[BZOJ2793] [Poi2012] Vouchers

2793: [poi2012]vouchers time limit:20 Sec Memory limit:64 MB
submit:316 solved:148
[Submit] [Status] [Discuss] Description

Considering the set of positive integers, there are now n groups of people to take the number, assuming that group I to x people, they each take the number of x must be a multiple, and is the smallest remaining x.
The number of M in a positive integer is labeled as a lucky number, asking who gets the lucky number.

Input

The first line is a positive integer m (m<=1,000,000), and the following m line has a positive integer x (x<=1,000,000) for each line, indicating that X is a lucky number.
The next line is a positive integer n (n<=1,000,000), and the following n lines a positive integer x (x<=1,000,000) for each line, indicating that this group comes in X person.

Output

The first line outputs a non-negative integer k, which indicates that the K-person has taken the lucky number, and the following K-line represents the number of the person who took the lucky number, which is numbered starting from 1 in the order in which they came.

Sample Input4
1
6
8
16
3
4
2
4

Sample Output3
2
4
6
HINT

Hint

A total of 10 people, the number they take away is 4 8 12 16 2 6 20 24 28 32.

2nd, 4, 6 people take the lucky number 8, 16, 6.

(don't think about it too hard, actually it's very water)

Source

Acknowledgement Oimaster

[Submit] [Status] [Discuss]

Maintain a next array for each x, representing the position of the next X, and then brute force processing.

Briefly estimate the upper limit complexity of $\sum\limits_{i=1}^n\frac mi=o (MLOGN) where m=\max{a_i}$

But it's definitely not going to be that complicated.

1#include <cstdio>2#include <algorithm>3 #defineN 10000104 #definell Long Long5 using namespacestd;6 intN,next[n],maxn,tot,m,a[n];7 ll Ans[n],sum,vis[n];8 intMain ()9 {Tenscanf"%d",&n); One      for(intI=1; i<=n;i++) Ascanf"%d", &a[i]), maxn=Max (A[I],MAXN); -scanf"%d",&m); -      while(m--) the     { -         intx; -scanf"%d",&x); -ll now=sum;sum+=x; +          for(inti=x;i&&next[x]+x<=MAXN;) -         if(!vis[next[x]+=x]) +vis[next[x]]=++now,i--; A     } at      for(intI=1; i<=n;i++) -     if(Vis[a[i]]) -ans[++tot]=Vis[a[i]]; -Sort (ans+1, ans+tot+1); -printf"%d\n", tot); -      for(intI=1; i<=tot;i++) inprintf"%lld\n", Ans[i]); -}

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[BZOJ2793] [Poi2012] Vouchers