Zoj 3791 an easy game [combined count]

Question address:

Http://acm.zju.edu.cn/onlinejudge/showProblem.do? Problemcode = 3791.

Description:

Given two 01 string S1 and S2 with a length of N, the K step is required. Each step is used to reverse the number of M positions in S1 (that is, 0 is changed to 0 ), ask how many methods can be used to convert S1 to S2.

1 <= n <= 100, 0 <= k <= 100, 0 <= m <= n.

Solution:

Typical combination counting problems.

1. First, compare S1 and S2. Note that there are N1 different positions in S1 and S2, and N2 in the same digital position.
2. Count. If S1 and S2 have different odd locations after step 1, the even locations are the same (odd + even = N) there are a [k1] [odd] methods. Retrieve I from odd, and retrieve J from even to reverse (I + J = m ), the method can be C [odd] [I] * C [even] [J. It is not difficult to see the counting equation:

A [k1 + 1] [Odd-I + J] + = A [k1] [odd] * C [odd] [I] * C [even] [J];

1. The final result is a [k] [0] (because there is no location digital difference)

Source code:

 

 1 //zoj3791_zzy_2014.07.19_AC_combinatorial counting 2 #include <iostream> 3  4 using namespace std; 5  6 typedef long long LL; 7  8 class EasyGame 9 {10 public:11     EasyGame(LL N, LL K, LL M): n(N), k(K), m(M) {12         ans = 0;13         ini();14         for(int i = 0; i < 2; ++i)15             for(int j = 0; j <= n; ++j)16                 a[i][j] = 0;17     }18     void getString();19     void solve();20     int getAns();21 private:22     static const LL mod = 1000000009;23     LL n, k, m, n1, n2, ans;24     string s1, s2;25     LL C[101][101];26     LL a[2][101];27     void ini();28 };29 30 void EasyGame::ini()31 {32     C[0][0] = 1;33     C[1][0] = 1;34     C[1][1] = 1;35     for(int i = 2; i <= n; ++i) {36         C[i][0] = 1;37         C[i][i] = 1;38         for(int j = 1; j < i; ++j) {39             C[i][j] = (C[i-1][j] + C[i-1][j-1]) % mod;40         }41     }42 }43 44 void EasyGame::getString()45 {46     cin >> s1;47     cin >> s2;48     n1 = n2 = 0;49     for(int idx = 0; idx < s1.size(); ++idx) {50         if(s1[idx] != s2[idx])51             n1++;52         else53             n2++;54     }55 }56 57 void EasyGame::solve()58 {59     a[0][n1] = 1;60     for(int step = 1; step <= k; ++step) {61         for(int idx = 0; idx <= n; ++idx)62             a[step % 2][idx] = 0;63         for(int odd = 0; odd <= n; ++odd) {64             int even = n - odd;65             for(int i = 0; i <= m; ++i) {66                 int j = m - i;67                 if(i > odd) break;68                 if(j > even) continue;69                 int odd2 = odd - i + j;70                 a[step % 2][odd2] += a[(step - 1) % 2][odd] * (C[odd][i] * C[even][j] % mod) % mod;71                 a[step % 2][odd2] %= mod;72             }73         }74     }75     ans = a[k % 2][0];76 }77 78 int EasyGame::getAns()79 {80     return ans;81 }82 83 int main()84 {85     LL N, K, M;86     while(cin >> N >> K >>M)87     {88         EasyGame *egp = new EasyGame(N, K, M);89         egp -> getString();90         egp -> solve();91         cout << egp -> getAns() << endl;92     }93     return 0;94 }