分析
- 给我们n个数(n<3000,每个数不超过5000),n个数中任意两个数,相加存在 n*(n-1)/2个结果,让我们输出前m大的数,我们将 相加的结果作为 下标(因为相加的结果<=10000),统计下标出现的次数, 在统计的完之后在 同 10000 向0 遍历 统计出前m大就行了
代码
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
void fre() { system("clear"), freopen("A.txt", "r", stdin); freopen("Ans.txt","w",stdout); }
void Fre() { system("clear"), freopen("A.txt", "r", stdin);}
#define ios ios::sync_with_stdio(false)
#define Pi acos(-1)
#define pb push_back
#define fi first
#define se second
#define ll long long
#define ull unsigned long long
#define db double
#define Pir pair
#define m_p make_pair
#define INF 0x3f3f3f3f
#define esp 1e-7
#define mod (ll)(1e9 + 7)
#define for_(i, s, e) for(int i = (ll)(s); i <= (ll)(e); i ++)
#define rep_(i, e, s) for(int i = (ll)(e); i >= (ll)(s); i --)
#define sc scanf
#define pr printf
#define sd(a) scanf("%d", &a)
#define ss(a) scanf("%s", a)
using namespace std;
const int mxn = 1e4 + 10;
int ba[mxn];
int ar[mxn];
int main()
{
int n, m;
while(~ sc("%d %d", &n, &m))
{
for_(i, 1, n) sd(ar[i]);
memset(ba, 0, sizeof(ba));
for_(i, 1, n)
for_(j, i + 1, n)
ba[ar[i] + ar[j]] ++;
rep_(i, mxn - 1, 0)
{
while(ba[i] && m)
{
pr(m == 1 ? "%d" :"%d ", i);
ba[i] --;
m --;
}
if(! m) break;
}
pr("\n");
}
return 0;
}