Some items--give you n integers, ask for the product of all the odd numbers in them.

Problem description gives you n integers, asking for the product of all the odd numbers in them. Input data contains multiple test instances, one row per test instance, and N for each row, indicating that there is a total of n for this group of data, followed by n integers, you can assume that each set of data must have at least one odd number. Output outputs the product of all the odd numbers in each set, and one row for the test instance. Sample Input

3 1 2 34 2 3 4 5

Sample Output

315

Code

#include <iostream>using namespace Std;int main () {    int n,m,i,s;    while (Cin>>n)    {        S=1;        for (i=0; i<n; i++)        {            cin>>m;            if (m%2!=0)                s=s*m;        }        cout<<s<<endl;    }    return 0;}

Learning sentiment: ashamed to say, because I did not examining, think that the number of data in each group is a meaning, not aware of the scope of the first digital limit, kept to how to end the cycle with Ctrl + Z, even the back with a double loop, know I find the answer from the Internet to analyze again ..... Alas, take this as a warning to yourself.

Some items--give you n integers, ask for the product of all the odd numbers in them.