HDU 4352 XHXJ's LIS HDU 题解

题目

#define xhxj (Xin Hang senior sister(学姐))

If you do not know xhxj, then carefully reading the entire description is very important.

As the strongest fighting force in UESTC, xhxj grew up in Jintang, a border town of Chengdu.

Like many god cattles, xhxj has a legendary life:

2010.04, had not yet begun to learn the algorithm, xhxj won the second prize in the university contest. And in this fall, xhxj got one gold medal and one silver medal of regional contest. In the next year's summer, xhxj was invited to Beijing to attend the astar onsite. A few months later, xhxj got two gold medals and was also qualified for world's final. However, xhxj was defeated by zhymaoiing in the competition that determined who would go to the world's final(there is only one team for every university to send to the world's final) .Now, xhxj is much more stronger than ever，and she will go to the dreaming country to compete in TCO final.

As you see, xhxj always keeps a short hair(reasons unknown), so she looks like a boy( I will not tell you she is actually a lovely girl), wearing yellow T-shirt. When she is not talking, her round face feels very lovely, attracting others to touch her face gently。Unlike God Luo's, another UESTC god cattle who has cool and noble charm, xhxj is quite approachable, lively, clever. On the other hand,xhxj is very sensitive to the beautiful properties, "this problem has a very good properties"，she always said that after ACing a very hard problem. She often helps in finding solutions, even though she is not good at the problems of that type.

Xhxj loves many games such as，Dota, ocg, mahjong, Starcraft 2, Diablo 3.etc，if you can beat her in any game above, you will get her admire and become a god cattle. She is very concerned with her younger schoolfellows, if she saw someone on a DOTA platform, she would say: "Why do not you go to improve your programming skill". When she receives sincere compliments from others, she would say modestly: "Please don’t flatter at me.(Please don't black)."As she will graduate after no more than one year, xhxj also wants to fall in love. However, the man in her dreams has not yet appeared, so she now prefers girls.

Another hobby of xhxj is yy(speculation) some magical problems to discover the special properties. For example, when she see a number, she would think whether the digits of a number are strictly increasing. If you consider the number as a string and can get a longest strictly increasing subsequence the length of which is equal to k, the power of this number is k.. It is very simple to determine a single number’s power, but is it also easy to solve this problem with the numbers within an interval? xhxj has a little tired，she want a god cattle to help her solve this problem,the problem is: Determine how many numbers have the power value k in \([L，R]\) in O(1)time.

For the first one to solve this problem，xhxj will upgrade 20 favorability rate。

给定L到R区间，和一个数字\(K\)

对\([L,R]\)每一个数字，把数字当作序列，数字的每一位当作序列的元素，求\(LIS\)长度为\(k\)的数字的个数

输入格式

First a integer \(T(T<=10000)\),then \(T\) lines follow, every line has three positive integer \(L,R,K(0<L \le R < 2 ^ {63}-1,1 \le K \le 10)\).

输出格式

For each query, print "Case #t: ans" in a line, in which \(t\) is the number of the test case starting from 1 and ans is the answer.

题解

依旧是我很不熟的数位dp，直接套板子，求\(LIS\)用\(O(n log n)\)的复杂度的方法，\([L,R]\)区间差分。

重点说一下状压下的\(LIS\)，由于最长长度为10，所以可以不用二分，这样求\(LIS\)的复杂度是\(O(n)\)。

注意这个存储方式不同，状态从右往左编号，依次0-9。如果在状态中为1，说明它出现过，因为是\(LIS\)，所以顺序肯定是递增的，所以只需要存下来出现过什么数字，就相当于原来的dp数组。

当更新状态时，从新的pos开始，往后面遍历，遇到的第一个1就是比自己大或者等于的，将其置0，新pos位置 置1，这就是一个更新操作。

举个例子:

原状态:

0001000010

表示长度为1的上升子序列结尾元素是1，长度为2的上升子序列结尾元素是6, LIS长度为2

那么现在发现，有一个长度为2的上升子序列结尾元素为4，需要更新一下

从第4位(实际上是第5位)开始

00010[0]0010
0001[0]00010
000[1]000010

找到一个1，把它置零

000[0]000010

然后把 第4位 置1(更新)

00000[1]0010

现在更新后的状态就是

0000010010

表示长度为1的上升子序列结尾元素是1，长度为2的上升子序列结尾元素是4, LIS长度为2

LIS长度是通过1的个数算出来的，可以用lowbit计算，我使用的是内置函数__builtin_popcountll，应该也是lowbit实现的

代码

#include <bits/stdc++.h>
#define ll long long
using namespace std;
int k, number[20];
ll dp[20][1 << 10][11];

ll newStatus(ll status, ll num) {
    for (int i = num; i <= 9; i++)
        if (status & (1 << i)) return (status ^ (1 << i)) | (1 << num);
    return status | (1 << num);
}
ll dfs(int position, ll status, bool lead, bool limit) {
    if (position == 0) return __builtin_popcountll(status) == k;
    if (!limit && dp[position][status][k] != -1) return dp[position][status][k];
    int up = limit ? number[position] : 9;
    ll ans = 0;
    for (int i = 0; i <= up; i++)
        ans += dfs(position - 1, (lead && i == 0) ? 0 : newStatus(status, i), lead && i == 0, limit && i == up);
    if (!limit) dp[position][status][k] = ans;
    return ans;
}
ll solve(ll x) {
    int length = 0;
    while (x) number[++length] = x % 10, x /= 10;
    return dfs(length, 0ll, true, true);
}

int main() {
    int t;
    scanf("%d", &t);
    memset(dp, -1, sizeof(dp));
    for (int Case = 1; Case <= t; Case++) {
        long long l, r;
        scanf("%lld%lld%d", &l, &r, &k);
        printf("Case #%d: %lld\n", Case, solve(r) - solve(l - 1));
    }
}