hdu - 5945 Fxx and game 【dp + 单调队列】

Fxx and game

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 413    Accepted Submission(s): 100

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

29 2 111 3 3

 

Sample Output

43

 

Source

BestCoder Round #89

 

ｄｐ方程还是比较好推的，　关键使用单调队列进行优化。

#include <map>#include <set>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <iostream>#include <stack>#include <cmath>#include <string>#include <vector>#include <cstdlib>//#include <bits/stdc++.h>//#define LOACL#define space " "using namespace std;typedef long long LL;//typedef __int64 Int;typedef pair<int, int> paii;const int INF = 0x3f3f3f3f;const double ESP = 1e-5;const double PI = acos(-1.0);const LL MOD = 1e9 + 7;const int MAXN = 1e6 + 10;int dp[MAXN], que[MAXN], loc[MAXN];int X, k, t;int main() {int T;scanf("%d", &T);while (T--) {scanf("%d%d%d", &X, &k, &t);dp[1] = 0; dp[0] = INF;int tail = 1, head = 1;que[head] = 0; loc[head] = 1;for (int i = 2; i <= X; i++) {dp[i] = INF;if (i%k == 0) dp[i] = dp[i/k] + 1;//找出队顶while (head <= tail && i - t > loc[head]) head++;dp[i] = min(dp[i], que[head] + 1);//新元素加入到队列while (head <= tail && dp[i] < que[tail]) tail--;que[++tail] = dp[i]; loc[tail] = i;}printf("%d\n", dp[X]);}return 0;}