CodeForces 663A Rebus
A. 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
? + ? - ? + ? + ? = 42output
Possible
9 + 13 - 39 + 28 + 31 = 42input
? - ? = 1output
Impossibleinput
? = 1000000output
Possible
1000000 = 1000000#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>
using namespace std;
char a[1000];
int n;
int num1,num2;
int b[1000];
int main()
{
gets(a);
int len=strlen(a);
num1=0;num2=0;
int cnt=1;int pos;int now=1;b[0]=1;
for(int i=0;i<len;i++)
{
if(a[i]=='+') {b[cnt++]=1;now++;}
if(a[i]=='-') {b[cnt++]=-1;now--;}
if(a[i]=='=')
{
pos=i;
break;
}
}
n=0;
for(int i=pos+1;i<len;i++)
{
if(a[i]<='9'&&a[i]>='0')
n=n*10+a[i]-'0';
}
for(int i=0;i<cnt;i++)
{
while(now<n&&b[i]<n&&b[i]>0)
now++,b[i]++;
while(now>n&&b[i]>-n&&b[i]<0)
now--,b[i]--;
}
if(now!=n)
{
printf("Impossible\n");
return 0;
}
printf("Possible\n");
int j=0;
for(int i=0;i<len;i++)
{
if(a[i]!='?')
printf("%c",a[i]);
else
{
printf("%d",abs(b[j++]));
}
}
cout<<endl;
return 0;
}腾讯云开发者
