UVa11059
2015-07-29 19:29
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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)
程序分析: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)
程序代码:
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)
程序分析: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)
程序代码:
#include <stdio.h> int a[20]; int main() { int n,N=0,c2=0; while(scanf("%d",&n)==1) { long long c,c2=0; ++N; for(int i=0; i<n; i++) scanf("%d",&a[i]); for(int q=0;q<n;q++) { c=1; for(int z=q;z<n;z++) { c*=a[z]; if(c>c2) c2=c; } } if(c2<=0) 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 0; }
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