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
[a,b] (a≤b). Please be fast to get the gold medal!
Input
The first line of the input is a single integer T, 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 1≤a≤b≤100000.
For each test case, there are two numbers a and b, as described in the statement. It is guaranteed that 1≤a≤b≤100000.
Output
For each testcase, print one line indicating the answer.
Sample Input
2
1 10
1 1000
Sample Output
10
738
Author
XJZX
Source
Recommend
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 }