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Pop-group “Pink elephant” entered on recording their debut album. In fact they have only two songs: “My love” and “I miss you”, but each of them has a large number of remixes.
The producer of the group said that the album should consist of n remixes. On second thoughts the musicians decided that the album will be of interest only if there are no more than a remixes on “My love” in a row and no more than b remixes on “I miss you” in a row. Otherwise, there is a risk that even the most devoted fans won’t listen to the disk up to the end.
How many different variants to record the album of interest from n remixes exist? A variant is a sequence of integers 1 and 2, where ones denote remixes on “My love” and twos denote remixes on “I miss you”. Two variants are considered different if for some i in one variant at i-th place stands one and in another variant at the same place stands two.
The only line contains integers n, a, b (1 ≤ a, b ≤ 300; max( a, b) + 1 ≤ n ≤ 50 000).
Output the number of different record variants modulo 10 9+7.
| input | output |
|---|---|
3 2 1 | 4 |
Notes
In the example there are the following record variants: 112, 121, 211, 212.
#include <iostream>
#include<cstring>
#include<cstdio>
using namespace std;const int maxn=50000+5;
const int INF=1e9+7;
int dp[maxn][5];
int main()
{int n,a,b;while(cin>>n>>a>>b){memset(dp,0,sizeof(dp));dp[0][1]=1;dp[0][2]=1;for(int i=0;i<=n;i++){for(int j=1; j+i<=n && j<=b; j++){dp[i+j][1] += dp[i][2];dp[i+j][1]%=INF;}for(int j=1; j+i <= n && j <= a; j++){dp[i+j][2]+=dp[i][1];dp[i+j][2]%=INF;}}cout<<((dp[n][1]+dp[n][2])%INF)<<endl;}return 0;
}
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