(打表或数学)HDU 6063 RXD and math

题意分析:

代码:

#include <iostream>
#include <cmath>

using namespace std;

#define ll long long

const ll mod = 1e9+7;

void seeTable(); // 打表找规律

ll quickpower(ll n, ll k)
{
ll ans = 1;
n %= mod;
while (k)
{
if(k & 1) ans = (ans * n) % mod;
k >>= 1;
n = (n*n) % mod;
}
return ans;
}

int main(int argc, char const *argv[])
{
// seeTable();  // 打表可知, 其实求的就是n^k

ll n, k;
int cnt = 1;
while (~scanf("%I64d %I64d", &n, &k))
{
printf("Case #%d: ", cnt++);
printf("%I64d\n", quickpower(n, k));
}
return 0;
}

void seeTable()
{
for(int n = 2; n < 6; ++n)
{
int k = 1;
while (k < 11)
{
int nk = pow(n, k);
int sum = 0;

for (int i = 1; i <= nk; ++i)
{
int tmp = i;
bool flag = true;
for (int i = 2; tmp != 1; ++i)
{
int cnt = 0; // 记录有多少个相同的素数
while (tmp % i == 0)
{
++cnt;
if(cnt > 1)
{
flag = false;
}
tmp /= i;
}
}
if(flag)
{ // 每个素数只出现一次
sum += (int)sqrt((double)nk/(double)i);
}
}
printf("n=%-3d k=%-3d \t n^k=%-8d \t result=%d\n", n, k, nk, sum);
printf("==============================================================\n");
k++;
}
}
}