HDU 4989

Summary

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 405    Accepted Submission(s): 247

Problem Description 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.

Input Multi test cases, every case occupies two lines, the first line contain n, then second line contain n numbers a 1, a 2, ……a n separated by exact one space. Process to the end of file.

[Technical Specification]

2 <= n <= 100

-1000000000 <= a i <= 1000000000

Output For each case, output the final sum.

Sample Input

4
1 2 3 4
2
5 5
        

Sample Output

25
10

   
    
     Hint
    Firstly 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.

   
    
        

Source BestCoder Round #8

題目的意思看了挺久的。，給你一個N，叫你輸入N個數。然後從N個數中任意取兩個數相加，得到一個新的數，将所有新的數求和就是答案。注意一點就好，相同的元素隻算一個。是以去掉重複的就好。上代碼

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int main()
{
	int n,i,j;
	while(~scanf("%d",&n))
	{
		int a[105];
		int b[10005];
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		__int64 sum=0;
		int l=0;
		for(i=0;i<n;i++)
			scanf("%d",&a[i]);
		for(i=0;i<n;i++)
			for(j=i+1;j<n;j++)
			{
				b[l++]=a[i]+a[j];
			}
			sort(b,b+l);
			int n1=unique(b,b+l)-b;  //去掉相同的。此時新數組個數n1.
			for(i=0;i<n1;i++)
				sum+=b[i];
			printf("%I64d\n",sum);
	}
	return 0;
}