hdu 4278 码表数字跳跃找规律

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Faulty Odometer

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 324    Accepted Submission(s): 236

Problem Description

　　You are given a car odometer which displays the miles traveled as an integer. The odometer has a defect, however: it proceeds from the digit 2 to the digit 4 and from the digit 7 to the digit 9, always skipping over the digit 3 and 8. This defect shows up in all positions (the one's, the ten's, the hundred's, etc.). For example, if the odometer displays 15229 and the car travels one mile, odometer reading changes to 15240 (instead of 15230).

 

Input

　　Each line of input contains a positive integer in the range 1..999999999 which represents an odometer reading. (Leading zeros will not appear in the input.) The end of input is indicated by a line containing a single 0. You may assume that no odometer reading will contain the digit 3 and 8.

 

Output

　　Each line of input will produce exactly one line of output, which will contain: the odometer reading from the input, a colon, one blank space, and the actual number of miles traveled by the car.

 

Sample Input

   
   
   
   

    
    
    
    15
2005
250
1500
999999
0
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    15: 12
2005: 1028
250: 160
1500: 768
999999: 262143
   
   
   
   

 

Source

2012 ACM/ICPC Asia Regional Tianjin Online

 

Recommend

liuyiding

 

题意 ：有个码表坏了  数字显示的时候  不会显示3和8  2直接到4    7直接到9 

输入一个码表显示的时间 问真正的时间是什么 

 

思路：

例如  2050

那么可以分成 2000+50  

对于50   也就是5个10  那么中间的3不算 也就是4个10

暴力出 10 100 1000  。。。。。的数据可以知道 10 对应为8  1000对应为512
那么就是  4个8 加上2个512 

 

如果是5050  就是4个512 加上4个8 

 

 

暴力打表程序：

#include<stdio.h>
int main()
{
 int i,j,cnt,n;
 while(scanf("%d",&n))
 {
  cnt=0;
  for(i=0;i<=n;i++)
  {
   j=i;
   while(j!=0)
   {
    if(j%10==3||j%10==8) { cnt++;break;}
     j=j/10;
   }
  }
     printf("%d\n",n-cnt);
 }
}

ac代码

 

：

#include<stdio.h>
int a[20]={1,8,64,512,4096,32768,262144,2097152,16777216,134217728};
int main()
{
	int n,k,pos;
	while(scanf("%d",&n)!=EOF)
	{
		if(!n) break;
		k=n;
		int cnt=0,ans=0;
		while(k!=0)
		{
                pos=k%10;
				if(pos>7)
					pos=pos-2;
				else if(pos>2)
					pos=pos-1;
				ans+=pos*a[cnt];
				cnt++; 
				k=k/10;
		}
		printf("%d: %d\n",n,ans);
	}
	return 0;
}