# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
22296 | Lazy Against The Machine (#40) | Fully Generate (KRIII5_FG) | C++14 | 0 ms | 952 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <stdio.h>
long long arr[40000000];
long long nuj[40000000];
long long pro[40000000];
long long pr2[40000000];
long long mod = 1000000007;
// calculate n^k % m
long long modpow(long long n, long long k)
{
long long ret = 1;
n %= mod;
while (k)
{
if (k & 1)
ret = (ret * n) % mod;
n = (n * n) % mod;
k /= 2;
}
return ret;
}
int main()
{
arr[1] = 1;
arr[2] = 2;
nuj[1] = 1;
nuj[2] = 3;
pro[1] = 1;
pro[2] = 2;
pr2[1] = 1;
pr2[2] = 2;
long long sum = 1;
long long start = 2;
long long end = 2;
int n;
scanf("%d", &n);
for (int i = 2;; i++)
{
for (int j = 0; j < arr[i] ; j++)
{
arr[start + j] = i;
nuj[start + j] = nuj[start + j - 1] + i;
pro[start + j] = (pro[start + j - 1] * i) % mod;
if (nuj[start + j] >= n)
{
end = start + j;
goto hell;
}
}
start += arr[i];
}
hell:
for (int i = 2; i < end; i++)
{
pr2[i] = (pr2[i - 1] * modpow(i, arr[i])) % mod;
}
printf("%lld", (pr2[end - 1] * modpow(end, n - nuj[end - 1])) % mod);
return -0;
}
Compilation message (stderr)
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