HDOJ 5400 Arithmetic Sequence 暴力枚举

Arithmetic Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 382    Accepted Submission(s): 196

Problem Description

A sequence b1,b2,⋯,bn are called (d1,d2)-arithmetic sequence if and only if there exist i(1≤i≤n) such that for every j(1≤j<i),bj+1=bj+d1and for every j(i≤j<n),bj+1=bj+d2.

Teacher Mai has a sequence a1,a2,⋯,an. He wants to know how many intervals [l,r](1≤l≤r≤n) there are that al,al+1,⋯,ar are (d1,d2)-arithmetic sequence.

 

Input

There are multiple test cases.

For each test case, the first line contains three numbers n,d1,d2(1≤n≤105,|d1|,|d2|≤1000), the next line contains n integers a1,a2,⋯,an(|ai|≤109).

 

Output

For each test case, print the answer.

 

Sample Input

5 2 -20 2 0 -2 05 2 32 3 3 3 3

 

Sample Output

125

 

Author

xudyh

 

Source

2015 Multi-University Training Contest 9

 

/* ***********************************************Author        :CKbossCreated Time  :2015年08月18日 星期二 12时21分22秒File Name     :1005.cpp************************************************ */#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <string>#include <cmath>#include <cstdlib>#include <vector>#include <queue>#include <set>#include <map>using namespace std;typedef pair<int,int> pII;typedef long long int LL;const int maxn=100100;const int INF=0x3f3f3f3f;int n,d1,d2;int a[maxn];LL ans;vector<pII> up,down;void bd2(){            LL aaa=n-1;    for(int i=0,sz=up.size();i<sz;i++)    {        pII pi=up[i];        LL dur=pi.second-pi.first+1;        aaa+=dur*(dur-1)/2;    }    cout<<aaa<<endl;}int main(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    while(scanf("%d%d%d",&n,&d1,&d2)!=EOF)    {        ans=n;        up.clear(); down.clear();        /// one duan        int left=-1,right=-1;        int Left=-1,Right=-1;        bool cl=false;        bool Cl=false;        for(int i=1;i<=n;i++)             scanf("%d",a+i);        a[n+1]=INF; n++;        for(int i=1;i<=n;i++)         {            if(i>1)            {                int ca=a[i]-a[i-1];                if(ca==d1)                {                    if(left==-1)                     {                        left=i-1; right=i; cl=true;                    }                    else right=i;                }                else                {                    if(cl==true)                    {                        up.push_back(make_pair(left,right));                        left=right=-1;                        cl=false;                    }                }            }            if(i>1)            {                int ca=a[i]-a[i-1];                if(ca==d2)                {                    if(Left==-1)                    {                        Left=i-1; Right=i; Cl=true;                    }                    else Right=i;                }                else                {                    if(Cl==true)                    {                        down.push_back(make_pair(Left,Right));                        Left=Right=-1;                        Cl=false;                    }                }            }        }        if(d1==d2)        {            bd2();            continue;        }        /// single duan        for(int i=0,sz=up.size();i<sz;i++)        {            pII pi=up[i];            LL dur=pi.second-pi.first+1;            ans+=dur*(dur-1)/2;        }        for(int i=0,sz=down.size();i<sz;i++)        {            pII pi=down[i];            LL dur=pi.second-pi.first+1;            ans+=dur*(dur-1)/2;        }        /// double duan        for(int i=0,j=0,sz1=up.size(),sz2=down.size();i<sz1&&j<sz2;)        {            if(up[i].second==down[j].first)            {                LL duan1=up[i].second-up[i].first;                LL duan2=down[j].second-down[j].first;                ans+=duan1*duan2;                i++; j++;            }            else            {                if(up[i].second>down[j].first) j++;else if(up[i].second<down[j].first) i++;            }        }printf("%lld\n",ans);    }        return 0;}