Bzoj 2764 [jloi2011] gene completion

Question: http://www.lydsy.com/JudgeOnline/problem.php? Id = 2764

Given a base sequence s with a length of N and a base sequence T with a length of M, we hope to add a certain base to the sequence T to pair the sequence s with the sequence T, the matching rule is:AAndTPairing,CAndGThe number and location of the base to be added are different. 0 <m ≤ n ≤ 2000 .

Question:
We can consider calculating the number of different solutions matching sequence T in sequence S. Using DP can easily solve this problem.
Ling F [I] [J] Indicates the number of schemes from t to J in the first I-bit matching sequence of sequence S. F [I] [J] If not S [I] Match T [J] , Is F [I? 1] [J] . F [I? 1] [J? 1] The final answer is F [N] [m] , DP can be rolled.
Roughly estimate the upper limit of the answer. You can find that the answer exceeds 264 But cannot exceed C (2000,1000) Or smaller, DP can be directly coupled with high precision.

Code:

#include <cstdio>const int maxn = 2001, maxl = 70, mod = 1000000000;struct BigInt{    int len, num[maxl];    void getint(const int &x) { num[len++] = x; }    void Print()    {        printf("%d", num[len - 1]);        for(int i = len - 2; i >= 0; --i)            printf("%9.9d", num[i]);        putchar(‘\n‘);    }    void operator += (const BigInt &x)    {        if(len < x.len) len = x.len;        for(int i = 0; i < len; ++i)        {            num[i] += x.num[i];            if(num[i] >= mod)            {                num[i] -= mod;                ++num[i + 1];            }        }        if(num[len]) ++len;    }} f[maxn];inline bool check(char a, char b){    return a == ‘A‘ && b == ‘T‘ || a == ‘G‘ && b == ‘C‘ || a == ‘C‘ && b == ‘G‘ || a == ‘T‘ && b == ‘A‘;}int n, m;char s[maxn], t[maxn];int main(){    scanf("%d%d%s%s", &n, &m, s, t);    f[0].getint(1);    for(int i = 1; i <= n; ++i)        for(int j = m; j; --j)            if(check(s[i - 1], t[j - 1])) f[j] += f[j - 1];    f[m].Print();    return 0;}

Bzoj 2764 [jloi2011] gene completion