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Question c:repeat number

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Title Description

definition:a+b = c, if all the digits of C being same (c is more than ten) , Then we call A and B is Repeat number. My question is what many Repeat Numbers in [x, Y].

Input

There is several test cases.

Each test cases contains the integers x, y (1<=x<=y<=1,000,000) described above.

Proceed to the end of file.

Output

For each test output, the number of couple of Repeat number in one line.

Sample input

1 10 10 12

Sample output

5 2

Tips

If a equals B, we can call a, b be Repeat Numbers too, and A is the Repeat Numbers for itself.

This problem, I was violent plus two points, the enumeration of the possible values of C can be. Then we can find the minimum value of the middle point and the boundary difference.

#include <iostream> #include <cstring> #include <cstdio> #include <algorithm> #include <    Cmath> #include <vector>using namespace std;vector<int> g;void init () {int k=1;        for (int i=0;i<6;i++) {k=k*10+1;    for (int j=1;j<=9;j++) g.push_back (K*J);    }}int Main () {int x, y;    Init ();        while (cin>>x>>y) {int p=lower_bound (G.begin (), G.end (), 2*x)-g.begin ();        int Q=lower_bound (G.begin (), G.end (), 2*y)-g.begin ();        int ans=0;        for (int i=0;i<10;i++)//cout<<g[i]<<endl;       if (g[q]>2*y) q--;        cout<<p<< "" <<q<<endl;            for (int i=p;i<=q;i++) {int mid = G[I]/2; if (mid* 2 = = G[i]) ans + = Mid-x < Y-mid?            mid-x+1:y-mid+1; else ans+= mid-x+1 < y-mid?        Mid-x+1:y-mid;    } cout<<ans<<endl; } return 0;}