6 solutions
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0
#include <stdio.h> void fun1(int n,int m,int a[32],int i); void fun2(int n,int m); int main() { int n,m,a[32],i=0; scanf("%d%d", &n,&m); fun1(n,m,a,i); fun2(n,m); return 0; }
void fun1(int n,int m,int a[32],int i) { while (n > 0) { a[i] = n % m; n = n / m; i++; } for (int j = i - 1; j >= 0; j--) { printf("%d", a[j]); }
}
void fun2(int n,int m) { printf("\n"); int x=n; while(x!=1) { if(x%m!=0) break; x=x/m; } if(x==1) printf("yes"); else printf("no");
}
Information
- ID
- 960
- Time
- 1000ms
- Memory
- 256MiB
- Difficulty
- 5
- Tags
- # Submissions
- 145
- Accepted
- 53
- Uploaded By