@[lwx20211103](/user/727008) `if (s[i - 1] == t[i - 1])`->`if (s[i - 1] == t[j - 1])`
and you can get an [AC](https://www.luogu.com.cn/record/98044442).
by Jerrlee✅ @ 2022-12-26 14:55:09
@[lwx20211103](/user/727008) 看楼上
by Jim_Franklin @ 2022-12-26 15:04:52
@[Jerrlee✅](/user/367652) thx
by lwx20211103 @ 2022-12-26 15:33:30
你的策略和我的不太一样，我是把它反过来，和原来的字符串求最长相同子序列，给你一份代码
```C++
#include<bits/stdc++.h>
using namespace std;
char s1[1005],s2[1005];
int dp[1005][1005];
int main(void)
{
cin>>s1+1;
s1[0]=' ';
int len=strlen(s1)-1;
for(int i=1,j=len;i<=len;i++,j--)s2[j]=s1[i];
for(int i=1;i<=len;i++)
{
for(int j=1;j<=len;j++)
{
if(s1[i]==s2[j])dp[i][j]=dp[i-1][j-1]+1;
else dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
}
}
cout<<len-dp[len][len];
return 0;
}
```
by 凤凰工作室 @ 2023-01-04 16:31:40