POJ 2785 4 Values whose Sum is 0（想法题）

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4 Values whose Sum is 0

Time Limit: 15000MS		Memory Limit: 228000K
Total Submissions: 20334		Accepted: 6100
Case Time Limit: 5000MS

Description

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 2²⁸ ) 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).

思路

题意：给定四个长度为n的数组A， B， C， D。 从每个数组中取一个数， 这样得到四个数， 并且这四个数的之和为0. 求这样组合的个数。

题解：直接算出组合数的话，复杂度太高，分成两堆来求，算出 A[i] + B[i] 的值，然后在A[i] + B[i]中找 等于 -C[i] - D[i] 的个数

 

+ View Code?

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#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int maxn = 4005; int cd[maxn*maxn];   int main() {     int N;     while (~scanf("%d",&N))     {         int a[maxn],b[maxn],c[maxn],d[maxn];         for (int i = 0;i < N;i++)    scanf("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]);         for (int i = 0;i < N;i++)    for (int j = 0;j < N;j++)    cd[i*N+j] = c[i] + d[j];         sort(cd,cd + N*N);         int res = 0;         for (int i = 0;i < N;i++)         {             for (int j = 0;j < N;j++)             {                 int tmp = -a[i] - b[j];                 int pos1 = lower_bound(cd,cd + N*N,tmp) - cd;                 int pos2 = upper_bound(cd,cd + N*N,tmp) - cd;                 res += pos2 - pos1;             }         }         printf("%d\n",res);     }     return 0; }