C

题目

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.
TechnicalSpecificationTechnicalSpecification
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.

题意

对任意一组数据，输入N，有N个整数，取其中每一对之和（排列组合），可以构成新的数，有N*(N - 1)/2个，求这些新的数之和。

解析

1.定义两个数组，一个存N个整数，一个存新生成的数（这个数组要开大，不然runtime，因为构成新的数的数量是排列组合得到了总数）。

2.元素的范围大， 故用long long保证不超出数据范围。

3.

for(i=0; i<n; i++)        {            for(j=i+1; j<n; j++)            {                b[k]=a[i]+a[j];                k++;            }        }

这里用双循环完成排列组合得到每一对之和。

代码

include

include

using namespace std;
int main()
{
int n;
while (scanf(“%d”,&n)!=EOF)
{
int i, j, k=0;
long long a[105];
long long b[10000];

    for(i=0; i<n; i++)    {        scanf("%I64d",&a[i]);    }    for(i=0; i<n; i++)    {        for(j=i+1; j<n; j++)        {            b[k]=a[i]+a[j];            k++;            //记录新数存入数组b中。        }    }    sort(b,b+k);            //给新数的数组排序，用下面的算法去除相同的元素。    for(i=0; i<k; i++)    {       if(i<=k-2)            if(b[i]==b[i+1])                b[i]=0;    }   long long  sum=0;    for(i=0; i<k; i++)    {        sum=sum+b[i];    }    printf("%I64d\n",sum);}return 0;

}