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\n",max);
}
else{
printf("0\n");
}
delete val;
}
return 0;
}