F(x)（逆向思维）_逆向思维 竞赛

F(x)

Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8583    Accepted Submission(s): 3386

 

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

Source

2013 ACM/ICPC Asia Regional Chengdu Online

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题意：题目给了个f(x)的定义：F(x) = An * 2n-1 + An-1 * 2n-2 + ... + A2 * 2 + A1 * 1，Ai是十进制数位，然后给出a,b求区间[0,b]内满足f(i)<=f(a)的i的个数。

 

常规想：这个f(x)计算就和数位计算是一样的，就是加了权值，所以dp[pos][sum]，这状态是基本的。a是题目给定的，f(a)是变化的不过f(a)最大好像是4600的样子。如果要memset优化就要加一维存f(a)的不同取值，那就是dp[10][4600][4600]，这显然不合法。

这个时候就要用减法了，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，

也就是说初始的是时候sum是f(a),枚举一位就减去这一位在计算f(i)的权值，那么最后枚举完所有位 sum>=0时就是满足的，后面的位数凑足sum位就可以了。

仔细想想这个状态是与f(a)无关的(新手似乎很难理解)，一个状态只有在sum>=0时才满足，如果我们按常规的思想求f(i)的话，那么最后sum>=f(a)才是满足的条件。

所以我们就逆向思维，考虑小于他的。

这题一定要用减法，这样最终的结果就会和f(A)无关(sum>=0)如果最后判断sum<=f(A)那么dp中还要记录f(A)

代码：

#include<bits/stdc++.h>
using namespace std;
int dp[20][200000];
int bit[20];
int dfs(int pos,int num,bool flag)
{
    if(pos == -1)
        return num >= 0;
    if(num < 0)///!!!
        return 0;
    if(!flag && dp[pos][num] != -1)
        return dp[pos][num];
    int ans = 0;
    int end = flag?bit[pos]:9;
    for(int i = 0;i <= end;i++)
    {
        ans += dfs(pos-1,num - i*(1<<pos),flag && i==end);
    }
    if(!flag)
        dp[pos][num] = ans;
    return ans;
}
int F(int x)
{
    int ret = 0;
    int len = 0;
    while(x)
    {
        ret += (x%10)*(1<<len);
        len++;
        x /= 10;
    }
    return ret;
}
int A,B;
int calc()
{
    int len = 0;
    while(B)
    {
        bit[len++] = B%10;
        B/=10;
    }
    return dfs(len-1,F(A),1);
}
int main()
{
    int T;
    int iCase = 0;
    scanf("%d",&T);
    memset(dp,-1,sizeof(dp));
    while(T--)
    {
        iCase++;
        scanf("%d%d",&A,&B);
        printf("Case #%d: %d\n",iCase,calc());
    }
    return 0;
}