【CUGBACM15级BC第8场 A】hdu 4989 Summary

Summary

[align=center]Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1135    Accepted Submission(s): 599
[/align]

[align=left]Problem Description[/align]

Small W is playing a summary game. Firstly, He takes N numbers. Secondly he takes out every pair of them and add this two numbers, thus he can get N*(N - 1)/2 new numbers. Thirdly he deletes the repeated
number of the new numbers. Finally he gets the sum of the left numbers. Now small W want you to tell him what is the final sum.
 

[align=left]Input[/align]

Multi test cases, every case occupies two lines, the first line contain n, then second line contain n numbers a1, a2, ……an separated by exact one space. Process to the
end of file.

[Technical Specification]

2 <= n <= 100

-1000000000 <= ai <= 1000000000
 

[align=left]Output[/align]

For each case, output the final sum.
 

[align=left]Sample Input[/align]

4
1 2 3 4
2
5 5

 

[align=left]Sample Output[/align]

25
10
HintFirstly small W takes any pair of 1 2 3 4 and add them, he will get 3 4 5 5 6 7. Then he deletes the repeated numbers, he will get 3 4 5 6 7, Finally he gets the sum=3+4+5+6+7=25.

 

题目大意：给定一个集合，从中任选两个数求和，要求输出所有组成可能的和（相同算一种）

解题思路：枚举两个数，用set去重后求和

#include <iostream>
#include <set>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <bitset>
#include <string>
#include <vector>
#include <iomanip>
#include <cstring>
#include <algorithm>
#include <functional>
#define PI acos(-1)
#define eps 1e-8
#define inf 0x3f3f3f3f
#define debug(x) cout<<"---"<<x<<"---"<<endl
typedef long long ll;
using namespace std;

long long a[110];
set<long long> gg;
set<long long>::iterator iter;
int main()
{
int n;
while (~scanf("%d", &n))
{
gg.clear();
for (int i = 0; i < n; i++)
{
scanf("%lld", &a[i]);
}
for (int i = 0; i < n; i++)
{
for (int j = i + 1; j < n; j++)
{
int xx = a[i] + a[j];
gg.insert(xx);
}
}
ll ans = 0;
for (iter = gg.begin(); iter != gg.end(); iter++)
{
ans += *iter;
}
cout << ans << endl;
}
return 0;
}