P2679 子串

#include <bits/stdc++.h>
#define read read()
#define up(i,l,r) for(register int i = (l);i <= (r);i++)
#define down(i,l,r) for(register int i = (l);i >= (r);i--)
#define traversal_vedge(i) for(register int i = head[u]; i ;i = e[i].nxt)
#define ll long long
using namespace std;
int read
{
    int x = 0, f = 1; char ch = getchar();
    while(ch < 48 || ch > 57) {if(ch == '-')f = -1; ch = getchar();}
    while(ch >=48 && ch <=57) {x = 10 * x + ch - 48;ch = getchar();}
    return x * f; 
}
const int mod = 1000000007;
int n,m,maxk,f[2][205][205][2];
char a[1005],b[205];
bool c = 0;

int main()
{
    freopen("input.txt","r",stdin);
    n = read; m = read; maxk = read;
    scanf("%s",a+1);
    scanf("%s",b+1);
    f[0][0][0][0] = f[1][0][0][0] = 1;// 两行0初始化为1; 
    up(i,1,n)
    {
        c ^=1;
        up(j,1,m)
            up(k,1,maxk)
            {
                if(a[i] == b[j])
                {
                    
                    f[c][j][k][0] = (f[c^1][j][k][1] + f[c^1][j][k][0])%mod;
                    f[c][j][k][1] = ((f[c^1][j-1][k][1] + 
                    f[c^1][j-1][k-1][0])%mod + f[c^1][j-1][k-1][1])%mod; 
                }
                else
                {
                    f[c][j][k][0] = (f[c^1][j][k][1] + f[c^1][j][k][0])%mod;
                    f[c][j][k][1] = 0;
                }
            }
    }
    printf("%d\n",(f[n&1][m][maxk][0] + f[n&1][m][maxk][1])%mod);
    return 0;
}