4 Values whose Sum is 0 (二分+排序)
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题面
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 2 28 ) 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:
给一个n*4的矩阵,输入n*4个数,在每一列找一个数,使得四个数的和为0;
分析:
先分别求出a和b,c和d两列任意两个数的和存放到相应的数组,将cd的和进行排序后,再用二分法进行查找;二分查找的时候注意,倘若中间的数据符合条件的话要再往两边进行查找,因为不能排除有多个数字相等的情况
注意:
求第二组数据的时候,根据提交的结果是不需要初始化total的;
题解:
#include<iostream> #include<cstring> #include<cstdio> #include<algorithm> using namespace std; const int N=4005; int a[N],b[N],c[N],d[N]; int ab[N*N],cd[N*N]; int main() { int n,total=0,i,j; while (cin>>n) { for (i=0;i<n;i++) cin>>a[i]>>b[i]>>c[i]>>d[i]; int num1=0,num2=0; for (i=0;i<n;i++) for (j=0;j<n;j++) { ab[num1++]=a[i]+b[j]; cd[num2++]=-(c[i]+d[j]); } sort (cd,cd+num2); for (i=0;i<num1;i++) { int mid,up=num2-1,low=0; while (low<=up) { mid=low+(up-low)/2; if (ab[i]==cd[mid]) { total++; for (j=mid+1;j<=up;j++) { if (ab[i]==cd[j]) total++; else break; } for (j=mid-1;j>=low;j--) { if (ab[i]==cd[j]) total++; else break; } break; } else { if (ab[i]>cd[mid]) low=mid+1; else up=mid-1; } } } cout << total << endl; } return 0; }
转载自https://www.cnblogs.com/lisijie/p/7289457.html
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