Multi-University Training Contest 4 — 1002题
Multi-University Training Contest 4 — 1002题
Problem Killer
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 13 Accepted Submission(s): 9
Problem Description
You are a "Problem Killer", you want to solve many problems.
Now you have n problems, the i-th problem's difficulty is represented by an integer ai (1≤ai≤109).
For some strange reason, you must choose some integer l and r (1≤l≤r≤n), and solve the problems between the l-th and the r-th, and these problems' difficulties must form an AP (Arithmetic Progression) or a GP (Geometric Progression).
So how many problems can you solve at most?
You can find the definitions of AP and GP by the following links:
https://en.wikipedia.org/wiki/Arithmetic_progression
https://en.wikipedia.org/wiki/Geometric_progression
Input
The first line contains a single integer T, indicating the number of cases.
For each test case, the first line contains a single integer n, the second line contains n integers a1,a2,⋯,an.
T≤104,∑n≤106
Output
For each test case, output one line with a single integer, representing the answer.
Sample Input
251 2 3 4 6101 1 1 1 1 1 2 3 4 5
Sample Output
46
题目大意：
给定一组数，求其中包含的最长等比级数或等差级数的项数。已知数据里边不包含0，这组数的长度范围是从1到10^9。
因为输入的数据较多，为防止超时，用scanf读取数据，然后从头开始遍历，每次计算后边两个数的差，与前两项的差对比，若相同则计数累加，若不同，计数器归2，从新开始计数。同理再次遍历得出最长的等比级数的项数，最后两相比较，输出结果。
需要注意的是，若用int型保存数据，则计算等比级数的公比时，在除的过程中，要注意强制数据类型转换，否则结果会出错。
#include
#include
#include
#include
using namespace std;
int data[1000005];
const double e = 0.00000001;
int main()
{
int t;
scanf("%d", &t);
while (t--)
{
int n;
scanf("%d", &n);
for (int i = 1; i <= n; i++)
{
scanf("%d", &data[i]);
}
int max_ap = 1, max_temp = 2;
int gongcha_first, gongcha_last;
gongcha_first = data[2] - data[1];
for (int i = 3; i <= n; i++)
{
gongcha_last = data[i] - data[i - 1];
if (gongcha_last == gongcha_first)
max_temp += 1;
else
{
gongcha_first = gongcha_last;
if (max_temp > max_ap)
max_ap = max_temp;
max_temp = 2;
}
}
if (max_temp > max_ap)
max_ap = max_temp;
int max_gp = 1, max_temp2 = 2;
double gongbi_first, gongbi_last;
gongbi_first = data[2] * 1.0 / data[1];
for (int i = 3; i <= n; i++)
{
gongbi_last = data[i] * 1.0 / data[i - 1];
if (fabs(gongbi_last - gongbi_first) < e)
max_temp2 += 1;
else
{
gongbi_first = gongbi_last;
if (max_temp2>max_gp)
max_gp = max_temp2;
max_temp2 = 2;
}
}
if (max_temp2>max_gp)
max_gp = max_temp2;
if (n == 1)
printf("%d\n", 1);
else
printf("%d\n", max_ap > max_gp ? max_ap : max_gp);
}
return 0;
}