Handstand 2——AT（先打表）

题目描述

Given is a positive integer N.Find the number of pairs (A,B) of positive integers not greater than
N that satisfy the following condition:
·When A and B are written in base ten without leading zeros, the last digit of A is equal to the first digit of B, and the first digit of A is equal to the last digit of B.
Constraints
·1≤N≤2×105
·All values in input are integers.

输入

Input is given from Standard Input in the following format:
N

输出

Print the answer.

样例输入 Copy

【样例1】
25
【样例2】
1
【样例3】
100
【样例4】
2020
【样例5】
200000

样例输出 Copy

【样例1】
17
【样例2】
1
【样例3】
108
【样例4】
40812
【样例5】
400000008

提示

样例1解释
The following 17 pairs satisfy the condition: (1,1), (1,11), (2,2), (2,22), (3,3), (4,4), (5,5),
(6,6), (7,7), (8,8), (9,9), (11,1), (11,11), (12,21), (21,12), (22,2), and (22,22).

#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize (2)
#pragma G++ optimize (2)
#include <bits/stdc++.h>
#include <algorithm>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
using namespace std;
#define wuyt main
typedef long long ll;
#define HEAP(...) priority_queue<__VA_ARGS__ >
#define heap(...) priority_queue<__VA_ARGS__,vector<__VA_ARGS__ >,greater<__VA_ARGS__ > >
template<class T> inline T min(T &x,const T &y){return x>y?y:x;}
template<class T> inline T max(T &x,const T &y){return x<y?y:x;}
//#define getchar()(p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 21) + 1], *p1 = buf, *p2 = buf;
ll read(){ll c = getchar(),Nig = 1,x = 0;while(!isdigit(c) && c!='-')c = getchar();
if(c == '-')Nig = -1,c = getchar();
while(isdigit(c))x = ((x<<1) + (x<<3)) + (c^'0'),c = getchar();
return Nig*x;}
#define read read()
const ll inf = 1e15;
const int maxn = 2e5 + 7;
const int mod = 1e9 + 7;
#define start int wuyt()
#define end return 0
int n;
ll num[10][10];
ll fun(int left, int right){
    ll res=0;
    for(int i=1;i<=n;i++){
        int teemp=i%10,temp=i;
        while(temp>9)
            temp/=10;
        if(temp==left&&teemp==right)
            res++;
    }
    return res;
}
int main(){
    n=read;ll res=0;
    for(int i=1;i<=9;i++)
        for(int j=1;j<=9;j++)
            num[i][j]=fun(i,j);
    for(int i=1;i<=9;i++)
        for(int j=1;j<=9;j++)
            res+=num[i][j]*num[j][i];
    cout<<res<<endl;
    return 0;
}
 
/**************************************************************
    Language: C++
    Result: 正确
    Time:110 ms
    Memory:2024 kb
****************************************************************/