算法:
對於第i個數,模擬計算出其置換週期,記爲ai,答案就是所有ai的最小公倍數。
1 2 3 4 5
4 1 5 2 3
次數:3 3 2 3 2,最小公倍數6.
#include <iostream>
#include <cstdio>
int gcd(int a, int b)
{
if (b == 0)
return a;
else
return gcd(b, a % b);
}
int lcm(int a, int b)
{
return a / gcd(a, b) * b;
}
int rota(int *f, int i)
{
int num = 1;
int t = f[i];
while (f[i] != f[t])
{
t = f[t];
++num;
}
return num;
}
int main()
{
int n;
scanf("%d", &n);
int f[n+1];
for (int i = 1; i <= n; ++i)
{
scanf("%d", &f[i]);
}
int ans = 1;
for (int i = 1; i <= n; ++i)
{
int num = rota(f, i);
ans = lcm(ans, num);
}
printf("%d\n", ans);
//system("pause");
return 0;
}