题目：10700 - Camel trading

题目大意：给出一些表达式，表达式由数字和加号乘号组成，数字范围【1，20】。这些表达式可能缺少了括号，问这样的表达式加上括号后能得到的最大值和最小值。

解题思路：因为这些数的都是正整数，所以可以用贪心。不然看出最大值就是先做完加法在做乘法，最小值就是先做乘法在做加法。注意这里的数值要用long long 因为比表达式的值可能会超过int。

代码：

#include 
  
   
#include 
   
     const int N = 15; char op[N]; char str[3 * N]; long long num[N]; long long caculate_max (int count) { long long temp[N]; for (int i = 0; i < count; i++) temp[i] = num[i]; for (int i = count - 2; i >= 0; i--) if (op[i] == '+') { temp[i] += temp[i + 1]; temp[i + 1] = temp[i]; } long long sum = temp[0]; for (int i = 0; i < count - 1; i++) { if (op[i] == '*') sum *= temp[i + 1]; } return sum; } long long caculate_min (int count) { long long temp[N]; for (int i = 0; i < count; i++) temp[i] = num[i]; for (int i = count - 2; i >= 0; i--) { if (op[i] == '*') temp[i] = temp[i] * temp[i + 1]; } long long sum = temp[0]; for (int i = 0; i <= count - 2; i++) { if (op[i] == '+') sum += temp[i + 1]; } return sum; } int init () { int t = 0; long long sum; scanf ("%s", str); for (int i = 0; i < strlen (str); i++) { if (str[i] == '+' || str[i] == '*') op[t++] = str[i]; else { sum = 0; while (str[i] >= '0' && str[i] <= '9') { sum = sum * 10 + str[i] - '0'; i++; } num[t] = sum; i--; } } return t + 1; } int main () { int t; int count; scanf ("%d", &t); while (t--) { count = init(); printf ("The maximum and minimum are %lld and %lld.\n", caculate_max(count), caculate_min(count)); } return 0; }