二分查找的应用
1 #include <iostream> 2 #include <algorithm> 3 #include <cstdio> 4 using namespace std; 5 6 int a1[4001],a2[4001]; 7 int b1[4001],b2[4001]; 8 int sum1[4001*4000],sum2[4001*4000]; 9 10 int binarySearch(int left,int right,int key) 11 { 12 __int64 ans=0; //64位整数 13 while(left<=right) 14 { 15 int mid = (left + right) / 2; 16 if(sum2[mid] == key) 17 { 18 ans++; 19 for(int i=mid-1; i>=left; i--) //搜到后可能相邻的左边有相等的也满足条件的数! 20 { 21 if(sum2[i]!=key) 22 { 23 break; 24 } 25 ans++; 26 } 27 for(int j=mid+1; j<=right; j++) 28 { 29 if(sum2[j]!=key) 30 { 31 break; 32 } 33 ans++; 34 } 35 return ans; 36 } 37 else if(sum2[mid] < key) left = mid+1; 38 else right = mid -1; 39 } 40 return 0; 41 } 42 43 int main() 44 { 45 int n; 46 cin>>n; 47 __int64 ans=0; 48 for(int i = 0; i < n; i++) 49 { 50 cin>>a1[i]>>a2[i]>>b1[i]>>b2[i]; 51 52 } 53 int times = 0; 54 for(int i = 0; i < n; i++) //四数相加转化为两数,通过两两随机相加,以便二分 55 for(int j = 0; j < n; j++) 56 { 57 sum1[times] = a1[i] + a2[j]; 58 sum2[times] = b1[i] + b2[j]; 59 times++; //加时的下表 60 } 61 sort(sum2,sum2 + times); 62 63 for(int i = 0; i < times; i++) 64 { 65 int key = -sum1[i]; 66 ans += binarySearch(0,times-1,key); 67 } 68 cout<<ans<<endl; 69 return 0; 70 }
二分 基础
Time Limit:15000MS Memory Limit:228000KB 64bit IO Format:%I64d & %I64uDescription
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).