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HDU 4731 Minimum palindrome 打表找规律
题目链接
虽然想到了可能有规律，但是比赛的时候没有去仔细推敲。
暴力打表找出可以得到对应的长度n和对应字符集m所对应的答案
10 1 1 1 a 2 2 aa 3 3 aaa 4 4 aaaa 5 5 aaaaa 6 6 aaaaaa 7 7 aaaaaaa 8 8 aaaaaaaa 9 9 aaaaaaaaa 10 10 aaaaaaaaaa 20 2 1 1 a 2 1 ab 3 2 aab 4 2 aabb 5 3 aaaba 6 3 aaabab 7 3 aaababb 8 3 aaababbb 9 4 aaaababba 10 4 aaaababbaa 11 4 aaaababbaaa 12 4 aaaababbaaaa 13 4 aaaababbaabab 14 4 aaaababbaababb 15 4 aaaababbaababba 16 4 aaaababbaababbaa 17 4 aaaababbaababbaaa 18 4 aaaababbaababbaaaa 19 4 aaaababbaababbaabab 20 4 aaaababbaababbaababb 10 3 1 1 a 2 1 ab 3 1 abc 4 1 abca 5 1 abcab 6 1 abcabc 7 1 abcabca 8 1 abcabcab 9 1 abcabcabc 10 1 abcabcabca 10 4 1 1 a 2 1 ab 3 1 abc 4 1 abca 5 1 abcab 6 1 abcabc 7 1 abcabca 8 1 abcabcab 9 1 abcabcabc 10 1 abcabcabca
然后很容易发现规律，m=1和m=3的时，循环节分别是a和abc，m=2的时候，从长度为9的时候开始循环，循环节为babbaa，然后前面8个打表就行了。
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1 #include <stdio.h> 2 const char ans[][9] = {"a", "ab", "aab", "aabb", "aaaba", "aaabab", "aaababb", "aaababbb"}; 3 int main() { 4 int T; 5 scanf("%d", &T); 6 for (int Tc = 1; Tc <= T; ++Tc) { 7 int n, m; 8 scanf("%d%d", &m, &n); 9 printf("Case #%d: ", Tc); 10 if (m == 1) { 11 for (int i = 0; i < n; ++i) 12 putchar(‘a‘); 13 } 14 else if (m >= 3) { 15 for (int i = 0; i < n; ++i) 16 putchar("abc"[i % 3]); 17 } 18 else { 19 if (n <= 8) 20 printf("%s", ans[n - 1]); 21 else { 22 printf("aaaa"); 23 for (int i = 4; i < n; ++i) 24 putchar("babbaa"[(i + 2) % 6]); 25 } 26 } 27 putchar(‘\n‘); 28 } 29 return 0; 30 }
HDU 4731 Minimum palindrome 打表找规律
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