HDU 4734中如何运用数位动态规划解决F(x)问题？

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题目：HDU 4734 F(x) 时间限制：1000/500 MS (Java/其他) 内存限制：32768/32768 K (Java/其他) 总提交次数：6341 接受提交次数：2438

问题描述：对于一个小数x，它有n位数字（A1, A2, A3, ..., An），我们需要计算以下表达式的值：x=A1 + A2 + A3 + ... + An

直接输出结果：x

HDU 4734

F(x)

Time Limit: 1000/500 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6341Accepted Submission(s): 2438

Problem Description

For a decimal number x with n digits (A_nA_n-1A_n-2 ... A₂A₁), we define its weight as F(x) = A_n * 2^n-1 + A_n-1 * 2^n-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

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关于数位dp详细的题解在 hdu2089的题解里：javascript:void(0)

本篇是我做的第二个数位dp练习题。

这题有个坑一直卡着我。dfs的第二个参数sum。本题限制不超过f(A)。令upnum=f(A)，

那我就可以用它来限制着dfs的递归，当递归过程的sum>upnum，就return，可是死活不能AC

然后看着网上的代码，改成给这个参数赋初值为upnum，然后递归时去减，减到<0就return，没想到就AC了。

大概是跟dp数组的记忆化有关。

：

#include <stdio.h> #include <string.h> #define mset(a,i) memset(a,i,sizeof(a)) int dp[26][15240];//dp[i][sum]表示第i位，和sum时的记忆。 int a[26]; int upnum; int dfs(int pos,int sum,bool limit) { if(sum<0)return 0; if(pos==-1)return sum>=0; if(!limit&&dp[pos][sum]!=-1)return dp[pos][sum]; int up=limit?a[pos]:9; int ans=0; for(int i=0;i<=up;i++) { ans+=dfs(pos-1,sum-i*(1<<pos),limit&&i==up); } if(!limit)dp[pos][sum]=ans; return ans; } int solve(int x) { int top=0; while(x) { a[top++]=x%10; x/=10; } return dfs(top-1,upnum,1); } int f(int A) { int ans=0; for(int j=0;A;j++,A/=10) ans+=(1<<j)*(A%10); return ans; } int main() { int T,A,B,r=1; scanf("%d",&T); memset(dp,-1,sizeof(dp)); while(T--) { scanf("%d%d",&A,&B); upnum=f(A); printf("Case #%d: %d\n",r++,solve(B)); } return 0; }