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13级春季校赛的热身题，但优化后我的代码也超时了，后来看了看学长的解法，觉得最简单的还是map，再一次感受到了map的强大。
题目描述如下
| Description |
|
There is an integer set A. How many couples of a and b, which can make a+b=0(a∈A, b∈A，a<=b). |
| Input |
|
There are multiple test cases. The first line is an integer T indicating for the number of test cases. The first line of each case is an integer n, standing for the number of integers in set A. The second line is n integers in set A separated by space. (1<=n<=100 000，|ai|<=1000 000 000) |
| Output |
|
For each test case, output all the unique couples of a and b by the order of a from small to large. If there is no such a situation, output "<empty>". Output a blank line after each case.
|
| Sample Input |
|
2 6 -1 0 1 4 -1 -4 3 1 1 2 |
| Sample Output |
|
-4 4 -1 1
<empty> |
map的遍历方式：使用迭代器，it->first指原集合的元素，it-second指映射集合的元素。这样仅用O(N)的复杂度就求出了答案，尽管map是c++封装好的函数，运行相对较慢。
对比数组来说，map的好处有很好的处理负值，能够容下更大的数字等等，代码如下：
#include<cstdio> #include<map> #include<cstring> #include<iostream> using namespace std; int main() { int n,t,b,mark; scanf("%d",&t); while(t--) { map<int,int>a; a.clear(); mark = 0; scanf("%d",&n); while(n--) { scanf("%d",&b); a[b]++; } map<int,int>::iterator it; for(it = a.begin(); it->first < 0 && it != a.end(); it++) { ///cout<<"first = "<<it->first<<endl; if(it->second) { ///cout<<"second = "<<it->second<<endl; if(a[-it->first]) { printf("%d %d\n",it->first,-it->first); mark = 1; } } } if(a[0] >= 2) { cout<<"0 0"<<endl; mark = 1; } if(mark == 0) printf("<empty>\n"); printf("\n"); } }
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原文地址：http://www.cnblogs.com/jifahu/p/5505891.html
