B. School Marks
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
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 test(s)
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：有n个数，先给出p个数，现在要求另外n-p个数，使得这n个数的中位数大于等于y，并且n个数之和不大于x。没有方案就输出-1.
代码：
#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;
}