Codeforces Round #371 (Div. 2) B. Filya and Homework(水题 分类讨论)
2016-09-14 11:13
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描述:
B. Filya and Homework
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Today, hedgehog Filya went to school for the very first time! Teacher gave him a homework which Filya was unable to complete without your help.
Filya is given an array of non-negative integers a1, a2, ..., an.
First, he pick an integer x and then he adds x to
some elements of the array (no more than once), subtract x from some other elements (also, no more than once) and do no change other
elements. He wants all elements of the array to be equal.
Now he wonders if it's possible to pick such integer x and change some elements of the array using this x in
order to make all elements equal.
Input
The first line of the input contains an integer n (1 ≤ n ≤ 100 000) —
the number of integers in the Filya's array. The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109) —
elements of the array.
Output
If it's impossible to make all elements of the array equal using the process given in the problem statement, then print "NO" (without quotes) in the only line
of the output. Otherwise print "YES" (without quotes).
Examples
input
output
input
output
Note
In the first sample Filya should select x = 1, then add it to the first and the last elements of the array and subtract from the second
and the third elements.
题意:
给定一个序列,对于每一个元素,只能 + 或者 - 一个数val。这个数一旦选定,就不能改。
问能否变成全部数字都一样。
思路:
先用set维护有几个不同数字,然后分类讨论,特别要注意有三个不同数字的时候一定要满足a[0]+a[2]==2*a[1]
代码:
#include <bits/stdc++.h>
using namespace std;
set<int>s;
int a[4];
int main(){
std::ios::sync_with_stdio(false);
std::cin.tie(0);
int n;
cin>>n;
int x;
for(int i=0; i<n ;i++){
cin>>x;
s.insert(x);
}
if(s.size()>3)puts("NO");
else if(s.size()==3){
int now=0;
for(set<int>::iterator it=s.begin(); it!=s.end(); it++){
a[now++]=*it;
}
sort(a, a+3);
if(a[0]+a[2]==2*a[1])puts("YES");
else puts("NO");
}
else puts("YES");
return 0;
}
描述:
B. Filya and Homework
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Today, hedgehog Filya went to school for the very first time! Teacher gave him a homework which Filya was unable to complete without your help.
Filya is given an array of non-negative integers a1, a2, ..., an.
First, he pick an integer x and then he adds x to
some elements of the array (no more than once), subtract x from some other elements (also, no more than once) and do no change other
elements. He wants all elements of the array to be equal.
Now he wonders if it's possible to pick such integer x and change some elements of the array using this x in
order to make all elements equal.
Input
The first line of the input contains an integer n (1 ≤ n ≤ 100 000) —
the number of integers in the Filya's array. The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109) —
elements of the array.
Output
If it's impossible to make all elements of the array equal using the process given in the problem statement, then print "NO" (without quotes) in the only line
of the output. Otherwise print "YES" (without quotes).
Examples
input
5 1 3 3 2 1
output
YES
input
5 1 2 3 4 5
output
NO
Note
In the first sample Filya should select x = 1, then add it to the first and the last elements of the array and subtract from the second
and the third elements.
题意:
给定一个序列,对于每一个元素,只能 + 或者 - 一个数val。这个数一旦选定,就不能改。
问能否变成全部数字都一样。
思路:
先用set维护有几个不同数字,然后分类讨论,特别要注意有三个不同数字的时候一定要满足a[0]+a[2]==2*a[1]
代码:
#include <bits/stdc++.h>
using namespace std;
set<int>s;
int a[4];
int main(){
std::ios::sync_with_stdio(false);
std::cin.tie(0);
int n;
cin>>n;
int x;
for(int i=0; i<n ;i++){
cin>>x;
s.insert(x);
}
if(s.size()>3)puts("NO");
else if(s.size()==3){
int now=0;
for(set<int>::iterator it=s.begin(); it!=s.end(); it++){
a[now++]=*it;
}
sort(a, a+3);
if(a[0]+a[2]==2*a[1])puts("YES");
else puts("NO");
}
else puts("YES");
return 0;
}
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