F(x)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4734

F(x)

Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8015    Accepted Submission(s): 3141

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

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

 

题目大意：t组测试样例，输入n，m,按上述方法计算出f(n),然后问你[0,m]之中有多少个数按上述方法求出来的值小于等于f(n)

个人思路：其实这道题就算加了权值的数位dp了，每位数按着方法乘以权值就行，然后还有一个优化，不然会超时。。。这道题按照一般的思路，就算直接求了，从最高位开始，一直往下

求，如果在这期间已经大于了就直接退出来，但是这里有一个优化dp[pos][sum],这里的sum 不是当前的前缀和(加了权值的）而是当前位离答案还差<=sum的数的个数，为什么是这样呢？

如果sum值是前缀和的话，因为目标是不一样的(也就是f(n))，所以即使计算出这个也没啥用，大概思路就算这个了

看代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<stdio.h>
#include<string.h>
#include<cmath>
#include<math.h>
#include<algorithm>
#include<set>
#include<queue>
#include<map>
typedef long long ll;
using namespace std;
const ll mod=1e9+7;
const int maxn=1e8+10;
const int maxk=100+10;
const int maxx=1e4+10;
const ll maxa=43200;
#define INF 0x3f3f3f3f3f3f
int a[15];
int dp[15][10000];
int edge;
int dfs(int pos,int sum,int limit)
{
    if(pos==-1)
        return sum<=edge?1:0;
     //  return sum<=edge;
     if(sum>edge)
        return 0;
    if(!limit&&dp[pos][edge-sum]!=-1) return dp[pos][edge-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==a[pos]);
    }
    if(!limit) dp[pos][edge-sum]=ans;
    return ans;
}
ll f(ll n)
{
    ll t=1,ans=0;
    while(n)
    {
        ans+=n%10*t;
        n=n/10;
        t*=2;
    }
    return ans;
}
ll solve(ll n)
{
    int pos=0;
    while(n)
    {
        a[pos++]=n%10;
        n=n/10;
    }
    return dfs(pos-1,0,1);
}
int main()
{
    int t;
    cin>>t;
    memset(dp,-1,sizeof(dp));
    for(int i=1;i<=t;i++)
    {
        ll n,m;
        cin>>n>>m;
        edge=f(n);
       // cout<<edge<<endl;
        printf("Case #%d: %I64d\n",i,solve(m));
    }
    return 0;
}