Codeforces 767D 贪心

Cartons of milk

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Olya likes milk very much. She drinks k cartons of milk each day if she has at least k and
drinks all of them if she doesn't. But there's an issue — expiration dates. Each carton has a date after which you can't drink it (you still can drink it exactly at the date written on the carton). Due to this, if Olya's fridge contains a carton past its expiry
date, she throws it away.

Olya hates throwing out cartons, so when she drinks a carton, she chooses the one which expires the fastest. It's easy to understand that this strategy minimizes the amount of cartons thrown out and lets her avoid it if it's even possible.


Milk.
Best before: 20.02.2017.

The main issue Olya has is the one of buying new cartons. Currently, there are n cartons of milk in Olya's fridge, for each one an
expiration date is known (how soon does it expire, measured in days). In the shop that Olya visited there are m cartons, and the expiration
date is known for each of those cartons as well.

Find the maximum number of cartons Olya can buy so that she wouldn't have to throw away any cartons. Assume that Olya drank no cartons today.

Input

In the first line there are three integers n, m, k (1 ≤ n, m ≤ 106, 1 ≤ k ≤ n + m) —
the amount of cartons in Olya's fridge, the amount of cartons in the shop and the number of cartons Olya drinks each day.

In the second line there are n integers f1, f2, ..., fn (0 ≤ fi ≤ 107) —
expiration dates of the cartons in Olya's fridge. The expiration date is expressed by the number of days the drinking of this carton can be delayed. For example, a 0 expiration
date means it must be drunk today, 1 — no later than tomorrow, etc.

In the third line there are m integers s1, s2, ..., sm (0 ≤ si ≤ 107) —
expiration dates of the cartons in the shop in a similar format.

Output

If there's no way for Olya to drink the cartons she already has in her fridge, print -1.

Otherwise, in the first line print the maximum number x of cartons which Olya can buy so that she wouldn't have to throw a carton away.
The next line should contain exactly x integers — the numbers of the cartons that should be bought (cartons are numbered in an order
in which they are written in the input, starting with 1). Numbers should not repeat, but can be in arbitrary order. If there are multiple correct
answers, print any of them.

Examples

input

3 6 2
1 0 1
2 0 2 0 0 2

output

3
1 2 3

input

3 1 2
0 0 0
1

output

-1

input

2 1 2
0 1
0

output

1
1

Note

In the first example k = 2 and Olya has three cartons with expiry dates 0, 1 and 1 (they
expire today, tomorrow and tomorrow), and the shop has 3 cartons with expiry date 0 and 3 cartons
with expiry date 2. Olya can buy three cartons, for example, one with the expiry date0 and
two with expiry date 2.

In the second example all three cartons Olya owns expire today and it means she would have to throw packets away regardless of whether she buys an extra one or not.

In the third example Olya would drink k = 2 cartons today (one she alreay has in her fridge and one from the shop) and the remaining
one tomorrow.

题意：冰箱里n瓶牛奶，商店里m瓶，每天最多喝k瓶，n瓶牛奶的保质期，商店里m瓶的保质期，问最多能买多少瓶牛奶而且使得所有牛奶都喝完且不过保质期

题解：贪心，先把冰箱的牛奶平摊完，然后贪心算出最多能买多少瓶

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
int s[10000005],c[1000005],a[1000005],b[1000005];
bool cmp(int i,int j){
return a[i]<a[j];
}
int main(){
int n,m,k,i,j;
scanf("%d%d%d",&n,&m,&k);
memset(s,0,sizeof(s));
for(i=0;i<n;i++){
scanf("%d",&j);
s[j]++;
}
for(i=10000000;i;i--){
if(s[i]>k){
s[i-1]+=s[i]-k;
s[i]=k;
}
}
if(s[0]>k){
printf("-1\n");
return 0;
}
for(i=0;i<m;i++){
scanf("%d",&a[i]);
b[i]=i;
}
sort(b,b+m,cmp);
int ans=0;
for(i=0,j=0;i<m;i++){
for(;s[j]==k;j++);
if(a[b[i]]>=j){
s[j]++;
c[ans++]=b[i];
}
}
printf("%d\n",ans);
for(int i=0;i<ans;i++){
printf("%d ",c[i]+1);
}
printf("\n");
return 0;
}