HDU 5400 Arithmetic Sequence(数学)——多校练习9

Arithmetic Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 0    Accepted Submission(s): 0

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

 

/*********************************************************************/

题意：给你一个序列b1,b2,⋯,bn，问有多少个区间[l,r]满足该区间存在一个i，使得[l,i]是以d1为公差的等差数列，[i,r]是以d2为公差的等差数列

解题思路：老办法，观察样例吧，第一个样例的取法是{0},{0,2},{0,2,0},{0,2,0,-2},{2},{2,0},{2,0,-2},{0},{0,-2},{-2},{-2,0},{0}共12种

由此可见我们只需先找出满足题意的区间[l,r]，知道里面序列的个数k，那么该区间的情况数为

但是有一点不能忽略，即一个元素bi，若它既属于前一个区间，又属于后一个区间的话，它本身会被多计算一次，比如样例1中的-2，它既属于区间[1,4]，又属于区间[4,5]，那么若单纯带公式再相加，那么取法{-2}会被计算两次，所以解题的时候判断一下，减去多算的即为结果。

放上一组样例以供参考

Input

5 2 2
0 2 0 2 4

Output

9

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<stdio.h>#include<string.h>#include<stdlib.h>#include<queue>#include<math.h>#include<vector>#include<map>#include<set>#include<stdlib.h>#include<cmath>#include<string>#include<algorithm>#include<iostream>#define exp 1e-10using namespace std;const int N = 100005;const int inf = 1000000000;const int mod = 258280327;int s[N];int main(){    int n,d1,d2,i,t,j,sum;    __int64 ans,k;    while(~scanf("%d%d%d",&n,&d1,&d2))    {        t=1;sum=j=0;k=1;ans=0;        for(i=1;i<=n;i++)            scanf("%d",&s[i]);        for(i=1;i<n;i++)        {            if(t)            {                if(s[i]+d1==s[i+1])                    k++;                else                    t=0;                j=0;            }            if(!t)            {                if(s[i]+d2==s[i+1])                {                    k++;                    j++;                }                else                {                    ans+=(k+1)*k/2;                    if(sum+k>i)                        ans--;                    //printf("%d\n",k);                    k=1;                    t=1;                    sum=i;                    if(j)                        i--;                }            }        }        ans+=(k+1)*k/2;        if(sum+k>i)            ans--;        //printf("%d\n",k);        printf("%I64d\n",ans);    }    return 0;}

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