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链接:http://acm.hust.edu.cn/vjudge/problem/27946
分析:Two Pointers搞搞。
1 #include <iostream> 2 #include <algorithm> 3 using namespace std; 4 5 typedef long long LL; 6 7 int n, a[20]; 8 9 LL gao() { 10 LL ans = 0; 11 for (int L = 0; L < n; L++) { 12 LL res = 1; 13 for (int R = L; R < n; R++) { 14 res *= a[R]; 15 ans = max(ans, res); 16 } 17 } 18 return ans; 19 } 20 21 int main() { 22 int kase = 0; 23 while (cin >> n) { 24 for (int i = 0; i < n; i++) cin >> a[i]; 25 printf("Case #%d: The maximum product is %lld.\n\n", ++kase, gao()); 26 } 27 return 0; 28 }
UVa11059 Maximum Product (Two Pointers)
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原文地址:http://www.cnblogs.com/XieWeida/p/5837895.html