UVA 10700 Camel trading

#include <iostream>
#include <cstdio>
#include <stack>
#define sc(x)    scanf("%d",&x)
#define sc1(x)    scanf("%lld",&x)
#define pf(x)    printf("%d
",x)
#define FOR(i,b,e)    for(int i=b;i<N;i++)
using namespace std;
stack <long long>Max;
stack <long long>Min;
int main()
{
    int N;
    sc(N);
    while(N--)
    {
        while(!Max.empty())
        Max.pop();
        while(!Min.empty())
        Min.pop();
        long long a, t, ans_max = 1, ans_min = 0;
        sc1(a);
        char ch;
        Min.push(a);
        Max.push(a);
        while((ch = getchar()) != '
')
       {
               sc1(a);
               if(ch == '+'){
               Min.push(a);
               t = Max.top();
               Max.pop();
               t += a;
               Max.push(t);
            }
            else if(ch=='*'){
               Max.push(a);
               t = Min.top();
               Min.pop();
               t *= a;
               Min.push(t);               
            }
       }
       while(!Min.empty())
       {
               ans_min += Min.top();
               Min.pop();
       }
       while(!Max.empty())
       {
               ans_max *= Max.top();
               Max.pop();
       }
        printf("The maximum and minimum are %lld and %lld.
",ans_max,ans_min);
    }
    return 0;
}

这个是数学思路，需要用栈解决，同时输入字符是getchar();

通过一个一个字符输入，避免了字符串的输入，那时还要考虑转换成数组问题。