Codeforces Round #224 (Div. 2) A. Ksenia and Pan Scales

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Ksenia has ordinary pan scales and several weights of an equal mass. Ksenia has already put some weights on the scales, while other weights are untouched. Ksenia is now wondering whether it is possible to put all the remaining weights on the scales so that the scales were in equilibrium.

The scales is in equilibrium if the total sum of weights on the left pan is equal to the total sum of weights on the right pan.

Input

The first line has a non-empty sequence of characters describing the scales. In this sequence, an uppercase English letter indicates a weight, and the symbol "|" indicates the delimiter (the character occurs in the sequence exactly once). All weights that are recorded in the sequence before the delimiter are initially on the left pan of the scale. All weights that are recorded in the sequence after the delimiter are initially on the right pan of the scale.

The second line contains a non-empty sequence containing uppercase English letters. Each letter indicates a weight which is not used yet.

It is guaranteed that all the English letters in the input data are different. It is guaranteed that the input does not contain any extra characters.

Output

If you cannot put all the weights on the scales so that the scales were in equilibrium, print string "Impossible". Otherwise, print the description of the resulting scales, copy the format of the input.

If there are multiple answers, print any of them.

Sample test(s)
input
AC|TL
output
AC|TL
input
|ABCXYZ
output
XYZ|ABC
input
W|TF
output
Impossible
input
ABC|D
output
Impossible

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题目大意：

第一行的字符串相当于天平，' | ' 将字符串分成左右两边，相当于天平的两边，左边两边的重量是字符的个数

第二行的字符串相当于一个个砝码

题目要求：把所有的砝码都放到天平上，判断天平能否平衡，能的话输出使它平衡的一种情况，否则输出Impossble

#include<bits/stdc++.h>#define M 100000using namespace std;char s1[M],s2[M];int main(){    int flag=0,ans1=0,ans2=0;    scanf("%s",s1);    int l1=strlen(s1);    for(int i=0;i<l1;++i){        if(!flag) ans1++;        if(s1[i]=='|') flag=1;        if(flag) ans2++;    }    scanf("%s",s2);    int l2=strlen(s2);    int tmp=(ans1+ans2+l2)/2;    if(ans1>tmp||ans2>tmp||((ans1+ans2+l2)%2)){        printf("Impossible\n");    }    else{        for(int i=0;i<abs(ans1-tmp);++i) cout<<s2[i];        for(int i=0;i<l1;++i) cout<<s1[i];        for(int i=abs(ans1-tmp);i<l2;++i) cout<<s2[i];        cout<<endl;    }    return 0;}

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