小弟不才，求完整代码

如图，交了十多次还不ac，求解！ac了就好 用c语言提交 在线等 求大神帮解

// 20151108csdnacm.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include "stdio.h"
#define MAX_N 100
#define MAX_COMB 100*99/2

int pair_add(int a[],int n,int b[]);
int repeat_del(int a[],int n);
int final_sum(int a[],int n);

int main(int argc, char* argv[])
{
int n = 1,pair_num,final_num,sum;
int i;
printf("Please input the number scale n (2=<n<=100):");
scanf("%d",&n);
while ( n != 0) {
int a[MAX_N];
int add[MAX_COMB];
for(i = 0;i < n;i++) { //Input the numbers (and we can add error handling if it's needed）.
scanf("%d",&a[i]); //cannot written as a[i] here.
}
pair_num = pair_add(a,n,add);
final_num = repeat_del(add,pair_num);
sum = final_sum(add,final_num);
printf("SUM is %d\n",sum);

    printf("\nPlease input the number scale n (2=<n<=100 and 0 to quit):");
    scanf("%d",&n);
}   
return 0;

}

int pair_add(int a[],int n,int b[]) {
int i,j,k = 0;
for (i = 0;i < n-1;i++) {
for (j = i+1;j < n;j++) {
b[k] = a[i] + a[j];
k++;
}
}
return k;
}

int repeat_del(int a[],int n) {
int i,j,k;
for (i = 0;i < n-1;i++) {
for (j = i+1;j < n;j++) {
if (a[i] == a[j]) {
for (k = j;k < n-1;k++)
a[k] = a[k+1];
n--;
j--;
}
}
}
return n;
}

int final_sum(int a[],int n) {
int sum = 0,i;
for(i = 0;i < n;i++)
sum += a[i];
return sum;
}

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
附上文字版

不是难题，但是我C语言有点生疏了，很多地方处理的不是很快，附上经过验证的代码给题主：

 #include "stdafx.h"
#include "stdio.h"
#define MAX_N 100
#define MAX_COMB 100*99/2

int pair_add(int a[],int n,int b[]);
int repeat_del(int a[],int n);
int final_sum(int a[],int n);

int main(int argc, char* argv[])
{
    int n = 1,pair_num,final_num,sum;
    int i;
    printf("Please input the number scale n (2=<n<=100):");
    scanf("%d",&n);
    while ( n != 0) {
        int a[MAX_N];
        int add[MAX_COMB];
        for(i = 0;i < n;i++) {      //Input the numbers (and we can add error handling if it's needed.
            scanf("%d",&a[i]);      //cannot written as a[i] here.
        }
        pair_num = pair_add(a,n,add);
        final_num = repeat_del(add,pair_num);
        sum = final_sum(add,final_num);
        printf("SUM is %d\n",sum);

        printf("\nPlease input the number scale n (2=<n<=100 and 0 to quit):");
        scanf("%d",&n);
    }   
    return 0;
}

int pair_add(int a[],int n,int b[]) {
    int i,j,k = 0;
    for (i = 0;i < n-1;i++) {
        for (j = i+1;j < n;j++) {
            b[k] = a[i] + a[j];
            k++;
        }
    }
    return k;
}

int repeat_del(int a[],int n) {
    int i,j,k;
    for (i = 0;i < n-1;i++) {
        for (j = i+1;j < n;j++) {
            if (a[i] == a[j]) {
                for (k = j;k < n-1;k++)
                    a[k] = a[k+1];
                n--;
                j--;
            }
        }
    }
    return n;
}

int final_sum(int a[],int n) {
    int sum = 0,i;
    for(i = 0;i < n;i++)
        sum += a[i];
    return sum;
}

你可以把printf语句去掉以符合题意要求，我是为了测试方便才做的while循环，你也可以删去。

这个是结果显示图。
题主如果觉得哪些地方费解或者冗余可以反馈给我，也是对我的一个提醒。
望采纳~