2015 Multi-University Training Contest 6 hdu 5362 Just A String

Just A String

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 643    Accepted Submission(s): 182

Problem Description
soda has a random string of length n which is generated by the following algorithm: each of n characters of the string is equiprobably chosen from the alphabet of size m.

For a string s, if we can reorder the letters in string s so as to get a palindrome, then we call s a good string.

soda wants to know the expected number of good substrings in the random string.

Input
There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

The first line contains two integers $n and m (1 \leq n,m \leq 2000)$.

Output
For each case, if the expected number is E, a single integer denotes$ E\dot mn mod 1000000007$.

Sample Input
3
2 2
3 2
10 3

Sample Output
10
40
1908021

Author
zimpha@zju

 

Source

2015 Multi-University Training Contest 6 1010

 

解题：动态规划

 

吗各级，T了一下午

 

 dp[i][j] 表示长度为i的有j种字母是奇数个的串的个数

 

dp[i][j]可以有两种方向转移过来

一种是dp[i-1][j-1]选那种个数是偶数的字符 既然有j-1种是奇数，那么剩下的 m - j + 1的种数的个数都是偶数，增加其中一个，就多出一种个数是奇数的种数，偶数的选择方式有m - j + 1种

 

另一种转移方向是 dp[i-1][j+1] 从j + 1这些个数是奇数的种数中选择任一一个，增加这种一个，就会少个奇数个数的种数，可以发现有j + 1种选择方式

 

#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;
const int maxn = 2002;
const int mod = 1000000007;
long long dp[maxn][maxn],PM[maxn];
int main() {
    PM[0] = dp[1][0] = 1;
    int kase,n,m;
    scanf("%d",&kase);
    while(kase--) {
        scanf("%d%d",&n,&m);
        dp[1][1] = m;
        for(int i = 1; i <= n; ++i) PM[i] = PM[i-1]*m%mod;
        for(int i = 2; i <= n; ++i) {
            for(int j = 0, k = min(i,m); j <= k; ++j) {
                dp[i][j] = 0;
                if(j) dp[i][j] += dp[i-1][j-1]*(m - j + 1);
                if(j + 1 <= min(i - 1,k)) dp[i][j] += dp[i-1][j+1]*(j + 1);
                dp[i][j] %= mod;
            }
        }
        long long ret = 0;
        for(int i = 1; i <= n; ++i)
            ret += dp[i][i&1]*(n - i + 1)%mod*PM[n-i]%mod;
        printf("%I64d\n",ret%mod);
    }
    return 0;
}