Test instructions: There are 5 sets, the size of the collection is N, and each one sets out a number, asking if you can find five numbers and 0. Data range: t<=50;n<=200

Analysis:

Brute force enumeration is n^5*t, time-out, then you need to use some tricks.

Here is the magic of a pointer: How to find a B in the complexity of O (n), so that A+b==c (a, B, respectively, belongs to a series, A, b). The practice is to first sort a, a, and then a pointer I point to the first, the pointer J points to the tail of B, the number of the pointer to determine whether the sum of ==c, if equal to the end of the search, if less than, then i++, if greater than J-, if there is a pointer has gone to the head has not found A+b==c It indicates that there is no such ab;

This problem re-deformation: There are three series a,b,c, how to find the complexity of O (n^2) a,b,c, making a+b==c. method or the above, is more than a loop to traverse C.

The topic is to use this technique, but also with a little division of the idea of treatment, in fact, is not divided, anyway, I remember this, that is, do not give a question to the problem of the whole solution, the topic said sum, we do not have to one step the sum ah, we can sum this problem into: a1+a2 A3+a4;a5 sum is (A1+A2) + (A3+A4) +a5, and because is to find equals 0, so is: (A1+A2) + (A3+A4) ==-a5, so that the problem is converted into the above model.

In addition, the sort + de-weight can be done with set, and the Set.insert () will be automatically weighed and sorted in ascending order.

Code:

#include <iostream> #include <set> #include <algorithm>using namespace Std;set<long long> s1,s2 ; int T,n;long long A[6][300];long long Sum[3][50000];int main () {cin>>t;while (t--) {cin>>n;s1.clear (); S2.clear (); for (int. i=0;i<5;i++) {for (int j=0;j<n;j++) {cin>>a[i][j];}} for (int i=0;i<n;i++) {for (int j=0;j<n;j++) {S1.insert (a[0][i]+a[1][j]); S2.insert (A[2][i]+a[3][j]);}} for (int i=0;i<n;i++) a[4][i]*= ( -1), int len1=0,len2=0;set<long long>::iterator it;for (It=s1.begin (); it!= S1.end (); it++) sum[1][len1++]= (*it); for (It=s2.begin (); It!=s2.end (); it++) sum[2][len2++]= (*it); int ok1=0;for (int i= 0;i<n;i++) {int ok=1;int j=0,k=len2-1;while (Sum[1][j]+sum[2][k]!=a[4][i]) {if (Sum[1][j]+sum[2][k]<a[4][i]) j+ +;else K--;if (j>=len1| | k<0) {ok=0;break;}} if (OK) {ok1=1;break;}} if (OK1) cout<< "Yes" <<endl;else cout<< "No" <<ENDL;}

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！ HDU 4,334 sets each one number and 0 whether there is-thinking, card time-(magic of the pointer)