1600: Twenty-four point
Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 490 Solved: 78
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Description
Given four numbers, can you get twenty-four through the addition, subtraction, multiplication, and division? Each number can be used only once.
Input
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).
Output
For each case, print the “Yes” or “No”. If twenty-four point can be get, print “Yes”, otherwise, print “No”.
Sample Input
2 2 3 9
1 1 1 1
5 5 5 1
Sample Output
Yes
No
Yes
HINT
For the first sample, (2/3+2)*9=24.
Source
思路：
dfs，把数组当参数，4个数先选2个算一下，然后变成三个数，传入当下一层，避免了麻烦的选数
1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 #include <algorithm> 5 #include <stack> 6 #include <cctype> 7 #include <vector> 8 #include <cmath> 9 #include <map> 10 #include <queue> 11 12 #define ll long long 13 #define eps 1e-8 14 15 using namespace std; 16 17 double a[5]; 18 19 int dfs(double p[],int len) 20 { 21 if(len==1){ 22 if(fabs(p[0]-24)<eps){ 23 return 1; 24 } 25 else{ 26 return 0; 27 } 28 } 29 double f[5]; 30 int cou; 31 int i,j,k; 32 for(i = 0;i <len-1;i++){ 33 for(j=i+1;j<len;j++){ 34 cou=0; 35 for(k=0;k<len;k++){ 36 if(k!=i && k!=j){ 37 f[cou]=p[k];cou++; 38 } 39 } 40 f[cou]=p[i]+p[j]; 41 if(dfs(f,cou+1)) return 1; 42 f[cou]=p[i]-p[j]; 43 if(dfs(f,cou+1)) return 1; 44 f[cou]=p[j]-p[i]; 45 if(dfs(f,cou+1)) return 1; 46 f[cou]=p[i]*p[j]; 47 if(dfs(f,cou+1)) return 1; 48 if(p[j]!=0){ 49 f[cou]=p[i]/p[j]; 50 if(dfs(f,cou+1)) return 1; 51 } 52 if(p[i]!=0){ 53 f[cou]=p[j]/p[i]; 54 if(dfs(f,cou+1)) return 1; 55 } 56 } 57 } 58 return 0; 59 } 60 61 int main() 62 { 63 //freopen("in.txt","r",stdin); 64 //scanf("%d",&T); 65 //for(int ccnt=1;ccnt<=T;ccnt++){ 66 while(scanf("%lf%lf%lf%lf",&a[0],&a[1],&a[2],&a[3])!=EOF){ 67 if(dfs(a,4) == 1){ 68 printf("Yes "); 69 } 70 else{ 71 printf("No "); 72 } 73 } 74 return 0; 75 }