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 ypoints (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, nis 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.
Examples
Input
5 3 5 18 4 3 5 4
Output
4 1
Input
5 3 5 16 4 5 5 5
Output
-1
Note
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：
给你n,k,p,x,y,五个数，分别表示小明要考n门课，现在已经考了k门课了，每门课的最高成绩为p,但小明有两个小要求。
1、小明考的n门课的总成绩不能超过x;
2、小明的n门课的成绩在排完序后，他的中位数要>=y。
问小明剩下的几门课要考多少分，若有多种情况，输出其中任意一种就行，如果不能满足两个条件，输出-1.
//思路：
首先通过输入的k门成绩求出小明剩余的几门课，几门是y以下的，几门是y以上的。
求出这两个数，下面的就简单了，分别对它们赋值为1，y就行了（保证总分最小）。
C++版本一
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
#define INF 0x3f3f3f3f
#define ll long long
#define N 10010
using namespace std;
int a[N];
int b[N];
int main()
{
int n,k,p,x,y;
while(scanf("%d%d%d%d%d",&n,&k,&p,&x,&y)!=EOF)
{
int i,j;
int sum=0;
for(i=1;i<=k;i++)
scanf("%d",&a[i]),sum+=a[i];
int sx=x-sum;
for(i=k+1;i<=n;i++)
{
if(sx-(n-i)>=y)
a[i]=y,sx-=y;
else
a[i]=1,sx-=1;
b[i]=a[i];
}
sort(a+1,a+n+1);
if(a[n/2+1]<y||sx<0)
printf("-1\n");
else
{
for(i=k+1;i<n;i++)
printf("%d ",b[i]);
printf("%d\n",b[n]);
}
}
return 0;
}
C++版本二
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 1005
#define MAXN 2005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b) for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b) for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v) memset ((t) , v, sizeof(t))
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf printf
#define DBG pf("Hi\n")
typedef long long ll;
using namespace std;
int n,p,k,x,y;
int a[maxn];
int ans[maxn];
int main()
{
// freopen("C:/Users/asus1/Desktop/IN.txt","r",stdin);
int i,j;
while (~scanf("%d%d%d%d%d",&n,&p,&k,&x,&y))
{
int s=0,l=0,r=0,num=0,cnt=0;
for (i=0;i<p;i++)
{
scanf("%d",&a[i]);
s+=a[i];
if (a[i]<y) l++; //记录比y小的数有多少
else if (a[i]>=y) r++;//记录大于等于y的数有多少
}
if (l>=n/2+1){ //如果比y小的数超过了一半，那么中位数不可能达到y了
printf("-1\n");
continue;
}
int xx=n/2+1-r;
while (xx>0){ //填充右边，尽量填y使得和最小
ans[cnt++]=y;
xx--;s+=y;
}
xx=n-r-l-cnt;
while (xx>0) //填充左边，填1使和尽量小
{
ans[cnt++]=1;
xx--;s++;
}
if (s>x){ //和超过x输出-1
printf("-1\n");
continue;
}
printf("%d",ans[0]);
for (i=1;i<cnt;i++)
printf(" %d",ans[i]);
printf("\n");
}
return 0;
}
京公网安备 11010502036488号