数位DP HDU3555

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 15025    Accepted Submission(s): 5427

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

3 1 50 500

 

Sample Output

0 1 15

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.

 

Author

fatboy_cw@WHU

 

Source

2010 ACM-ICPC Multi-University Training Contest（12）——Host by WHU

 

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题意：

  1~nz中包含49的数有几个。

代码：

/*
本题与HDU2089相似，把那道题的代码改了一下找出了不含49的，然后用总数减去，简单粗暴，当然有更好的方法。
*/
#include<iostream>
#include<string>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<vector>
#include<iomanip>
#include<queue>
#include<stack>
using namespace std;
int t;
long long dp[40][10],n;
void init()
{
    for(int i=1;i<22;i++)
    {
        for(int j=0;j<10;j++)
        {
            for(int k=0;k<10;k++)
            {
                if(j==4&&k==9) continue;
                dp[i][j]+=dp[i-1][k];
            }
        }
    }
}
long long insum(long long n)
{
    long long sum=0;
    int lne=1,c[30]={0};
    while(n)
    {
        c[lne++]=n%10;
        n/=10;
    }
    for(int i=lne-1;i>0;i--)
    {
        for(int j=0;j<c[i];j++)   //j不能等于c[i],因为dp[i][j]中的j代表第i位取j时的总数，而j后面的数
                                  //要全部遍历一遍，这里j后面并非取全部数而是取到给出的数的后几位
        {
            if(j==9&&c[i+1]==4) continue;
            sum+=dp[i][j];
        }
        if(c[i]==9&&c[i+1]==4) break;
    }
    return sum;
}
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        memset(dp,0,sizeof(dp));
        dp[0][0]=1;
        cin>>n;
        n+=1;
        init();
        long long a=insum(n);
        cout<<n-a<<endl;
    }
    return 0;
}

/*
别人的一个更好的方法。很难想到。
*/
#include<iostream>
#include<string>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<vector>
#include<iomanip>
#include<queue>
#include<stack>
using namespace std;
int t;
long long dp[25][3],n; //dp[i][0]表示到第i位没有49的个数，dp[i][1]表示到第i位没有49但第i位是9的个数,
                       //dp[i]][2]表示到第i位包含49的个数。
void init()
{
    dp[0][0]=1;        //初始化
    dp[0][1]=0;
    dp[0][2]=0;
    for(int i=1;i<23;i++)
    {
        dp[i][0]=10*dp[i-1][0]-dp[i-1][1];  //到第i位没有49的个数等于第i位依次取0~9连上后面没有49的，
                                            //然后去掉第i位取4，i-1位为9的情况。
        dp[i][1]=dp[i-1][0];                //第i位是9时
        dp[i][2]=10*dp[i-1][2]+dp[i-1][1];  //到第i位包含49的个数等于第i位依次取0~9连上后面包含49的,
                                            //再加上第i位取4时，i-1位是9的情况。
    }
}
long long insum(long long n)
{
    long long sum=0;
    int c[23]={0};
    int cnt=1;
    while(n)
    {
        c[cnt++]=n%10;
        n/=10;
    }
    bool flag=false;
    for(int i=cnt-1;i>0;i--)   //从高位到低位依次枚举
    {
        sum+=dp[i-1][2]*c[i];           //到第i-1位包含49，就加上0~c[i]个
        if(flag) sum+=dp[i-1][0]*c[i];
        else
        {
            if(c[i]>4) sum+=dp[i-1][1];  //如果第i位大于4了，并且i-1是9，则一定包含49.
        }
        if(c[i+1]==4&&c[i]==9) flag=true; //当从高位出现49之后后面dp[i][0]就无意义了，全加上
    }
    return sum;
}
int main()
{
    scanf("%d",&t);
    {
        while(t--)
        {
            scanf("%lld",&n);
            init();
            printf("%lld
",insum(n+1));  //计算结果不包含n本身
        }
    }
    return 0;
}

//记忆化搜索法
#include<iostream>
#include<string>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<vector>
#include<iomanip>
#include<queue>
#include<stack>
using namespace std;
long long n;
int t;
long long dp[40][3];
int c[40];
long long dfs(int lne,int have,int lim)
{
    if(lne<=0)
    return have==2;
    if(!lim&&dp[lne][have]!=-1)
    return dp[lne][have];
    long long ans=0;
    int nnum=lim?c[lne]:9;
    for(int i=0;i<=nnum;i++)
    {
        int nhave=have;
        if(have==0&&i==4) nhave=1;
        if(have==1&&i!=4&&i!=9) nhave=0;
        if(have==1&&i==9) nhave=2;
        ans+=dfs(lne-1,nhave,lim&&i==nnum);
    }
    if(!lim)
    dp[lne][have]=ans;
    return ans;
}
int main()
{
        scanf("%d",&t);
        while(t--)
        {
            scanf("%lld",&n);
            memset(dp,-1,sizeof(dp));
            int cnt=0;
            while(n)
            {
                c[++cnt]=n%10;
                n/=10;
            }
            c[cnt+1]=0;
            printf("%lld
",dfs(cnt,0,1));
        }
    return 0;
}