Fxx and game

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

2 9 2 1 11 3 3

 

Sample Output

4 3

分析：dp+单调队列（需手写，否则难以卡过去）；

代码：

手写：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <unordered_map>
#include <queue>
#include <stack>
#include <ctime>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, ls[rt]
#define Rson mid+1, R, rs[rt]
#define sys system("pause")
#define freopen freopen("in.txt","r",stdin)
const int maxn=1e6+10;
using namespace std;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
inline ll read()
{
    ll x=0;int f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,m,k,t,d[maxn],f[maxn],head,tail;
int main()
{
    int i,j;
    scanf("%d",&t);
    while(t--)
    {
        int a,b,c;
        scanf("%d%d%d",&a,&b,&c);
        memset(d,inf,sizeof(d));
        d[a]=0;
        head=tail=a;
        f[head]=a;
        for(i=a-1;i>=1;i--)
        {
            if((ll)i*b<=a)d[i]=d[i*b]+1;
            while(head<=tail&&f[tail]-i>c)tail--;
            if(head<=tail)d[i]=min(d[i],d[f[tail]]+1);
            f[--head]=i;
            while(head<tail&&d[f[head+1]]>d[f[head]])f[head+1]=f[head],head++;
        }
        printf("%d
",d[1]);
    }
    //system("Pause");
    return 0;
}

stl：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <unordered_map>
#include <queue>
#include <stack>
#include <ctime>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, ls[rt]
#define Rson mid+1, R, rs[rt]
#define sys system("pause")
#define freopen freopen("in.txt","r",stdin)
const int maxn=1e6+10;
using namespace std;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
inline ll read()
{
    ll x=0;int f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,m,k,t,dp[maxn];
struct node
{
    int c,id;
    node(int _c,int _id):c(_c),id(_id){}
    bool operator<(const node&p)const
    {
        return c>p.c;
    }
};
priority_queue<node>pq;
int main()
{
    int i,j;
    scanf("%d",&t);
    while(t--)
    {
        int a,b,c;
        while(!pq.empty())pq.pop();
        scanf("%d%d%d",&a,&b,&c);
        memset(dp,inf,sizeof(dp));
        pq.push(node(0,a));
        dp[a]=0;
        for(i=a-1;i>=1;i--)
        {
            while(!pq.empty())
            {
                node x=pq.top();
                if(x.id-i>c)pq.pop();
                else
                {
                    dp[i]=x.c+1;
                    break;
                }
            }
            if((ll)b*i<=a)dp[i]=min(dp[i],dp[i*b]+1);
            pq.push(node(dp[i],i));
        }
        printf("%d
",dp[1]);
    }
    //system("Pause");
    return 0;
}