大数取模 模板
1 scanf("%s%d",a,&b); 2 int len=strlen(a); 3 int ans=0; 4 for(int i=st;i<len;i++) 5 ans=(int)(((long long)ans*10+a[i]-'0')%b); 6 printf("%d\n",ans);
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
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include<cmath> 6 using namespace std; 7 #define maxn 10000000 8 int t,b; 9 char a[205]; 10 int main() 11 { 12 scanf("%d",&t); 13 int cnt=0; 14 while(t--) 15 { 16 scanf("%s%d",a,&b); 17 int len=strlen(a); 18 int ans=0; 19 int st=0; 20 printf("Case %d: ",++cnt); 21 if(a[0]=='-') 22 st=1; 23 for(int i=st;i<len;i++) 24 ans=(int)(((long long)ans*10+a[i]-'0')%b); 25 if(ans==0) 26 printf("divisible\n"); 27 else 28 printf("not divisible\n"); 29 30 } 31 32 return 0; 33 } 34 35 //6 36 // 37 //101 101 38 // 39 //0 67 40 // 41 //-101 101 42 // 43 //7678123668327637674887634 101 44 // 45 //11010000000000000000 256 46 // 47 //-202202202202000202202202 -101
