UVA - 11059:Maximum Product
5438 ワード
Maximum Product
出典:UVA
ラベル:
参考資料:
類似タイトル:
タイトル
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.
入力
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).
しゅつりょく
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.
入力サンプル
3 2 4 -3
5 2 5 -1 2 -1
出力サンプル
Case #1: The maximum product is 8. Case #2: The maximum product is 20.
リファレンスコード
出典:UVA
ラベル:
参考資料:
類似タイトル:
タイトル
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.
入力
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).
しゅつりょく
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.
入力サンプル
3 2 4 -3
5 2 5 -1 2 -1
出力サンプル
Case #1: The maximum product is 8. Case #2: The maximum product is 20.
リファレンスコード
#include
int main()
{
int n,kase=0;
while(scanf("%d",&n)!=EOF)
{
int arr[n];
int i,j;
for(i=0;i<n;i++)
scanf("%d",&arr[i]);
long long maxNum=0;
for(i=0;i<n;i++)
{
long long res=1;
for(j=i;j<n;j++)
{
res*=arr[j];
if(res>maxNum)maxNum=res;
}
}
printf("Case #%d: The maximum product is %lld.
",++kase,maxNum);
}
return 0;
}