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hdu 5945 Fxx and game 单调队列优化dp

2024-09-06 16:31:27   217人阅读

Fxx and game

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)



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
#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<cstdio>#include<cmath>#include<string>#include<queue>#include<algorithm>#include<stack>#include<cstring>#include<vector>#include<list>#include<set>#include<map>using namespace std;#define ll long long#define pi (4*atan(1.0))#define eps 1e-14#define bug(x)  cout<<"bug"<<x<<endl;const int N=1e6+10,M=1e6+10,inf=2e9;const ll INF=1e18+10,mod=2147493647;int dp[N];int q[N];int main(){    int T;    scanf("%d",&T);    while(T--)    {        memset(dp,0,sizeof(dp));        int n,k,t;        scanf("%d%d%d",&n,&k,&t);        int l=1,r=1;        dp[1]=0;        q[r++]=1;        for(int i=2;i<=n;i++)        {            dp[i]=inf;            while(l<r&&q[l]<i-t)l++;            if(l<r)dp[i]=dp[q[l]]+1;            if(i%k==0)dp[i]=min(dp[i/k]+1,dp[i]);            while(l<r&&dp[q[r-1]]>=dp[i])r--;            q[r++]=i;        }        //for(int i=1;i<=n;i++)         //   printf("%d ",dp[i]);        //printf("\n");        printf("%d\n",dp[n]);    }    return 0;}

 

hdu 5945 Fxx and game 单调队列优化dp


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