HDU6053 前缀和+容斥

#include
#include
#include
#include
#include
using namespace std;
typedef long long ll;
const int MAXN=2e5+10;
const ll mod=1e9+7;
ll dp[MAXN];
ll cnt[MAXN],a[MAXN],n;//cnt[MAXN]1到i的a[i]的个数 
ll fsm(ll x,ll n)
{
	ll s=1;
	while(n)
	{
		if(n&1)
			s=(s*x)%mod;
		n>>=1;
		x=(x*x)%mod;
	}
	return s%mod;
}
int main()
{
	int t;
	int cas=1;
	ll ans;
	ll res;
	scanf("%d",&t);
	while(t--)
	{
		memset(dp,0,sizeof(dp));
		memset(cnt,0,sizeof(cnt));
		ll mxn=0;
		scanf("%d",&n);
		for(int i=0;i=2;i--)	
		{
			res=1;
			if(cnt[i-1])
			{
				dp[i]=0;
				continue;
			}
			for(int j=i;j<=mxn;j+=i)
			{
				ll num=cnt[min(mxn,(ll)j+i-1)]-cnt[j-1];
				ll x=j/i;
				res=(res*fsm(x,num))%mod;
			} 
			dp[i]=res;
		}
		for(int i=mxn;i>=2;i--)
		{
			for(int j=i+i;j<=mxn;j+=i)
			{
				dp[i]=(dp[i]-dp[j]+mod)%mod;
			}
			ans=(ans+dp[i])%mod;	
		}	
		 printf("Case #%d: %lld\n",cas,ans);
		 cas++;
	}
	return 0;
}