CodeCraft-19 and Codeforces Round #537 (Div. 2)

Problem  CodeCraft-19 and Codeforces Round #537 (Div. 2) - D. Destroy the Colony

Time Limit: 2000 mSec

Problem Description

Input

 

Output

 For each question output the number of arrangements possible modulo 10^9+7.

Sample Input

abba
2
1 4
1 2

Sample Output

2
0

题解：这个题的每一部分都比较简单，加在一起就有难度了。关于组合数学的部分就不说了，比较简单，这个题的精华在于dp的部分，对于每次的询问，相当于强制让两种字符不在左边（或者不在右边），也就是在原始的dp递推方程中要跳过这两个字符的方案数，如果每次都重新递推，递推的复杂度是O(n * k)，一共有k^2个组合，复杂度会炸，这时就要考虑利用之前dp的信息来减少递推的计算量，新的tmp_dp[j]的状态定义长度为j的，没有第i种字符的方案数，这其实就是长度为j的总方案数减去强制有第i种字符的方案数，而强制有i的方案数就应该是没有i的长度为j - cnt[i]的方案数，这恰好就是tmp_dp[j - cnt[i]]，因此tmp[j] = dp[j] - tmp_dp[j - cnt[i]]，可以在O(n)的时间内完成递推，一共有k^2个组合，因此复杂度时O(n * k^2)，可以接受。

#include <bits/stdc++.h>

using namespace std;

#define REP(i, n) for (int i = 1; i <= (n); i++)
#define sqr(x) ((x) * (x))

const int maxn = 100000 + 10;
const int maxm = 200000 + 100;
const int maxs = 52;

typedef long long LL;
typedef pair<int, int> pii;
typedef pair<double, double> pdd;

const LL unit = 1LL;
const int INF = 0x3f3f3f3f;
const LL Inf = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-14;
const double inf = 1e15;
const double pi = acos(-1.0);
const LL mod = 1000000007;

inline int cal(char ch)
{
    if (isupper(ch))
        return ch - 'A' + 26;
    return ch - 'a';
}

LL pow_mod(LL x, LL n, LL mod)
{
    LL ans = 1, base = x;
    while (n)
    {
        if (n & 1)
            ans = (ans * base) % mod;
        base = (base * base) % mod;
        n >>= 1;
    }
    return ans;
}

void Add(LL &a, LL b)
{
    a += b;
    if (a > mod)
        a -= mod;
}

void Sub(LL &a, LL b)
{
    a -= b;
    if (a < 0)
        a += mod;
}

string str;
LL n;
LL cnt[maxs];
LL fact[maxn], inv[maxn];
LL dp[maxn], tmp_dp[maxn];
LL ans[maxs][maxs];

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    cin >> str;
    n = str.size();
    for (int i = 0; i < n; i++)
    {
        cnt[cal(str[i])]++;
    }
    fact[0] = fact[1] = 1;
    for (LL i = 2; i <= n; i++)
    {
        fact[i] = (fact[i - 1] * i) % mod;
    }
    inv[n] = pow_mod(fact[n], mod - 2, mod);
    for (LL i = n - 1; i >= 0; i--)
    {
        inv[i] = (inv[i + 1] * (i + 1)) % mod;
    }
    LL num = (fact[n / 2] * fact[n / 2]) % mod;
    for (int i = 0; i < maxs; i++)
    {
        num = (num * inv[cnt[i]]) % mod;
    }
    dp[0] = 1;
    for (int i = 0; i < maxs; i++)
    {
        if (!cnt[i])
            continue;
        for (int j = n; j >= cnt[i]; j--)
        {
            Add(dp[j], dp[j - cnt[i]]);
        }
    }
    for (int i = 0; i < maxs; i++)
    {
        ans[i][i] = dp[n / 2];
    }
    for (int i = 0; i < maxs; i++)
    {
        if (!cnt[i])
            continue;
        for (int i = 0; i <= n; i++)
            tmp_dp[i] = dp[i];
        for (int t = cnt[i]; t <= n; t++)
        {
            Sub(tmp_dp[t], tmp_dp[t - cnt[i]]);
        }
        for (int j = i + 1; j < maxs; j++)
        {
            if (!cnt[j])
                continue;
            for (int t = cnt[j]; t <= n; t++)
            {
                Sub(tmp_dp[t], tmp_dp[t - cnt[j]]);
                ans[i][j] = (2LL * tmp_dp[n / 2]) % mod;
                ans[j][i] = ans[i][j];
            }
            for (int t = n; t >= cnt[j]; t--)
                Add(tmp_dp[t], tmp_dp[t - cnt[j]]);
        }
    }
    int q;
    cin >> q;
    int x, y;
    while (q--)
    {
        cin >> x >> y;
        int l = cal(str[x - 1]);
        int r = cal(str[y - 1]);
        cout << (num * ans[l][r]) % mod << "
";
    }
    return 0;
}