2017-2018 ACM-ICPC Southeastern European(SEERC 2017) A concerts

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题目大意：字符串只由A~Z组成，给定s字符串长度k(k<=300)，p字符串长度是n(<=1e5)，再给定字母A~Z匹配成功后需要间隔的次数a[27]，再给出字符串s, p。求s匹配p的方案数

dp[i][j]:匹配到i位置时已经匹配了j个的方案数。
(1) s[j]==p[i]：dp[i][j] = dp[i-1][j] + dp[i-a[id]-1][j-1];
(2) s[j]!=p[i]：dp[i][j] = dp[i-1][j]；
喜提新技能:列出所有情况，找规律得转移方程

#include <cstdio>
#include <cstring>

using namespace std;
typedef long long ll;
const int mod = 1e9+7;
const int N = 1e5+10;
const int K = 310;
int dp[N][K];
char s[K], p[N];
int a[27];
int k, n;

int main()
{
    scanf("%d %d", &k, &n);
    for(int i = 0; i < 26; ++i)
        scanf("%d", &a[i]);
    scanf("%s", s+1);
    scanf("%s", p+1);
    memset(dp, 0, sizeof(dp));
    for(int i = 1; i <= n; ++i)
    {
        dp[i][1] = dp[i-1][1];
        if(s[1]==p[i]) ++dp[i][1];
        //printf("%d\n", sum[i][1]);
    }
    for(int i = 1; i <= n; ++i)
        for(int j = 2; j <= k; ++j)
        {
            dp[i][j] = dp[i-1][j];
            if(s[j]==p[i])
            {
                int id = s[j-1]-'A';
                if(i-a[id]-1<1) continue;
                dp[i][j] = (dp[i][j]+dp[i-a[id]-1][j-1])%mod;
            }
        }
    printf("%d\n", dp[n][k]);
    return 0;
}