"Hdu 5945" Fxx and Game (recursive |dp+ monotone queue)

Fxx and Game Time limit:3000/1500 MS (java/others) Memory limit:131072/65536 K (java/others)
Total submission (s): 347 Accepted Submission (s):
Problem Description Young theoretical computer scientist Fxx designed a game for his students.
In each game, you'll get three integers x,k,t.in each step and you can have only do one of the following moves:

1.x=x−i (0<=i<=t).

2.if k| X,x=x/k.
Now Fxx wants-him the minimum steps to make X become 1.
Input in the first line, there are an integer T (1≤t≤20) indicating the number of test cases.

As for the following T lines, each line contains three integers x,k,t (0≤t≤106,1≤x,k≤106)

For the text Case,we assure that it's possible to make X become 1.

Output for each test case, output the answer.

Sample Input

2 9 2 1 11 3 3
Sample Output

4 3
Source Bestcoder Round #89
Recommend wange2014 | We have carefully selected several similar problems for you:5947 5946 5943 5942 5941 "Solving" " recursive |dp+ monotone queue" "from small to large processing: from 1 to n,f[i] means I at least a few transformations can change to 1" "F[i]=max (F[j] (max (i-t,1) <=j<i), [k|i]f[i/k]) +1, each time first to determine whether the current number divisible by K, and then F[j] (max (i-t,1) <=j<i) in each of the comparison, Maximum Value " "Then, we can find that the actual maximum value of f[i/k] (I can divide K) and F[j] (max (i-t,1) <=j<i) is a large one. Then, we can use the monotone queue to maintain the maximum value of F[j] (max (i-t,1) <=j<i), so that we can dispense with a heavy cycle, it is possible to AC. " [during the exam, use DFS to implement the above process, resulting in a stack overflow. ]

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int t,k,n,t;
int f[1000010],que[1000010],minn[1000010];
int main ()
{
//	freopen ("Int.txt", "R", stdin);	freopen ("My.txt", "w", stdout);
	int i,j;
	scanf ("%d", &t);
	while (t--)
	 {
	 	memset (f,127,sizeof (f));
	 	int head,tail; head=tail=1;
	 	f[1]=0; minn[tail]=0; Que[tail]=1;
	 	scanf ("%d%d%d", &n,&k,&t);
	 	for (I=2;i<=n;++i)
	     {
	     	if (!) ( i%k)) f[i]=f[i/k]+1;
	     	int M=max (i-t,1);
	     	while (tail>head&&que[head]<m) head++;
	     	F[i]=min (f[i],minn[head]+1);
	     	while (Tail>=head&&f[i]<=minn[tail]) tail--;
			tail++; Minn[tail]=f[i]; que[tail]=i; 
		 }
		printf ("%d\n", F[n]);
	 }
	return 0;