time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Not so long ago, Vlad had a birthday, for which he was presented with a package of candies. There were nn types of candies, there are aiai candies of the type ii (1≤i≤n1≤i≤n).
Vlad decided to eat exactly one candy every time, choosing any of the candies of a type that is currently the most frequent (if there are several such types, he can choose any of them). To get the maximum pleasure from eating, Vlad does not want to eat two candies of the same type in a row.
Help him figure out if he can eat all the candies without eating two identical candies in a row.
Input
The first line of input data contains an integer tt (1≤t≤1041≤t≤104) — the number of input test cases.
The following is a description of tt test cases of input, two lines for each.
The first line of the case contains the single number nn (1≤n≤2⋅1051≤n≤2⋅105) — the number of types of candies in the package.
The second line of the case contains nn integers aiai (1≤ai≤1091≤ai≤109) — the number of candies of the type ii.
It is guaranteed that the sum of nn for all cases does not exceed 2⋅1052⋅105.
Output
Output tt lines, each of which contains the answer to the corresponding test case of input. As an answer, output "YES" if Vlad can eat candy as planned, and "NO" otherwise.
You can output the answer in any case (for example, the strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive answer).
Example
input
Copy
6 2 2 3 1 2 5 1 6 2 4 3 4 2 2 2 1 3 1 1000000000 999999999 1 1
output
Copy
YES NO NO YES YES YES
Note
In the first example, it is necessary to eat sweets in this order:
In the second example, all the candies are of the same type and it is impossible to eat them without eating two identical ones in a row.
In the third example, first of all, a candy of the type 22 will be eaten, after which this kind will remain the only kind that is the most frequent, and you will have to eat a candy of the type 22 again.
解题说明:此题是一道模拟题,只需要判断每次不从相同的糖果中取即可。
#include <stdio.h>
int main()
{
int nCases;
int n;
int most, premost;
int buf;
scanf("%d", &nCases);
while (nCases-- > 0)
{
scanf("%d", &n);
most = 0;
premost = 0;
while (n-- > 0)
{
scanf("%d", &buf);
if (buf >= most)
{
premost = most;
most = buf;
}
else if (buf > premost)
{
premost = buf;
}
}
if (most - premost <= 1)
{
puts("YES");
}
else
{
puts("NO");
}
}
return 0;
}