POJ-2785 4 Values whose Sum is 0 (binary enumeration sort + dichotomy)

Title Link:http://poj.org/problem?id=2785

Test instructions is giving you 4 numbers. To take a number from each column, make the sum of four number 0, and find out the number of cases of such a combination.

When one of the series has multiple identical numbers, consider them as different numbers.

The practice is to put the first two columns and values in an array (a), the next two columns and the existence of another array (B), the array is n^2. Then the B array is sort, the a array is traversed, and the second finds the number of elements in the B array and the array of a and 0. One of the points to note is that in case the A array is all 0, and the B array is 0 (there are other cases), the two points can only find a match. So it is possible to find exactly the number that matches the number, subtract their position, and add up, that's the answer.

Here can use two functions Lower_bound () and Upper_bound (), all using two points to find, the previous function is to return an array of greater than or equal to a number of positions, followed by the return is greater than this number of positions (can not understand Google or Baidu How to use these two functions).

The code is as follows:

1#include <iostream>2#include <cstdio>3#include <algorithm>4 5 using namespacestd;6 Const intMAXN =4010;7typedefLong LongLL;8 9 LL A[MAXN], B[MAXN], C[MAXN], D[MAXN];TenLL NUM1[MAXN * MAXN], NUM2[MAXN *MAXN]; One  A intMain () - { -     intN; the      while(~SCANF ("%d", &N)) { -          for(inti =0; I < n; i++) { -scanf"%lld%lld%lld%lld", A + I, B + i, C + i, D +i); -         } +         intf =0; -          for(inti =0; I < n; i++) { +              for(intj =0; J < N; J + +) { ANUM1[F] = A[i] +B[j]; atnum2[f++] = C[i] +D[j]; -             } -         } -LL res =0; -         inttemp =0; -Sort (num2, num2 +f); in          for(inti =0; i < F; i++) { -temp = Lower_bound (num2, num2 + F,-num1[i])-num2; to             if(Temp < F && Num2[temp] + num1[i] = =0) { +Res + = Upper_bound (num2, num2 + F,-num1[i])-num2-temp; -             } the         } *cout << Res <<Endl; $     }Panax Notoginseng}

POJ-2785 4 Values whose Sum is 0 (binary enumeration sort + dichotomy)