HDU-4734 F(x) （数位DP）

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F(x)

http://acm.hdu.edu.cn/showproblem.php?pid=4734

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

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. 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

题目大意：统计区间[0,b]中的数x满足f(x)<=f(a)的个数？

好不容易遇见一个较为简单的数位DP，结果忘了求f(a)，直接传入a，还debug半天...

看数据范围可以知道，max(f(x))=f(999999999)=4599，所以设dp[25][4615]即够

dp[i][j]表示长度为i的数，f(x)大小不超过j的数的个数

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int n,len,bit[25];int a,b;int dp[25][4615];//dp[i][j]表示长度为i的数，f(x)大小不超过j的数的个数int dfs(int pos,int rem,bool limit) {//pos表示当前考虑的位，rem表示还剩余多少数，limit表示当前位是否有限制    if(pos<=0) {        return rem>=0?1:0;    }    if(rem<0) {//如果剩余的数小于0，则剪去        return 0;    }    if(!limit&&dp[pos][rem]!=-1) {        return dp[pos][rem];    }    int mx=limit?bit[pos]:9;    int res=0;    for(int i=0;i<=mx;++i) {        res+=dfs(pos-1,rem-i*(1<<(pos-1)),limit&&i==mx);    }    if(!limit&&dp[pos][rem]==-1) {        dp[pos][rem]=res;    }    return res;}int f(int x) {    int res=0,w=1;    while(x>0) {        res+=(x%10)*w;        w<<=1;        x/=10;    }    return res;}int getCnt(int x) {//返回[0,x]中满足题意的数的答案    len=0;    while(x>0) {        bit[++len]=x%10;        x/=10;    }    return dfs(len,f(a),true);}int main() {    int T,kase=0;    memset(dp,-1,sizeof(dp));    scanf("%d",&T);    while(kase<T) {        scanf("%d%d",&a,&b);        printf("Case #%d: %d\n",++kase,getCnt(b));    }    return 0;}

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