51nod 1042 Number 0-9 (digital DP)

Title Description

Give an interval of a-B to count the number of occurrences in 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 a total of 10 rows, respectively, 0-9 occurrences of the input example

10 19

Output Example

1
1
1
1
1
1
1
1
1

Thinking of solving problems

Solution One:
Considering the feasible value of each digit from low to High, dp[i] represents the number of feasible values for the current number of digits. Take 235678 as an example, assuming that the now=5 is currently traversed, this high=23,after=678,tmp=1000:
1. When I < now, the high value range is [0,23], then dp[i]+= (high+1) *tmp;
2. When I > now, the high value range is [0,22], then dp[i]+=high*tmp;
3. When i = Now, corresponds to whether the high value is equal to two cases. When the high =high, the low value can not exceed after, that is, there are after+1 (plus a low value of 0 this case); When the high value is [0,22], the corresponding case is high* tmp. Then dp[i]+=high*tmp+after+1.
PS: When i=0, it is necessary to subtract the high value of 0, that is, dp[0]-=tmp.

Solution Two:
Implemented in recursive mode, Dp[i] represents the number of possible values for the current number of digits. Also take 235678 as an example, assuming that the current traversal to 5 this digit, because in the recursive process, 235*** this situation in the previous layer has been calculated, so the current layer x=234, corresponding Now=4,before =23=t, tmp=1000.
1. Consider that the value range of the _,# position is [0,x], the dp[0~now]+=tmp, and for the high 2, 3, corresponds to appear (now+1) *tmp times, namely dp[2,3]+= (now+1) *tmp;
2. Consider _ _ I _ _ _ (i∈[0,9]), high-level corresponding to the value [0,22], then dp[i]+=before*tmp. Also need to subtract i=0 and high value is also 0 of the case. Code Implementation

Solution One

#include <bits/stdc++.h>
using namespace std;
#define IO Ios::sync_with_stdio (false); \
cin.tie (0); \
cout.tie (0)
; typedef long long LL;
ll num[10]= {0};
void Solve (ll x,int mark)
{
    ll tmp=1,after=0,high,now;
    while (x)
    {
        now=x%10;
        HIGH=X/10;
        for (int i=0; i<=9; i++)
        {
            if (now<i)
                num[i]=num[i]+mark*high*tmp;
            else if (now>i)
                num[i]=num[i]+mark* (high+1) *tmp;
            else
                num[i]=num[i]+mark* (high*tmp+after+1);
        }
        num[0]=num[0]-mark*tmp;
        x/=10;
        after+=now*tmp;
        tmp*=10;
    }
}
int main ()
{
    IO;
    ll A, B;
    cin>>a>>b;
    Solve (b,1);
    Solve (a-1,-1);
    for (int i=0; i<10; i++)
        cout<<num[i]<<endl;
    return 0;
}

Solution Two:

 #include <bits/stdc++.h> using namespace std; #define IO Ios::sync_with_stdio (false); \
Cin.tie (0); \ cout.tie (0);
typedef long Long LL;

ll num[10]= {0};
    void Solve (ll x,ll tmp,int Mark) {ll now=x%10,before=x/10,t=before;
    for (int i=0;i<=now;i++) num[i]+=mark*tmp;
    for (int i=0;i<10;i++) num[i]+=mark*before*tmp;
    num[0]-=mark*tmp;
        while (t) {num[t%10]+=mark*tmp* (now+1);
    t/=10;
} if (before) solve (Before-1,tmp*10,mark);
    } int main () {IO;
    ll A, B;
    cin>>a>>b;
    Solve (b,1,1);
    Solve (a-1,1,-1);
    for (int i=0; i<10; i++) cout<<num[i]<<endl;
return 0; }