HDU 5400(Arithmetic Sequence-暴力找区间)

Arithmetic Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 937    Accepted Submission(s): 411

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+d1 and 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

 

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暴力找区间

#include<bits/stdc++.h> using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  #define Lson (x<<1)#define Rson ((x<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (100000007)#define MAXN (1000000+10)typedef long long ll;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}int n;ll d1,d2,a[MAXN],b[MAXN]; int main(){//freopen("E.in","r",stdin);while(cin>>n>>d1>>d2){For(i,n) scanf("%lld",&a[i]);For(i,n-1) b[i]=a[i+1]-a[i];//For(i,n-1) cout<<b[i]<<' ';cout<<endl;if (d1==d2){ll sz=1,ans=0;Fork(i,2,n) if (a[i]-a[i-1]==d1) sz++; else ans+=sz*(sz+1)/2,sz=1;if (sz) ans+=sz*(sz+1)/2;cout<<ans<<endl; } else {ll sz=1,ans=0; bool flag=0;Fork(i,2,n){if (!flag){if (a[i]-a[i-1]==d1) sz++;else if (a[i]-a[i-1]==d2) sz++,flag=1;else ans+=sz*(sz+1)/2,sz=1,flag=0;}else{if (a[i]-a[i-1]==d2) sz++;else if (a[i]-a[i-1]==d1) ans+=sz*(sz+1)/2-1,sz=2,flag=0;else ans+=sz*(sz+1)/2,sz=1,flag=0;}}if (sz) ans+=sz*(sz+1)/2;cout<<ans<<endl; } }return 0;}