Fxx and game
Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 2101 Accepted Submission(s): 573
Problem Description
Young theoretical computer scientist Fxx designed a game for his students.
In each game, you will get three integers X,k,t.In each step, you can only do one of the following moves:
1.X=X−i(0<=i<=t).
2.if k|X,X=X/k.
Now Fxx wants you to tell him the minimum steps to make X become 1.
Input
In the first line, there is 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 each 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
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哇哈哈哈哈这题上数学课的时候发现忘记把第一个数存入队列了(上课的时候没事干思忖了一下单调队列的原理还有单调队列和单调栈的用途)，然后下课飞奔至姬房，顺带发现忘记压入后面的数了，反正一个下课就把这题搞好咯~ 不虚此行，撤！
1 #include "bits/stdc++.h"
2 using namespace std;
3 typedef long long LL;
4 const int MAX=1e6+5;
5 int T,n,k,t;
6 int dp[MAX];
7 deque <int> q;
8 int main(){
9 freopen ("fxx.in","r",stdin);
10 freopen ("fxx.out","w",stdout);
11 int i,j;
12 scanf("%d",&T);
13 int an;
14 while (T--){
15 scanf("%d%d%d",&n,&k,&t);
16 if (t==0){
17 an=0;
18 while (n!=1){n/=k;an++;}
19 printf("%d\n",an);
20 continue;
21 }
22 if (k==0){
23 printf("%d\n",(n-2)/t+1);
24 continue;
25 }
26 dp[1]=0;
27 while (q.size()) q.pop_back();
28 q.push_back(1);
29 for (i=2;i<=n;i++){
30 dp[i]=999999999;
31 if (i%k==0) dp[i]=min(dp[i],dp[i/k]+1);
32 while (q.size() && (i-t)>q.front()) q.pop_front();
33 if (q.size()) dp[i]=min(dp[i],dp[q.front()]+1);
34 while (q.size() && dp[i]<dp[q.back()]) q.pop_back();
35 q.push_back(i);
36 }
37 printf("%d\n",dp[n]);
38 }
39 return 0;
40
