Arithmetic Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1445    Accepted Submission(s): 632

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 -2
0 2 0 -2 0
5 2 3
2 3 3 3 3

Sample Output

12
5

Author

xudyh

对每个数先预处理出左右能够达到的最远距离。然后当d1!=d2时,用乘法原理得到区间数(左边有l[i]个区间,右边有r[i]个区间,左右就是l[i]*r[i]个区间)

当d1==d2 时,左右会算重，只算左边就好了。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <queue>
#include <algorithm>
using namespace std;
typedef long long LL;
const int N = 100005;
int a[N];
LL l[N],r[N],ans; ///l[i]记录第i个数左边与其成方差为d1的数的个数,r[i]记录第i个数右边与其方差为
                  ///d2的数的个数(包括自身)
int main()
{
    int n,d1,d2;
    while(scanf("%d%d%d",&n,&d1,&d2)!=EOF){
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
        }
        l[1] = 1,r[n]=1;
        ans = 0;
        for(int i=2;i<=n;i++){ ///预处理
            if(a[i]==a[i-1]+d1){
                l[i] = l[i-1]+1;
            }else l[i] = 1;
        }
        for(int i=n-1;i>=0;i--){
            if(a[i]+d2==a[i+1]){
                r[i] = r[i+1]+1;
            }else r[i]=1;
        }
        if(d1==d2){
            for(int i=1;i<=n;i++){
            ans+=l[i];
        }
        }
        else
            for(int i=1;i<=n;i++){ ///乘法原理
           // printf("%lld %lld\n",l[i],r[i]);
            ans+=l[i]*r[i];
        }
        printf("%lld\n",ans);
    }
}