Educational Codeforces Round 14 A. Fashion in Berland

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最新推荐文章于 2023-03-13 16:03:25 发布

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本文链接：https://blog.youkuaiyun.com/sdut_jk17_zhangming/article/details/79562115

本文介绍了一个基于Berland时尚规则的夹克扣法判断算法。该算法接收一个包含n个按钮状态的输入，并根据Berland时尚规则判断夹克是否正确扣好。如果夹克只有一个按钮，则必须扣上；如果有多个按钮，则除了一个可以不扣以外，其余都必须扣上。

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According to rules of the Berland fashion, a jacket should be fastened by all the buttons except only one, but not necessarily it should be the last one. Also if the jacket has only one button, it should be fastened, so the jacket will not swinging open.

You are given a jacket with n buttons. Determine if it is fastened in a right way.

Input

The first line contains integer n (1 ≤ n ≤ 1000) — the number of buttons on the jacket.

The second line contains n integers a_i (0 ≤ a_i ≤ 1). The number a_i = 0 if the i-th button is not fastened. Otherwise a_i = 1.

Output

In the only line print the word "YES" if the jacket is fastened in a right way. Otherwise print the word "NO".

Example

Input

3
1 0 1

Output

YES

Input

3
1 0 0

Output

NO

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>

int main()
{
    int n,i,a[1010],sum = 0;
    scanf("%d",&n);
    for(i = 0;i < n;i++)
    {
        scanf("%d",&a[i]);
    }
    if(n == 1)
    {
        if(a[0] == 1)
            printf("YES\n");
        else
            printf("NO\n");
    }
    else
    {
        for(i = 0;i <n;i++)
        {
            if(a[i] == 0)
                sum++;
        }
    if(sum == 1)
        printf("YES\n");
    else
        printf("NO\n");
    }
    return 0;
}