Codeforces 1092C Prefixes and Suffixes【字符串+思维】
2020-11-19 09:53
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题目链接:点这里
题意:理解错了题意导致WA好几发,QAQ暴击
题意是判断给你的2*n-2个字符串是前缀还是后缀,不是判断这个字符串的内容...我真的欲哭无泪,理解能力太菜了
思路:将两个n-1长的字符串取出,先判断第一个取出的字符串和给出的字符串前缀的匹配程度。如果匹配程度大于半数,则这个为所需字符串-1,否则就是另外一个。同时要注意回文串的情况,所以开了一个cnt数组进行标记,其中p和s的数量要刚好等于n-1···
#include <bits/stdc++.h> using namespace std; int cnt[300], n, f; int main() { string mx1 = "", mx2 = "", pr, s[300]; scanf("%d", &n); getchar(); for (int i = 1; i <= 2 * n - 2; i++) { cin >> s[i]; if (s[i].size() == n - 1) { if (mx1.size() == 0) mx1 = s[i]; else mx2 = s[i]; } } for (int i = 1; i <= 2 * n - 2; i++) if (mx1.substr(0, s[i].size()) == s[i]) f++; if (f >= (2 * n - 2) / 2 && mx1.substr(1, n - 2) == mx2.substr(0, n - 2)) pr = mx1; else pr = mx2; for (int i = 1; i <= n * 2 - 2; i++) if (s[i] == pr.substr(0, s[i].size()) && cnt[s[i].size()] != 1) { cnt[s[i].size()] = 1; printf("P"); } else { printf("S"); } }
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