Codeforces 664 B Rebus【思维】_pair<cont int ,int>啥意思-优快云博客

本文介绍了解决特定形式的Rebus谜题的算法。该算法的目标是在问号位置填入1到n之间的数字，使等式成立。文章提供了一个示例代码，详细解释了如何通过逐步调整每个问号处的数值来找到合适的解决方案。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

题目大意：

在？处填入1-n的数字使得等式成立，如果不成立，输出Impossible。

思路：

先将所有？置为1.然后对应每个数进行调整。

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
char a[1050];
int num[1050];
int abs(int aaa)
{
    if(aaa<0)return -aaa;
    else return aaa;
}
int main()
{
    while(gets(a))
    {
        int cont=0;
        int d=1;
        int val=0;
        int n=strlen(a);
        for(int i=0;i<n;i++)
        {
            if(a[i]=='+')d=1;
            if(a[i]=='-')d=-1;
            if(a[i]=='?')
            {
                num[cont++]=d;
                val+=d;
            }
        }
        int contz=1;
        int sum=0;
        for(int i=n-1;i>=0;i--)
        {
            if(a[i]>='0'&&a[i]<='9')
            {
                sum+=(a[i]-'0')*contz;
                contz*=10;
            }
            else break;
        }
        for(int i=0;i<cont;i++)
        {
            while(val<sum&&num[i]>0&&num[i]<sum)num[i]++,val++;
            while(val>sum&&num[i]<0&&num[i]>-sum)num[i]--,val--;
        }
        if(val!=sum)
        {
            printf("Impossible\n");
            continue;
        }
        printf("Possible\n");
        printf("%d",num[0]);
        contz=1;
        for(int i=1;i<n;i++)
        {
            if(a[i]!='?')printf("%c",a[i]);
            else
            {
                printf("%d",abs(num[contz++]));
            }
        }
        printf("\n");
    }
}