hdu3555数位dp

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 4063    Accepted Submission(s): 1403

Problem Description

The counter-terrorists found a time bomb in the dust. But this time the terrorists improve on the time bomb. The number sequence of the time bomb counts from 1 to N. If the current number sequence includes the sub-sequence "49", the power of the blast would add one point.
Now the counter-terrorist knows the number N. They want to know the final points of the power. Can you help them?

 

Input

The first line of input consists of an integer T (1 <= T <= 10000), indicating the number of test cases. For each test case, there will be an integer N (1 <= N <= 2^63-1) as the description.

The input terminates by end of file marker.

 

Output

For each test case, output an integer indicating the final points of the power.

 

Sample Input

3150500

 

Sample Output

0115
Hint
From 1 to 500, the numbers that include the sub-sequence "49" are "49","149","249","349","449","490","491","492","493","494","495","496","497","498","499",so the answer is 15.
题意：求0-n中出现了多少个含有49的数字。
小结：第一次做数位dp的题，对数位dp的一般模式有了一些了解，不过还是有很多地方没搞懂，dp的初始化，这里为什么n要++，为什么以i开始的时候，后面呢能取到所有出现49的情况等等。
代码:
#include<iostream>#include<vector>#include<string>#include<queue>#include<map>#include<cstdio>#include<cstring>#define maxn 2005#define ll unsigned long long#define INF 0xfffffffusing namespace std;int digit[25];ll dp[25][3];//dp[i][0]表示长度为i不出现49的数字个数//dp[i][1]表示长度为i以9开头不出现49的数字个数//dp[i][2]表示长度为i出现49的数字个数void init(){    memset(dp,0,sizeof(0));    dp[0][0]=1;//这里一开始还是不理解，不过从后面推的话是正确的    for(int i=1;i<=20;i++)    {        dp[i][0]=10*dp[i-1][0]-dp[i-1][1];        dp[i][1]=dp[i-1][0];        dp[i][2]=dp[i-1][1]+10*dp[i-1][2];    }}int main(){    int t;    ll n,ans;    init();    cin>>t;    while(t--)    {        ans=0;        cin>>n;        n++;//不理解为什么要++        //求出n的每个数位        int cnt=0;        memset(digit,0,sizeof(digit));        while(n)        {            digit[++cnt]=n%10;            n/=10;        }        int flag=0;//标记是否前面出现过49        for(int i=cnt;i>=1;i--)        {            ans+=digit[i]*dp[i-1][2];//i-1位含有49   不理解这里有可能以digit[i]开头的，后面的所有dp[i-1][2]情况不是都取的到            if(flag)            {                ans+=digit[i]*dp[i-1][0];//i-1位不含有49    这里也是            }            if(flag==0&&digit[i]>4)            {                ans+=dp[i-1][1];            }            if(digit[i]==9&&digit[i+1]==4)            flag=1;        }        cout<<ans<<endl;    }return 0;}
第二种做法：标准的记忆化搜索
今天大强给我讲了下划状态图，说dp的题都可以这种方法来做，一般状态图画出来的话就好些状态转移方程。
这题在花状态图的时候用了四个标记变量：pos记录当前的位置,pre记录上一个位置的数字,flag记录是否出现过49,limit记录是否受限制，就是能否取到9.
代码：
#include<iostream>#include<cstring>#define ll __int64using namespace std;int digit[25];ll dp[25][10][3];//dp记录的是没有限制的情况下的状态ll dfs(int pos,int pre,int flag,int limit){    if(pos==-1) return flag==1;//如果已经遍历完n，最后一个数字是否为含有49的数字看flag    if(!limit&&dp[pos][pre][flag]!=-1) return dp[pos][pre][flag];    ll sum=0;    int end=(limit?digit[pos]:9);    for(int i=0;i<=end;i++)    {        if(pre==4&&i==9)        sum+=dfs(pos-1,i,1,limit&&i==end);//受限制的条件是前面的所有位都是取最大的        else        sum+=dfs(pos-1,i,flag,limit&&i==end);    }    if(!limit) dp[pos][pre][flag]=sum;    return sum;}ll work(ll n){    int cnt=0;    while(n)    {        digit[cnt++]=n%10;        n/=10;    }    memset(dp,-1,sizeof(dp));    ll ans=dfs(cnt-1,-1,0,1);    return ans;}int main(){    int t;    ll n;    cin>>t;    while(t--)    {        cin>>n;        cout<<work(n)<<endl;    }    return 0;}