简单的dp 但是一个大数加法  套用了末位大牛的类模板

#include 
    
     #include 
     
      #include 
      
       #define maxs 10005#define maxn 110using namespace std;char X[maxs], Z[maxn];struct bign{    int len, s[maxn];    bign operator = (const char *num)    {        len = strlen(num);        for (int i = 0; i < len; i++)            s[i] = num[len-i-1]-'0';        return *this;    }    bign operator + (const bign &b) const    {        bign c; c.len = 0;        for (int i = 0, g = 0; g || i < max(len, b.len); i++)        {            int x = g;            if (i < len) x += s[i];            if (i < b.len) x += b.s[i];            c.s[c.len++] = x%10;            g = x/10;        }        return c;    }    bign operator += (const bign &b)    {        *this = *this + b;        return *this;    }    void print()    {        for (int i = len-1; i >= 0; i--)            printf("%d", s[i]);    }};bign dp[maxn][maxs];int main(){    int T;    scanf("%d", &T);    while (T--)    {        scanf("%s%s", X+1, Z+1);        int len1 = strlen(X+1), len2 = strlen(Z+1);        for (int i = 0; i <= len1; i++)            dp[0][i] = "1";        for (int i = 1; i <= len2; i++)            for (int j = i; j <= len1; j++)            {                dp[i][j] = dp[i][j-1];                if (Z[i] == X[j])                    dp[i][j] += dp[i-1][j-1];            }        dp[len2][len1].print();//printf("%d",dp[len2][len1]);        puts("");    }    return 0;}