poj 2785 4 Values whose Sum is 0(hash)

4 Values whose Sum is 0

Time Limit: 15000MS Memory Limit: 228000KTotal Submissions: 14140 Accepted: 4002Case 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 -4626 -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).

Source

Southwestern Europe 2005

题意：

给你四列数。每列里找一个数使得他们的和为0.问一共有多少这样的数。

思路：

把前两列的数任意组合的和hash。在任意组合后两列。然后去查hash表看能否使值为0.

详细见代码：

#include<algorithm>#include<iostream>#include<string.h>#include<sstream>#include<stdio.h>#include<math.h>#include<vector>#include<string>#include<queue>#include<set>#include<map>//#pragma comment(linker,"/STACK:1024000000,1024000000")using namespace std;const int INF=0x3f3f3f3f;const double eps=1e-8;const double PI=acos(-1.0);const int maxn=4010;const int up=536870912;const int mod=13000007;const int sz=maxn*maxn;typedef __int64 ll;int maze[5][maxn];int h[sz],f[sz];void Insert(int x){    int key=(x+up)%mod;    while(f[key]&&h[key]!=x)        key=(key+1)%sz;    if(!f[key])        h[key]=x;    f[key]++;}int getf(int x){    int key=(x+up)%mod;    while(f[key]&&h[key]!=x)        key=(key+1)%sz;    return f[key];}int main(){    int n,i,j;    ll ans;    while(~scanf("%d",&n))    {        memset(f,0,sizeof f);        for(j=0;j<n;j++)            for(i=0;i<4;i++)                scanf("%d",&maze[i][j]);        for(i=0;i<n;i++)            for(j=0;j<n;j++)                Insert(maze[0][i]+maze[1][j]);        ans=0;        for(i=0;i<n;i++)            for(j=0;j<n;j++)                ans+=getf(-maze[2][i]-maze[3][j]);        printf("%I64d\n",ans);    }    return 0;}