#include <iostream>
using namespace std;
int main()
{
int num1=0, num2=0, num3=0, num4=0, num5=0, num6=0, num7=0, num8=0, num9=0, num10=0,even,odd,sum;
cout<<"Enter (10) Integers : ";
cin>> num1>>num2>>num3>>num4>>num5>>num6>>num7>>num8>>num9>>num10;
if(num1 %2==0)
even=even + num1;
else
odd=odd+ num1;
if(num2 %2==0)
even=even + num2;
else
odd=odd+ num2;
if(num3 %2==0)
even=even + num3;
else
odd=odd+ num3;
if(num4 %2==0)
even=even + num4;
else
odd=odd+ num4;
if(num5 %2==0)
even=even + num5;
else
odd=odd+ num5;
if(num6%2==0)
even=even + num6;
else
odd=odd+ num6;
if(num7 %2==0)
even=even + num7;
else
odd=odd+ num7;
if(num8%2==0)
even=even + num8;
else
odd=odd+ num8;
if(num9 %2==0)
even=even + num9;
else
odd=odd+ num9;
if(num10 %2==0)
even=even + num10;
else
odd=odd+ num10;
sum = num1 + num2 + num3 + num4 + num5 + num6 + num7 + num8 + num9 + num10;
cout<<"\nThe sum of ODD numbers is "<<odd;
cout<<" \nThe sum of EVEN numbers is "<<even;
cout<<" \nThe sum of all input numbers is "<<sum;
return 0;
}
答案 0 :(得分:0)
您应该使用此代码:
int numarr[10];
int sum = 0, even = 0, odd = 0;
for (int i = 0; i < 10; i++)
{
cin >> numarr[i];
if (numarr[i] % 2 == 0)
even = even + numarr[i];
else
odd = odd + numarr[i];
sum = sum + numarr[i];
}
那些多个if-else
陈述令人不悦。