4 Values whose Sum is 0 poj2785
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The SUM problem can be formulated as follows: given four lists A, B, C, D of integer values, compute how many quadruplet (a, b, c, d ) ∈ A x B x C x D are such that a + b + c + d = 0 . In the following, we assume that all lists have the same size n .
Input
The first line of the input file contains the size of the lists n (this value can be as large as 4000). We then have n lines containing four integer values (with absolute value as large as 2 28 ) that belong respectively to A, B, C and D .
Output
For each input file, your program has to write the number quadruplets whose sum is zero.
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
Sample Explanation: Indeed, the sum of the five following quadruplets is zero: (-45, -27, 42, 30), (26, 30, -10, -46), (-32, 22, 56, -46),(-32, 30, -75, 77), (-32, -54, 56, 30).
这道题刚拿到手的时候完全没思路,但是暴力绝逼超时,先是想着用set后来发现不可行,二分尽然过了,,,,,
View Code
1 #include <iostream> 2 using namespace std; 3 #include<string.h> 4 #include<set> 5 #include<stdio.h> 6 #include<math.h> 7 #include<queue> 8 #include<map> 9 #include<algorithm> 10 #include<cstdio> 11 #include<cmath> 12 #include<cstring> 13 #include <cstdio> 14 #include <cstdlib> 15 #include<cstring> 16 int a[4010],b[4010],c[4010],d[4010]; 17 int a1[16000000],b1[16000000]; 18 set<int>TM; 19 int main() 20 { 21 int t; 22 cin>>t; 23 for(int i=0;i<t;i++) 24 cin>>a[i]>>b[i]>>c[i]>>d[i]; 25 int add=0; 26 for(int i=0;i<t;i++) 27 for(int j=0;j<t;j++) 28 { 29 b1[add]=(a[i]+b[j]); 30 a1[add++]=c[i]+d[j]; 31 } 32 sort(b1,b1+add); 33 int sum=0; 34 for(int i=0;i<add;i++) 35 { 36 int tou=0,wei=add,zhongjian; 37 while(tou<=wei) 38 { 39 zhongjian=(tou+wei)/2; 40 if(a1[i]+b1[zhongjian]==0) 41 { 42 sum++; 43 for(int j=zhongjian-1;j>=0;j--) 44 { 45 if(b1[j]!=b1[j+1]) 46 break; 47 else 48 sum++; 49 } 50 for(int j=zhongjian+1;j<add;j++) 51 { 52 if(b1[j]!=b1[j-1]) 53 break; 54 else 55 sum++; 56 } 57 break; 58 } 59 else if(a1[i]+b1[zhongjian]<0) 60 tou=zhongjian+1; 61 else 62 wei=zhongjian-1; 63 } 64 } 65 cout<<sum<<endl; 66 return 0; 67 }
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4 Values whose Sum is 0 (二分+排序)
poj 2785 4 Values whose Sum is 0