CodeForces 47B Coins

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解决一个有趣的问题：通过三次两两比较，确定三枚不同重量硬币的价值排序。使用简单的输入输出方法，根据比较结果输出硬币按价值从低到高的字母排序。

One day Vasya came across three Berland coins. They didn't have any numbers that's why Vasya didn't understand how their denominations differ. He supposed that if one coin is heavier than the other one, then it should be worth more. Vasya weighed all the three pairs of coins on pan balance scales and told you the results. Find out how the deminations of the coins differ or if Vasya has a mistake in the weighting results. No two coins are equal.

Input

The input data contains the results of all the weighting, one result on each line. It is guaranteed that every coin pair was weighted exactly once. Vasya labelled the coins with letters «A», «B» and «C». Each result is a line that appears as (letter)(> or < sign)(letter). For example, if coin "A" proved lighter than coin "B", the result of the weighting is A<B.

Output

It the results are contradictory, print Impossible. Otherwise, print without spaces the rearrangement of letters «A», «B» and «C» which represent the coins in the increasing order of their weights.

Sample test(s)

Input

A>B
C<B
A>C

Output

CBA

Input

A<B
B>C
C>A

Output

ACB

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;

int main(){
    int num[3];
    int min,mid,max;
    char temp[10];
    memset(num,0,sizeof(num));
    for(int i=0;i<3;i++){
        cin>>temp;
        if(temp[1]=='>'){
            num[temp[0]-'A']++;
        }
        if(temp[1]=='<'){
            num[temp[2]-'A']++;
        }
    }
    if(num[0]==num[1]||num[1]==num[2]||num[2]==num[0]){
        cout<<"Impossible"<<endl;
    }
    else{
        for(int i=0;i<3;i++){
            if(num[i]==2){
                max=i;
            }
            if(num[i]==1){
                mid=i;
            }
            if(num[i]==0){
                min=i;
            }
        }
        printf("%c%c%c\n",min+'A',mid+'A',max+'A');
    }
    return 0;
}