51nod 1042 number of digits 0-9 digits DP

Title Link: https://www.51nod.com/onlineJudge/questionCode.html#!problemId=1042

Title: Give an interval of a-B to count the number of occurrences within this interval of 0-9. For example, 10-19,1 appears 11 times (10,11,12,13,14,15,16,17,18,19, of which 11 includes 2 1), and the remaining numbers appear 1 times. Input

Two number A, a (1 <= a <= b <= 10^18)

Output

Output total 10 rows, respectively, 0-9 occurrences

This problem feels very troublesome, tangled for a long time.

We use recursion to consider each one. Into the already calculated low, current bit, not yet calculated high.

There are only a few cases when dealing with the current bit:

1, the current position and low-level relationship

2, the relationship between high and low

3, the current relationship with the high

Three cases can be processed. Note that 0 of special cases, such as 01 in fact, do not have 0.

#include <iostream>
#include <bits/stdc++.h>
#define LL Long long
using namespace std;

void Dfs (LL a,ll b,ll c[])
{
    ll n=a/10,m=a%10,t=n;
    for (int i=0;i<=m;i++)   c[i]+=b;//Current bit-to-low effect for
    (int i=0;i<10;i++)   c[i]+=b*n;//High-to-low effect
    c[0] -=b;//0 special treatment, the number of 0 minus while
    (t)//The current bit of the effect of the high
        (c[t%10]+=b* (m+1);//Plus 0
        t/=10;
    }
    if (n)   dfs (N-1,B*10,C);//n has been processed, so to process n-1
}

ll x[20],y[20];

int main ()
{
    ll A, B;
    cin>>a>>b;
    DFS (a-1,1,x);
    DFS (b,1,y);
    for (int i=0;i<10;i++)   cout<<y[i]-x[i]<<endl;
}