[kuangbin带我飞]数位DP F(x)

最新推荐文章于 2021-06-07 20:39:12 发布

原创最新推荐文章于 2021-06-07 20:39:12 发布 · 490 阅读

0 ·

CC 4.0 BY-SA版权

本文介绍了一道关于位运算与数位DP的算法题，详细解析了如何利用位运算特性优化数位DP算法，避免重复计算，提高算法效率。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

New~ 欢迎参加2016多校联合训练的同学们~

F(x)

Time Limit: 1000/500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3768 Accepted Submission(s): 1397

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

Sample Output

      
       Case #1: 1
Case #2: 2
Case #3: 13

Source

2013 ACM/ICPC Asia Regional Chengdu Online

Recommend

liuyiding | We have carefully selected several similar problems for you: 5722 5721 5720 5719 5718

Statistic | Submit | Discuss | Note

先学习一个位运算

<<是左移，比如1<<n，表示1往左移n位，即数值大小2的n次方
>>右移，类似左移，数值大小除以2的n次方

题目意思:对于一个数a，一个数b；

求在[0,b]的范围内有多少个数的F函数的值要小于等于FA；

输出数的个数！

解题思路:

对于每一对a，b所求出的FA 显然都是不一样的!

有错误的做法是:用数位DP把每一个数的	F函数值求出来并和FA 比较，好像这样的思路也是可以的,但是你会发现没有办法保存DP[pos][sum],因为这个DP里面保存的res是针对一个FA所得到的结论，在第二组样例中这样的DP显然就不能重复利用了！如果对每一组样例的DP初始化重新计算，这样你的答案会正确。只是显然这样会TLE。

有正确的做法是：每一次数位DP,其sum值保存的为FA-FX,这样就可以重复利用你的DP啦！

错误代码：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<iostream>
    using namespace std;
#define ll long long
int fa;
int DP[20][6000];
int str[20];
int multi(int x){
    int count=1;
    int ans=0;
    while(x){
        ans+=(x%10)*count;
        count<<=1;
        x=x/10;
    }
    return ans;
}
int dfs(int pos,int sum,int limit){
    if(pos<0){
        return sum-fa<=0;
    }
    if(sum>fa)
        return 0;
    if(!limit&&DP[pos][sum]!=-1)
        return DP[pos][sum];
    int res=0;
    int end=limit?str[pos]:9;
    for(int i=0;i<=end;i++){
        res+=dfs(pos-1,sum+i*(1<<pos),limit&&(i==end));
    }

    return res;
}
int solve(int x){
    int len=0;
    while(x){
        str[len++]=x%10;
        x=x/10;
    }
    len--;
    return dfs(len,0,1);
}
int main(){
    int cas;
    int a,b;
    scanf("%d",&cas);
    memset(DP,-1,sizeof(DP));
    for(int ii=1;ii<=cas;ii++){

        scanf("%d%d",&a,&b);
        fa=multi(a);

        printf("Case #%d: %d\n",ii,solve(b));
    }
    return 0;
}<span style="color:#ff0000;">
</span>

正确代码：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<iostream>
    using namespace std;
#define ll long long
int fa;
int DP[20][6000];
int str[20];
int multi(int x){
    int count=1;
    int ans=0;
    while(x){
        ans+=(x%10)*count;
        count<<=1;
        x=x/10;
    }
    return ans;
}
int dfs(int pos,int sum,int limit){
    if(pos<0){
        return sum>=0;
    }
    if(sum<0)
        return 0;
    if(!limit&&DP[pos][sum]!=-1)
        return DP[pos][sum];
    int res=0;
    int end=limit?str[pos]:9;
    for(int i=0;i<=end;i++){
        res+=dfs(pos-1,sum-i*(1<<pos),limit&&(i==end));
    }
    if(!limit)
        DP[pos][sum]=res;
    return res;
}
int solve(int x){
    int len=0;
    while(x){
        str[len++]=x%10;
        x=x/10;
    }
    len--;
    return dfs(len,fa,1);
}
int main(){
    int cas;
    int a,b;
    scanf("%d",&cas);
    memset(DP,-1,sizeof(DP));
    for(int ii=1;ii<=cas;ii++){

        scanf("%d%d",&a,&b);
        fa=multi(a);
        //cout<<fa<<endl;
        printf("Case #%d: %d\n",ii,solve(b));
    }
    return 0;
}