Feed the monkey_dp_递推

最新推荐文章于 2018-05-05 08:45:37 发布

原创最新推荐文章于 2018-05-05 08:45:37 发布 · 316 阅读

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本文介绍了一个经典的编程竞赛问题——喂猴策略问题。该问题要求在考虑猴子对不同水果连续食用天数限制的情况下，计算出所有可能的喂食方案数量。通过动态规划的方法，有效地解决了这一问题，并提供了一个具体的代码实现案例。

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Feed the monkey

Time Limit: 2000 msMemory Limit: 131072 KiB

Problem Description

Alice has a monkey, she must feed fruit to the monkey every day.She has three kinds of fruits, bananas,
peaches and apples. Every day, she chooses one in three, and pick one of this to feed the monkey.
But the monkey is picky, it doesn’t want bananas for more than D1 consecutive days, peaches for more than D2
consecutive days, or apples for more than D3 consecutive days. Now Alice has N1 bananas, N2 peaches and N3
apples, please help her calculate the number of schemes to feed the monkey.

Input

Multiple test cases. The first line contains an integer T (T<=20), indicating the number of test case.
Each test case is a line containing 6 integers N1, N2, N3, D1, D2, D3 (N1, N2, N3, D1, D2, D3<=50).

Output

One line per case. The number of schemes to feed the monkey during (N1+N2+N3) days.
The answer is too large so you should mod 1000000007.

Sample Input

1
2 1 1 1 1 1

Sample Output

Hint

Answers are BPBA, BPAB, BABP, BAPB, PBAB, and ABPB(B-banana P-peach A-apple)

Source

“浪潮杯”山东省第七届ACM大学生程序设计竞赛

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

const int N=200;
ll dp[N][N][N][3];

int n1,n2,n3,d1,d2,d3;


int main()
{
    //freopen("in.txt","r",stdin);
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin>>t;
    while(t--){
        cin>>n1>>n2>>n3>>d1>>d2>>d3;
        memset(dp,0,sizeof(dp));

        for(int i=n1;i>=0;i--)
        for(int j=n2;j>=0;j--)
        for(int k=n3;k>=0;k--){
            for(int s=1;s<=min(d1,i);s++){
                if(i==n1&&j==n2&&k==n3){
                    dp[i-s][j][k][0]=1;
                }else{
                    dp[i-s][j][k][0]=((dp[i-s][j][k][0]+dp[i][j][k][1])%mod+dp[i][j][k][2])%mod;
                }
            }
            for(int s=1;s<=min(d2,j);s++){
                if(i==n1&&j==n2&&k==n3){
                    dp[i][j-s][k][1]=1;
                }else{
                    dp[i][j-s][k][1]=((dp[i][j-s][k][1]+dp[i][j][k][0])%mod+dp[i][j][k][2])%mod;
                }
            }
            for(int s=1;s<=min(d3,k);s++){
                if(i==n1&&j==n2&&k==n3){
                    dp[i][j][k-s][2]=1;
                }else{
                    dp[i][j][k-s][2]=((dp[i][j][k-s][2]+dp[i][j][k][1])%mod+dp[i][j][k][0])%mod;
                }
            }
        }

        ll ans=((dp[0][0][0][0]+dp[0][0][0][1])%mod+dp[0][0][0][2])%mod;
        cout<<ans<<endl;
    }
    return 0;
}