Submission #19711

#	Time	Username	Problem	Language	Result	Execution time	Memory
19711		emppu	팔찌 (kriii4_V)	C++14	100 / 100	226 ms	60316 KiB

V

#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cstring>
#include <cassert>

#include <algorithm>
#include <vector>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <deque>
#include <list>
#include <bitset>
#include <tuple>

using namespace std;

#define f0(_X,_Y) for(int (_X)=0;(_X)<(_Y);++(_X))
#define f1(_X,_Y) for(int (_X)=1;(_X)<=(_Y);++(_X))
#define ff(_X,_Y,_Z) for(int (_X)=(_Y);(_X)<=(_Z);++(_X))
#define fF(_X,_Y,_Z) for(int (_X)=(_Y);(_X)<(_Z);++(_X))
#define rf0(_X,_Y) for(int _X=(_Y)-1;(_X)>=0;--(_X))
#define rf1(_X,_Y) for(int _X=(_Y);(_X)>0;--(_X))
#define rff(_X,_Y,_Z) for(int _X=(_Y);(_X)>=(_Z);--(_X))
#define rfF(_X,_Y,_Z) for(int _X=(_Y);(_X)>(_Z);--(_X))
#define PRT(_X) std::cout<< #_X ": "<<_X<<std::endl;

#define scan1(_X) scanf("%d",&_X);
#define scan2(_X,_Y) scanf("%d%d",&_X,&_Y);
#define scan3(_X,_Y,_Z) scanf("%d%d%d",&_X,&_Y,&_Z);
#define define1(_1) int _1; scan1(_1)
#define define2(_1,_2) int _1,_2; scan2(_1,_2)
#define define3(_1,_2,_3) int _1,_2,_3; scan3(_1,_2,_3)
#define EXPAND(_1) _1
#define SELECT(_1,_2,_3,_4,NAME,...) NAME
#define scan(...) EXPAND(SELECT(__VA_ARGS__, scan4, scan3, scan2, scan1)(__VA_ARGS__))
#define define(...) EXPAND(SELECT(__VA_ARGS__, define4, define3, define2, define1)(__VA_ARGS__))
#define print(_X) printf("%d\n",_X)
#define PAIR_STRUCT(_T,_X,_Y,...) struct _T{int _X,_Y,##__VA_ARGS__; bool friend operator < (const _T &p, const _T &q){if(p._X!=q._X) return p._X<q._X; return p._Y<q._Y;}}

typedef long long ll;

const int MOD = 1000000007;
ll inv(ll a, ll b=MOD)
{
	ll s = 0;  ll  old_s = 1;
	ll r = b;  ll  old_r = a;
	while (r!=0)
	{
		ll quotient = old_r / r;
		tie(old_r, r) = make_tuple(r, old_r - quotient * r);
		tie(old_s, s) = make_tuple(s, old_s - quotient * s);
	}
	ll inv = old_s%MOD;
	if(inv<0) inv += MOD;
	return inv;
}

const int N = 1000005;
ll pi[N];
ll power[N];
ll s[N];
ll in[N];
int dv[N];
vector<int> d[N];

ll mul(ll a, ll b){return a*b%MOD;}
ll add(ll a, ll b){return (a+b)%MOD;}

int main()
{
	define(n,color);
	// divisors
	ff(i,2,n)
	{
		for(int j=i;j<=n;j+=i) if(!dv[j]) dv[j]=i;
	}
	// inverse
	in[1]=1;
	ff(i,2,n)
	{
		if(dv[i]!=i)
			in[i]=  mul(in[dv[i]],in[i/dv[i]]);
		else
			in[i]= inv(i);
	}
	// euler
	pi[1]=1;
	ff(i,2,n)
	{
		int num = i;
		while(num%dv[i]==0) num/=dv[i];

		if(num==1)
			pi[i]=i/dv[i]*(dv[i]-1);
		else
			pi[i] = pi[num]*pi[i/num];
	}

	// a^?
	power[0]=1;
	f1(i,n)
		power[i]=mul(power[i-1],color);

	const ll two = inv(2);

	ll ans=1;
	f1(divisor,n)
	{
		for(int i=divisor;i<=n;i+=divisor)
			s[i]+=pi[divisor]*power[i/divisor];
	}
	f1(i,n)
	{
		ll sum = 0;
		sum = s[i]%MOD;

		if(i&1)
			sum=add(sum,mul(i,power[(i+1)/2]));
		else
		{
			ll term = mul(two,i);
			term = mul(term, 1+color);
			term = mul(term, power[i/2]);
			sum=add(sum,term);
		}
		ans+=mul(sum,mul(two,in[i]));
	}
	ans %= MOD;
	printf("%lld\n",ans);
	return 0;
}

• Subtask 1: 6 / 6
• Subtask 2: 94 / 94