PAT 1019. General Palindromic Number (20)

1496 단어 number
Input Specification:
Each input file contains one test case. Each case consists of two non-negative numbers N and b, where 0 <= N <= 109 is the decimal number and 2 <= b <= 109 is the base. The numbers are separated by a space.
Output Specification:
For each test case, first print in one line "Yes"if N is a palindromic number in base b, or "No"if not. Then in the next line, print N as the number in base b in the form "ak ak-1 ... a0". Notice that there must be no extra space at the end of output.
Sample Input 1:
27 2

Sample Output 1:
Yes
1 1 0 1 1

Sample Input 2:
121 5

Sample Output 2:
No
4 4 1
#include<stdio.h>
int main(){
	long long n, b;
	int a[1000];
	scanf("%Ld%Ld
", &n,&b); int index = 0; while(n != 0){ a[index++] = n%b; n /= b; } int i = 0; int j = index - 1; int flag = 0; while(i<j){ if(a[i] != a[j]){ flag = 1; break; } i++; j--; } if(flag) printf("No
"); else printf("Yes
"); for(i =index-1; i> 0; i--){ printf("%d ", a[i]); } printf("%d
", a[0]); return 0; }

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