传送门
题面：
Trouble
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6560 Accepted Submission(s): 1847
Problem Description
Hassan is in trouble. His mathematics teacher has given him a very difficult problem called 5-sum. Please help him.
The 5-sum problem is defined as follows: Given 5 sets S_1,...,S_5 of n integer numbers each, is there a_1 in S_1,...,a_5 in S_5 such that a_1+...+a_5=0?
The 5-sum problem is defined as follows: Given 5 sets S_1,...,S_5 of n integer numbers each, is there a_1 in S_1,...,a_5 in S_5 such that a_1+...+a_5=0?
Input
First line of input contains a single integer N (1≤N≤50). N test-cases follow. First line of each test-case contains a single integer n (1<=n<=200). 5 lines follow each containing n integer numbers in range [-10^15, 1 0^15]. I-th line denotes set S_i for 1<=i<=5.
Output
For each test-case output "Yes" (without quotes) if there are a_1 in S_1,...,a_5 in S_5 such that a_1+...+a_5=0, otherwise output "No".
Sample Input
221 -11 -11 -11 -11 -131 2 3-1 -2 -34 5 6-1 3 2-4 -10 -1
Sample Output
NoYes
Source
2012 Multi-University Training Contest 4
题目描述：有5个集合，每个集合有n个数，问你从每个集合中取出一个数，取出的五个数的和能否为0。
题目分析：试着分析下这道题的时间复杂度。如果我们暴力去解题的话，时间复杂度为N*n^5，故极限时间必定超过1e9。故大暴力是绝对不可行的。
而因为题目让我们求五个数的和，因此我们可以适当进行拆分。
我们可以先使前两个集合求和，第3、4集合求和，并将这两个新的集合排序，此时，对于最后一个集合，我们就可以通过二分答案的方法去查找是否存在解。
因为新集合A和新集合B是具有单调性的，因此，我们只需要先从头开始对集合5进行枚举，同时对新集合A进行枚举，最后再对新集合B从最后开始枚举，根据单调性，如果满足e[i]+A[j]+B[k]>0则令B的下标-1，直到满足小于或者等于停止。（等于0则输出yes即可）
ps：这一题事实上是hdu-1895的简化版，在hdu1895这题中还要求求出满足等于0的个数。
具体看代码：
#include
#define maxn 2550
using namespace std;
typedef long long ll;
ll a[maxn],b[maxn],c[maxn],d[maxn],e[maxn];
vectortmp1,tmp2,tmp3;
int main()
{
int t;
scanf("%d",&t);
while(t--){
int n;
tmp1.clear(),tmp2.clear();
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%lld",&a[i]);
for(int i=1;i<=n;i++) scanf("%lld",&b[i]);
for(int i=1;i<=n;i++) scanf("%lld",&c[i]);
for(int i=1;i<=n;i++) scanf("%lld",&d[i]);
for(int i=1;i<=n;i++) scanf("%lld",&e[i]);
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
tmp1.push_back(a[i]+b[j]);
tmp2.push_back(c[i]+d[j]);
}
}
sort(tmp1.begin(),tmp1.end());
sort(tmp2.begin(),tmp2.end());
bool vis=true;
for(int i=1;i<=n;i++){
int p=tmp2.size()-1;
for(int j=0;j=0;j++){
while(e[i]+tmp1[j]+tmp2[p]>0) p--;
if(e[i]+tmp1[j]+tmp2[p]==0){
vis=false;
puts("Yes");
break;
}
}
if(!vis) break;
}
if(vis) puts("No");
}
return 0;
}