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CF 301 B School Marks

栏目:htmlcss时间:2015-06-05 09:02:48

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: nkpx and y (1?≤?n?≤?999n is odd, 0?≤?k?<?n1?≤?p?≤?1000n?≤?x?≤?np1?≤?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 "".

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
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 "".


思路:先找出比y小的数填1,然后剩余的数填y,如果这个条件没法满足,那输出⑴




#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<queue> #include<stack> #include<vector> #include<set> #include<map> #define L(x) (x<<1) #define R(x) (x<<1|1) #define MID(x,y) ((x+y)>>1) #define eps 1e⑻ typedef __int64 ll; #define fre(i,a,b) for(i = a; i <b; i++) #define free(i,b,a) for(i = b; i >= a;i--) #define mem(t, v) memset ((t) , v, sizeof(t)) #define ssf(n) scanf("%s", n) #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 bug pf("Hi ") using namespace std; #define INF 0x3f3f3f3f #define N 1005 int n,k,p,x,y; int lem,rim; int a[N]; int num; int le,ri; void solve() { int t=n/2; if(le>t) //小的数目超过1半 { printf("⑴ "); return ; } int i,flag=0; int lnum=min(t-le,t-(ri-t⑴)-le); //右侧大的数目可能多于1半 x-=lnum; int hh=le+lnum+ri; //添加lnum个数后还要几个数 hh=n-hh; if(x>=hh*y) { for(i=0;i<lnum;i++) { if(flag) pf(" "); flag=1; printf("1"); } for(i=0;i<hh;i++) { if(flag) pf(" "); printf("%d",y); flag=1; } } else printf("⑴ "); } int main() { int i,j; scanf("%d%d%d%d%d",&n,&k,&p,&x,&y); int s; le=ri=0; //记录比y小,大 for(i=0;i<k;i++) { scanf("%d",&s); if(s>=y) ri++; else le++; x-=s; } if(p<y) { printf("⑴ "); return 0; } solve(); return 0; } /* 5 3 5 25 4 3 3 3 9 7 2 14 1 2 2 2 1 1 2 2 9 7 2 14 1 2 2 2 1 1 2 2 */






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