UVA 10700 Camel trading

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <stack>
 4 #define sc(x)    scanf("%d",&x)
 5 #define sc1(x)    scanf("%lld",&x)
 6 #define pf(x)    printf("%d\n",x)
 7 #define FOR(i,b,e)    for(int i=b;i<N;i++)
 8 using namespace std;
 9 stack <long long>Max;
10 stack <long long>Min;
11 int main()
12 {
13     int N;
14     sc(N);
15     while(N--)
16     {
17         while(!Max.empty())
18         Max.pop();
19         while(!Min.empty())
20         Min.pop();
21         long long a, t, ans_max = 1, ans_min = 0;
22         sc1(a);
23         char ch;
24         Min.push(a);
25         Max.push(a);
26         while((ch = getchar()) != ‘\n‘)
27        {
28                sc1(a);
29                if(ch == ‘+‘){
30                Min.push(a);
31                t = Max.top();
32                Max.pop();
33                t += a;
34                Max.push(t);
35             }
36             else if(ch==‘*‘){
37                Max.push(a);
38                t = Min.top();
39                Min.pop();
40                t *= a;
41                Min.push(t);
42             }
43        }
44        while(!Min.empty())
45        {
46                ans_min += Min.top();
47                Min.pop();
48        }
49        while(!Max.empty())
50        {
51                ans_max *= Max.top();
52                Max.pop();
53        }
54         printf("The maximum and minimum are %lld and %lld.\n",ans_max,ans_min);
55     }
56     return 0;
57 }

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。