POJ2785:4 Values whose Sum is 0(二分)
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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).
题意:每列找一个数,得到和为0的序列,有几种不同的方案
对1,2列的数求一个和,3,4列的数求一个和,然后进行二分查找
#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;}
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