Uva11059
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.
1 #include <cstdio> 2 #include <iostream> 3 #include <queue> 4 #include <vector> 5 #include<string.h> 6 #include<map> 7 #include<bits/stdc++.h> 8 #define LL long long 9 using namespace std; 10 LL ay[20]; 11 int main() 12 { 13 int n,no=0; 14 LL ans,sum; 15 while(~scanf("%d",&n)) 16 { 17 no++; 18 ans=-1; 19 for(int i=0;i<n;i++) 20 scanf("%lld",&ay[i]); 21 for(int i=0;i<n;i++) 22 for(int j=i;j<n;j++){ 23 sum=1; 24 for(int k=i;k<=j;k++) 25 { 26 sum*=ay[k]; 27 ans=max(ans,sum); 28 } 29 } 30 if(ans<0)ans=0; 31 printf("Case #%d: The maximum product is %lld.\n\n",no,ans); 32 } 33 return 0; 34 }
题意:
求最大的连续子列的乘积。如果乘积为负数,则输出0。
注意点:
枚举起点终点求解。ans为最大值,sum为动态过程值。
1.ans的初值应该设为负数,不然如果设为正数的话,当乘积为负数的话,ans=max(ans,sum)>0,无法得到0。