【Lightoj】1214 - 能否整除（同餘定理）

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 whethera is divisible by b or not. We know that an integera is divisible by an integer b if and only if there exists an integerc 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) andb (|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' ifa 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

簡單同餘定理。

代碼如下（這是初學的代碼，比較繁瑣，不推薦，建議看下面的）：

#include <stdio.h>#include <string.h>int main(){int u;char t[222];int a[222];long int b;int l;//數字總位元 long int y;//餘數 long int ty;//臨時存放餘數 int i,j;scanf ("%d",&u);getchar();for (int p=1;p<=u;p++){memset (a,0,sizeof (a));memset (t,'0',sizeof (t));scanf ("%s",t);scanf ("%ld",&b);printf ("Case %d: ",p);if (b<0)b=-b;l=strlen(t);if (t[0]=='-'){l--;for (i=1,j=l;i<=l;i++,j--){a[i]=t[j]-48;}}else{for (i=1,j=l-1;i<=l;i++,j--){a[i]=t[j]-48;}}y=a[1]%b;for (i=2;i<=l;i++){if (a[i]==0)continue;ty=a[i]%b;for (j=2;j<=i;j++){ty=(ty*10)%b;}y=(y+ty)%b;}if (y==0){printf ("divisible\n");}else{printf ("not divisible\n");}}return 0;}

改進的代碼（注意取餘的時候用longlong型）：

#include <cstdio>#include <cstring>int main(){int u;int num = 1;char a[222];long long b;int l;long long ans;// __int64scanf ("%d",&u);while (u--){scanf ("%s %lld",a,&b);l = strlen(a);//if (b < 0)//不取絕對值也能AC，為什麼。 //b = -b;ans = 0;for (int i = 0 ; i < l ; i++){if (a[i] == '-')continue;ans = (ans * 10 + (a[i] - '0')) % b;}if (ans == 0)printf ("Case %d: divisible\n",num++);elseprintf ("Case %d: not divisible\n",num++);}return 0;}