Description
The Little Elephant very much loves sums on intervals.
This time he has a pair of integers l and r (l ≤ r). The Little Elephant has to find the number of such integers x (l ≤ x ≤ r), that the first digit of integer x equals the last one (in decimal notation). For example, such numbers as 101, 477474 or 9 will be included in the answer and47, 253 or 1020 will not.
Help him and count the number of described numbers x for a given pair l and r.
The single line contains a pair of integers l and r (1 ≤ l ≤ r ≤ 1018) — the boundaries of the interval.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64dspecifier.
On a single line print a single integer — the answer to the problem.
2 47
12
47 1024
98
In the first sample the answer includes integers 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44.
题意：有这种数字，第一位和最后一位相同，问你两个数字范围内有多少符合要求的数字~
解法：当然是找规律啊，很容易看出规律是9*10^n，然后考虑两个端点的情况，他们是不是也是符合要求的（这里的代码写的不清楚QAQ，最有价值的只有cmd函数了）
#include<bits/stdc++.h>
using namespace std;
long long a[100005];
int pos;
int n;
int d;
long long cmd(string s)
{
long long sum1=0,sum2=0;
if(s.length()==1)
{
sum1=0;
}
else if(s.length()==2)
{
sum1=9;
}
else
{
sum1=10;
for(int i=1;i<s.length()-2;i++)
{
sum1*=10;
}
// cout<<sum1<<endl;
sum1=sum1-1;
// cout<<sum1<<endl;
sum1+=9;
}
int ans1=s[0]-'0'-1;
long long sum3=0;
long long pos1=1;
if(s.length()>=3)
{
for(int i=1; i<=s.length()-2; i++)
{
sum3=sum3*10+(s[i]-'0');
}
// sum3=sum3+1;
for(int i=0; i<s.length()-2; i++)
{
pos1*=10;
}
if((s[0]-'0')<=(s[s.length()-1]-'0'))
{
sum1=sum1+ans1*pos1+sum3+1;
}
else
{
sum1=sum1+ans1*pos1+sum3;
}
}
else
{
if(s[0]<=s[s.length()-1])
{
sum1=sum1+ans1*pos1+sum3+1;
}
else
{
sum1=sum1+ans1*pos1+sum3;
}
}
return sum1;
}
int main()
{
string s1,s2;
cin>>s1>>s2;
// cout<<cmd(s1)<<endl;
// cout<<cmd(s2)<<endl;
if(s1[0]==s1[s1.length()-1]&&s2[0]<=s2[s2.length()-1])
{
cout<<cmd(s2)-cmd(s1)+1<<endl;
}
else if(s1[0]==s1[s1.length()-1]&&s2[0]>s2[s2.length()-1])
{
cout<<cmd(s2)-cmd(s1)+1<<endl;
}
else if(s1[0]<=s1[s1.length()-1]&&s2[0]<=s2[s2.length()-1])
{
cout<<cmd(s2)-cmd(s1)<<endl;
}
else if(s1[0]<=s1[s1.length()-1]&&s2[0]>s2[s2.length()-1])
{
cout<<cmd(s2)-cmd(s1)<<endl;
}
else if(s1[0]>s1[s1.length()-1]&&s2[0]<=s2[s2.length()-1])
{
cout<<cmd(s2)-cmd(s1)<<endl;
}
else if(s1[0]>s1[s1.length()-1]&&s2[0]>s2[s2.length()-1])
{
cout<<cmd(s2)-cmd(s1)<<endl;
}
else if(s1[0]>s1[s1.length()-1]&&s2[0]<=s2[s2.length()-1])
{
cout<<cmd(s2)-cmd(s1)<<endl;
}
return 0;
}