
很简单的一道题,唯一的难点就在于能不能想到当a和b和c三个相等时,应该提前结束,如果是暴力的话只有百分之50的分

#include <iostream>
using namespace std;
int main()
{
int T;
cin >> T;
while (T—)
{
long long A, B, C;
int K;
cin >> A >> B >> C >> K;
for (int i = 0; i < K; ++i)
{
long long na = (B + C) / 2;
long long nb = (A + C) / 2;
long long nc = (A + B) / 2;
// 如果数值不再变化,提前退出循环
if (na == A && nb == B && nc == C)
break;
A = na;
B = nb;
C = nc;
}
cout << A << " " << B << " " << C << endl;
}
return 0;
}




