UVA 11059

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.

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，就输出0（表示无解）。 1<=n<=18 -10<=S<=10

输出格式每输出一组案例就空一行（注意）

题目分析：连续子序列有两个要素：起点和重点。所以只要枚举起点和终点就好。由于每个元素与的最大值不会超过10,且不超过18个元素，最大乘积不会超过10的18次方。所以可以用long long储存（试了一下，用int的话，输出18个10的结果不对，int存不下）

代码如下：（妈的，刚开始想到3重循环去了，还傻逼的循环了一个len表示子序列的长度，尽管试了很多案例都对了，就是不能过。重想了一下，抱着试一试的心态，写了，然后TM就过了，也是RLGL.....）

 #include <stdio.h>

 int a[];

 int main()

 {

     int n,N=,c2=;

     while(scanf("%d",&n)==)

     {

         long long c,c2=;

         ++N;

         for(int i=; i<n; i++)

             scanf("%d",&a[i]);

         for(int q=;q<n;q++)

         {

             c=;

             for(int z=q;z<n;z++)

             {

                 c*=a[z];

                 if(c>c2)

                     c2=c;

             }

         }

         if(c2<=)

             printf("Case #%d: The maximum product is 0.\n\n",N);

         else

             printf("Case #%d: The maximum product is %lld.\n\n",N,c2);

     }

     return ;

 }

巴特西

UVA 11059

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