hdu 1895 Sum Zero hash

时间：2017-09-29 00:33:44 阅读：168 评论：0 收藏：0 [点我收藏+]

标签：problem algo eps search groups scan tle ack oid

Sum Zero

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)

Problem Description

There are 5 Integer Arrays and each of them contains no more than 300 integers whose value are between -100,000,000 and 100,000,000, You are to find how many such groups (i,j,k,l,m) can make A[0][i]+A[1][j]+A[2][k]+A[3][l]+A[4][m]＝0. Maybe the result is too large, you only need tell me the remainder after divided by 1000000007.

Input

In the first line, there is an Integer T(0<T<20), means the test cases in the input file, then followed by T test cases.
For each test case, there are 5 lines Integers, In each line, the first one is the number of integers in its array.

Output

For each test case, just output the result, followed by a newline character.

Sample Input

1
4 -2 3
-5 -1 -7 -10 -1
-10 2 4 -6 2
-4 -1
-7 -7 -1 -4 -6

Sample Output

11

Author

Sempr|CrazyBird|hust07p43

Source

HDU 2008-4 Programming Contest

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<bitset>
#include<set>
#include<map>
#include<time.h>
using namespace std;
#define LL long long
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=5e4+10,M=1e5+10,inf=1e9+10;
const LL INF=1e18+10,mod=1e9+7;
const double eps=(1e-8),pi=(4*atan(1.0));

int si[10];
int a[6][310];
struct handhash
{
    const static int side=1e5+7;
    vector<int>v[M];
    vector<int>nu[M];
    void init()
    {
        for(int i=0;i<side;i++)
        v[i].clear(),nu[i].clear();
    }
    void add(int x)
    {
        int z=(abs(x))%side;
        for(int i=0;i<v[z].size();i++)
        if(v[z][i]==x)
        {
            nu[z][i]++;
            return;
        }
        v[z].push_back(x);
        nu[z].push_back(1);
    }
    int query(int x)
    {
        int z=(abs(x))%side;
        for(int i=0;i<v[z].size();i++)
            if(v[z][i]==x)return nu[z][i];
        return 0;
    }
}mp;
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        mp.init();
        for(int i=1;i<=5;i++)
        {
            scanf("%d",&si[i]);
            for(int j=1;j<=si[i];j++)
                scanf("%d",&a[i][j]);
        }
        for(int i=1;i<=si[1];i++)
            for(int j=1;j<=si[2];j++)
                mp.add(a[1][i]+a[2][j]);
        LL ans=0;
        for(int k=1;k<=si[3];k++)
        for(int i=1;i<=si[4];i++)
        for(int j=1;j<=si[5];j++)
            ans+=mp.query(-a[3][k]-a[4][i]-a[5][j]);
        printf("%lld\n",ans%mod);
    }
    return 0;
}