#include <iostream>
using namespace std;
//快速幂,计算a^b%29
int quickPower(int a, int b)
{
int ans = 1;
a %= 29;
while (b)
{
if (b & 1)
ans = (ans * a) % 29;
b >>= 1;
a = (a * a) % 29;
}
return ans;
}
int main()
{
int X;
int a, b, c;
while (cin >> X)
{
if (X == 0)
break;
a = (quickPower(2, 2 * X + 1) - 1);
b = (quickPower(3, X + 1) - 1) * 15;
c = (quickPower(22, X + 1) - 1) * 18;
cout << a * b * c % 29 << endl;
}
return 0;
}
HDU 1452(Happy 2004)
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转载自blog.csdn.net/Intelligence1028/article/details/105388381
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