# ZOJ-3962 Seven Segment Display（数位DP）

### Seven Segment Display

#### Time limit:2000ms Memory limit:65536kB

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.

Digit Energy Cost
(units/s)
0 6
1 2
2 5
3 5
4 4
5 5
6 6
7 3
Digit Energy Cost
(units/s)
8 7
9 6
A 6
B 5
C 4
D 5
E 5
F 4

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.

3
5 89ABCDEF
3 FFFFFFFF
7 00000000

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.

#### 题意

现在有一块8位的7段数字显像板，每个16进制数都有自己的能量消耗。给出当前的位置和经历的时间，问在当前序列下经历了t秒后一共消耗了多少的能量。

#### 思路

这个题用数位DP来解决比较好。首先找状态事件，一个是位数，第二个是已经枚举的位的能量消耗；其次是找子状态，这个题的很简单就是一个已经枚举过的位的能量消耗；那么最后就可以找状态转移了，显然有当前的消耗的等于下一位时枚举总能量为sum+cost[i]时的消耗。方程写一下： 。然后就是随便判一下越没越过最大值的问题了。（题目数据很水 – -）

#### 代码

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

ll m;
ll dp;
int dit;
int cost = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6, 6, 5, 4, 5, 5, 4};

ll dfs (int pos, ll sum, bool lead) {
if (pos == -1) {
return sum;
}
if (!lead && dp[pos][sum] != -1) {
return dp[pos][sum];
}
ll ans = 0;
for (int i = 0; i <= nd; ++i) {
}
dp[pos][sum] = ans;
}
return ans;
}

ll solve (ll v) {
int pos = 0;
for (int i = 0; i < 8; ++i) {
dit[i] = v%16;
v /= 16;
}
return dfs(7, 0, 1);
}

int main() {
int t;
m = (ll)0xffffffff+1;
memset(dp, -1, sizeof(dp));
scanf("%d", &t);
while (t--) {
ll n;
char str;
scanf("%lld", &n);
--n;
scanf("%s", str);
ll v = 0;
for (int i = 0; i < 8; ++i) {
v *= 16;
if (str[i] <= '9' && str[i] >= '0') {
v += str[i] - '0';
}
else {
v += str[i] - 'A' + 10;
}
}
ll t = n%m;
if (t+v >= m) {
ll tmp = (t+v)%m;
cout << (n/m+1)*solve(m-1)-solve(v-1)+solve(tmp) << '\n';
}
else {
cout << (n/m)*solve(m) + solve(t+v) - solve(v-1) << '\n';
}
}
return 0;
}