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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.
题意：
在n个位置填充1 or 2. 要求连续的1的个数不超过a，连续的2的个数不超过b. 求方案数.
思路：
用一个二维数组，来进行存储方案数，其实只需要第i个位置后连续的j（<a||<b）个位置填1或2,那么我们可以dp[i+j][1] 表示第i个位置填1，在i位置后面连续填j个2的方案个数 dp[i+j][2] 表示第i个位置填2，在i位置后面连续填j个1的方案个数 !!!!!
代码：
#include <bits/stdc++.h>
using namespace std;
const int N=1e6+2;
const int mod=1e9+7;
int main()
{long long dp[N][3];memset(dp,0,sizeof(dp));int i,j,a,b,n;cin>>n>>a>>b;dp[0][1]=dp[0][2]=1;for(i=0;i<=n;i++){for(j=1;j<=a&&i+j<=n;j++)dp[i+j][1]=(dp[i+j][1]+dp[i][2])%mod;for(j=1;j<=b&&i+j<=n;j++)dp[i+j][2]=(dp[i+j][2]+dp[i][1])%mod;}cout<<(dp[n][1]+dp[n][2])%mod<<endl;return 0;} 提问：：：
求哪个大神告诉我，为什么对它进行求余1e9+7就能AC？？？？好困惑，菜鸟一个
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