转载请注明出处: http://www.cnblogs.com/fraud/          ——by fraud

Just A String

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 320    Accepted Submission(s): 62

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≤n,m≤2000).

 



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



Sample Input
3
2 2
3 2
10 3
 
Sample Output
10
40
1908021

当时状态真是见鬼,其实这题还是比较容易的一个dp

dp[i][j]表示长度为i时,j种字符是奇数个的字符串种数

从而dp[i][j] = dp[i-1][j+1]*(j+1) + dp[i-1][j-1]*(m-j+1)

最后Σdp[i][i&1]*(n-i+1)*(m^(n-i))

 /**
* code generated by JHelper
* More info: https://github.com/AlexeyDmitriev/JHelper
* @author xyiyy @https://github.com/xyiyy
*/ #include <iostream>
#include <fstream> //#####################
//Author:fraud
//Blog: http://www.cnblogs.com/fraud/
//#####################
//#pragma comment(linker, "/STACK:102400000,102400000")
#include <iostream>
#include <sstream>
#include <ios>
#include <iomanip>
#include <functional>
#include <algorithm>
#include <vector>
#include <string>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <set>
#include <map>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <climits>
#include <cctype> using namespace std;
#define rep2(X, L, R) for(int X=L;X<=R;X++)
typedef long long ll; int dp[][];
int dp2[];
const int mod = ; class hdu5362 {
public:
void solve(std::istream &in, std::ostream &out) {
int n, m;
in >> n >> m;
dp[][] = ;
rep2(i, , n) {
for (int j = (i & ); j <= m && j <= i; j++) {
if (!j)dp[i][j] = dp[i - ][j + ];
else if (j == i || j == m)dp[i][j] = (ll) dp[i - ][j - ] * (m - j + ) % mod;
else dp[i][j] = ((ll) dp[i - ][j - ] * (m - j + ) + (ll) dp[i - ][j + ] * (j + )) % mod;
}
}
dp2[] = ;
rep2(i, , n) {
dp2[i] = (ll) dp2[i - ] * m % mod;
}
int ans = ;
rep2(i, , n) {
ans = (ans + (ll) dp[i][i & ] * (n - i + ) % mod * dp2[n - i]) % mod;
}
out << ans << endl;
}
}; int main() {
std::ios::sync_with_stdio(false);
std::cin.tie();
hdu5362 solver;
std::istream &in(std::cin);
std::ostream &out(std::cout);
int n;
in >> n;
for (int i = ; i < n; ++i) {
solver.solve(in, out);
} return ;
}

hdu5362 Just A String(dp)的更多相关文章

  1. codeforces 710E E. Generate a String(dp)

    题目链接: E. Generate a String time limit per test 2 seconds memory limit per test 512 megabytes input s ...

  2. bnuoj 34985 Elegant String DP+矩阵快速幂

    题目链接:http://acm.bnu.edu.cn/bnuoj/problem_show.php?pid=34985 We define a kind of strings as elegant s ...

  3. CodeForces 710E Generate a String (DP)

    题意:给定 n,x,y,表示你要建立一个长度为 n的字符串,如果你加一个字符要花费 x时间,如果你复制前面的字符要花费y时间,问你最小时间. 析:这个题,很明显的DP,dp[i]表示长度为 i 的字符 ...

  4. LeetCode-Interleaving String[dp]

    Interleaving String Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2. Fo ...

  5. PAT 1040 Longest Symmetric String[dp][难]

    1040 Longest Symmetric String (25)(25 分) Given a string, you are supposed to output the length of th ...

  6. 87. Scramble String (String; DP)

    Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrin ...

  7. 115. Distinct Subsequences (String; DP)

    Given a string S and a string T, count the number of distinct subsequences of T in S. A subsequence ...

  8. 72. Edit Distance (String; DP)

    Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2 ...

  9. 97. Interleaving String (String; DP)

    Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2. For example,Given:s1 = ...

随机推荐

  1. switch case加条件语句(非等值) php

    <?php $s_level=rand(1,6); echo $s_level.'<br/>'; switch(true){ case $s_level<3 : echo 'l ...

  2. C++是怎么实现多态性的

    C++是怎么实现多态性的,C++中多态实现的原理, 当一个类中有虚函数时,系统会为该类构造一个虚函数表vtable,他是一个指针数组,存放每个虚函数的入口地址,编译器还会在此类中隐含插入一个指针vpt ...

  3. Jasper_filter data_pass field data from main to sub to filter some data

    main report: 1 add variable <variable name="Variable_rule" class="java.lang.String ...

  4. arm linux上的第一个应用程序 BOA移植

    1. 首先, linux在开发板上能跑起来了. 包括网络驱动也有了, ifconfig之后, 能看到在rcS里面设置的IP, 也能ping通windows主机了, 当然, 也要window关掉防火墙才 ...

  5. Facebook和Google如何激发工程师的创造力

    http://taiwen.lofter.com/post/664ff_ad8a15 今天终于“朝圣”了两个伟大的公司——Facebook和Google,对创造力和驱动力的来源有了更多的理解,尤其是对 ...

  6. 转:BZERO()等的区别

    BZERO()等的区别 bzero  原型: extern void bzero(void *s, int n); 用法: #include <string.h> 功能:置字节字符串s的前 ...

  7. 转:AM335X 启动流程

    链接: http://blog.csdn.net/hudaweikevin/article/details/10376585  作者:David_Hu 启动顺序(针对TI OMA3 EVM) linu ...

  8. css 选择器 (学习笔记)

    参考 http://zachary-guo.iteye.com/blog/605116 1. div+p  选择紧接在 <div> 元素之后的所有 <p> 元素.解释 : fi ...

  9. POSTGRESQL小玩

    因为CDH上需要用它来建HIVE的元库... 参考: http://www.cnblogs.com/mchina/archive/2012/06/06/2539003.html 一.简介 Postgr ...

  10. QT下int与QByteArray的转换

    int转QByteArray QByteArray intToByte(int i) { QByteArray abyte0; abyte0.resize(4); abyte0[0] = (uchar ...