hdu 5400 Arithmetic Sequence

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                         ***Arithmetic Sequence***

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

题目大意：就是给你n个数，然后一个d1和d2，求：
1：这个区间是一个等差数列，且公差为d1或d2；

2：若区间的下标范围为[l,r]，应有l<=i<=r，使得[l,i]范围是公差为d1的等差数列，[i,r]范围是公差为d2的等差数列，就是找一共有几种排列方法
吧

解题思路：首先由至少 n 个，然后根据数据推出公式就行了，
直接给出代码吧。。。。

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 1e5+5;

int data[maxn];
int main()
{
    int d1,d2,n,sum,j,i,t;
    __int64 ans, k;
    while(~scanf("%d%d%d",&n,&d1,&d2))
    {
        t = 1;
        sum = j = 0;
        ans = 0;
        k = 1;
        for(i=1; i<=n; i++)
            scanf("%d",&data[i]);
        for(i=1; i<n; i++)
        {
            if(t)
            {
                if(data[i]+d1 == data[i+1])
                    k++;
                else
                    t = 0;
                j = 0;
            }
            if(!t)
            {
                if(data[i]+d2 == data[i+1])
                {
                    k++;
                    j++;
                }
                else
                {
                    ans += (k+1)*k/2;
                    if(sum+k > i)
                        ans--;
                    k = 1;
                    t = 1;
                    sum = i;
                    if(j)
                    i--;
                }
            }
        }
        ans += (k+1)*k/2;
        if(sum+k > i)
            ans--;
        printf("%I64d\n",ans);
    }
    return 0;
}