第十四届浙江省赛 ZOJ 3962 Seven Segment Display（数位DP）

Seven Segment Display

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Time Limit: 1 Second      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 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

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：数位DP。n秒后如果没超过0xFFFFFFFF， 直接solve(r)-solve(l-1), 如果超过的话可以用一整段减去中间那段

不经过的，即solve(0xFFFFFFFF) - ( solve(r-1) - solve(l) ). ( r = (l+n-1)%(0xFFFFFFF)),

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<map>#include<string>#include<set>#include<cctype>using namespace std;typedef long long ll;const ll mod = (ll)0xffffffff+1;int val[16] = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6, 6, 5, 4, 5, 5, 4};int a[20];ll dp[20][1005];ll change(char *str){    ll tmp = 0;    for(int i = 0; i < 8; i++)    {        if(str[i]>= '0' && str[i] <= '9')            tmp = tmp*16+str[i]-'0';        else            tmp = tmp*16+str[i]-'A'+10;    }    return tmp;}ll dfs(int pos, ll sum, int limit){    if(pos < 0) return sum;    if(!limit && dp[pos][sum] != -1) return dp[pos][sum];    int up = limit ? a[pos] : 15;    ll tmp = 0;    for(int i = 0; i <= up; i++)        tmp += dfs(pos-1, sum+val[i], limit&&(i==a[pos]));    if(!limit) dp[pos][sum] = tmp;    return tmp;}ll solve(ll x){    int pos = 0;    memset(a, 0, sizeof(a));    while(x)    {        a[pos++] = x%16;        x /= 16;    }    return dfs(7, 0, 1);//因为前导0也要花费，所以不管怎么样都从第七位开始}int main(void){    ll setp;    int t;    char str[20];    memset(dp, -1, sizeof(dp));    cin >> t;    while(t--)    {        scanf("%lld %s", &setp, str);        setp--;        ll l = change(str);        ll bei = setp/mod;  //一开始做以为n会很大，还处理了n是0xFFFFFFFF的倍数的情况，其实不需要。        setp = setp%mod;        if(l+setp >= mod)        {            ll r = (l+setp)%mod;            printf("%lld\n", (bei+1)*solve((ll)0xffffffff)-(solve(l-1)-solve(r)));        }        else        {            ll r = l+setp;            printf("%lld\n", bei*solve((ll)0xffffffff)+solve(r)-solve(l-1));        }    }    return 0;}