poj 1023

来源：互联网发布：android反编译apk mac 编辑：程序博客网时间：2024/05/17 17:17

// Memory: 720K  Time: 16MS

#include <iostream>#include <cmath>#include <cstdio> using namespace std;void strrev(char str[], int n){    for(int i=0; i < n/2; i++)    {       char t = str[i];       str[i] = str[n-i-1];       str[n-i-1] = t;    }} char data[70];char re[70];int main(){    intn;    scanf("%d",&n);    while(n--)    {       long long  len, num;       scanf("%lld", &len);       scanf("%s", data);       scanf("%lld", &num);       strrev(data, len);       for (int i=1; i <= len; i++)       {           //cout << num%(int)pow(2.0,i+0.0);           if (num % (long long)pow(2.0, i+0.0) != 0)           {                re[i-1] = '1';                if (data[i-1] == 'p')                {                    num -= (long long)pow(2.0,i-1.0);               }                else                {                    num += (long long)pow(2.0,i-1.0);                }           }           else                re[i-1] = '0';       }       if (num != 0)           printf("Impossible\n");        else       {           re[len] = 0;           strrev(re, len);           printf("%s\n", re);       }    }     return0;}

刚开始加减那个地方没想通，后来发现人家是用num来减，减成0了就算成功