Hdu2476string painter (interval DP)

Problem descriptionthere are two strings A and B with equal length. both strings are made up of lower case letters. now you have a powerful string painter. with the help of the painter, you can change a segment of characters of a string to any other character you want. that is, after using the painter, the segment is made up of only one kind of character. now your task is to change A to B using STR Ing painter. What's the minimum number of operations?
Inputinput contains multiple cases. Each case consists of two lines: the first line contains string a. The second line contains string B. The length of both strings will not be greater than 100.
Outputa single line contains one integer representing the answer.
Sample Input

zzzzzfzzzzzabcdefedcbaababababababcdcdcdcdcdcd

Sample output

67. Given two strings a and B, evaluate the minimum number of operations required for a to change a to B. During each operation, any part of a is converted into any letter.
#include<stdio.h>#include<string.h>int min(int a,int b){    return a>b?b:a;}int main(){    int dp[105][105],ans[105];    char a[105],b[105];    while(scanf("%s%s",a,b)>0)    {        int len=strlen(a);        memset(dp,0,sizeof(dp));        for(int r=0;r<len;r++)        for(int i=0;i<len-r;i++)        {            int j=i+r;            dp[i][j]=dp[i+1][j]+1;            for(int k=i+1;k<=j;k++)            if(b[i]==b[k])            dp[i][j]=min(dp[i][j],dp[i+1][k]+dp[k+1][j]);        }        for(int i=0;i<len; i++)        ans[i]=dp[0][i];        for(int i=0;i<len; i++)        if(a[i]==b[i])        {            if(i==0)ans[i]=0;            else ans[i]=ans[i-1];        }        else{            for(int k=0;k<i;k++)            ans[i]=min(ans[i],ans[k]+dp[k+1][i]);        }        printf("%d\n",ans[len-1]);    }}

Hdu2476string painter (interval DP)