nyoj 43 24 Point game 【dfs&&递归】
2015-09-16 18:42
429 查看
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
来源经典改编上传者张云聪
分析:
这道题是一个不错的题,运用深搜加递归,佷6666666666.就是把每一种可能暴力枚举出来 如果能算出24则返回1.
代码:
#include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<cmath> #define e 1e-6 using namespace std; double num[7],v; int n; int dfs(int t) { if(t==n) { if(fabs(v-num )<e) return 1; return 0; } double left,right; for(int i=t;i<n;i++) for(int j=i+1;j<=n;j++) { left=num[i]; right=num[j]; num[i]=num[t]; num[j]=left+right; if(dfs(t+1)) return 1; num[j]=left-right; if(dfs(t+1)) return 1; num[j]=left*right; if(dfs(t+1)) return 1; num[j]=right-left; if(dfs(t+1)) return true; if(left) num[j]=right/left;if(dfs(t+1)) return 1; if(right) num[j]=left/right; if(dfs(t+1)) return 1; num[i]=left; num[j]=right; } return 0; } int main() { int T; scanf("%d",&T); while(T--) { scanf("%d%lf",&n,&v); for(int i=1;i<=n;i++) scanf("%lf",&num[i]); if(dfs(1)) printf("Yes\n"); else printf("No\n"); } return 0; }
相关文章推荐
- 论文研读《JOTS: Joint Online Tracking and Segmentation》
- svn 125005 问题
- Android动画效果——X、Y轴抖动
- IBM jni 专栏
- java SE复习笔记7
- Android Studio导入github下载的工程
- bzoj100题纪念。。。
- 移动web最简洁的滑动效果Swipe JS(适合初学者)
- 版本库控制 - git git git
- java.lang.InstantiationException
- 逆波兰运算器浮点求值(c++版本)
- k-means聚类方法的简单java实现
- A - ACM Computer Factory - poj 3436(最大流)
- 毕业一年总结
- Debain下源码安装OpenVAS
- iOS中几种数据永久存储方式
- JSP中的9大内置对象
- UI06_LTView
- Linux 网络配置
- Android 保持键盘隐藏