[2016-04-08] [Codeforces] [628] B [New Skateboard]

• Time: 2016-04-08-13:24:57
• Title Number: [2016-04-08][codeforces][628][b][new skateboard]
• The main topic: given a length of the most 3x5 3 x 5 String, asking how many substrings can be divisible by 4,
• Analysis:
  • Two digits after each number can be divisible by 4, then this number will be divisible by 4.
  • Dp[i] Indicates the number of words RP divisible by 4 from 0~i, then dp[i] = dp[i-1] + (a[i]%4 = 0) + ((a[i-1] * + a[i])%4?0:i)
    • That is, if the potential of the first can be divisible by 4, then the answer is +1, if the second bit is added, the next two bits are divisible by 4, then the answer is to add I, because the string containing the next two bits have I

  
 

   
  

  
#include <cstring>
   
  

  
#include <cstdio>
   
  

  
using namespace std;
   
  

  
typedef long long LL;
   
  

  
const int maxn = 3 * 1E5 + 10;
   
  

  
char str[maxn];
   
  

  
int main(){
   
  

  
 scanf("%s",str);
   
  

  
 int n = strlen(str);
   
  

  
 LL ans = 0;int tmp;
   
  

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

  
 int tmp = str[i] - ‘0‘;
   
  

  
 if(tmp % 4 == 0) ++ans;
   
  

  
 if(i){
   
  

  
 tmp  =  Span class= "PLN" >  ( str  [ i  - 1      -    ' 0 '   10    +   tmp    
   
  

  
 if(tmp % 4 == 0) ans += i;
   
  

  
 } 
   
  

  
 }
   
  

  
 printf("%I64d\n",ans);
   
  

  
 return 0;
   
  

  
}
  
 

From for notes (Wiz)

[2016-04-08] [Codeforces] [628] B [New Skateboard]