Codeforces Round #339 (Div. 2) B. Gena's Code

最新推荐文章于 2018-08-11 09:46:55 发布

AnICoo1

最新推荐文章于 2018-08-11 09:46:55 发布

阅读量711

点赞数

CC 4.0 BY-SA版权

分类专栏：数学问题

本文链接：https://blog.youkuaiyun.com/helloiamclh/article/details/50525862

数学问题专栏收录该内容

61 篇文章

订阅专栏

本文介绍了一个关于计算坦克游戏中各国坦克数量乘积的问题，其中涉及“美丽数”的概念，即仅由1和0组成的数字且最多只有一个1。文章提供了一种高效的算法实现，用于快速计算这一乘积。

It's the year 4527 and the tanks game that we all know and love still exists. There also exists Great Gena's code, written in 2016. The problem this code solves is: given the number of tanks that go into the battle from each country, find their product. If it is turns to be too large, then the servers might have not enough time to assign tanks into teams and the whole game will collapse!

There are exactly n distinct countries in the world and the i-th country added a_i tanks to the game. As the developers of the game are perfectionists, the number of tanks from each country is beautiful. A beautiful number, according to the developers, is such number that its decimal representation consists only of digits '1' and '0', moreover it contains at most one digit '1'. However, due to complaints from players, some number of tanks of one country was removed from the game, hence the number of tanks of this country may not remain beautiful.

Your task is to write the program that solves exactly the same problem in order to verify Gena's code correctness. Just in case.

Input

The first line of the input contains the number of countries n (1 ≤ n ≤ 100 000). The second line contains n non-negative integers a_iwithout leading zeroes — the number of tanks of the i-th country.

It is guaranteed that the second line contains at least n - 1 beautiful numbers and the total length of all these number's representations doesn't exceed 100 000.

Output

Print a single number without leading zeroes — the product of the number of tanks presented by each country.

Sample test(s)

input
3
5 10 1

output
50

input
4
1 1 10 11

output
110

input
5
0 3 1 100 1

output
0

Note

In sample 1 numbers 10 and 1 are beautiful, number 5 is not not.

In sample 2 number 11 is not beautiful (contains two '1's), all others are beautiful.

In sample 3 number 3 is not beautiful, all others are beautiful.

题意：给你n个数，求他们的乘积，其中最多有一个不美丽数，如果一个数字的10进制数是由
1和0组成且1的个数为1，就称其为美丽数

思路：查找美丽数，因为美丽数最多含有一个1，所以说最后的输出一定有好多0，所以说只需要记录0的数量就行了，不美丽的数，也要记录下来，以为最后要乘，当输入的数有0的时候，最后的结果一定是0，所以说提前结束，最后输出一个不美丽数，然后输出记录的0数

ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 101000
#define LL long long
#define ll __int64
#define INF 0xfffffff
#define MINF 1<<30
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
using namespace std;
char s[MAXN];
char ans[MAXN];
int main()
{
	int n;
	int i,j;
	while(scanf("%d",&n)!=EOF)
	{
		ll num=0;
		int flag=0;
		for(i=0;i<n;i++)
		{
			scanf("%s",s);
			int len=strlen(s);
			int cnt=0;
			int bz=0;
			for(j=0;j<len;j++)
			{
				if(s[j]=='1')
				cnt++;
				if(s[j]!='1'&&s[j]!='0')
				{
					bz=1;
					break;
				}
			}
			if(bz||cnt>1)
			{
				flag=1;
				strcpy(ans,s);
			}
			else if(cnt)
			num+=len-1;
			else
			{
				printf("0\n");
				return 0;
			}
		}
		if(flag)
		printf("%s",ans);
		else
		printf("1");
		for(i=0;i<num;i++)
		printf("0");
		printf("\n");
	}
	return 0;
}