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cf 540b School Marks 貪心

來源：程序員人生發布時間：2015-07-01 08:03:21 閱讀次數：3697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".

鏈接：http://codeforces.com/problemset/problem/540/B

題意：1共要寫n個數，已寫了k個數，要求寫出接下來的n-k個數。要求每一個數小于等于p，大于等于1。所有數總和小于等于x。中位數必須是大于等于y。

做法：貪心，由于有上限要求，又要改變中位數，所以只需要用到1和y就能夠了。統計大于等于y的個數，不夠n/2+1，就用y補，剩下的全放1就能夠了。

#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(⑴.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("⑴"); } } 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("⑴"); } } } return 0; }

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cf 540b School Marks 貪心