BNU34058——干了这桶冰红茶!——————【递推】
干了这桶冰红茶!
64-bit integer IO format: %lld Java class name: Main
None
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Tag it!
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
#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;
}
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