• HDU-5327 Olympiad


    http://acm.hdu.edu.cn/showproblem.php?pid=5327

    Olympiad

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 182    Accepted Submission(s): 130

    Problem Description
    You are one of the competitors of the Olympiad in numbers. The problem of this year relates to beatiful numbers. One integer is called beautiful if and only if all of its digitals are different (i.e. 12345 is beautiful, 11 is not beautiful and 100 is not beautiful). Every time you are asked to count how many beautiful numbers there are in the interval [a,b] (ab)
    . Please be fast to get the gold medal!
     
    Input
    The first line of the input is a single integer T (T1000)
    , indicating the number of testcases.
    For each test case, there are two numbers a
    and b
    , as described in the statement. It is guaranteed that 1ab100000
    .
     
    Output
    For each testcase, print one line indicating the answer.
     
    Sample Input
    2
    1 10
    1 1000
     
    Sample Output
    10
    738
    #include<iostream>
    #include<cstdio>
    using namespace std;
    int a[100000];
    int g(int n)
    {
         int a[10],r=0,q,i,flag=0;;
        while(n)
        {
            q=n%10;
            a[r++]=q;
            for(i=0;i<r-1;i++)
            {
                if(q==a[i])
                    flag=1;
            }
            n=n/10;
            if(flag==1)
                break;
        }
         if(flag==0)
             return 0;
         else
           return 1;
    
    }
    void fun()
    {
        int i;
        a[0]=0;
        for(i=1;i<=100000;i++)
        {
            if(g(i)==0)
               a[i]=a[i-1]+1;
            else
              a[i]=a[i-1];
    
        }
    }
     int main()
     {
         fun();
         int t,n,m;
         scanf("%d",&t);
         while(t--)
         {
    
             scanf("%d%d",&n,&m);
             if(a[n]>a[n-1])
               printf("%d
    ",a[m]-a[n]+1);
            else
               printf("%d
    ",a[m]-a[n]);
    
         }
         return 0;
    
     }
     
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  • 原文地址:https://www.cnblogs.com/cancangood/p/4690493.html
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