F(x)

来源：互联网发布：ubuntu如何安装flash 编辑：程序博客网时间：2024/05/08 14:15

F(x)

Time Limit: 1000/500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5055 Accepted Submission(s): 1880

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
题意：找出i在0到b之间的f(i)小于等于f(a)的数的个数。
思路：数位dp。主要在于状态转移不好想。dp[i][j]表示i位数比j小的数的个数。用递归完成的话就只需要思考边界和状态转移。
边界：
dp[i][j]如果j小于0，显然是dp[i][j]=0的，如果i==0，说明就是0，显然任何数都比0大，所以dp[i][j]对于j>=0的时候dp[i][j]=1,否则dp[i][j]=0。
状态转移：
dp[i][j]+=dp[i-1][j-k*(1<<(i))]；j的初始值是sum
完成上述两步推导就能开始写这题了。
#include <bits/stdc++.h>using namespace std;typedef long long LL;const int MOD = 1000000007;const int maxn = 200010;int  num[30],len;LL  dp[30][10000];int a[50];int sum;LL dfs(int pos,LL zhi,int last)//zhi是差值{    if(pos<0) return zhi>=0;//当差值大于等于0时，说明这个数符合要求，+1，否则+0    if(zhi<0) return 0;//当差值小于0时就+0，    if(!last&&dp[pos][zhi]!=-1)    {        return dp[pos][zhi];    }    int len=last?num[pos]:9;    LL res = 0;    for(int i=0; i<=len; i++)    {        res += dfs(pos-1,(zhi-i*(1<<pos)),last&&(i==len));    }    if(!last)dp[pos][zhi]= res;    return res;}LL solve(LL n){    len = 0;    while(n)    {        num[len++] = n%10;        n /= 10;    }    return dfs(len-1,sum,1);//用sum一直去减}int main(){    memset(dp,-1,sizeof(dp));    int _;    scanf("%d",&_);    int g=0;    while(_--)    {        LL n,m;        ++g;        scanf("%I64d%I64d",&m,&n);        sum=0;        int l=0;        while(m)//算sum        {            sum+=(m%10)*(1<<l);            l++;            m=m/10;        }        printf("Case #%d: ",g);        if(sum==0)            printf("%d\n",1);        else        printf("%I64d\n",solve(n));    }    return 0;}

0 0