POJ 2785 4 Values whose Sum is 0 [二分]

传送门

13773503

njczy2010

2785

Accepted

25248K

7079MS

G++

1423B

2015-01-11 10:26:48

4 Values whose Sum is 0

Time Limit: 15000MS		Memory Limit: 228000K
Total Submissions: 16102		Accepted: 4659
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).

Source

Southwestern Europe 2005

 

 

sign，抱着脑袋想了好久怎么用hash做，就是没想着用二分，，，，，，

#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<string>

#define N 4002
#define M 10
#define mod 1000000007
//#define p 10000007
#define mod2 1000000000
#define ll long long
#define LL long long
#define eps 1e-9
#define inf 0x3fffffff
#define maxi(a,b) (a)>(b)? (a) : (b)
#define mini(a,b) (a)<(b)? (a) : (b)

using namespace std;

int x[N*N];
int a[N],b[N],c[N],d[N];
ll ans;
int k;
int n;

void ini()
{
    int i,j;
    ans=0;
    for(i=1;i<=n;i++){
        scanf("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]);
    }
    k=0;
    for(i=1;i<=n;i++){
        for(j=1;j<=n;j++){
            x[k]=a[i]+b[j];k++;
        }
    }
    sort(x,x+k);
}

void solve()
{
    int i,j;
    int te;
    int t1,t2;
    for(i=1;i<=n;i++){
        for(j=1;j<=n;j++){
            te=-c[i]-d[j];
            t1=lower_bound(x,x+k,te)-x;
            t2=upper_bound(x,x+k,te)-x;
            ans+=(ll)(t2-t1);
        }
    }
}

void out()
{
    printf("%I64d
",ans);
}

int main()
{
    //freopen("data.in","r",stdin);
   // freopen("data.out","w",stdout);
    //scanf("%d",&T);
    //for(int ccnt=1;ccnt<=T;ccnt++)
    //while(T--)
    while(scanf("%d",&n)!=EOF)
    {
        ini();
        solve();
        out();
    }

    return 0;
}