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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。
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原文地址:http://www.cnblogs.com/tyx0604/p/4392667.html