题目: http://acm.hdu.edu.cn/showproblem.php?pid=4734
F(x)
Time Limit: 1000/500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3347 Accepted Submission(s): 1258
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)
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
Recommend
题意：对于一个n位的十进制数x（AnAn-1……A1），我们定义它的权重为：
F(x)=An*2n-1+An-1*2n-2+……+A1*20
现在，给你两个十进制数A和B，请计算出在闭区间[0, B]之中，有多少个数x的权重不大于A的权重，即F(x)<=F(A)。
F(x)=An*2n-1+An-1*2n-2+……+A1*20
现在，给你两个十进制数A和B，请计算出在闭区间[0, B]之中，有多少个数x的权重不大于A的权重，即F(x)<=F(A)。
题解：
数位dp
预处理出A的权重。f[i][j]为到第i位比j小的方案数。
1 #include<bits/stdc++.h> 2 using namespace std; 3 int DP[12][4616],digit[12]; 4 int read() 5 { 6 int s=0,fh=1;char ch=getchar(); 7 while(ch<'0'||ch>'9'){if(ch=='-')fh=-1;ch=getchar();} 8 while(ch>='0'&&ch<='9'){s=s*10+(ch-'0');ch=getchar();} 9 return s*fh; 10 } 11 int GetFA(int k) 12 { 13 int sum=0,tmp=1; 14 while(k!=0) 15 { 16 sum+=(k%10)*tmp;k/=10; 17 tmp*=2; 18 } 19 return sum; 20 } 21 int Dfs(int len,int FA,int limit) 22 { 23 if(FA<0)return 0; 24 if(len<=0)return 1; 25 if(limit==0&&DP[len][FA]!=-1)return DP[len][FA]; 26 int end=limit?digit[len]:9,ret=0,i; 27 for(i=0;i<=end;i++)ret+=Dfs(len-1,FA-i*(1<<(len-1)),limit&&(i==end)); 28 if(limit==0)DP[len][FA]=ret; 29 return ret; 30 } 31 int Calc(int FA,int B) 32 { 33 int len=0; 34 memset(digit,0,sizeof(digit)); 35 while(B!=0){digit[++len]=B%10;B/=10;} 36 return Dfs(len,FA,1); 37 } 38 int main() 39 { 40 int case1=0,A,B,FA,T; 41 T=read(); 42 memset(DP,-1,sizeof(DP)); 43 while(T--) 44 { 45 A=read();B=read(); 46 FA=GetFA(A); 47 //memset(DP,-1,sizeof(DP)); 48 printf("Case #%d: %d ",++case1,Calc(FA,B)); 49 } 50 fclose(stdin); 51 fclose(stdout); 52 return 0; 53 }