Codeforces - 339B(div2) - Gena's Code（模拟）

B. Gena's Code

time limit per test

0.5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 ai 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 aiwithout 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.

做这道题可把我恶心死了，bug一个接一个，绝壁是个模拟！！！本来就有点困想赶紧写完洗洗睡，尼玛debug完清醒了= =

题意：给你N个数，由0与1组成且仅含一个1的数称为“美丽数”，保证最多只会给出一个不美丽数，求N个数的乘积

乍一看好简单，范围看错以为才10^5，答案也才10^10，拿int，longlong交了一发直接wa233333 仔细一看尼玛10^5个数每个10^5  QAQ 乘方一下那是多少个0 啊T T蠢哭了

思路：字符串输入！！！【别学我不看懂题意瞎设int T T】找出不美丽数存起来，数其他数一共有多少0，cnt计数一下，最后统一输出。如果输入有0， 直接输出0就好。需要注意的是，如果你像我一样用1的个数来判断cnt，那就要注意出现12345跟10000的差别。还有就是，如果N个数都是美丽数，本来这点应该考虑到的，脑子糊了又wa了一发，蓝过

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

char s2[100005]; //存不美丽数
char s[100005]; //输入
int main()
{
  int n;
  while(scanf("%d", &n) != EOF){
    bool flag = 0, flag2 = 0; //flag标记是否出现0， flag2标记是否N个数都是美丽数
    int cnt = 0, cnt2, cnt3;
    int k;
    for(int i = 0; i < n; i++){
      scanf("%s", s);
      cnt2 = 0;
      cnt3 = 0;
      int len = strlen(s);
      for(int j = 0; j < len; j++){
        if(s[j] == '1') cnt2++;
        if(s[j] == '0') cnt3++;
      }
      if(s[0] == '0' && len == 1) flag = 1;
      else if(len == 1 && s[0] == '1') continue;
      else if(len > 1 && cnt2 == 1 && cnt2 + cnt3 == len){  // cnt2+cnt3=len ！！！ 论计算123和100的区别，白白wa了一发
        cnt = cnt + len - 1;
      }
      else{
        k = len;
        flag2 = 1;
        strcpy(s2, s);
      }
    }
    if(flag == 1)
      printf("0\n");
    else{
      if(flag2 == 1){  //N个数都是美丽数，flag2标记
        printf("%s", s2);
        while(cnt--) printf("0");
        printf("\n");
      }
      else{
        printf("1");
        while(cnt--) printf("0");
        printf("\n");
      }
    }
  }
  return 0;
}