HDUOJ-----F(x)

F(x)

Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 468    Accepted Submission(s): 171

Problem Description
For a decimal number x with n digits (AnAn-1An-2 ... A2A1), we define its weight as F(x) = An * 2n-1 + An-1 * 2n-2 + ... + A2 * 2 + A1 * 1. Now you are given two numbers A and B, please calculate how many numbers are there between 0 and B, inclusive, whose weight is no more than F(A).
 

 

Input
The first line has a number T (T <= 10000) , indicating the number of test cases. For each test case, there are two numbers A and B (0 <= A,B < 109)
 

 

Output
For every case,you should output "Case #t: " at first, without quotes. The t is the case number starting from 1. Then output the answer.
 

 

Sample Input
3 0 100 1 10 5 100
 

 

Sample Output
Case #1: 1 Case #2: 2 Case #3: 13
 

 

Source
2013 ACM/ICPC Asia Regional Chengdu Online
 

 

Recommend
liuyiding
 
 
代码：
 1 #include<stdio.h>
 2 #include <string.h>
 3 int sty[10][5000];
 4 int pow1[10]={1,2,4,8,16,32,64,128,256,512};
 5 int pow2[10]={1,10,100,1000,10000,100000,1000000,10000000,100000000,1000000000};
 6 
 7 int cal(int ca,int b);
 8 int f(int a);
 9 int main()
10 {
11     int i,j,a,b,l,number,I,count;
12      memset(sty,0xff,5000*10*sizeof(int));
13     while(scanf("%d",&number)!=EOF)
14     {
15       for(I=1;I<=number;I++)
16       {
17         scanf("%d%d",&a,&b);
18         a=f(a);
19         b++; 
20         count=0;
21        for(i=9;i>=0;i--)
22        {
23          l=b/pow2[i];
24          b-=l*pow2[i];
25          for(j=0;j<l;j++)
26            count+=cal(i,a-j * pow1[i]); 
27           a-=l*pow1[i];
28        }
29       printf("Case #%d: %d\n",I,count);
30       }
31     }
32    return 0;
33 }
34 
35 int cal(int ca,int b)
36 {
37    if(b<0)
38         return 0;
39    if(ca==0)
40          return 1;
41    if(sty[ca][b]!=-1)
42          return sty[ca][b];
43     int n = 0,i;
44     for(i = 0;i < 10;i++)
45       n += cal(ca - 1,b - i * pow1[ca-1]);
46       sty[ca][b] = n;
47     return n;
48 }
49 int f(int a)
50 {
51    int n=0,l,i;
52     for(i=9;i>=0;i--)
53     {
54     l=a/pow2[i];
55     n+=l*pow1[i];
56     a-=l*pow2[i];
57     }
58    return n;
59 }
View Code

看了别人的思路，写的

posted @ 2013-09-15 19:49  龚细军  阅读(190)  评论(0)  编辑  收藏  举报