[2016-03-31] [Codeforces] [659C] [Tanya and Toys]

• Time: 2016-03-31-23:49:13 Thursday
• Title Number: [2016-03-31][codeforces][659c][tanya and Toys].md
• The main topic: There are $10^9$ species of goods, the value of the item I, has used n items, to M-yuan, ask the most can buy the number of items not owned
• Analysis: Greedy, from the lowest start to buy, assuming $m = 10^9$, then also buy items not more than $10^6$, because $\frac{(1+k) K}{k} < 10^9$
• Problems encountered: The answer could be 0

  
 

   
  

  
#include <algorithm>
   
  

  
#include <cstdio>
   
  

  
using namespace std;
   
  

  
int a[100000 + 10];
   
  

  
int ans[1000000];
   
  

  
int main(){
   
  

  
 int n,m;
   
  

  
 scanf("%d%d",&n,&m);
   
  

  
 for(int i = 0;i < n ; ++i){
   
  

  
 scanf("%d",&a[i]);
   
  

  
 }
   
  

  
 a[n] = 0;
   
  

  
 a[n + 1] = 1E9 + 10;
   
  

  
 sort(a,a+n+2); 
   
  

  
  int   =    0  ;  
   
  

  
  for   ( int   i  =    0   i  <   n  +  2    &&   m  >    0     ++  i  ) {  
   
  

  
  for   ( int   J  =   a  [ i  ]    +    1   j  <   a  [ i  +    1  ]    &&  Span class= "PLN" > m  >    0  ;++  j  ) {  
   
  

  
 m -= j;
   
  

  
 if(m < 0) break;
   
  

  
 ans[cnt++] = j;
   
  

  
 }
   
  

  
 }
   
  

  
 printf("%d\n",cnt);
   
  

  
 for(int i = 0;i < cnt;++i){
   
  

  
 printf("%d ",ans[i]);
   
  

  
 }
   
  

  
 printf("\n");
   
  

  
 return 0;
   
  

  
}
  
 

From for notes (Wiz)

[2016-03-31] [Codeforces] [659C] [Tanya and Toys]