Arithmetic Sequence
2024-10-07 12:38:53
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 -2
0 2 0 -2 0
5 2 3
2 3 3 3 3
Sample Output
12
5
题意:给定一序列和d1,d2.问有多少间隔可以保证存在i,对于每一个j,有j(1≤j<i),bj+1=bj+d1, j(i≤j<n),bj+1=bj+d2.成立。
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
const int N = ;
const int oo = 0x3f3f3f3f;
long long al[N], ar[N], as[N], n; // al数组存从左边到这个点一直满足的+d1的数有几个,ar是往右边数满足+d2的有几个。
int main()
{
long long d1, d2, i, ans;
while(~scanf("%lld %lld %lld", &n, &d1, &d2))
{
for(i = ; i <= n; i++)
scanf("%lld", &as[i]);
al[] = ar[n] = ;
for(i = ; i <= n; i++)
if(as[i] == as[i-]+d1)
al[i] = al[i-]+;
else al[i] = ; // 只要间隔,不满足了那么连续的就是1个,它自身
for(i = n-; i >= ; i--)
{
if(as[i]+d2 == as[i+])
ar[i] = ar[i+]+;
else ar[i] = ;
}
ans = ;
for(i = ; i <= n; i++)
{
if(d1 == d2) ans += al[i]; // 如果d1,d2相等,al直接相加
else ans += al[i]*ar[i]; // if不等,就等于两者相乘,即间隔种类数
}
printf("%lld\n", ans);
}
return ;
}
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