Time limit: 1.0s Memory limit: 256.0MB Total points for this question: 20 points
【Problem Description】
The math teacher gave Xiao Ming a problem to sum arithmetic series. But the careless Xiao Ming forgot part of the sequence, and only remembered the N integers.
Now given these N integers, Xiaoming wants to know how many terms are there in the shortest arithmetic sequence containing these N integers?
【Input format】
The first line of input contains an integer N.
The second row contains N integers A 1 , A 2 , . . . , AN . (Note that A 1 ∼ AN is not necessarily given in the order in the arithmetic sequence)
【Output format】
Output an integer representing the answer.
【Sample input】
5
2 6 4 10 20
【Example output】
10
【Example description】
The shortest arithmetic progression that contains 2, 6, 4, 10, 20 is 2, 4, 6, 8, 10, 12, 14, 16, 18, 20.
[Scale and conventions of evaluation cases]
For all evaluation cases, 2 ≤ N ≤ 100000, 0 ≤ A i ≤ 109 10^910 9.
The reason why I wrote this problem solution is because I saw the same problem solution on the Internet and entered a misunderstanding, that is, "the minimum value of the difference between the two is the tolerance"
Then there is the following code
#include <bits/stdc++.h>
using namespace std;
int main()
{
int n;cin>>n;
int a[n+10];
for(int i=0;i<n;i++)
cin>>a[i];
sort(a,a+n);
int x = 99999999999;
for(int i=1;i<n;i++){
x = min(a[i]-a[i-1],x);
}
if(x==0) cout<<n<<endl;
else {
int ans = (a[n-1]-a[0])/x+1;
cout<<ans<<endl;
}
return 0;
}
Because of this, I have been depressed for a long time. I really can't think of the principle of this code. This code is still AC! ! !
Suppose a sequence of 1 3 8 is given, so the tolerance is 2? Obviously not
The H question of the Blue Bridge Cup will not be so simple
correct answer:
To understand that for a sequence of
2 6 4 10 20 given by the title,
first sort the sequence: 2, 4, 6, 10, 20 (there should be no doubt about this)
to make the sequence the shortest, a[0]=2, a[n ]=20;
By the arithmetic sequence, the general term: an=a1+(n-1)*d→ n=(an-a1)/d+1
At this time, we only need to find the value of the tolerance d.
The following paragraph needs to be detailed! ! !
d={greatest common divisor of the differences of all adjacent integers}
All integer differences of an ordered sequence must be a multiple of the tolerance to form an arithmetic sequence, that is, it (a[n]-a[n-1])/d==0must be established for an ordered sequence. To make the sequence the shortest, the tolerance as large as possible, so just ask for the greatest common divisor of all differences
#include<bits/stdc++.h>
using namespace std;
int N,a[100005];
int gcd(int a,int b){
return b==0?a:gcd(b,a%b);
}
int main(){
cin>>N;
for(int i=0;i<N;i++){
cin>>a[i];
}
sort(a,a+N);
//求差值最大公约数
int d=a[1]-a[0];
for(int i=2;i<N;i++){
d=gcd(d,a[i]-a[i-1]);
}
if(d==0)printf("%d\n",N);
else{
printf("%d ",(a[N-1]-a[0])/d+1);
}
return 0;
}