2017-2018 ACM-ICPC Southeastern European Regional Programming Contest (SEERC 2017)

全靠 wxh的博客 补完这套。wxhtxdy！

$f[i][k]$ 表示在第 $i$ 个位置刚好匹配了 $k$ 个字符。转移方程 $$ f[i][k] = \sum_{i - j > h[s[k - 1]]} f[j][k - 1] $$

前缀和优化加滚动就行了。

好像可以直接用 $f[i][k]$ 表示前缀和，就没这么多事了。

#include <bits/stdc++.h>

const int N = 1e5 + 7;
const int MOD = 1e9 + 7;

int dp[2][N], sum[2][N], h[N];
char s[N], t[N];

void M(int &a) {
    if (a >= MOD) a -= MOD;
}

int main() {
    freopen("in.txt", "r", stdin);
    int k, n;
    scanf("%d%d", &k, &n);
    for (int i = 0; i < 26; i++)
        scanf("%d", h + i);
    scanf("%s%s", t + 1, s + 1);
    for (int i = 1; i <= n; i++) {
        if (s[i] == t[1])
            dp[1][i] = 1;
        sum[1][i] =  sum[1][i - 1] + dp[1][i];
    }
    for (int j = 2; j <= k; j++) {
        int cur = j & 1, pre = cur ^ 1;
        memset(dp[cur], 0, sizeof(dp[cur]));
        memset(sum[cur], 0, sizeof(sum[cur]));
        for (int i = 1; i <= n; i++) {
            if (s[i] == t[j]) {
                int where = i - h[t[j - 1] - 'A'] - 1;
                if (where > 0)
                    M(dp[cur][i] += sum[pre][where]);
            }
            //if (i == 2) printf("%d\n", dp[cur][i]);
            M(sum[cur][i] += sum[cur][i - 1]);
            M(sum[cur][i] += dp[cur][i]);
        }
    }
    printf("%d\n", sum[k & 1][n]);
    return 0;
}

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