HDU 4278 Faulty Odometer 8进制转10进制

Faulty Odometer

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=4278

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

HINT

题意

有辆车的里程表坏了，3和8都不能显示，会直接跳过去，给你汽车里程表，输出实际跑了多远

题解：

8进制转10进制

代码

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
const int maxn=2501;
#define mod 1000000009
#define eps 1e-7
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
//**************************************************************************************

int d[8]={0,1,2,4,5,6,7,9};
map<int,int> H;
ll pow(ll x,ll y)
{
    if(y==0)
        return 1;
    ll ans=1;
    for(int i=0;i<y;i++)
        ans*=x;
    return ans;
}
int main()
{
    ll n;
    H[0]=0,H[1]=1,H[2]=2,H[4]=3,H[5]=4,H[6]=5,H[7]=6,H[9]=7;
    while(cin>>n)
    {
        if(n==0)
            break;
        printf("%lld: ",n);
        ll ans=0;
        ll t=0;
        while(n)
        {
            ans+=(H[n%10])*(pow(8,t++));
            n/=10;
        }
        cout<<ans<<endl;
    }
}