hdoj5400Arithmetic Sequence【等差数列】

Arithmetic Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1315    Accepted Submission(s): 569

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+d1and 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

 

题意：给定一个序列求在这个序列中左边满足公差为d1的公差序列右边满足公差为d2的区间个数区间只有一个数或只有两个数满足任一个公差均可

枚举每一个数左边满足d1的元素个数a和右边满足公差为d2的元素个数b则以这个数为分界线总长度为a+b+1的序列满足条件的区间个数即为（a+1）*（b+1）

列如 公差为 2 -2 序列为0 2 0 则以2位分界线 a=1；b=1；则区间个数为4 {0,0},{0,2,0},{2,2},{2,0};

#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>#include<queue>using namespace std;const int maxn=100010;long long l[maxn];long long r[maxn];long long num[maxn];int main(){int n,d1,d2,i,j,k;while(scanf("%d%d%d",&n,&d1,&d2)!=EOF){for(i=1;i<=n;++i){l[i]=r[i]=1;}int left,right;long long ans1=0,ans2=0;for(i=1;i<=n;++i){scanf("%lld",&num[i]);if(i!=1&&num[i]-num[i-1]==d1){l[i]+=(i-left);ans1+=l[i];}else {left=i;ans1+=l[i];}}for(i=n;i>=1;--i){if(i!=n&&num[i+1]-num[i]==d2){r[i]+=(right-i);ans2+=l[i]*r[i];}else {right=i;ans2+=l[i]*r[i];}}printf("%lld\n",d1==d2?ans1:ans2);}return 0;}