hdu1404 Digital Deletions

Digital Deletions

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1689 Accepted Submission(s): 605

Problem Description

Digital deletions is a two-player game. The rule of the game is as following.

Begin by writing down a string of digits (numbers) that's as long or as short as you like. The digits can be 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 and appear in any combinations that you like. You don't have to use them all. Here is an example:



On a turn a player may either:
Change any one of the digits to a value less than the number that it is. (No negative numbers are allowed.) For example, you could change a 5 into a 4, 3, 2, 1, or 0.
Erase a zero and all the digits to the right of it.


The player who removes the last digit wins.


The game that begins with the string of numbers above could proceed like this:



Now, given a initial string, try to determine can the first player win if the two players play optimally both.

Input

The input consists of several test cases. For each case, there is a string in one line.

The length of string will be in the range of [1,6]. The string contains only digit characters.

Proceed to the end of file.

Output

Output Yes in a line if the first player can win the game, otherwise output No.

Sample Input

   
   
   
   

    
    
    
    0
00
1
20
   
   
   
   

Sample Output

   
   
   
   

    
    
    
    Yes
Yes
No
No
   
   
   
   
   
   
   
   

    
    
    
    由于,数字是很小的,所以可以,先把表打出来,这样,就可以 了!
   
   
   
   
   
   
   
   

    
    
    
    
#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
#define M 1000000
int pri[M],ten[]={1,10,100,1000,10000,100000,1000000};
char str[10];
int getlen(int n){
    if(n/100000)return 6;
    if(n/10000)return 5;
    if(n/1000)return 4;
    if(n/100)return 3;
    if(n/10)return 2;
    return 1;
}
int getsg(int n){
    int m,len,i,j,k,low;
    len=getlen(n);
    for(i=0;i<len;i++){
        m=n;low=(m-m/ten[i+1]*ten[i+1])/ten[i];
        for(j=low;j<9;j++){
          m+=ten[i];pri[m]=1;
        }
    }
    if(len!=6){
        for(i=len,j=1;i<6;i++,j++){
            m=n*ten[j];
            for( k=0;k<ten[j-1];k++)
            pri[m+k]=1;
        }
    }
}
int main()
{
    int i,ans;
    pri[0]=1;
    for(int i=1;i<M;i++)
    if(!pri[i])getsg(i);
    while(scanf("%s",str)!=EOF){
        if(str[0]=='0'){
            printf("Yes\n");continue;
        }
        for(i=0,ans=0;str[i]!='\0';i++){
            ans=ans*10+str[i]-'0';
        }
        if(pri[ans])printf("Yes\n");
        else printf("No\n");
    }
    return 0;
}