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;

}

巴特西

PAT A1019 General Palindromic Number (20 分)

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