2017浙江省赛 E - Seven Segment Display ZOJ - 3962

地址：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3962

题目：

A seven segment display, or seven segment indicator, is a form of electronic display device for displaying decimal numerals that is an alternative to the more complex dot matrix displays. Seven segment displays are widely used in digital clocks, electronic meters, basic calculators, and other electronic devices that display numerical information.

Edward, a student in Marjar University, is studying the course "Logic and Computer Design Fundamentals" this semester. He bought an eight-digit seven segment display component to make a hexadecimal counter for his course project.

In order to display a hexadecimal number, the seven segment display component needs to consume some electrical energy. The total energy cost for display a hexadecimal number on the component is the sum of the energy cost for displaying each digit of the number. Edward found the following table on the Internet, which describes the energy cost for display each kind of digit.

For example, in order to display the hexadecimal number "5A8BEF67" on the component for one second, 5 + 6 + 7 + 5 + 5 + 4 + 6 + 3 = 41 units of energy will be consumed.

Edward's hexadecimal counter works as follows:

• The counter will only work for n seconds. After n seconds the counter will stop displaying.
• At the beginning of the 1st second, the counter will begin to display a previously configured eight-digit hexadecimal number m.
• At the end of the i-th second (1 ≤ i < n), the number displayed will be increased by 1. If the number displayed will be larger than the hexadecimal number "FFFFFFFF" after increasing, the counter will set the number to 0 and continue displaying.

Given n and m, Edward is interested in the total units of energy consumed by the seven segment display component. Can you help him by working out this problem?

Input

There are multiple test cases. The first line of input contains an integer T (1 ≤ T ≤ 10⁵), indicating the number of test cases. For each test case:

The first and only line contains an integer n (1 ≤ n ≤ 10⁹) and a capitalized eight-digit hexadecimal number m (00000000 ≤ m ≤ FFFFFFFF), their meanings are described above.

We kindly remind you that this problem contains large I/O file, so it's recommended to use a faster I/O method. For example, you can use scanf/printf instead of cin/cout in C++.

Output

For each test case output one line, indicating the total units of energy consumed by the eight-digit seven segment display component.

Sample Input

3
5 89ABCDEF
3 FFFFFFFF
7 00000000

Sample Output

208
124
327

Hint

For the first test case, the counter will display 5 hexadecimal numbers (89ABCDEF, 89ABCDF0, 89ABCDF1, 89ABCDF2, 89ABCDF3) in 5 seconds. The total units of energy cost is (7 + 6 + 6 + 5 + 4 + 5 + 5 + 4) + (7 + 6 + 6 + 5 + 4 + 5 + 4 + 6) + (7 + 6 + 6 + 5 + 4 + 5 + 4 + 2) + (7 + 6 + 6 + 5 + 4 + 5 + 4 + 5) + (7 + 6 + 6 + 5 + 4 + 5 + 4 + 5) = 208.

For the second test case, the counter will display 3 hexadecimal numbers (FFFFFFFF, 00000000, 00000001) in 3 seconds. The total units of energy cost is (4 + 4 + 4 + 4 + 4 + 4 + 4 + 4) + (6 + 6 + 6 + 6 + 6 + 6 + 6 + 6) + (6 + 6 + 6 + 6 + 6 + 6 + 6 + 2) = 124.

思路：

　　对于第i位x，先算出从x到x+1要k秒，然后设lf=n-k，则这一位从x+1开始进行lf/(16^i-1)次变化，剩下lf%(16^i-1)的时间第i为的数值保持不变。

　　

 #include <bits/stdc++.h>

 using namespace std;

 #define MP make_pair
 #define PB push_back
 #define ll first
 #define rr second
 typedef long long LL;
 typedef pair<int,int> PII;
 const double eps=1e-;
 const double pi=acos(-1.0);
 const int K=1e6+;
 const int mod=1e9+;

 LL n,num[],w[]= {,,,,,,,,,,,,,,,};
 LL pr[];
 char ss[];

 int main(void)
 {
     std::ios::sync_with_stdio(false);
     std::cin.tie();
     pr[]=;
     for(int i=; i<=; i++)
         pr[i]=pr[i-]*;
     int t;
     cin>>t;
     while(t--)
     {
         LL ans=,sum=;
         cin>>n>>(ss+);
         for(int i=; i<=; i++)
             if(ss[-i]<='' && ss[-i]>='')
                 num[i]=ss[-i]-'';
             else
                 num[i]=ss[-i]-'A'+;
         for(int i=; i<=; i++)
         {
             LL ta,tb,lf;
             lf=n-pr[i-]+sum;
             ta=lf/pr[i-],tb=lf%pr[i-];
             sum+=num[i]*pr[i-];
             if(lf<)
             {
                 ans+=n*w[num[i]];
                 continue;
             }
             ans+=(n-lf)*w[num[i]];
             while(num[i]+< && ta>)
                 ans+=w[num[i]+]*pr[i-],ta--,num[i]++;
             if(ta==)
             {
                 ans+=tb*w[(num[i]+)%];
                 continue;
             }
             ans+=*(ta/)*pr[i-];
             int j;
             for(j=; j<ta%; j++)
                 ans+=w[j]*pr[i-];
             ans+=tb*w[j];
         }
         cout<<ans<<"\n";
     }
     return ;
 }

再贴一份刚改的简短的代码，前面的那个太啰嗦了

 #include <bits/stdc++.h>

 using namespace std;

 #define MP make_pair
 #define PB push_back
 #define ll first
 #define rr second
 typedef long long LL;
 typedef pair<int,int> PII;
 const double eps=1e-;
 const double pi=acos(-1.0);
 const int K=1e6+;
 const int mod=1e9+;

 LL n,num[],w[]= {,,,,,,,,,,,,,,,};
 LL pr[];
 char ss[];

 int main(void)
 {
     std::ios::sync_with_stdio(false);
     std::cin.tie();
     pr[]=;
     for(int i=; i<=; i++)
         pr[i]=pr[i-]*;
     int t;
     cin>>t;
     while(t--)
     {
         LL ans=,sum=;
         cin>>n>>(ss+);
         for(int i=; i<=; i++)
             if(ss[-i]<='' && ss[-i]>='')
                 num[i]=ss[-i]-'';
             else
                 num[i]=ss[-i]-'A'+;
         for(int i=; i<=; i++)
         {
             LL lf=n-pr[i-]+sum;
             sum+=num[i]*pr[i-];
             if(lf<)
             {
                 ans+=n*w[num[i]];
                 continue;
             }
             ans+=(n-lf)*w[num[i]];
             for(int j=;j<;j++)
                 ans+=pr[i-]*w[(num[i]+j+)%]*(lf/pr[i]+(j+<=(lf/pr[i-])%?:));
             ans+=(lf%pr[i-])*w[(num[i]++lf/pr[i-])%];
         }
         cout<<ans<<"\n";
     }
     return ;
 }