PAT A1019 General Palindromic Number (20 分)

AC代码

#include <cstdio>
const int max_n = 1000;
long long ans[max_n];
int num = 0;
void change(long long a, long long d) {
do {
ans[num++] = a % d;
a /= d;
} while(a != 0);
}

int main() {
#ifdef ONLINE_JUDGE
#else
freopen("1.txt", "r", stdin);
#endif // ONLINE_JUDGE}
long long a, d;
scanf("%lld %lld", &a, &d);
change(a, d);
bool flag = 1;
for(int i = 0; i < num - 1- i; i++) {
if(ans[i] != ans[num - 1 - i]) flag = false;
}
if(flag == true) printf("Yes\n");
else if(flag == false) printf("No\n");
for(int i = num - 1; i >= 0; i--) {
if(i != num - 1) printf(" ");
printf("%d", ans[i]);
}
return 0;
}

转载于:https://www.cnblogs.com/isChenJY/p/11300606.html