poj 2785 4 Values whose Sum is 0

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

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【分析】
呼呼呼，又是一道数字哈希
让我们两段两段的考虑吧
就考虑 a[i]+b[j] 和 c[p]+d[q] 是否相等
将上述两个值分别哈希
然后查询就好咯hohoho

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【代码】

//poj 2785 零和游戏 #include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<iomanip>#include<cmath>#define fo(i,j,k) for(i=j;i<=k;i++)using namespace std;const int N=4001;const int H=20345677;const int MAX=1000000000;const int key=1;int a[N],b[N],c[N],d[N];int hash[H+1],sum[H+1];void insert(int num){    int tmp=num;    num=(num+MAX)%H;    while(hash[num]!=MAX && hash[num]!=tmp)      num=(num+key)%H;    hash[num]=tmp;    sum[num]++;}int find(int num){    int tmp=num;    num=(num+MAX)%H;    while(hash[num]!=MAX && hash[num]!=tmp)      num=(num+key)%H;    if(hash[num]==MAX) return 0;    return sum[num];}int main(){    int n,i,j,k,tot=0;    fo(i,0,H) hash[i]=MAX;    scanf("%d",&n);    fo(i,1,n)      scanf("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]);    fo(i,1,n)      fo(j,1,n)        insert(a[i]+b[j]);    fo(i,1,n)      fo(j,1,n)        tot+=find(-c[i]-d[j]);    printf("%d\n",tot);    return 0;}