题目链接
https://www.nowcoder.com/acm/contest/85/J
思路 用 MAP 标记 哪一个 打败哪一个 输出的时候 输出对应的 就可以了
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> using namespace std; typedef long long ll; typedef unsigned long long ull; typedef pair <int, int> pii; typedef pair <ll, ll> pll; const double PI = 3.14159265358979323846264338327; const double E = 2.718281828459; const double eps = 1e-6; const int INF = 0x3f3f3f3f; const int maxn = 1e5 + 5; const int MOD = 1e9 + 7; int main() { map <string, string> m; map <string, int> vis; string s1, s2; for (int i = 0; i < 3; i++) { cin >> s1 >> s2; m[s2] = s1; vis[s1] = 1; vis[s2] = 1; } int n; scanf("%d", &n); while (n--) { cin >> s1; if (vis[s1]) cout << m[s1] << endl; else cout << "Fake\n"; } }转载于:https://www.cnblogs.com/Dup4/p/9433249.html
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