hdu 4734 F(x) （数位dp）

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本文介绍了一种计算F(x)权值的方法，并提供了一个完整的C++代码实现。F(x)是对于一个十进制数x，其权值定义为各数字乘以对应位置的2的幂次之和。文章通过动态规划方法解决给定两个数A和B时，求解0到B之间所有数中权值不超过F(A)的数量。

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注意：dp[pos][sum]，sum不是存当前枚举的数的前缀和(加权的)，而是枚举到当前pos

位，后面还需要凑sum的权值和的个数

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 10;
const int maxm = 1e4+5;
int dp[maxn][maxm], a[maxn], A, B, sumA;

int dfs(int pos, int sum, int limit)
{
    if(pos == -1) return sum <= sumA;
    if(sum > sumA) return 0;
    if(!limit && dp[pos][sumA-sum] != -1) return dp[pos][sumA-sum];
    int up = limit ? a[pos] : 9;
    int tmp = 0;
    for(int i = 0; i <= up; i++)
        tmp += dfs(pos-1, sum+i*(1<<pos), limit && a[pos] == i);
    if(!limit) dp[pos][sumA-sum] = tmp;
    return tmp;
}

int solve()
{
    int pos = 0, p = 1;
    sumA = 0;
    while(A)
    {
        sumA += A%10*p;
        p *= 2;
        A /= 10;
    }
    while(B)
    {
        a[pos++] = B%10;
        B /= 10;
    }
    return dfs(pos-1, 0, 1);
}

int main(void)
{
    int ca = 1, t;
    memset(dp, -1, sizeof(dp));
    cin >> t;
    while(t--)
    {
        scanf("%d%d", &A, &B);
        printf("Case #%d: %d\n", ca++, solve());
    }
    return 0;
}

F(x)

Time Limit: 1000/500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5013 Accepted Submission(s): 1866

Problem Description

For a decimal number x with n digits (A _nA _n-1A _n-2 ... A ₂A ₁), we define its weight as F(x) = A _n * 2 ^n-1 + A _n-1 * 2 ^n-2 + ... + A ₂ * 2 + A ₁ * 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 < 10 ⁹)

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