HDU-2017 多校训练赛2-1009-TrickGCD

描述

题解

代码

#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;

typedef long long ll;

const int MAXN = 1e5 + 10;
const ll MOD = 1e9 + 7;

int mu[MAXN];

void mobius(int x)
{
mu[1] = 1;
for (int i = 1; i <= x; i++)
{
for (int j = i + i; j <= x; j += i)
{
mu[j] -= mu[i];
}
}
}

int n;
int cnt[MAXN << 1];

ll QPow(ll x, int y)
{
ll ret = 1;
while (y > 0)
{
if (y & 1)
{
ret = ret * x % MOD;
}
x = x * x % MOD;
y >>= 1;
}

return ret;
}

template <class T>
inline void scan_d(T &ret)
{
char c;
ret = 0;
while ((c = getchar()) < '0' || c > '9');
while (c >= '0' && c <= '9')
{
ret = ret * 10 + (c - '0'), c = getchar();
}
}

int main()
{
mobius(MAXN);

int T, ce = 1, mn;

scan_d(T);
while (T--)
{
mn = MAXN;
memset(cnt, 0, sizeof(cnt));

scan_d(n);

int x;
while (n--)
{
scan_d(x);
cnt[x]++;
mn = min(mn, x);
}

ll ans = 0;
int t = MAXN << 1;
for (int i = 1; i <= t; i++)
{
cnt[i] += cnt[i - 1];
}
for (int i = 2; i <= mn; i++)
{
ll temp = 1;
for (int j = 1; i * j <= MAXN; j++)
{
temp = (temp * QPow((ll)j, cnt[i * j + i - 1] - cnt[i * j - 1])) % MOD;
}
ans = (ans - temp * mu[i] + MOD) % MOD;
}

printf("Case #%d: %lld\n", ce++, ans);
}

return 0;
}