cf 540b School Marks 贪心
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.
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.
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.
5 3 5 18 4 3 5 4
4 1
5 3 5 16 4 5 5 5
-1
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".
链接：http://codeforces.com/problemset/problem/540/B
题意：一共要写n个数，已经写了k个数，要求写出接下来的n-k个数。要求每个数小于等于p，大于等于1。所有数总和小于等于x。中位数必须是大于等于y。
做法：贪心，因为有上限要求，又要改变中位数，所以只需要用到1和y就可以了。统计大于等于y的个数，不够n/2+1，就用y补，剩下的全放1就可以了。
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <malloc.h>
#include <ctype.h>
#include <math.h>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#include <stack>
#include <queue>
#include <vector>
#include <deque>
#include <set>
#include <map>
#define INF 999999999
#define eps 0.00001
#define LL __int64d
#define pi acos(-1.0)
int num[1010];
int main()//n, k, p, x and y (1 ≤
{
int n,k,p,x,y;
while(scanf("%d%d%d%d%d",&n,&k,&p,&x,&y)!=EOF)
{
int ji=0;
int sum=0;
for(int i=0;i<k;i++)
{
scanf("%d",&num[i]);
sum+=num[i];
if(num[i]>=y)
ji++;
}
int fir=1;
if(ji>=(n/2+1))
{//quan1
if(sum+(n-k)<=x)
{
for(int i=0;i<n-k;i++)
{
if(fir)
{
printf("1");
fir=0;
}
else
printf(" 1");
}
puts("");
}
else
{
puts("-1");
}
}
else
{
if(sum+((n/2+1)-ji)*y+(n-k-((n/2+1)-ji))<=x&&n-k>=((n/2+1)-ji))
{
for(int i=0;i<((n/2+1)-ji);i++)
{
if(fir)
{
fir=0;
printf("%d",y);
}
else
printf(" %d",y);
}
for(int i=0;i<(n-k-((n/2+1)-ji));i++)
{
if(fir)
{
fir=0;
printf("1");
}
else
printf(" 1");
}
puts("");
}
else
{
puts("-1");
}
}
}
return 0;
}