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2015
05-23

HDU 4352-XHXJ’s LIS-动态规划-[解题报告]HOJ

XHXJ’s LIS

问题描述 :

#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。

输入:

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

输出:

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

样例输入:

1
123 321 2

样例输出:

Case #1: 139 

题意:

给出L和R找出在[L,R]中满足最长递增子序列长度等于K的个数。

题解:

状压想不到,看了kuangbin的才明白,1<<10状压存数出现了没有。利用了nlogn递增字序列算法的思想。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<math.h>
using namespace std;
typedef long long lld;
#define oo 0x3f3f3f3f
#define mod 1000000007
#define maxn 30+5
lld dp[maxn][1<<10][15];
int bit[maxn];
lld L,R,K;

int NextSt(int st,int x)
{
    //找到第一个大于x的数并且替换他,nlogn求最长递增子序列的思想
    for(int i=x;i<10;i++)
        if(st&(1<<i)) return ((st^(1<<i))|(1<<x));
    return st|(1<<x);
}

int GetLen(int st)
{
    int cnt=0;
    while(st)
    {
        if(st&1) cnt++;
        st>>=1;
    }
    return cnt;
}

lld dfs(int pos,int st,int zero,int f)
{
    if(pos<1) return GetLen(st)==K;
    if(!f&&dp[pos][st][K]!=-1) return dp[pos][st][K];
    int last= f ? bit[pos] : 9;
    lld res=0;
    for(int i=0;i<=last;i++)
    {
        res+=dfs(pos-1,(zero==0&&i==0)?0:NextSt(st,i),zero||i,f&&(i==last));
    }
    if(!f) dp[pos][st][K]=res;
    return res;
}

lld GetAns(lld n)
{
    int len=0;
    while(n)
    {
        bit[++len]=n%10;
        n/=10;
    }
    return dfs(len,0,0,1);
}

int main()
{
    int T;
    memset(dp,-1,sizeof dp);
    scanf("%d",&T);
    for(int cas=1;cas<=T;cas++)
    {
        cin>>L>>R>>K;
        cout<<"Case #"<<cas<<": "
            <<GetAns(R)-GetAns(L-1)<<endl;
    }
	return 0;
}

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参考:http://blog.csdn.net/my_acm_dream/article/details/41885523