欢迎您访问程序员文章站本站旨在为大家提供分享程序员计算机编程知识!
您现在的位置是: 首页  >  IT编程

4 Values whose Sum is 0 POJ - 2785

程序员文章站 2022-05-25 16:25:26
4 Values whose Sum is 0 Time Limit: 15000MS Memory Limit: 228000K Total Submissions: 29243 Accepted: 8887 Case Time Limit: 5000MS Description The SUM ......
4 values whose sum is 0
time limit: 15000ms   memory limit: 228000k
total submissions: 29243   accepted: 8887
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). 
 
题意:abcd四个数列,每个数列抽一个数字,使得和为0,问一共有几组。当一个数列中有多个相同的数字时,把他们作为不同的数字看待。
思路:直接爆搜肯定会超时,将它们对半分成ab和cd再考虑就可以解决。比如在ab中取出a,b后,cd中要取出-a-b,因此先把cd中所有取数字的情况n*n种给纪录下来,然后排序,用二分查找次数。
 
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<set>
#include<vector>
using namespace std;
#define inf 0x3f3f3f3f
#define eps 1e-10
typedef long long ll;
const int maxn = 4002;
const int mod = 1e9 + 7;
int gcd(int a, int b) {
    if (b == 0) return a;  return gcd(b, a % b);
}

int n;
int a[maxn],b[maxn],c[maxn],d[maxn];
int cd[maxn*maxn];

void solve()
{
    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);
    ll res=0;
    for(int i=0;i<n;i++)
        for(int j=0;j<n;j++)
        {
            int ab=-(a[i]+b[j]);
            res+=upper_bound(cd,cd+n*n,ab)-lower_bound(cd,cd+n*n,ab);
        }
    cout<<res<<endl;
}
int main()
{
    scanf("%d",&n);
    for(int i=0;i<n;i++)
        scanf("%d %d %d %d",&a[i],&b[i],&c[i],&d[i]);
    
    solve();
}