hdu4734——F(x) (数位DP)

F(x)

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

Total Submission(s): 5174    Accepted Submission(s): 1932


[align=left]Problem Description[/align]
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).
 

[align=left]Input[/align]
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)
 

[align=left]Output[/align]
For every case,you should output "Case #t: " at first, without quotes. Thet is the case number starting from 1. Then output the answer.
 

[align=left]Sample Input[/align]

3
0 100
1 10
5 100

 

[align=left]Sample Output[/align]

Case #1: 1
Case #2: 2
Case #3: 13

 

[align=left]Source[/align]
2013 ACM/ICPC Asia Regional Chengdu Online 

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-1))]；

完成上述两步推导就能开始写这题了。

#include <iostream>

#include <cstdio>

#include <cstring>

#include <queue>

#include <cmath>

#include <algorithm>

#include <vector>

#include <map>

#include <string>

#include <stack>

using namespace std;

typedef long long ll;

#define PI 3.1415926535897932

#define E 2.718281828459045

#define INF 0x3f3f3f3f

#define mod 100000007

const int M=1005;

int n,m;

int cnt;

/*int sx,sy,sz;

int mp[1000][1000];

int pa[M*10],rankk[M];

int head[M*6],vis[M*100];

int dis[M*100];

ll prime[M*1000];

bool isprime[M*1000];

int lowcost[M],closet[M];

char st1[5050],st2[5050];

int len[M*6];

typedef pair<int ,int> ac;

//vector<int> g[M*10];

int has[10500];

int month[13]= {0,31,59,90,120,151,181,212,243,273,304,334,0};

int dir[8][2]= {{0,1},{0,-1},{-1,0},{1,0},{1,1},{1,-1},{-1,1},{-1,-1}};

void getpri()

{

    ll i;

    int j;

    cnt=0;

    memset(isprime,false,sizeof(isprime));

    for(i=2; i<1000000LL; i++)

    {

        if(!isprime[i])prime[cnt++]=i;

        for(j=0; j<cnt&&prime[j]*i<1000000LL; j++)

        {

            isprime[i*prime[j]]=1;

            if(i%prime[j]==0)break;

        }

    }

}

struct node

{

    int v,w;

    node(int vv,int ww)

    {

        v=vv;

        w=ww;

    }

};

vector<int> g[M*100];

string str[1000];

*/

int dp[20][200000];

int bit[50];

int dfs(int cur,int s,int e,int z){

    if(s<0)return 0;//不能光相信模板，这么实际的情况都米想到

    if(cur<0) return s>=0;//s的初值是F（n）,递归到头s>=0就是当前数小于F（n）return check(s);

    if(!e&&!z&&dp[cur][s]!=-1) return dp[cur][s];

    int endx=e?bit[cur]:9;

    int ans=0;

    for(int i=0;i<=endx;i++){

            //if(i==4||s&&i==2)continue;

        if(z&&!i) ans+=dfs(cur-1,s-i*(1<<cur),e&&i==endx,1);

        else ans+=dfs(cur-1,s-i*(1<<cur),e&&i==endx,0);

            //ans+=dfs(cur-1,get_news(s,i),e&&i==endx,0);

    }

    if(!e&&!z) dp[cur][s]=ans;

    return ans;

}

int F(int a){

    int ans=0,len=0;

    while(a){

        ans+=(a%10)*(1<<len);

        len++;

        a/=10;

    }

    return ans;

}

int solve(int x){

    int len=0;

    while(x){

        bit[len++]=x%10;

        x/=10;

    }

    return dfs(len-1,F(n),1,1);

}

int main()

{

    int i,j,k,t;

    int l,r,cas=0;

    scanf("%d",&t);

    memset(dp,-1,sizeof(dp));//对于数位DP，memset拿到外边就行，因为统计的数位信息会有被重用的可能

    while(t--)

    {

        scanf("%d%d",&n,&m);

        printf("Case #%d: %d\n",++cas,solve(m));

    }

    return 0;

}