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);
}