Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
Description
Little Vova studies programming in an elite school. Vova and his classmates are supposed to write n progress tests, for each test they will get a mark from 1 to p. Vova is very smart and he can write every test for any mark, but he doesn‘t want to stand out from the crowd too much. If the sum of his marks for all tests exceeds value x, then his classmates notice how smart he is and start distracting him asking to let them copy his homework. And if the median of his marks will be lower than y points (the definition of a median is given in the notes), then his mom will decide that he gets too many bad marks and forbid him to play computer games.
Vova has already wrote k tests and got marks a1, ..., ak. He doesn‘t want to get into the first or the second situation described above and now he needs to determine which marks he needs to get for the remaining tests. Help him do that.
Input
The first line contains 5 space-separated integers: n, k, p, x and y (1 ≤ n ≤ 999, n is odd, 0 ≤ k < n, 1 ≤ p ≤ 1000, n ≤ x ≤ n·p, 1 ≤ y ≤ p). Here n is the number of tests that Vova is planned to write, k is the number of tests he has already written, p is the maximum possible mark for a test, x is the maximum total number of points so that the classmates don‘t yet disturb Vova, y is the minimum median point so that mom still lets him play computer games.
The second line contains k space-separated integers: a1, ..., ak (1 ≤ ai ≤ p) — the marks that Vova got for the tests he has already written.
Output
If Vova cannot achieve the desired result, print "-1".
Otherwise, print n - k space-separated integers — the marks that Vova should get for the remaining tests. If there are multiple possible solutions, print any of them.
Sample Input
Input
5 3 5 18 43 5 4Output
4 1Input
5 3 5 16 45 5 5Output
-1Hint
The median of sequence a1, ..., an where n is odd (in this problem n is always odd) is the element staying on (n + 1) / 2 position in the sorted list of ai.
In the first sample the sum of marks equals 3 + 5 + 4 + 4 + 1 = 17, what doesn‘t exceed 18, that means that Vova won‘t be disturbed by his classmates. And the median point of the sequence {1, 3, 4, 4, 5} equals to 4, that isn‘t less than 4, so his mom lets him play computer games.
Please note that you do not have to maximize the sum of marks or the median mark. Any of the answers: "4 2", "2 4", "5 1", "1 5", "4 1", "1 4" for the first test is correct.
In the second sample Vova got three ‘5‘ marks, so even if he gets two ‘1‘ marks, the sum of marks will be 17, that is more than the required value of 16. So, the answer to this test is "-1".
题意：
共有n个数字，已知k个，所有数字不大于p，求剩余未知数字使数字总和不大于x且中位数不小于y。
按y将已知的k个数字分为左和右两边，再在两侧添加n-k个数字，左侧添1，右侧添y以保证总和尽可能的小。
附AC代码：
1 #include<iostream>
2 #include<cstring>
3 #include<algorithm>
4 #include<cmath>
5 using namespace std;
6
7 int n,k,p,x,y,xx,mid;
8 int a[1100],ans[1100];
9
10 int main(){
11 while(cin>>n>>k>>p>>x>>y){
12 memset(a,0,sizeof(a));
13 memset(ans,0,sizeof(ans));
14 int sum=0,cnt=0;
15 int l=0,r=0;
16 for(int i=0;i<k;i++){//划分左右
17 cin>>a[i];
18 sum+=a[i];
19 if(a[i]<y)
20 l++;
21 else if(a[i]>=y)
22 r++;
23 }
24 mid=n/2+1;
25 if(l>=mid){//"=" 当左边数目大于或等于一半+1时，中位数不可能为>=y
26 cout<<"-1"<<endl;
27 continue;
28 }
29 if(r<=mid){//下方第二组数据时不判定的话xx会为负数。。
30 xx=mid-r;//右侧个数
31 while(xx--){//取右侧最小值y
32 ans[cnt++]=y;
33 sum+=y;
34 }
35 }
36
37 xx=n-r-l-cnt;//剩余数量
38 while(xx--){//取左侧最小值1
39 ans[cnt++]=1;
40 sum++;
41 }
42 if(sum>x){//大于x时同样不符合要求
43 cout<<"-1"<<endl;
44 continue;
45 }
46 else{
47 cout<<ans[0];//保证输出格式
48 for(int i=1;i<cnt;i++){
49 cout<<" "<<ans[i];
50 }
51 cout<<endl;
52 }
53 }
54 return 0;
55 }
56 /*
57 5 3 5 25 4
58 3 3 3
59 */
60 /*
61 9 7 2 13 1
62 2 2 2 1 1 2 2
63 */