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  • 最大乘积(Maximum Product,UVA 11059)

    Problem D - Maximum Product

    Time Limit: 1 second

    Given a sequence of integers S = {S1, S2, ..., Sn}, you should determine what is the value of the maximum positive product involving consecutive terms ofS. 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 elementSi is an integer such that-10 ≤ Si ≤ 10. Next line will haveN 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., whereM is the number of the test case, starting from1, andP 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.
    
    



    #include <stdio.h>
    #include <set>
    
    using namespace std;
    
    int main(){
    	int n;
    	int * val = NULL;
    	set<long long> s;
    	while((scanf("%d",&n) == 1) && n != 0){
    		val =  new int[n];
    		int i = 0;
    		for (;i  < n; i++) {
    			scanf("%d",&val[i]);
    		}
    		s.clear();
    		int j ;
    		for(i = 0;i < n -1;i++){	//枚举起点
    
    			for (j = i; j < n; ++j) {//枚举终点
    				int k;long long ji = 1;
    				for(k = i; k <= j;k++){
    					ji *= val[k];
    				}
    				s.insert(ji);
    			}
    
    		}
    
    		long long max = *(s.rbegin());
    		if(max > 0){
    			printf("%ld
    ",max);
    		}
    		else{
    			printf("0
    ");
    		}
                    delete val;
            }
    
    
    	return 0;
    
    }
    
    




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  • 原文地址:https://www.cnblogs.com/lvyahui/p/4009958.html
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