POJ2785：4 Values whose Sum is 0(二分)

標籤：poj   二分   

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).
題意：每列找一個數，得到和為0的序列，有幾種不同的方案對1，2列的數求一個和，3,4列的數求一個和，然後進行二分尋找

#include <stdio.h>#include <string.h>#include <algorithm>#include <map>using namespace std;#define up(i,x,y) for(i=x;i<=y;i++)#define down(i,x,y) for(i=x;i>=y;i--)#define mem(a,b) memset(a,b,sizeof(a))#define w(a) while(a)#define ll long longint n,a[4005],b[4005],c[4005],d[4005],sum1[16000005],sum2[16000005],len;int main(){    int i,j,ans,l,r,mid;    w(~scanf("%d",&n))    {        up(i,0,n-1)        scanf("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]);        len=0;        up(i,0,n-1)        up(j,0,n-1)        sum1[len++]=a[i]+b[j];        len=0;        up(i,0,n-1)        up(j,0,n-1)        sum2[len++]=c[i]+d[j];        ans=0;        sort(sum2,sum2+len);        up(i,0,len-1)        {            l=0,r=len-1;            w(l<r)            {                mid=(l+r)>>1;                if(sum2[mid]<-sum1[i])                    l=mid+1;                else                    r=mid;            }            while(sum2[l]==-sum1[i]&&l<len)            {                ans++;                l++;            }        }        printf("%d\n",ans);    }    return 0;}