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： 0-9分别如图所示  每个数字led管由7部分组成 根据不同的数字亮不同的地方
 给一个区间 问这段区间内所有数 得有多少部分亮
题解： 每位每位的计算

我的代码998ms
#include<bits/stdc++.h>
using namespace std;
#define LL __int64
map<int,int>mp;
map<int,int>mpp;
int n,m;
int main()
{
    mp[0]=6;
    mp[1]=2;
    mp[2]=5;
    mp[3]=5;
    mp[4]=4;
    mp[5]=5;
    mp[6]=6;
    mp[7]=3;
    mp[8]=7;
    mp[9]=6;
    mpp[0]=0;
    int sum=0;
    for(int i=1;i<=1000000;i++)
    {
        int exm=i;
        mpp[i]=sum;
      while(exm!=0)
     {
         mpp[i]+=mp[exm%10];
         sum+=mp[exm%10];
         exm=exm/10;
     }
    }
    scanf("%d%d",&n,&m);
    printf("%d\n",mpp[m]-mpp[n-1]);
    return 0;
}

　　

我晏的代码62ms

#include <cstdio>
#include<bits/stdc++.h>
#include <cstring>
#include <ctime>
#include <algorithm>
#include <vector>
#include <queue>
#include <map>
#include <set>
using namespace std;
typedef long long ll;
const int N = 1000001;
ll H[N][11],a,b;
int M[20] = {6,2,5,5,4,5,6,3,7,6};
int main() {
    for(int i = 1; i <= 1000000; i++) {
        int tmp = i;
        while(tmp) H[i][tmp%10]++,tmp/=10;
    }
    ll ans = 0 ;
    scanf("%I64d%I64d",&a,&b);
    for(int i = a; i <= b; i++) {
        for(int j = 0; j <= 9 ; j ++) ans+= M[j] * H[i][j];
    }
    printf("%I64d\n",ans);
    return 0;
}

　　

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