hdu 5400 Arithmetic Sequence

http://acm.hdu.edu.cn/showproblem.php?pid=5400

Arithmetic Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1151    Accepted Submission(s): 501

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：有一种区间[l,r],存在i使得当l<=j<=i时,a[j+1]=a[j]+d1,当i<=j<=r时a[j+1]=a[j]+d2;

其实这种区间就是[l,i]区间a中元素是以d1位公差的等差数列,[i,r]区间a中的元素是以d2为公差的等差数列

求这样的区间有多少种

 

枚举i（1<=i<=n），预处理出以i为中心，左侧是满足以d1为公差的等差数列的最大长度al[ i ]，

右侧是满足以d2为公差的等差数列的最大长度ar[ i ]；

 

1、如果d1=d2,  那么以i为一个解的区间个数为 ans +=al[i](或ar[i]);

2、如果d1!=d2, 那么以i为一个解的区间个数为al[ i ]*ar[ i ]

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<algorithm>
#define N 100010

using namespace std;

long long a[N], al[N], ar[N];//al[i]表示从左边开始（以d1为公差的等差数列）记录元素的个数，ar[i]表示从右边开始（以d2为公差的等差数列）记录元素的个数

int main()
{
    long long n, d1, d2, i;
    while(~scanf("%lld%lld%lld", &n, &d1, &d2))
    {
        for(i =  ; i <= n ; i++)
            scanf("%lld", &a[i]);
        al[] = ar[n] = ;//初始化，最开始从左边开始i=1和从右边开始i=n时等差数列中的元素个数都为1
        for(i =  ; i <= n ; i++)
        {
            if(a[i] == a[i - ] + d1)
                al[i] = al[i - ] + ;//如果满足条件即a[i]是以d1为公差的等差数列里的元素，所以元素个数在本来的基础上+1
            else
                al[i] = ;//不满足条件则a[i]不是以d1为公差的等差数列里的元素，此时a[i]是另一个数列。其元素个数为1
        }
        for(i = n - ; i >=  ; i--)
        {
            if(a[i] == a[i + ] - d2)//由a[i+1]=a[i]+d2转化而来
                ar[i] = ar[i + ] + ;
            else
                ar[i] = ;
        }
        long long ans = ;
        for(i =  ; i <= n ; i++)
        {
            if(d1 == d2)
                ans += al[i];
            else
                ans += al[i] * ar[i];
        }
        printf("%lld\n", ans);
    }
    return ;
}