Given four numbers, can you get twenty-four through the addition, subtraction, multiplication, and division? Each number can be used only once.
标签：计算24点
Given four numbers, can you get twenty-four through the addition, subtraction, multiplication, and division? Each number can be used only once.
The input consists of multiple test cases. Each test case contains 4 integers A, B, C, D in a single line (1 <= A, B, C, D <= 13).
For each case, print the “Yes” or “No”. If twenty-four point can be get, print “Yes”, otherwise, print “No”.
2 2 3 9
1 1 1 1
5 5 5 1
Yes
No
Yes
For the first sample, (2/3+2)*9=24.
题目链接：http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1600
题目大意：给四个数判断能否算出24
题目分析：直接DFS爆搜，每次计算两个数字再放进数组，知道数组里只有一个数为止
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
double EPS = 1e-10;
double num[4];
bool flag;
bool equel(double a, double b)
{
if(fabs(a - b) <= EPS)
return true;
return false;
}
void DFS(int n)
{
if(flag || (n == 1 && equel(num[0], 24)))
{
flag = true;
return;
}
for(int i = 0; i < n; i++)
{
for(int j = i + 1; j < n; j++)
{
double c1 = num[i], c2 = num[j];
num[j] = num[n - 1];
num[i] = c1 + c2;
DFS(n - 1);
num[i] = c1 - c2;
DFS(n - 1);
num[i] = c2 - c1;
DFS(n - 1);
num[i] = c1 * c2;
DFS(n - 1);
if(!equel(c2, 0))
{
num[i] = c1 / c2;
DFS(n - 1);
}
if(!equel(c1, 0))
{
num[i] = c2 / c1;
DFS(n - 1);
}
num[i] = c1;
num[j] = c2;
}
}
}
int main()
{
while(scanf("%lf %lf %lf %lf", &num[0], &num[1], &num[2], &num[3]) != EOF)
{
flag = false;
DFS(4);
printf("%s\n", flag ? "Yes" : "No");
}
}
另外考虑，输出解的个数和所有解的方法，其实只要用字符串记录一下就可以了
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <iostream>
#include <string>
using namespace std;
double EPS = 1e-10;
double num[4];
bool flag;
int cnt;
string re[4];
bool equel(double a, double b)
{
if(fabs(a - b) <= EPS)
return true;
return false;
}
void DFS(int n)
{
if(n == 1 && equel(num[0], 24))
{
cnt ++;
cout << re[0] << endl;
return;
}
for(int i = 0; i < n; i++)
{
for(int j = i + 1; j < n; j++)
{
double c1 = num[i], c2 = num[j];
string re1, re2;
num[j] = num[n - 1];
num[i] = c1 + c2;
re1 = re[i];
re2 = re[j];
re[j] = re[n - 1];
re[i] = '(' + re1 + '+' + re2 + ')';
DFS(n - 1);
num[i] = c1 - c2;
re[i] = '(' + re1 + '-' + re2 + ')';
DFS(n - 1);
num[i] = c2 - c1;
re[i] = '(' + re2 + '+' + re1 + ')';
DFS(n - 1);
num[i] = c1 * c2;
re[i] = '(' + re1 + '*' + re2 + ')';
DFS(n - 1);
if(!equel(c2, 0))
{
num[i] = c1 / c2;
re[i] = '(' + re1 + '/' + re2 + ')';
DFS(n - 1);
}
if(!equel(c1, 0))
{
num[i] = c2 / c1;
re[i] = '(' + re2 + '/' + re1 + ')';
DFS(n - 1);
}
num[i] = c1;
num[j] = c2;
re[i] = re1;
re[j] = re2;
}
}
}
int main()
{
while(scanf("%lf %lf %lf %lf", &num[0], &num[1], &num[2], &num[3]) != EOF)
{
re[0] = num[0] + '0';
re[1] = num[1] + '0';
re[2] = num[2] + '0';
re[3] = num[3] + '0';
cnt = 0;
DFS(4);
printf("Answer num: %d\n", cnt);
}
}CSUOJ 1600 Twenty-four point (判断24点，另附给出表达式版)
标签：计算24点
原文地址：http://blog.csdn.net/tc_to_top/article/details/45556031
