ZOJ 3962 Seven Segment Display

Seven Segment Display

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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.

DigitEnergy Cost
(units/s)0612253544556673DigitEnergy Cost
(units/s)8796A6B5C4D5E5F4

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 numberm.
• 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 numberm (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

35 89ABCDEF3 FFFFFFFF7 00000000

Sample Output

208124327

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: 找规律, 分别统计每一位上各个数字出现的次数, 然后求和.

      (1): 最后一位上 从0-F, 每个数字都会出现n / 16次, 剩余n % 16次, 可以直接模拟

      (2): 非最后一位, 大致与(1)类似, 先统计当前数字出现的次数, 把这些数字先从n中减去, 然后每个数字都会出现的次数为 n / (16 ^ t) (从后往前, t依次取1, 2, 3, ...)

    剩余 n % (16 ^ t) 按 n / (16 ^ (t - 1)) 为一组依次加到该位置下一位应出现的数字的个数中

      (3): 求和即是所求







#include<cstdio>#include<cstring>#include<string>#include<iostream>#include<queue>#include<set>#include<vector>#include<algorithm>typedef long long ll;using namespace std;int f[20] = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6, 6, 5, 4, 5, 5, 4};ll num[20];int main() {    int t, n;    char s[20];    scanf("%d", &t);    while(t--) {        scanf("%d %s", &n, s);        memset(num, 0, sizeof num);        ll c = 16, d = 1;        ll k = 0;        for(int i = 7; i >= 0; i--) {            ll nn = n;            ll kk1, kk2;            if(s[i] >= 'A' && s[i] <= 'Z') kk1 = s[i] - 'A' + 10;            else kk1 = s[i] - '0';            kk2 = kk1;            if(k >= n)  {num[kk1] += n; continue; }            nn -= k;            num[kk1] += k;            ll t1 = nn / c;            ll t2 = nn % c;            for(int j = 0; j < 16; j++) {                num[j] += d * t1;            }            if(i != 7) kk1++;            for(ll j = t2; j > 0; j -= d) {                num[kk1 % 16] += min(j, d);                kk1++;            }            if(i == 7) k += 16 - kk2;            else k += (15 - kk2) * d;            d *= 16;            c *= 16;        }        ll ans = 0;        for(int i = 0; i < 16; i++) {            ans += f[i] * num[i];        }        printf("%lld\n", ans);    }    return 0;}