Basic 分类： POJ 2015-08-03 15:49 3人阅读 评论(0) 收藏

Basic
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 905 Accepted: 228

Description
The programming language Ada has integer constants that look like this: 123, 8#123#, 16#abc#. These constants represent the integers 123, 83 (123 base 8) and 2739 (abc base 16). More precisely, an integer may be a decimal integer given as a sequence of one or more digits less than 10, or it may be an integer to some specific base, given as the base followed by a sequence of one or more digits less than the base enclosed by # symbols. Lower case letters from a through f are used as the digits representing 10 through 15. In Ada, the base, if specified, must be a sequence of decimal digits. For this problem, however, the base may be of any form described above so long as it represents an integer between 2 and 16 inclusive.

Input
The first line of input contains a positive integer n. n lines follow.Input lines contain no spaces and are between 1 and 80 characters in length.

Output
For each line of input, output a line “yes” if it is a valid integer constant according to the above rules; otherwise output a line containing “no”.

Sample Input

5
2#101#
2#101##123#
17#abc#
16#123456789abcdef#
16#123456789abcdef#123456789abcdef#

Sample Output

yes
yes
no
yes
no

Source
Waterloo local 2003.01.25
奇葩题，看了好几遍，把百度，谷歌都用上只能说，不懂
题意就是：
给一串字符串，判断是否合法。合法情况为：第一个数字在2至16间，表示进制的基底，然后是一个用两个#包含在内的数字，表示该进制下的数字，用a到f表示10到15，新算出的值可以作为下一个数的基底（如果后面还有数的话）。
只是奇怪为什么放在排序专题里????

#include <map>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <string>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define LL long long
#define eps 1e-9
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define CRR fclose(stdin)
#define CWW fclose(stdout)
#define WW freopen("output.txt","w",stdout)
#define RR freopen("input.txt","r",stdin)

using namespace std;

const int MAX=100100;

char s[110];

int Oper(int sum,int i)
{
    if(sum<2||sum>16)
    {
        return 0;
    }
    if(s[i]=='#')
    {
        return 0;
    }
    double  ans=0;//这里要用double，用long long 都过不了，可能有的字符串太长会损失精度
    int j;
    for(j=i;s[j];j++)
    {
        if(s[j]=='#')
        {
            break;
        }
        if(sum>10)
        {
            if(s[j]>='0'&&s[j]<='9')
            {
                ans=ans*sum+s[j]-'0';
            }
            else if(s[j]>='a'&&s[j]<=sum-11+'a')
            {
                ans=ans *sum+10+s[j]-'a';
            }
            else
            {
                return 0;
            }
        }
        else if(sum<=10)
        {
            if(s[j]>='0'&&s[j]<=sum-1+'0')
            {
                ans=ans*sum+s[j]-'0';
            }
            else
            {
                return 0;
            }
        }
    }
    if((s[j]=='#'&&s[j+1]=='')||(s[j]=='#'&&s[j+1]=='#'&&Oper(ans,j+2)))
    {
        return 1;
    }
    else
    {
        return 0;
    }
}

int main()
{
    int n,i;
    scanf("%d",&n);
    while(n--)
    {
        scanf("%s",s);
        int ans=0;
        for(i=0;;i++)
        {
            if(s[i]>='0'&&s[i]<='9')
            {
                ans=ans*10+s[i]-'0';
            }
            else
            {
                break;
            }
        }
        if((s[i]=='#'&&Oper(ans,i+1))||(s[i]==''&&ans==0))
        {
            printf("yes
");
        }
        else
        {
            printf("no
");
        }
    }
    return 0;
}

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