Lets define a simple recursive function F(n), where
F(n) = p(x) =
n%10, if (n%10) > 0
0, if n = 0
F(n/10), Otherwise
Lets define another function S(p, q),
S(p, q) = ∑q
i=p
F(i)
In this problem you have to Calculate S(p, q) on given value of p and q.
Input
The input file contains several lines of inputs. Each line contains two non negative integers p and q
(p ≤ q) separated by a single space. p and q will fit in 32 bit signed integer. In put is terminated by a
line which contains two negative integers. This line should not be processed.
Output
For each set of input print a single line of the value of S(p, q).
Sample Input
1 10
10 20
30 40
-1 -1
Sample Output
46
48
5
•例如，求S(2,53)，将范围划分为3个区间：[2,9]，[10,50]和[51,
53]。
•对于第1个区间[2,9]，个位数之和2+3+4+…...+9=44；对于第2个区
间[10,50]，个位数之和(1+2+…...+9)4=454=180；对于第3个区
间[51,53]，个位数之和1+2+3=6。所以，第一轮，个位数的总和
为44+180+6=230。
•在[10,50]中，10,20,30,40和50的个位数是0，将这些数除以10后
得1,2,3,4,5，产生新区间[1,5]；进入第二轮，区间[1,5]中的数
只有个位数，个位数之和1+2+3+4+5=15。
•最后，两轮结果相加，得S(2,53)=230+15=245。
#include <bits/stdc++.h>
#define ll long long
ll ans;
ll f(ll x) {
if (x == 0) return 0;
else if (x % 10)
return x % 10;
else
return f(x / 10);
}
void solve(ll l, ll r) {
if (r - l < 9) {
for (int i = l; i <= r; i++)
ans += f(i);
return;
}
while (l % 10) {
ans += f(l);
l++;
}
while (r % 10) {
ans += f(r);
r--;
}
ans += 45 * (r - l) / 10;
solve(l / 10, r / 10);
}
int main () {
ll l, r;
while (scanf("%lld%lld", &l, &r), l >= 0 || r >= 0) {
ans = 0;
solve(l, r);
printf("%lld\n", ans);
}
return 0;
}