hdu6279 Circular Coloring

Circular Coloring

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others)
Total Submission(s): 28    Accepted Submission(s): 10

Problem Description

Bobo considers  (n+m) balls arranged in a circle.
The balls are numbered with  0,1,…,(n+m−1) where the ball  i and the ball  (i+1)mod(n+m) are adjacent.

Bobo would like to color  n of his balls black and  m of his balls white.
Bobo groups adjacent balls with same colors, and he determines the weight of the coloring as the product of the lengths of groups.

He would like to know the sum of the weight of the possible colorings, modulo  (109+7).

 

Input

The input consists of several test cases and is terminated by end-of-file.

Each test case contains two integers  n and  m.

 

Output

For each test case, print an integer which denotes the result.

## Constraint

*  1≤n,m≤5000
* The number of test cases does not exceed  5000.

 

Sample Input

1 22 35000 5000

 

Sample Output

640975597525

Hint

For the second sample, there are $10$ possible colorings (listed below).The number followed is the corresponding weight.* `BBWWW` ($6$)* `BWBWW` ($2$)* `BWWBW` ($2$)* `BWWWB` ($6$)* `WBBWW` ($6$)* `WBWBW` ($2$)* `WBWWB` ($2$)* `WWBBW` ($6$)* `WWBWB` ($2$)* `WWWBB` ($6$)

 

Source

CCPC2018-湖南全国邀请赛-重现赛（感谢湘潭大学）

 

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解法：dp，注意将（头部和尾部）从整体中分开来讨论，则剩下部分为一条链,用两次前缀和优化即可。

以第一维为起始颜色，now[0]相当于以第一维的颜色结束，now[1]相当于以第二维的颜色结束。

#include"bits/stdc++.h"
using namespace std;
const int mod = 1e9+7;
const int MX = 5002;
typedef long long LL;
int dp[MX][MX],now[2],si1[MX],si2[MX],sj1[MX],sj2[MX];
void init()
{
    int n = 5000, m = 5000;
    for(int i = 0; i <= n; i++)
        dp[i][0] = dp[0][i]= sj1[i] = sj2[i] = i;
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= m; j++){
            now[0] = si2[j]%mod;
            now[1] = sj2[i]%mod;
            dp[i][j] = now[0];

            (si1[j] += now[1]) %= mod;
            (sj1[i] += now[0]) %= mod;
            
            (si2[j] += si1[j]) %= mod;
            (sj2[i] += sj1[i]) %= mod;
        }
    }
}

LL work(int n, int m){
    LL ans = 0;
    for(int i = 1; i <= n; i++)
        (ans += (LL) i*i*dp[m][n-i]) %= mod;
    for(int j = 1; j <= m; j++)
        (ans += (LL) j*j*dp[n][m-j]) %= mod;
    cout<