HDU4734 F(x)

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

30 1001 105 100

Sample Output

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

人家说是入门级的，可我怎么想也没想出来，我知道优化的方向就是dp的初始化应该只有一次就好了，然而就是没想到转化一下思维。

一开始第二维表示的是当前数的价值的和，然后dp数组记忆化搜索的作用就显得不是那么大了。因为你存和跟你的A的价值的限制是有关系的，每次dp都初始化的话，作用就大打折扣了。

这时候就需要转化一下思维，第二维表示，价值不超过这个数的数的个数，这样，下一次相同的A出现的时候可以直接用。A变得时候不会影响到dp的状态的。

//#include <bits/stdc++.h>#include <cstdio>#include <cstring>using namespace std;typedef long long LL;int a,b;int dp[20][4600];int digit[20];int f(int x){    int sum = 0;    int p = 1;    while(x)    {        sum += (x%10)*p;        x /= 10;        p <<= 1;    }    return sum;}int dfs(int len,int sum,bool up){    if(sum < 0)return 0;    if(len == -1)return sum >= 0;    if(!up && dp[len][sum] != -1)return dp[len][sum];    int res = 0;    int n = up?digit[len] : 9;    for(int i = 0 ; i <= n ; ++i)    {        res += dfs(len - 1,sum - i*(1<<len),up && i==n);    }    if(!up)dp[len][sum] = res;    return res;}int cal(int x){    int len = 0;    while(x)    {        digit[len++] = x % 10;        x /= 10;    }    return dfs(len - 1,f(a),1);}int main(){    int t;    scanf("%d",&t);    memset(dp,-1,sizeof dp);    for(int tt = 1 ; tt <= t ; ++tt)    {        scanf("%d%d",&a,&b);        printf("Case #%d: %d\n",tt,cal(b));    }    return 0;}