UVa 11509 Maximum Product(最大乘积)

题目链接：UVa 11509

题意：

输入n个元素组成的序列S，你需要找出一个乘积最大的连续子序列。如果这个最大的乘积不是正数，应输出0（表示无解）。1<=n<=18,-10<=S[i]<=10。

分析：

两层循环对每个s[i]求最大连续子序列乘积，逐个取最大即可。

注意：

• C++中 long long 不能用%I64d输出。

CODE:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <algorithm>
#include <queue>
using namespace std;int n, a[20],cases=0;
long long ans, tmp;int main()
{
#ifdef LOCALfreopen("in.txt", "r", stdin);//freopen("out.txt", "w", stdout);
#endifwhile (cin >> n){for (int i = 0; i < n; i++)cin >> a[i];ans = 0;for (int i = 0; i < n; i++){tmp = 1;for (int j = i; j < n; j++){tmp *= a[j];ans = max(ans, tmp);}}printf("Case #%d: The maximum product is %lld.\n\n", ++cases, ans);}return 0;
}

附一个O(n)的解法。原博地址：UVa 11059

#include <iostream>
#include <cstdlib>
#include <cstring>using namespace std;int data[20];int main()
{int n,t = 1;while (cin >> n) {for (int i = 1 ; i <= n ; ++ i)cin >> data[i];long long max = 0,neg = 0,pos = 0,tem;for (int i = 1 ; i <= n ; ++ i) {if (data[i] == 0) neg = pos = 0;else if (data[i] > 0) {pos = pos*data[i];neg = neg*data[i];if (!pos) pos = data[i];}else if (data[i] < 0) {tem = pos*data[i];pos = neg*data[i];neg = tem;if (!neg) neg = data[i];}if (max < pos && data[i])max = pos;}cout << "Case #" << t ++ << ": The maximum product is " << max << ".\n\n";	}return 0;
}