hdu 4352 hdu_4352_XHXJ's LIS (数位DP+最长上升子序列+未压缩)

XHXJ's LIS

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 14   Accepted Submission(s) : 5

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<2[sup]63[/sup]-1 and 1<=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

1123 321 2

 

Sample Output

Case #1: 139 

 

Author

peterae86@UESTC03

 

Source

2012 Multi-University Training Contest 6

 

题意：

    问在[L,R]中有多少数，他的最长上升子序列的长度为k、（注意上升子序列不一定连续）

解析：

    数位DP中用位压缩，来保存上升子序列的状态。

    dp[i][j][k]，在输入k的边界下，第i位下，状态位j(j的二进制展开表示0-9，第几位为1时，说明这个数此时在上升子序列的状态数组中g[])

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;typedef long long int ll;int a[25];ll dp[25][1<<10][15];int k;int setsum(int x,int s){    for(int i=x;i<10;i++)                         //当前s中标为1的数，就是现在在g[]数组中的数    {                                            //1 7 8 2 4 5 6 就是在2替代7的位置（2标为1，7标为0），4替代8，这样使得后面的5，6加入到子序列中        if(s&(1<<i)) return (s^(1<<i))|(1<<x);   //找在它之前插入的，并且比他大的数i。用现在这个数x替代i,（把i标为0，x标为1）    }                                         //此时这个x就是更有潜力的在当前长度下i所处位置上的值，为了使未来的序列长度尽可能大    return s|(1<<x); }int getsum(int s){    int cnt=0;    while(s)    {        if(s&1)cnt++;        s>>=1;    }    return cnt;}ll dfs(int pos,int s,bool limit,bool lead){    ll ans=0;    if(pos==-1)    {if(getsum(s)==k)return 1;elsereturn 0;    }    if(!limit/*&&!lead*/&&dp[pos][s][k]!=-1) return dp[pos][s][k];    int up=limit?a[pos]:9;    int now;    for(int i=0;i<=up;i++)    {         ans+=dfs(pos-1,(lead&&i==0)?0:setsum(i,s),limit&&i==up,lead&&i==0);    }    if(!limit/*&&!lead*/) dp[pos][s][k]=ans;    return ans;}ll solve(ll n){    ll ans=0;    int pos=0;    while(n)    {        a[pos++]=n%10;        n=n/10;    }    //memset(dp,-1,sizeof(dp));    ans+=dfs(pos-1,0,true,true);    return ans;}int main(){    int t;    ll zuo,you;    scanf("%d",&t);int case1=0;memset(dp,-1,sizeof(dp));    while(t--)    {case1++;        scanf("%lld%lld%d",&zuo,&you,&k);        printf("Case #%d: %lld\n",case1,solve(you)-solve(zuo-1));    }    return 0;}