 algorithm: 大数整除 source : Light Oj 1214 explain : 从ans初始化为0,最低位开始ans=(ans+s[i]-'0')%b; 计算到最后,如果ans 等于0,可以整除; 否则,不可以整除; **********************************************/ #include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<algorithm> #define INF 0x3fffffff using namespace std; char s[2000002]; int b; long long ans; int t; int main() { scanf("%d",&t); for(int i=1;i<=t;i++) { memset(s,0,sizeof(s)); scanf("%s%d",&s,&b); if(b<0) b=-b; int l=strlen(s); ans=0; for(int j=0;j<l;j++) { if(s[j]=='-') continue; ans=(ans*10+(s[j]-'0'))%b; } if(ans==0) printf("Case %d: divisible\n",i); else printf("Case %d: not divisible\n",i); } return 0; } </span>