hdu4734---F(x)(数位dp)

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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设dp[i][j]表示i位数,F值小于等于j的数的个数

/*************************************************************************    > File Name: hdu4734.cpp    > Author: ALex    > Mail: [e-mail]     > Created Time: 2015年02月26日 星期四 13时17分59秒 ************************************************************************/#include <map>#include <set>#include <queue>#include <stack>#include <vector>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const double pi = acos(-1);const int inf = 0x3f3f3f3f;const double eps = 1e-15;typedef long long LL;typedef pair <int, int> PLL;int FA;int dp[15][10000];int bit[15];int dfs (int cur, int e, int k, bool flag, bool zero){    if (k < 0)    {        return 0;    }    if (cur == -1)    {        return 1;    }    if (!flag && ~dp[cur][k])    {        return dp[cur][k];    }    int ans = 0;    int end = flag ? bit[cur] : 9;    for (int i = 0; i <= end; ++i)    {        if (zero && !i)        {            ans += dfs (cur - 1, 0, k, flag && (i == end), 1);        }        else        {            ans += dfs (cur - 1, i, k - i * (1 << cur), flag && (i == end), 0);        }    }    if (!flag)    {        dp[cur][k] = ans;    }    return ans;}int calc (int n){    int cnt = 0;    while (n)    {        bit[cnt++] = n % 10;        n /= 10;    }    return dfs (cnt - 1, 0, FA, 1, 1);}int main (){    int t;    int icase = 1;    scanf("%d", &t);    memset (dp, -1, sizeof(dp));    while (t--)    {        int cnt = 0;        int a, b;        scanf("%d%d", &a, &b);        while (a)        {            bit[cnt++] = a % 10;            a /= 10;        }        FA = 0;        for (int i = 0; i < cnt; ++i)        {            FA += bit[i] * (1 << i);        }        printf("Case #%d: ", icase++);        printf("%d\n", calc (b));    }    return 0;}