F(x) HDU

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F(x) HDU - 4734

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

大致题意：就是问你0到b的范围中不大于F（a）的值的个数

思路：简单数位dp，
dp[i][j]表示第i位不大于j的个数。

代码如下

#include<iostream>#include<cstdio>#include<cstdlib>#include<algorithm>#include<cstring>#include<string>#include<queue>#include<map>using namespace std;int dp[20][200000];int bit[20];int a,b;int dfs(int pos,int num,bool flag){    if(pos==-1) return num>=0;    if(num<0) return 0;    if(!flag&&dp[pos][num]!=-1)    return dp[pos][num];    int ans=0;    int end=flag?bit[pos]:9;    for(int i=0;i<=end;i++)    {        ans+=dfs(pos-1,num-i*(1<<pos),flag&&i==end);    }    if(!flag) dp[pos][num]=ans;    return ans;}int F(int x){    int ret=0;    int len=0;    while(x)    {        ret+=(x%10)*(1<<len);        len++;        x/=10;    }    return ret;}int calc(){    int len=0;    while(b)    {        bit[len++]=b%10;        b/=10;    }    return dfs(len-1,F(a),1);}int main(){     int T;     int Icase=0;     cin>>T;     memset(dp,-1,sizeof(dp));     while(T--)     {        Icase++;        cin>>a>>b;        printf("Case #%d: %d\n",Icase,calc());     }    return 0;}

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