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

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题目链接：点击打开链接

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
Possible9 + 13 - 39 + 28 + 31 = 42
input
? - ? = 1
output
Impossible
input
? = 1000000
output
Possible1000000 = 1000000

题意：给你 ? + ? - ? + ? = n，包含多个?和+、-，其中?是1-n的整数，问是否存在这样的等式，如果存在输出“Possible"和任意一组，不存在输出”Impossile"。

思路：个人觉得字符串的输入是个很头疼的问题，最终用s[1010][1010]来解决，s[i]就等于"+"或“-”或“=”或“?"或"n"，最后一个字符串是n，将它转换为数字n，遍历一个s数组，得到“+”的个数cnt1和“-”的个数cnt2;

如果cnt1==0&&cnt2==0，意味着等式是? = n，输出n = n，即可；

第一个数肯定是被加起来的，所以cnt1+1个数被加起来，有cnt2个数被减掉，如果（cnt1+1)*n-cnt2*1<n的话肯定无解，同理(cnt1+1)-cnt2*n>n的话也无解；

反之一定有解，将cnt1+1个数保存在a[1010]中，将cnt2个数保存在b[1010]中，接下来是贪心的做法，先初始化a[1~cnt1+1]=1，b[1~cnt2]=1;

如果cnt1+1-cnt2>n，证明要减的更多，将他们尽可能均分给每一个减数：

cnt3=(cnt1+1-cnt2-n)/cnt2;            for(h=1;h<=cnt2;h++)            b[h]+=cnt3;            cnt3=(cnt1+1-cnt2-n)%cnt2;            if(cnt3>=1)            for(h=1;h<=cnt3;h++) b[h]++;

如果 cnt1+1-cnt2<n，证明要加的更多，同样的处理方法，这样可以保证每一个数都大于等于1且不会超过n：

  cnt3=(n-(cnt1+1-cnt2))/(cnt1+1);            for(h=1;h<=cnt1+1;h++)            a[h]+=cnt3;            cnt3=n-(cnt1+1-cnt2)-cnt3*(cnt1+1);            for(h=1;h<=cnt3;h++) a[h]++;

下面附上AC代码：这题毕竟卡了很久，凌晨两点半睡不着又下床想了这种写法才过掉的

#include<stdio.h>#include<string.h>#include<math.h>int main(){    char s[1010][1010];int i=0,j,n=0,cnt1=0,cnt2=0,cnt3=1,a[1010],b[1010];    while(scanf("%s",s[i])!=EOF)    {        if(strcmp(s[i],"?")==0||strcmp(s[i],"+")==0||strcmp(s[i],"-")==0||strcmp(s[i],"=")==0) i++;        else break;    }    for(j=0;j<i;j++)    {        if(strcmp(s[j],"+")==0) cnt1++;        if(strcmp(s[j],"-")==0) cnt2++;    }    int len=strlen(s[i]);    for(j=len-1;j>=0;j--)    {        n+=(s[i][j]-'0')*cnt3;        cnt3*=10;    }    //printf("%d %d\n",cnt1,cnt2);    if(cnt1==0&&cnt2==0) printf("Possible\n%d = %d\n",n,n);    else if(n+n*cnt1-cnt2<n||1+cnt1-cnt2*n>n) printf("Impossible\n");    else    {        int h;        printf("Possible\n");        for(h=1;h<=cnt1+1;h++)        a[h]=1;        for(h=1;h<=cnt2;h++)        b[h]=1;        if(cnt1+1-cnt2>n)        {            cnt3=(cnt1+1-cnt2-n)/cnt2;            for(h=1;h<=cnt2;h++)            b[h]+=cnt3;            cnt3=(cnt1+1-cnt2-n)%cnt2;            if(cnt3>=1)            for(h=1;h<=cnt3;h++) b[h]++;        }        else if(cnt1+1-cnt2<n)        {            cnt3=(n-(cnt1+1-cnt2))/(cnt1+1);            for(h=1;h<=cnt1+1;h++)            a[h]+=cnt3;            cnt3=n-(cnt1+1-cnt2)-cnt3*(cnt1+1);            for(h=1;h<=cnt3;h++) a[h]++;        }        //for(h=1;h<=cnt1+1;h++) printf("%d ",a[h]);        printf("%d ",a[1]);        int m,x;m=2;x=1;        for(h=1;h<i-1;h++)        {            if(strcmp(s[h],"+")==0) printf("+ %d ",a[m++]);            if(strcmp(s[h],"-")==0) printf("- %d ",b[x++]);        }        printf("= %d\n",n);    }}//? + ? - ? - ? - ? - ? - ? = 5//? + ? + ? + ? + ? + ? + ? + ? - ? = 5

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