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  • PAT 1019 General Palindromic Number

    #include <cstdio>
    #include <iostream>
    #include <cstdlib>
    
    using namespace std;
    
    void print(int d[], int s, int e, const char* msg) {
        printf("%s
    ", msg);
        if (s == e) {
            printf("0");
            return;
        }
        printf("%d", d[s]);
        for (int i=s+1; i<e; i++) {
            printf(" %d", d[i]);
        }
    }
    
    int main() {
        int n = 0, b = 0;
        scanf("%d%d", &n, &b);
        int ds[32] = {0};
        
        int e = 31;
        while(n) {
            ds[e--] = n % b;
            n /= b;
        }
        
        int p = e + 1, q = 31;
        while (p < q) {
            if (ds[p] != ds[q]) break;
            p++, q--;
        }
        if (p < q) {
            print(ds, e + 1, 32, "No");
        } else {
            print(ds, e + 1, 32, "Yes");
        }
        return 0;
    }

    存在感

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  • 原文地址:https://www.cnblogs.com/lailailai/p/4063901.html
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