POJ 2719（模拟+进制转换）

思路：把数按个、十、百……拆开，如果大于3，就减1，再按9进制加回去。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
using namespace std;
#define inf 0x3f3f3f3f
#define ll long long
const int maxn=200005;
const double eps=1e-8;
const double PI = acos(-1.0);
int main()
{
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    std::cout.tie(0);
    ll ans,n;
    while(cin>>n&&n)
    {
        ll u=n;
        ll jin=1;
        ans=0;
        while(n)
        {
            int t=n%10;
            if(t>3) t--;
            ans+=t*jin;
            n/=10;
            jin*=9;
        }
        cout<<u<<": "<<ans<<endl;
    }
    return 0;
}