Codeforces 360C Levko and Strings dp (看题解）

#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 2000 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}

int n, m;
int dp[N][N];
int sum[N];
char s[N];

int main() {
    scanf("%d%d%s", &n, &m, s + 1);
    dp[0][0] = 1;
    sum[0] = 1;
    for(int i = 1; i <= n; i++) {
        for(int j = 0; j <= m; j++) {
            dp[i][j] = 1LL * sum[j] * (s[i] - ‘a‘) % mod;
            for(int k = i - 1; k >= 0 && (i - k) * (n - i + 1) <= j; k--)
                add(dp[i][j], 1LL * dp[k][j - (i - k) * (n - i + 1)] * (‘z‘ - s[i]) % mod);
            add(sum[j], dp[i][j]);
        }
    }
    int ans = 0;
    for(int i = 0; i <= n; i++) add(ans, dp[i][m]);
    printf("%d\n", ans);
    return 0;
}

/*
*/