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】给出一个n*4的矩阵,每列上选一个数使得最后加起来为0,问有多少种取法
【思路】先用ab数组存a+b的所有组合,同理,存储cd数组,然后对cd数组进行排序,然后用upper_bound,lower_bound查找是否存在-ab[i],正好两者只差为1,即多了一种组合方式
#include<iostream> #include<stdio.h> #include<algorithm> #include<string.h> using namespace std; const int N=4000+10; int n; int a[N],b[N],c[N],d[N]; int ab[N*N],cd[N*N]; int main() { scanf("%d",&n); for(int i=1;i<=n;i++) { scanf("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]); } int k=0; for(int i=1;i<=n;i++) { for(int j=1;j<=n;j++) { ab[k]=a[i]+b[j]; cd[k]=c[i]+d[j]; k++; } } sort(cd,cd+k); long long ans=0; for(int i=0;i<k;i++) { int tmp=-ab[i]; ans+=(long long )(upper_bound(cd,cd+k,tmp)-lower_bound(cd,cd+k,tmp)); } printf("%I64d\n",ans); return 0; }