zoj 3818 Pretty Poem(暴力枚举)

本文主要是介绍zoj 3818 Pretty Poem(暴力枚举)，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

题目链接：zoj 3818 Pretty Poem

题目大意：给定一个字符串，忽略标点符号，考虑是否押韵，即为ABABA或者ABABCAB形式。

解题思路：暴力枚举A，B的长度，判断即可， 注意A，B，C非空，并且不相同。

#include <cstdio>
#include <cstring>
#include <algorithm>using namespace std;int n;
char s[105];bool check (int a, int b, int l) {for (int i = 0; i < l; i++)if (s[a + i] != s[b + i])return false;return true;
}bool judge (int l, int r, int flag) {int a = l + r;if (flag == 1) {if (!check(0, a, l) || !check(0, a*2, l))return false;if (!check(l, l + a, r))return false;} else {int c = n - 3 * a;if (!check(0, a, a) || !check(0, a*2+c, a))return false;if (l == c && check(0, 2*a, c))return false;if (r == c && check(l, 2*a, c))return false;}if (l == r && l == 1 && s[0] == s[1])return false;return true;
}bool solve () {scanf("%s", s);n = 0;int len = strlen(s);for (int i = 0; i < len; i++)if ((s[i] >= 'a' && s[i] <= 'z') || (s[i] >= 'A' && s[i] <= 'Z'))s[n++] = s[i];s[n] = '\0';for (int i = 1; i < n; i++) {for (int j = 1; j < n; j++) {if (3 * i + 2 * j == n && judge(i, j, 1))return true;if (3 * (i + j) < n && judge(i, j, 2))return true;}}return false;
}int main () {int cas;scanf("%d", &cas);while (cas--) {printf("%s\n", solve() ? "Yes" : "No");}return 0;
}

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