F(x) [数位DP]

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).
题意：给出对于数字x， f(x)的计算方法，问一个区间[0, B]内有多少个数字的f（i）值是不大于f(A)的

数位dp的思路很简单，用dp[pos][sum]表示状态，由于每一次询问的结果与输入f（A）有关，所以直接计算sum是不能记忆化的，必须每次询问去memset，这样子就会超出时间限制。如果在dp状态上记录一个f(A),这样空间也会受到限制。 所以在计算sum的时候，我们表示后面还差多少值，也就是给sum一个初始的F(A)，然后去减每一位的权值。

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
LL a;
LL dp[20][5555];
int dig[15];
int z[15];
LL dfs(int pos, int sum, int limit)
{
    if(pos < 1) return sum >= 0;
    if(sum < 0) return 0;
    if(!limit && dp[pos][sum] != -1)
        return dp[pos][sum];
    int n = limit ? dig[pos] : 9;
    LL res = 0;
    for(int i = 0; i <= n; ++i)
    {
        res += dfs(pos-1, sum-z[pos]*i, limit&&i==n);
    }
    if(!limit) dp[pos][sum] = res;
    return res;
}
LL f(LL a)
{
    int F = 0;
    int len = 1;
    while(a)
    {
        F += len*(a%10);
        a /= 10;
        len *= 2;
    }
    return F;
}
LL call(LL x)
{
    int len = 0;
    while(x)
    {
        dig[++len] = x%10;
        x /= 10;
    }
    return dfs(len, f(a), 1);
}

int main()
{
    ios::sync_with_stdio(false);
    memset(dp, -1, sizeof(dp));
    z[1] = 1;
    for(int i = 2; i < 15; ++i)
        z[i] = z[i-1]*2;
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    #endif // ONLION_JUDGE
    int Case = 1;
    int T;
    cin >> T;
    while(T--)
    {
        LL b;
        cin >> a >> b;
        cout << "Case #" << Case++ << ": " << call(b) << endl;
    }
    return 0;
}