The 3n + 1 problem （C++代码）

数字本身也正在运行次数类，所以最后的max要加1

每次循环结束记得要k归0，

最大值也要归零

#include<iostream>

using namespace std;

int main() 

{

long long i,n, m, t, s, max=0,k = 0;

while (cin >> n >> m)

{

cout << n << ' ' << m << ' ';

t = n;

if (n > m) n = m, m = t;

for ( i = n; i <=m; i++)

{

s = i;

while (s != 1)

{

if (s % 2 == 0)s = s / 2;

else s = 3 * s + 1;

k++;

}

if (max < k) max = k;

k = 0;

}

cout << max+1 << endl;

max = 0;

}

return 0;

}