Educational Codeforces Round 11 E. Different Subsets For All Tuples 动态规划

E. Different Subsets For All Tuples

题目连接：

http://www.codeforces.com/contest/660/problem/E

Description

For a sequence a of n integers between 1 and m, inclusive, denote f(a) as the number of distinct subsequences of a (including the empty subsequence).

You are given two positive integers n and m. Let S be the set of all sequences of length n consisting of numbers from 1 to m. Compute the sum f(a) over all a in S modulo 109 + 7.

Input

The only line contains two integers n and m (1 ≤ n, m ≤ 106) — the number of elements in arrays and the upper bound for elements.

　　

Output

Print the only integer c — the desired sum modulo 109 + 7.

Sample Input

1 3

Sample Output

6

Hint

题意

现在定义f(a)表示这个a串里面所有不相同的子序列的个数

现在给你n,m，让你用字符集为m，去构造出长度为n的串

然后让你算出所有f(a)的累加

题解：

考虑dp

dp[i][j]表示长度为i,以字符j结尾的答案是多少

dp[i][j]=sigma(dp[i-1][k]*2-dp[pre[j]-1][k])

然后这个玩意儿显然对于任意的j的都是一样的，而且pre[j]前面的每个位置都是可能的，这里的dp是个前缀和，所以直接扣除就可以了

那么直接化简为：dp[i]=dp[i-1]*(2m-1)

但是这个dp是没有考虑空串的

那么在加上空串就好了，所以答案就是

dp[i] = dp[i-1]*(2m-1)+m^(i-1)

代码

#include<bits/stdc++.h>
using namespace std;
const int mod = 1e9+7;

int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    long long ans = 2*m;
    long long tmp = 1;
    for(int i=2;i<=n;i++)
    {
        tmp = tmp * m % mod;
        ans = (ans * (2 * m - 1) % mod + tmp + mod) % mod;
    }
    cout<<ans<<endl;
}