Codeforces Round #347 (Div.2)_B. Rebus

题目链接：http://codeforces.com/contest/664/problem/B

B. Rebus

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a rebus of form ? + ? - ? + ? = n, consisting of only question marks, separated by arithmetic operation '+' and '-', equality and positive integer n. The goal is to replace each question mark with some positive integer from 1 to n, such that equality holds.

Input

The only line of the input contains a rebus. It's guaranteed that it contains no more than 100 question marks, integer n is positive and doesn't exceed 1 000 000, all letters and integers are separated by spaces, arithmetic operations are located only between question marks.

Output

The first line of the output should contain "Possible" (without quotes) if rebus has a solution and "Impossible" (without quotes) otherwise.

If the answer exists, the second line should contain any valid rebus with question marks replaced by integers from 1 to n. Follow the format given in the samples.

Examples

input

? + ? - ? + ? + ? = 42

output

Possible
9 + 13 - 39 + 28 + 31 = 42

input

? - ? = 1

output

Impossible

input

? = 1000000

output

Possible
1000000 = 1000000

 

解题报告：

看到大神们的代码，简直碉堡了。

1、这里的输入要掌控好。

2、思路：先置每个位置为1，记录下加上的数的个数sa,减去的个数sb,这样，p(离目标结果)只差p-n了。

3、根据每个问号前面的符号，自增，达到结果。

#include <cstdio>

using namespace std;

int a[1005];    ///结果
int v[1005];    ///记录每个符号

char c[2];  ///符号命令

int main()
{
    scanf("%s",c);  ///输入问号
    scanf("%s",c);  ///输入符号
    int m=1;    ///?的个数，即要填的数字的个数
    int sa=1;   ///加数的个数
    int sb=0;   ///减数的个数
    v[1]=1;

    while(c[0]!='=')
    {
        if(c[0]=='+')
        {
            v[++m]=1;
            sa++;
        }
        else {
            v[++m]=-1;
            sb++;
        }
        scanf("%s",c);///输入问号
        scanf("%s",c);///输入符号
    }

    int n;
    scanf("%d",&n); ///目标结果

    for(int i=1;i<=m;i++)   ///全部填上1
        a[i]=1;

    int p=sa-sb;

    for(int i=1;i<=m;i++)   ///一个一个处理各个数
    {
        while((p<n)&&(v[i]==1)&&(a[i]<n))
            a[i]++,p++;
        while((p>n)&&(v[i]==-1)&&(a[i]<n))
            a[i]++,p--;
    }

    if(p!=n)
    {
        printf("Impossible\n");
        return 0;
    }
    printf("Possible\n");
    printf("%d ",a[1]);

    for(int i=2;i<=m;i++)
    {
        if(v[i]==1) printf("+ ");
        else printf("- ");
        printf("%d ",a[i]);
    }
    printf("= %d\n",n);
}