题意：...
析：我们可以知道，a1+a2=b1,那么我们可以枚举a1，那么a2就有了，并且a1+a3=b2，所以a3就有了，我们再从把里面的剩下的数两两相加，并从b数组中去掉，
那么剩下的最小的就是a4，然后依次可以求出a5，a6....由于a最大才是5000，并且保证有唯一解，那么找到一个就直接退出。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <unordered_map>
#include <unordered_set>
#define debug() puts("++++");
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e4 + 5;
const int mod = 2000;
const int dr[] = {-1, 1, 0, 0};
const int dc[] = {0, 0, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
int b[maxn], c[maxn];
int a[maxn];
bool judge(int x){
a[1] = x; a[2] = b[1] - a[1];
int cnt = 2;
for(int i = 2; i <= m; ++i){
if(b[i]) a[++cnt] = b[i] - a[1];
else continue;
for(int j = 2; j < cnt; ++j){
bool ok = false;
for(int k = i+1; k <= m; ++k) if(a[j] + a[cnt] == b[k]){
b[k] = 0;
ok = true;
break;
}
if(!ok) return ok;
}
}
return true;
}
int main(){
while(scanf("%d", &n) == 1 && n){
m = n * (n-1) / 2;
for(int i = 1; i <= m; ++i) scanf("%d", c+i);
for(int i = 1; ; ++i){
memcpy(b+1, c+1, 4*m);
if(judge(i)) break;
}
for(int i = 1; i <= n; ++i)
if(i == 1) printf("%d", a[i]);
else printf(" %d", a[i]);
printf("
");
}
return 0;
}