UVa 353 - Pesky Palindromes

称号：字符串统计回文子的数量。

分析：dp，暴力。因为数据是小，直接暴力可以解决。

说明：（UVa最终评出800该）。

#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>using namespace std;char str[82];
char ans[3200][82];int main()
{while (~scanf("%s",str)) {int count = 0,len = strlen(str);for (int i = 1 ; i <= len ; ++ i)for (int s = 0 ; s < len ; ++ s) {int flag = 1;for (int t = s+i-1 ; t >= s ; -- t)if (str[s+s+i-1-t] != str[t]) {flag = 0;break;}if (flag) {for (int j = 0 ; j < i; ++ j)ans[count][j] = str[s+j];ans[count][i] = 0;int same = 0;for (int j = 0 ; j < count ; ++ j) if (!strcmp(ans[j], ans[count])) {same = 1;break;}if (!same) count ++;}}printf("The string \'%s\' contains %d palindromes.\n",str,count);/*printf("The %d unique palindromes in \'boy\' are",count);for (int i = 0 ; i < count-1 ; ++ i) {printf(" \'%s\'",ans[i]);if (i < count-2)printf(",");else printf(" and ");}printf("\'%s\'.\n\n",ans[count-1]);*/}return 0;
}

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