B - Little Elephant and Interval CodeForces - 205C（数位，模拟）

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 and 47, 253 or 1020 will not.

Help him and count the number of described numbers x for a given pair l and r.

Input
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 %I64d specifier.

Output
On a single line print a single integer — the answer to the problem.

Examples
Input
2 47
Output
12
Input
47 1024
Output
98
Note
In the first sample the answer includes integers 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44.

题意：
求l~r有多少个数字，满足首位和末位相同。

思路：
稍微有点恶心的数位模拟（学弟秒了，tql）

假设当前数字为 axxxxxb，假设a≤b，那么中间的xxxxx都可以算上，如果a>b，则中间的xxxxx可以算上，但是要减一。

最后再枚举其他首尾位置的数字，保证枚举的要小于原来的数字。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
#include <map>
#include <unordered_map>

using namespace std;
typedef unsigned long long ll;

const int maxn = 2e5 + 7;

ll P[20];
int num[20];

ll get(ll x) {
    if(x == 0) return 0;
    
    ll res = 0;
    int cnt = 0;
    ll tmp = x;
    while(tmp) {
        num[++cnt] = tmp % 10;
        tmp /= 10;
    }
    if(cnt == 1) {
        return x;
    }
    
    tmp = 0;
    for(int i = cnt - 1;i >= 2;i--) {
        tmp = tmp * 10 + num[i];
    }
    if(num[cnt] <= num[1]) {
        res += tmp + 1;
        x -= num[1] - num[cnt];
    } else {
        if(tmp != 0) {
            res += tmp;
        }
    }
    
    for(ll a = 1;a <= 9;a++) { //首位末位
        for(ll k = 0;k <= cnt - 2;k++) { //中间位数
            ll now = a * P[k + 1] + P[k + 1] - 10 + a;
            if(now < x) {
                res += P[k];
            }
        }
    }
    res += 9;
    return res;
}

int main() {
    P[0] = 1;
    for(int i = 1;i <= 19;i++) P[i] = P[i - 1] * 10;
    ll l,r;scanf("%llu%llu",&l,&r);
    printf("%llu\n",get(r) - get(l - 1));
    return 0;
}
//76843