Codeforces Round #129 (Div. 2) C

Source: Internet
Author: User


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 have to find the number of such integers  x   ( l x R ), that the first digit of Integer  x & Nbsp;equals the last one (in decimal notation). For example, such numbers as 101, 477474 or < Span class= "Tex-span" >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≤ lR ≤10 18)-the boundaries of the interval.

%lld specifier to read or write 64-bit integers inс++. It is preferred to use cin, cout streams or the %i64dspecifier.


On a single line print a single integer-the answer to the problem.

Examples input
2 47
47 1024

In the first sample, the answer includes integers 2, 3, 4, 5, 6, 7, 8, 9, one,,.

Test instructions: With this number, the first and the last one are the same, asking you how many numbers meet the required number in the two number range ~

Solution: Of course, is to find the law, it is easy to see the law is 9*10^n, and then consider the situation of two endpoints, they are not also meet the requirements (here the code is not clear Qaq, the most valuable only the CMD function)

#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) <&    Lt;endl; } else if (S1[0]>s1[s1.length () -1]&&s2[0]<=s2[s2.length ()-1]) {cout<<cmd (s2)-cmd (S1) <&    Lt;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) <&    Lt;endl; } return 0;}


Codeforces Round #129 (Div. 2) C

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