We say a sequence of characters is a palindrome if it is the same written forwards and backwards. For example, ‘racecar’ is a palindrome, but ‘fastcar’ is not. A partition of a sequence of characters is a list of one or more disjoint non-empty groups of consecutive characters whose concatenation yields the initial sequence. For example, (‘race’, ‘car’) is a partition of ‘racecar’ into two groups. Given a sequence of characters, we can always create a partition of these characters such that each group in the partition is a palindrome! Given this observation it is natural to ask: what is the minimum number of groups needed for a given string such that every group is a palindrome?

For example: • ‘racecar’ is already a palindrome, therefore it can be partitioned into one group. • ‘fastcar’ does not contain any non-trivial palindromes, so it must be partitioned as (‘f’, ‘a’, ‘s’, ‘t’, ‘c’, ‘a’, ‘r’). • ‘aaadbccb’ can be partitioned as (‘aaa’, ‘d’, ‘bccb’).

Input

Input begins with the number n of test cases. Each test case consists of a single line of between 1 and 1000 lowercase letters, with no whitespace within.

Output

For each test case, output a line containing the minimum number of groups required to partition the input into groups of palindromes.

Sample Input

3

racecar

fastcar

aaadbccb

Sample Output

1

7

3

题目大意：问最少的回文串的总数。

思路：开始我想用manacher算法。只需要到时候处理一下就可以。但是还是想简单了。折腾了两个小时都没搞好。果断弃坑。ps：按理说应该是可以的，可能是我太马虎了。

赛后发现原来是dp，真是藏的好深啊。。。

因为题目数据是1000。时间限制3s。我们就用比较好理解的O(n3)的复杂度的方法吧。

dp[ i ]表示前i个字符能构成的最小回文串总数。

代码：

#include
     
      #include
      
       #include
       
        #include
        
         #define INF 999999using namespace std;char a[1005];bool IsHuiWen(int l,int r){    for(int j=l,i=r;j<=i;j++,i--){        if(a[j]!=a[i])            return false;    }    return true;}int main(){    int T;    cin>>T;    while(T--){        int dp[1005];        scanf("%s",a+1);        int len=strlen(a+1);        for(int i=1;i<=len;i++){            dp[i]=INF;        }        dp[0]=0;        for(int i=1;i<=len;i++){            for(int j=1;j<=i;j++){                if(IsHuiWen(j,i)){                    dp[i]=min(dp[i],dp[j-1]+1); //如果j到i是回文串，那么我们就可以利用dp[j-1]的最少回文串来知道dp[i]的最少回文串。                }            }        }        cout<
         
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