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 }看了别人的思路，写的
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