hdu3555（数位dp）-优快云博客

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.

数位dp基础题，和网上的方法不大一样，用dp[i][j]记录以j开头的i位数里面含有49的有多少个。

此题和那个杭电的不要‘62’dp记录表示的正好相反，本来还以为计算方法一样的，结果代码一敲出来答案全是0。。。

所以自己又仔细思考了一下，这个dp记录方式如果相反的话，得到dp值是需要我们思考的！

typedef long long ll;
ll dp[25][25];

void init()
{
    memset(dp,0,sizeof(dp));
    ll hh=1;
    for(int i=2;i<=18;i++)
    {
       for(int j=0;j<10;j++)
          for(int k=0;k<10;k++)
        {
            if(j==4&&k==9)
            dp[i][j]+=hh;
            else
        dp[i][j]+=dp[i-1][k];
        }
        hh*=10;
    }
}

如果理解dp值的含义并自己推到一下其实我的代码很好理解的。

再就是最后的处理了。dp的最终处理绝对比算出dp值要难。。。。

将数字分解为每一位，从高位到低位枚举，不过，需要注意的是若遇到494949这个数字，我们枚举的步骤也得要理解！首先ans+=dp[6][0~3],然后ans+=[5][0~8],接着！看到49了！也就意味着49****后面4个数字怎么变都会含有49，那么我们在此直接ans+=4950!(因为从0到4949有4950个数)

？然后怎么办呢？继续？，这就是我说的要理解枚举的步骤，我们首先分别枚举了以0,1,2,3开头的6位数含有49的个数，剩下400000到494949含有49的个数，然后我们枚举了400000到489999的含有49的个数，剩下490000到494949含有49的个数，接着，我们就直接加上了4950，因为490000到494949所有的数字都有49，直接加上就ok！然后我们不就枚举完毕了吗？！！我们还需要继续往下用ans去加dp[0~4]的值吗？？我觉得已经说得比较明了了吧！所以枚举这个过程一定要自己理解！

#include <iostream>
#include <stdio.h>
#include<string.h>
#include<math.h>
using namespace std;
typedef long long ll;
ll dp[25][25];

void init()
{
    memset(dp,0,sizeof(dp));
    ll hh=1;
    for(int i=2;i<=18;i++)
    {
       for(int j=0;j<10;j++)
          for(int k=0;k<10;k++)
        {
            if(j==4&&k==9)
            dp[i][j]+=hh;
            else
        dp[i][j]+=dp[i-1][k];
        }
        hh*=10;
    }
}
int main()
{
    init();
    int t;
    //for(int i=1;i<=18;i++){for(int j=0;j<10;j++)cout<<dp[i][j]<<' ';cout<<endl;}
    scanf("%d",&t);
        ll n;
    while(t--)
    {
        int digit[25]={0},bj=0;
        scanf("%lld",&n);ll nn=n,ans=0;
        while(n)
        {digit[++bj]=n%10;n/=10;}
        bool flag=0;
        for(int i=bj;i>0;i--)
        {
            for(int j=0;j<digit[i];j++)
                ans+=dp[i][j];
               if(flag==1)
                break;
            if(digit[i]==4&&digit[i-1]==9&&flag==0)
            {
                    ll tem=0;
                    if(bj>2)
                    {
                    for(int mm=i-2;mm>0;mm--)
                        tem*=10,tem+=digit[mm];
                    ans+=tem+1;
                    }else
                    ans++;
                   flag=1;
            }
        }

        cout<<ans<<endl;
    }

    return 0;
}

转载于:https://www.cnblogs.com/martinue/p/5490509.html