BNU34058——干了这桶冰红茶！——————【递推】

干了这桶冰红茶！

Time Limit: 1000ms

Memory Limit: 65536KB

64-bit integer IO format: %lld      Java class name: Main

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BNUCIST的HWQ大神特别钟爱冰红茶这种神棍的饮料，有一天打Dota暴虐他寝室的WL后，决定大喝一顿庆祝一下。他决定用一种神棍的方式来喝冰红茶，那就是每口只喝1升，或者2升，或者3升（PS：HWQ大神真的能喝这么多= =）。爱思考的HWQ突然想知道，对于一桶整数升的冰红茶，他可以有多少种方案喝光，但似乎他不能马上想出解决的办法，纠结的他不知道答案他就喝不下去了。聪明的你快帮帮他吧。

 

Input

    输入一个整数T,代表数据组数。

    对于每一组数据，输入一个整数N，1<=N<=30，表示这桶冰红茶有N升。

 

Output

对于每个N，输出一个整数，代表方案数。

 

Sample Input

1
3

Sample Output

4

Hint

对于样例，3升的冰红茶，他可以（1）每次喝1升，连喝3口；（2）第一口喝1升，第二口喝2升；（3）第一口喝2升，第二口喝1升；（4）一口就喝掉3升。所以共有4种方案。

 

Source

2013年北京师范大学新生程序设计竞赛

 

 

解题思路：通过前面的递推出后边的。对于任意n可分为{{1，n-1}，{2，n-2}，{3，n-3}}，所以sum[i]=sum[i-1]+sum[i-2]+sum[i-3]；

 

#include<bits/stdc++.h>
using namespace std;
int sum[50];
int main(){
    int T;
    scanf("%d",&T);
    sum[0]=1;sum[1]=1;sum[2]=2;
    for(int i=3;i<32;i++){
        sum[i]=sum[i-1]+sum[i-2]+sum[i-3];
    }
    while(T--){

        int n;
        scanf("%d",&n);
        printf("%d\n",sum[n]);
    }
    return 0;
}