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B. School Marks (CF #301 (Div. 2))

來源：程序員人生發(fā)布時間：2015-06-16 08:59:05 閱讀次數(shù)：3196次

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 "⑴".

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
⑴

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 "⑴".

題意：有n個數(shù)，先給出p個數(shù)，現(xiàn)在要求另外n-p個數(shù)，使得這n個數(shù)的中位數(shù)大于等于y，并且n個數(shù)之和不大于x。沒有方案就輸出⑴.

代碼：

#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(⑴.0) #define eps 1e⑹ #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 ") 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++; //記錄比y小的數(shù)有多少 else if (a[i]>=y) r++;//記錄大于等于y的數(shù)有多少 } if (l>=n/2+1){ //如果比y小的數(shù)超過了1半，那末中位數(shù)不可能到達(dá)y了 printf("⑴ "); continue; } int xx=n/2+1-r; while (xx>0){ //填充右側(cè)，盡可能填y使得和最小 ans[cnt++]=y; xx--;s+=y; } xx=n-r-l-cnt; while (xx>0) //填充左側(cè)，填1使和盡可能小 { ans[cnt++]=1; xx--;s++; } if (s>x){ //和超過x輸出⑴ printf("⑴ "); continue; } printf("%d",ans[0]); for (i=1;i<cnt;i++) printf(" %d",ans[i]); printf(" "); } return 0; }

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B. School Marks (CF #301 (Div. 2))