首页 > 代码库 > NYOJ-43 24 Point game-DFS
NYOJ-43 24 Point game-DFS
24 Point game
时间限制:3000 ms | 内存限制:65535 KB
难度:5
- 描述
There is a game which is called 24 Point game.
In this game , you will be given some numbers. Your task is to find an expression which have all the given numbers and the value of the expression should be 24 .The expression mustn‘t have any other operator except plus,minus,multiply,divide and the brackets.
e.g. If the numbers you are given is "3 3 8 8", you can give "8/(3-8/3)" as an answer. All the numbers should be used and the bracktes can be nested.
Your task in this problem is only to judge whether the given numbers can be used to find a expression whose value is the given number。
- 输入
- The input has multicases and each case contains one line
The first line of the input is an non-negative integer C(C<=100),which indicates the number of the cases.
Each line has some integers,the first integer M(0<=M<=5) is the total number of the given numbers to consist the expression,the second integers N(0<=N<=100) is the number which the value of the expression should be.
Then,the followed M integer is the given numbers. All the given numbers is non-negative and less than 100 - 输出
- For each test-cases,output "Yes" if there is an expression which fit all the demands,otherwise output "No" instead.
- 样例输入
2 4 24 3 3 8 8 3 24 8 3 3
- 样例输出
Yes No
#include<cstdio> #include<cstring> #include<cmath> #include<cstdlib> #include<iostream> #include<algorithm> #include<queue> #include<stack> //#define MIN 1e-10 using namespace std; const double MIN=1e-10;//const double,习惯性的写成const int 导致搞了一个晚上 double a[10]; int n; double num; bool DFS(int x) { if(x==n) { if(fabs(num-a[n])<MIN) return true; return false; } double now,next; for(int i=x;i<n;i++) { for(int j=i+1;j<=n;j++) { now=a[i]; next=a[j]; a[i]=a[x]; a[j]=now+next;if(DFS(x+1)) return true; a[j]=now-next;if(DFS(x+1)) return true; a[j]=next-now;if(DFS(x+1)) return true; a[j]=now*next;if(DFS(x+1)) return true; if(next) //分母不为 0 { a[j]=now/next;if(DFS(x+1)) return true; } if(now) //分母不为 0 { a[j]=next/now;if(DFS(x+1)) return true; } a[i]=now; a[j]=next; } } return false; } int main() { int T; while(~scanf("%d",&T)) { while(T--) { memset(a,0,sizeof(a)); scanf("%d %lf",&n,&num); for(int i=1;i<=n;i++) { scanf("%lf",&a[i]); } if(DFS(1)) cout<<"Yes"<<endl; else cout<<"No"<<endl; } } return 0; }
NYOJ-43 24 Point game-DFS
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。