练习题目 1

 Little Elephant and Interval

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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 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 %I64dspecifier.

Output

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

Sample Input

Input

2 47

Output

12

Input

47 1024

Output

98

简单题意：　　给你两个数字， 判断两个数字之间有多少个首位与末位相同的数字个数，，范围1 -- 10^18

思路分析：　　　本来想用字符串，，发现__int64可以存储数字，。。 给出两个数字，先求从（0 -- 小数）的数字个数， 再求（0 -- 大树）的数字个数，然后相减。。

#include<stdio.h>
#include<string.h>
#include<math.h>
__int64 l, r, sum1, sum2;
__int64 a[20];
int lnn[22]; //保存每一位；
int getlen(__int64 n)
{
    int len = 0;
    while(n)
    {
        lnn[len] = n % 10;
        n = n / 10;
        len++;
    }
    return len;
}
//注意一下这里求阶乘，不要用那个pow函数，因为会有误差；
__int64 fac(int    s){
    int i;
    __int64 ans=1;
    for(i = 1; i <= s; i++) ans = ans * 10;
    return ans;
}
//这里是为了求出中间的那个数字；
__int64 getn(int ll,int rr)
{
    int i;
    __int64 j = 1, ans = 0;
    for(i = ll; i <= rr; i++)
    {
        ans += lnn[i] * j;
        j = j * 10;
    }

    return ans;
}

int main()
{
    int i, j, k;
    int len1 = 0,len2 = 0;
    scanf("%I64d%I64d", &l,&r);
    a[1] = 10; a[2] = 9;
    a[3] = 90;
    for(i = 4; i <= 18; i++) a[i] = a[i - 1] * 10;
    sum1 = sum2 = 0;
    len2 = getlen(r);
    //要单独讨论长度只有1的情况；
    if(len2 == 1)
        sum2 += r + 1;
    for(i = 1; i < len2; i++)
        sum2 += a[i];
    __int64 s2 = fac(len2 - 1);
    __int64 ans = 0;
     if(r / s2 > 1 && len2 != 1)
     {
         sum2 += a[len2] / 9 * (r / s2 - 1);
     }
     sum2 += getn(1, len2 - 2);
     //这里是首位小于等于末尾的情况，那么要加1；
     if(r / s2 <= r % 10 && len2 != 1)
        sum2++;

     memset(lnn, 0, sizeof(lnn));
     len1 = getlen(l - 1);
     if(len1 == 1)
        sum1 += l;
     for(i = 1;i < len1; i++)
        sum1 += a[i];
     __int64 s1 = fac(len1 - 1);
     if((l - 1) / s1 > 1 && len1 != 1)
     {
         sum1 += a[len1] / 9 * ((l - 1) / s1 - 1);
     }
    sum1 += getn(1, len1 - 2);
    if((l - 1) / s1 <= (l - 1) % 10 && len1 != 1)
        sum1++;

     printf("%I64d\n",sum2 - sum1);
}