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  • Uva

    Given a sequence of integers S = {S1, S2, ..., Sn}, you should determine what is the value of the maximum positive product involving consecutive terms of S. If you cannot find a positive sequence, you should consider 0 as the value of the maximum product.

    Input

    Each test case starts with 1 ≤ N ≤ 18, the number of elements in a sequence. Each element Si is an integer such that -10 ≤ Si ≤ 10. Next line will have N integers, representing the value of each element in the sequence. There is a blank line after each test case. The input is terminated by end of file (EOF).

    Output

    For each test case you must print the message: Case #M: The maximum product is P., where M is the number of the test case, starting from 1, and P is the value of the maximum product. After each test case you must print a blank line.

    Sample Input

    3
    2 4 -3
    
    5
    2 5 -1 2 -1
    
    

    Sample Output

    Case #1: The maximum product is 8.
    
    Case #2: The maximum product is 20.

    二重循环遍历过去就搞定了,因为每个数绝对值不超过10,不超过18个数,最大乘积不会超过10的18次方,这样就可以把乘积用long long存放。

    不知道为什么用printf打印结果不管怎么弄都是WA,都快疯了,最后用了cout却AC了,真是各种无语。

    AC代码:

    #include <iostream>
    #include <cstdio>
    #include <cstdlib>
    #include <cctype>
    #include <cstring>
    #include <string>
    #include <sstream>
    #include <vector>
    #include <set>
    #include <map>
    #include <algorithm>
    #include <stack>
    #include <queue>
    #include <bitset> 
    #include <cassert> 
    
    using namespace std;
    
    int s[20];
    
    int main()
    {
    	ios::sync_with_stdio(false);
    	int n;
    	int kase = 0;
    	while (cin >> n && n) {
    		for (int i = 0; i < n; i++) {
    			cin >> s[i];
    		}
    		long long pro = 0;
    		for (int star = 0; star < n; star++) {
    			long long proTem = 1;
    			for (int end = star; end < n; end++) {
    				proTem = proTem * (long long)s[end];
    				if (proTem > pro) {
    					pro = proTem;
    				}
    			}
    		}
    		//printf("Case #%d: The maximum product is %d.
    
    ", ++kase, pro);
    		cout << "Case #" << ++kase << ": The maximum product is " << pro << ".
    
    ";
    	}
    
    	return 0;
    }



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  • 原文地址:https://www.cnblogs.com/zhangyaoqi/p/4591552.html
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