HDU 5327 Olympiad (多校）

Olympiad

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 631 Accepted Submission(s): 436

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

Input

The first line of the input is a single integer

Output

For each testcase, print one line indicating the answer.

Sample Input

2
1 10
1 1000

Sample Output

10
738

Author

XJZX

Source

2015 Multi-University Training Contest 4

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开始想用字符串做，推出每一个数之前的美丽数的总和，再将a,b的总和数相减就是它们之间的美丽数之和了。但是做了半天还是没做出来，太麻烦。题解的做法是预处理，把每一个数是不是美丽数都判一遍，判的过程中用一个数组f记录这之前的美丽数之和，最后也是两个相减，也就是f[b]-f[a-1].

 1 #include<queue>
 2 #include<math.h>
 3 #include<stdio.h>
 4 #include<string.h>
 5 #include<iostream>
 6 #include<algorithm>
 7 using namespace std;
 8 #define N 123456
 9 #define M 12
10 
11 int a,b;
12 int ha[M],f[N];
13 int judge(int x)
14 {
15     int n;
16     memset(ha,0,sizeof(ha));
17     while(x)
18     {
19         n=x%10;
20         if(ha[n])return 0;
21         else ha[n]=1;
22         x=x/10;
23     }
24     return 1;
25 }
26 
27 int main()
28 {
29     f[0]=1;
30     for(int i=1;i<100005;i++)
31     {
32         f[i]=f[i-1];
33         if( judge(i) )
34             f[i]++;
35     }
36     int t;cin>>t;
37     while(t--)
38     {
39         scanf("%d%d",&a,&b);
40         cout<<f[b]-f[a-1]<<endl;
41     }
42     return 0;
43 }