我正在执行程序,我正在检查数组是否与数组一样平衡
int a5[] = {2, 1, 4, 3}; // balance array because i got even number on even position so program return 1
int a5[] = {3, 1, 4, 3}; // un balance array because i got odd number on even position so program return 0
这是我正在尝试的程序
int araay(int arg[], int length);
int main()
{
int a6[] = {3, 3, 4, 4};
int a7[] = {2, 2, 3, 4};
int a8[] = {4, 1, 2, 3};
int a9[] = {1, 1};
araay (a7,sizeof(a7));
}
int araay (int arg[], int length)
{
int sumEven = 0;
int sumOdd = 0;
for (int i=0; i<=length; i=i+2)
{
if (arg[i]%2 == 0)
{
sumEven++;
}
else
sumOdd++;
}
for (int i=1; i<=length; i=i+2)
{
if (arg[i]%2 == 0)
{
sumEven++;
}
else
sumOdd++;
}
return 0;
}
作为回报,它总是每次都给我00000
返回零值
答案 0 :(得分:3)
以下可能有所帮助:(http://ideone.com/NttqbY)
bool is_balanced(const std::vector<std::size_t>& v)
{
for (std::size_t i = 0; i != v.size(); ++i) {
if ((i % 2) != (v[i] % 2)) {
return false;
}
}
return true;
}
答案 1 :(得分:2)
感谢大家的意见和帮助
这是我试过的
int araay(int arg[], int length);
int main()
{
int a6[] = {3, 3, 4, 4};
int a7[] = {2, 3, 2, 3};
int a8[] = {4, 1, 2, 3};
int a9[] = {1, 1};
araay (a7,3);
}
int araay (int arg[], int length)
{
int sumEven = 0;
int sumOdd = 0;
for (int i=0; i<=length; i+=2)
{
if (arg[i]%2 != 0)
{
cout<<"unbalanced"<<endl;
// return 0;
}
else
{
sumEven++;
}
}
for (int i=1; i<=length; i=i+2)
{
if (arg[i]%2 == 0)
{
cout<<"unbalanced"<<endl;
sumEven++;
}
else
{
sumOdd++;
// return 0;
}
}
return 0;
}
但@jarod的答案看起来更合适,更轻松
答案 2 :(得分:0)
这是你可以做的
#include <iostream>
int check(int arg[])
{
for (auto i = 0; i < sizeof(arg); ++i)
{
if ((i % 2) != (arg[i] % 2))
{
return 0;
}
}
return 1;
}
void main()
{
int a[] = { 1, 2, 3, 4 };
int b[] = { 2, 3, 4, 5 };
int c[] = { 2, 2, 3, 3 };
std::cout << "a = " << check(a) << std::endl;
std::cout << "b = " << check(b) << std::endl;
std::cout << "c = " << check(c) << std::endl;
getchar();
}