1 #define _CRT_SECURE_NO_WARNINGS  
 2 #include <stdio.h>
 3 #include <math.h>
 4 #include <algorithm>
 5 #include <stdlib.h>
 6 #include <vector>
 7 #include <map>
 8 #include <queue>
 9 #include <string>
10 #include <iostream>
11 #include <ctype.h>
12
 
 #include <string.h>
13 #include <set>
14 #include <stack>
15 #include<functional>
16 using namespace std;
17 #define Size 55
18 #define maxn  1<<30
19 #define minn  1e-6
20 double a[5]; 
21 /*
22 遞歸能求解的問題是能縮小規模的問題,比如本題,先拿出兩個數字算一下,然後數字數減少了1
23 規模縮小了,所以能用遞歸來求解,,但是要定義遞歸的意義,本題是,在大小為n 的a[] 數組中通過一定的計算能否得到24    遞歸的出口就是當n=1 時候如果剩下的數字是24 那麽就可以
 
24 
25 要註意  用的函數的是fabs  不是abs
26 同時註意a-b  與b-a  a/b  與b/a  是不同的要分開來算
27 */
28 bool isZero(double x){
29     return fabs(x) <= minn;
30 }
31 bool solve(double b[], int n){
32     if (n == 1){
33         if (isZero(b[0] - 24)) return true;
34         else return false;
35     }
36     double tmp[5];
37     for
 
 (int i = 0; i < n - 1; i++)
38         for (int j = i + 1; j < n; j++){
39             int pos = 0;
40             double numLeft = b[i];//double
41             double numRight = b[j];//double
42             for (int k = 0; k < n; k++)
43                 if(k!=i&&k!=j) tmp[pos++] = b[k];
44             tmp[pos] = numLeft + numRight;
45             if (solve(tmp, pos + 1))  return true;
46             tmp[pos] = numLeft - numRight;
47             if (solve(tmp, pos + 1)) return true;
48             tmp[pos] = numRight-numLeft;//a-b  b-a
49             if (solve(tmp, pos + 1)) return true;
50             tmp[pos] = numLeft*numRight;
51             if (solve(tmp, pos + 1)) return true;
52             if (!isZero(numRight)){
53                 tmp[pos] = numLeft / numRight;
54                 if (solve(tmp, pos + 1)) return true;
55             }
56             if (!isZero(numLeft)){//a/b  b/a
57                 tmp[pos] = numRight / numLeft;
58                 if (solve(tmp, pos + 1)) return true;
59             }
60         }
61     return false;
62 }
63 
64 int main(){
65     while (1){
66         for (int i = 0; i < 4; i++)
67             cin >> a[i];
68         if (isZero(a[0])) break;
69         if (solve(a, 4)) cout << "YES" << endl;
70         else cout << "NO" << endl;
71     }
72     return 0;
73 }

遞歸_百煉 2787 算24