PAT 天梯赛 L1-054. 福到了 【字符串】

题目链接

https://www.patest.cn/contests/gplt/L1-054

思路

可以先将字符串用字符串数组 输入

然后用另一个字符串数组 从 n - 1 -> 0 保存 其反转的字符串

然后每一行比较一下 这两个字符串数组有没有什么不同 如果没有 就要输出 bu yong dao le

然后最后 输出“到了” 的字符串 注意 字符替换

AC代码

#include <cstdio>
#include <cstring>
#include <ctype.h>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <map>
#include <stack>
#include <set>
#include <numeric>
#include <sstream>
#include <iomanip>
#include <limits>

#define CLR(a) memset(a, 0, sizeof(a))
#define pb push_back

using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;

const double PI  = 3.14159265358979323846264338327;
const double E   = exp(1);
const double eps = 1e-6;

const int INF  = 0x3f3f3f3f;
const int maxn = 1e2 + 5;
const int MOD  = 1e9 + 7;

string in[maxn], tran[maxn];

string Reverse(string s)
{
    string ans = "";
    int len = s.size();
    for (int i = len - 1; i >= 0; i--)
        ans += s[i];
    return ans;
}

int main()
{
    char c;
    int n;
    scanf(" %c%d", &c, &n);
    getchar();
    for (int i = 0; i < n; i++)
    {
        getline(cin, in[i]);
        tran[n - 1 - i] = Reverse(in[i]);
    }
    int flag = 1;
    for (int i = 0; i < n; i++)
    {
        if (in[i] != tran[i])
        {
            flag = 0;
            break;
        }
    }
    if (flag)
        printf("bu yong dao le\n");
    for (int i = 0; i < n; i++)
    {
        int len = tran[i].size();
        for (int j = 0; j < len; j++)
        {
            if (tran[i][j] != ' ')
                printf("%c", c);
            else
                printf(" ");
        }
        printf("\n");
    }
}