HDU-4734 F(x) 【数位DP】

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

 

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思路：数位DP

• dp(i, j)表示枚举到第i位剩余数位权值和不超过j时的数的数量。
• 记忆化搜索即可。

 

Code:

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
long long dp[15][maxn];
int a[15], temp;

int f(int x) {
    int t = 0, ans = 0;
    while(x) {
        ans += x % 10 * (1 << t);
        x /= 10;
        ++t;
    }
    return ans;
} 

long long dfs(int pos, int res, bool limit) {
    if (res < 0) return 0;
    if (pos == 0) return 1;
    if (!limit && ~dp[pos][res]) return dp[pos][res];
    int up = limit ? a[pos] : 9;
    long long ans = 0;
    for (int i = 0; i <= up; ++i)
        ans += dfs(pos-1, res-i*(1<<(pos-1)), limit && a[pos] == i);
    if (!limit) dp[pos][res] = ans;
    return ans;
}

long long cal(int x) {
    int len = 0;
    while(x) {
        a[++len] = x % 10;
        x /= 10;
    }
    return dfs(len, temp, 1);
}

int main() {
    int T, a, b, kase = 0;
    memset(dp, -1, sizeof(dp));
    cin>>T;
    while(T--) {
        cin>>a>>b;
        temp = f(a);
        cout<<"Case #"<<++kase<<": "<<cal(b)<<endl;
    }

    return 0;
}

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