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"); } }转载于:https://www.cnblogs.com/Dup4/p/9433168.html
