Educational Codeforces Round 6 B. Grandfather Dovlet’s calculator （分解数字个位）

B. Grandfather Dovlet’s calculator

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Once Max found an electronic calculator from his grandfather Dovlet's chest. He noticed that the numbers were written with seven-segment indicators (https://en.wikipedia.org/wiki/Seven-segment_display).

Max starts to type all the values from a to b. After typing each number Max resets the calculator. Find the total number of segments printed on the calculator.

For example if a = 1 and b = 3 then at first the calculator will print 2 segments, then — 5 segments and at last it will print 5 segments. So the total number of printed segments is 12.

Input

The only line contains two integers a, b (1 ≤ a ≤ b ≤ 10⁶) — the first and the last number typed by Max.

Output

Print the only integer a — the total number of printed segments.

Sample test(s)

Input

1 3

Output

12

Input

10 15

Output
39


题意：求用棍子摆[l,r]的数字所需要的棍子数

思路：直接暴力分解个位数字就行了，然后相加得到数量，没想到暴力能过= =


ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 101000
#define LL long long
#define ll __int64
#define INF 0xfffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
using namespace std;
int a[10]={6,2,5,5,4,5,6,3,7,6};
int main()
{
	int l,r;
	while(scanf("%d%d",&l,&r)!=EOF)
	{
		ll ans=0;
		for(int i=l;i<=r;i++)
		{
			int num=i;
			while(num)
			{
				ans+=a[num%10];
				num/=10;
			}
		}
		printf("%I64d\n",ans);
	}
	return 0;
}