hdu 4352 - XHXJ's LIS (数位dp)多校联合

XHXJ’s LIS

Problem Description

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。

Input

First a integer T(T<=10000),then T lines follow, every line has three positive integer L,R,K.(
0<L<=R<263−1and1<=K<=10).

Output

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.

Sample Input

1
123 321 2

Sample Output

Case #1: 139

题意：先定义一个数的power value，把这个数看成一个字符串，他的最长上升子序列的长度就是他的power value，求某个区间内power value等于k的数的个数。
思路：dp[cur][S][K]表示处理到当前为止，最长上升子序列的状态集合为S，要求的最长上升子序列长度为K的有多少。
这个题跟上一个4507不一样，上一个我刚开始向的是直接dp保存在当前这个状态下，跟7没关系的数的平方和，这样必然会导致重复计算，想想肯定也是不对的。而这个题表示是到当前这个状态，最长上升子序列状态为S的，要求的是K，那么也就是说，后面的状态正好和S组成K，也就是当前这个数，可以加上后面的状态组成新的数，满足条件。所以都加上。

#include<iostream>#include<cstdio>#include<string>#include<cstring>#include<vector>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#include<algorithm>using namespace std;typedef long long LL;const int maxn=70;LL L,R;int K;int dig[maxn];LL dp[maxn][1024][11];int getnum(int S){    int sum=0;    for(int i=0;i<10;i++)        if(S&(1<<i))sum++;    return sum;}int getnew(int S,int x){    for(int i=x;i<10;i++)        if(S&(1<<i))return (S^(1<<i))|(1<<x);    return S|(1<<x);}LL dfs(int cur,int e,int z,int S){    if(cur<0)return getnum(S)==K;    if(!e&&dp[cur][S][K]!=-1)        return dp[cur][S][K];    int end=e?dig[cur]:9;    LL ans=0;    for(int i=0;i<=end;i++)    {        if(z&&!i)ans+=dfs(cur-1,e&&i==end,1,0);        else ans+=dfs(cur-1,e&&i==end,0,getnew(S,i));    }    if(!e)dp[cur][S][K]=ans;    return ans;}LL solve(LL x){    int len=0;    while(x)    {        dig[len++]=x%10;        x/=10;    }    return dfs(len-1,1,1,0);}int main(){    int T,cas=1;    scanf("%d",&T);    memset(dp,-1,sizeof(dp));    while(T--)    {        scanf("%I64d%I64d%d",&L,&R,&K);        printf("Case #%d: %I64d\n",cas++,solve(R)-solve(L-1));    }    return 0;}