hdu 4734 解题报告

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题目：

F(x)
Time Limit: 1000/500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9119 Accepted Submission(s): 3607

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

---

代码：

/* hdu-4734 */
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstring>
using namespace std;
const int N=1e5+5;
int digits[10];
int dp[10][N];
int a,b;

/*
参数：'
dp[pos][pre] - pre最初等于f(a),比如pos为是3，那么执行下一个pos位时，pre就要减去3*(1<<pos),
pos - 位置
limit - 判断是否是最大值,为真表示是最大值
*/
int f(int x)
{
    if(x==0) return 0;
    int ans=f(x/10);
    return ans*2+(x%10);
}
int dfs(int pos,int pre, int limit)
{
    if(pos==-1) return pre>=0;
    if(pre<0 ) return 0;

    if(!limit&&dp[pos][pre]) return dp[pos][pre];
    int maxd=limit?digits[pos]:9;
    int ans=0;//计数
    for(int i=0;i<=maxd;i++)
    {
        ans+=dfs(pos-1,pre-i*(1<<pos),limit&&i==digits[pos]);
    }
    if(!limit)
        dp[pos][pre]=ans;

    return ans;
}
int solve(int n)
{
    int len=0;

    while(n)
    {
        digits[len++]=n%10;
        n/=10;
    }
    return dfs(len-1,f(a),true);
}

int main()
{


    int t;
    scanf("%d",&t);
    for(int i=1;i<=t;i++)
    {
        scanf("%d%d",&a,&b);

        printf("Case #%d: %d\n",i,solve(b));
    }

    return 0;
}