Input

Output

Sample Input

6
-45 22 42 -16
-41 -27 56 30
-36 53 -37 77
-36 30 -75 -46
26 -38 -10 62
-32 -54 -6 45

Sample Output

5

Hint

題目意思:有四列數,從每一列中選一個數使4個數之和為0;一個有幾種情況;

解題思路:先將4列變成2堆數,再用二分快速查找:

 1 #include <iostream>
 2 #include <algorithm>
 3 
 4 using namespace std;
 5 const int MAX = 4000 + 300;
 6 const int MAX1 = 4000*4000 + 300;
 7 int N,ti,TTT;
 8 int a[4][MAX];
 9 int sum1[MAX1];
10 int sum2[MAX1];
11 
12 void twofe(int
 
 x)
13 {
14     //cout<<x<<"  x  "<<endl;
15     x = - x;
16     int tou = 0;
17     int wei = ti;
18 
19     while(tou <= wei)
20     {
21         int temp =( tou + wei )/2;
22 
23         if(x < sum2[temp])
24         {
25             wei = temp - 1;
26         }
27         else if ( x > sum2[temp])
28         {
29             tou = temp + 1;
30         }
31         else if( x == sum2[temp])
32         {
33           //  cout<<sum2[temp]<<" temp "<<endl;
34             TTT++;
35             for(int i = temp -1;i>=0;i--)
36                 if(sum2[i]==x)
37                     TTT++;
38                 else
39                     break;
40             for(int i = temp +1;i < ti;i++)
41                 if(sum2[i]==x)
42                     TTT++;
43                 else
44                     break;
45             return;
46         }
47     }
48 
49 
50 }
51 
52 int main()
53 {
54 
55     cin>>N;
56     for(int i = 0;i < N;i++)
57     {
58         for(int  j= 0;j < 4;j++)
59             cin>>a[j][i];
60     }
61 
62     ti = 0;
63     TTT =  0;
64     for(int i = 0;i<N;i++)
65         for(int j = 0;j<N;j++)
66         {
67             sum1[ti] = a[0][i] + a[1][j];
68             sum2[ti++] = a[2][i] + a[3][j];
69         }
70     sort(sum2,sum2+ti);
71   //  for(int i = 0;i < ti;i++)
72   //      cout<<sum2[i]<<‘ ‘;
73 //cout<<endl;
74 
75 
76     for(int i = 0;i < ti;i++)
77     {
78         twofe(sum1[i]);
79     }
80 
81     cout<<TTT<<endl;
82 
83 
84     return 0;
85 }

4 Values whose Sum is 0 :POJ - 2785