Codeforces Round #339 (Div. 2) B. Gena's Code
2016-01-15 20:08
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B. Gena's Code
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.
/*beautiful number是0,和1后面跟着一大串0的数,如果碰到0直接输出0即可,每组测试数据肯定只有1个not beautiful number,另外存下来就可以, 然后把beautiful word后面0的个数用strcat串起来就行。*/ #include<cstdio> #include<cstring> #include<stack> #include<iterator> #include<queue> #include<cmath> #include<set> #include<vector> #include<iostream> #include<map> #include<string> #include<algorithm> using namespace std; typedef long long ll; typedef unsigned long long ull; const int maxn=1e6+5; char ans[maxn],s[maxn]; int check(char *s) //是“0”,返回0,是0以外的beautiful word返回1,不是返回2. { int len=strlen(s); if(len==1&&s[0]=='0')return 0; int i,cnt=0; for(i=0;i<len;i++) { if(s[i]=='1')cnt++; if(cnt>=2)break; if(s[i]!='1'&&s[i]!='0')break; } if(i==len)return 1; else return 2; } int main() { int n,i; string tt="1"; scanf("%d",&n); for(i=0;i<n;i++) { scanf("%s",s); int key=check(s); if(key==0)break; else if(key==1) strcat(ans,s+1); //将0串起来,从s[1]开始. else tt=s; } if(i<n)puts("0"); else cout<<tt<<ans<<endl; return 0; }
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