Description
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。
First a integer T(T<=10000),then T lines follow, every line has three
positive integer L,R,K.(0<L<=R<2^63-1 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.
Examples
1
123 321 2
Output
Case
Problem Description
吐槽：一堆废话。
如果一个数的LIS的长度为k，它就有k的能量值，让你求[L,R]区间内，能量为k的数
的数目。
比如 15246 能量为4:1 2 4 6.
Solution
数位dp，二进制状态压缩，LIS O(nlogn)算法思想
基本的数位dp，就不说了。
先说LIS O(nlogn)算法思想：
定义f[len]数组，表示LIS的长度为len时，最后一位为f[len]。要求得数组为
a[i]={4 2 6 3 1 5};初始len=0，i=1;
if(a[i]>f[len]) ++len,f[len]=a[i];
else 从1~len中找到一个位置k，满足f[k-1]<a[i]<f[k],令f[k]=a[i];
---总的来说就是找到第一个比自己大或相等的数，将它删除，把自己放上去。
---如果没有比自己大的数，就把自己放到最后
f数组里保存的值是具有单调性的，所以查找k的过程我们就可以用二分的方法得到
所以复杂度就为O(nlogn).
整个过程就为：
4
2
2 6
2 3
1 3
1 3 5
最终答案为len=3，但要注意的是f数组里保存的数并不是LIS。
2进制状态压缩求LIS：
0~9只有十个数，LIS最大长度不过为10。
我们用10位二进制数来表示9~0的数字是否出现过。
比如上一个pos(数位dp中，枚举第pos个位置）的状态是0101000101表示8，6，2，0，出现过。
到当前pos时，枚举0~9。（在最初的状态的基础上改变）
0：状态不改变
1：0101000011，将2位置置0，1 位置置1
2: 状态不改变
3：0100001101，将6位置置0，3 位置置1
.
.
.
9:1101000101, 将9位置置1
---总的来说就是找到第一个比当前枚举的数大或相等的位置，将它置零，自己置1.
---如果没有就把自己的位置置1.
二进制中1的个数就是LIS的长度。
定义dp[20][1024][10]数组，分别为，当前枚举的pos，LIS的二进制状态十进制表示，
题目要求的K。
结束。具体看代码吧。
Code
#include<bits/stdc++.h>
using namespace std;
typedef int Int;
#define int long long
#define INF 0x3f3f3f3f
#define maxn 105
int a[maxn];
int dp[maxn][1<<10][12];
int l, r, k;
int gets(int s)
{
int cnt=0;
while(s)
{
if(s&1) cnt++;
s>>=1;
}
return cnt;
}
int news(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));
}
int dfs(int pos,int s,bool flag,bool limit)
{
if(pos==0) return gets(s)==k;
if(!limit&&dp[pos][s][k]!=-1) return dp[pos][s][k];
int up=limit?a[pos]:9;
int tmp=0;
for(int i=0;i<=up;i++)
{
tmp+=dfs(pos-1,flag&&i==0?0:news(s,i),i==0&&flag,limit&&i==up);
}
if(!limit) dp[pos][s][k]=tmp;
return tmp;
}
int solve(int x)
{
memset(a,0,sizeof(a));
int pos=0;
while(x)
{
a[++pos]=x%10;
x/=10;
}
return dfs(pos,0,1,1);
}
Int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
memset(dp,-1,sizeof(dp));
int t;
cin>>t;
for(int ca=1;ca<=t;ca++)
{
cout<<"Case #"<<ca<<": ";
cin>>l>>r>>k;
cout<<solve(r)-solve(l-1)<<endl;
}
cin.get(),cin.get();
return 0;
}