hdu 4734 F(x) （数位dp）

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F(x)

Time Limit: 1000/500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1140 Accepted Submission(s): 459

Problem Description

For a decimal number x with n digits (A_nA_n-1A_n-2 ... A₂A₁), we define its weight as F(x) = A_n * 2^n-1 + A_n-1 * 2^n-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. Thet 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

Source

2013 ACM/ICPC Asia Regional Chengdu Online

题意：

看懂f(x)函数之后，求[0,B]内的f(x)的值小于f(a)的有多少个。

思路：

数位dp，由于f(x)的值不大，可以考虑把f(x)放入状态中。

dp[i][j][k] - i 位 j 开头f(x)值为k的数的个数。

那么有dp[i][j][k]=dp[i-1][p][k-j*2^(i-1)] 。

然后算一个区间内的个数时，一位一位往下算就够了。

考虑到时间的原因，采用了一个sum数组纪录前缀和。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <map>#include <stack>#include <vector>#include <set>#include <queue>#pragma comment (linker,"/STACK:102400000,102400000")#define maxn 10005#define MAXN 100005#define mod 100000000#define INF 0x3f3f3f3f#define pi acos(-1.0)#define eps 1e-6typedef long long ll;using namespace std;int n,m,ans,cnt,tot,flag;int a,b,fac[10];int dp[10][10][5500],sum[10][10][5500];void solve(){    int i,j,t=0,u,v;    for(i=9;i>=1;i--)    {        t+=((a/fac[i-1])%10)*(1<<i-1);    }    a=t;    ans=0;    for(i=9;i>=1;i--)    {        u=(b/fac[i-1])%10;        for(j=0;j<u;j++)        {            ans+=sum[i][j][a];        }        a-=u*(1<<i-1);        if(a<0) return ;    }    ans++;}int main(){    int i,j,k,p,q,t,test=0;    fac[0]=1;    for(i=1;i<10;i++)    {        fac[i]=fac[i-1]*10;    }    dp[0][0][0]=1;    for(i=1;i<=9;i++)    {        for(j=0;j<=9;j++)        {            for(k=0;k<=5000;k++)            {                t=0;                for(p=0;p<=9;p++)                {                    if((k-(1<<i-1)*j)>=0) t+=dp[i-1][p][k-(1<<i-1)*j];                }                dp[i][j][k]=t;                if(k>0) sum[i][j][k]=sum[i][j][k-1]+t;                else sum[i][j][k]=t;            }        }    }    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&a,&b);        solve();        printf("Case #%d: %d\n",++test,ans);    }    return 0;}/*30 1001 105 100Case #1: 1Case #2: 2Case #3: 13*/

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