Submission #2636236

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#include <bits/stdc++.h>
using namespace std;

#define NDEBUG
#ifdef DEBUG
#include "../cout11.h"
#undef NDEBUG
#endif
#include <cassert>

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> ii;
typedef pair<ll,ll> llll;
typedef pair<double,double> dd;

typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef vector<ii> vii;
typedef vector<vector<ii>> vvii;
typedef vector<ll> vll;

#define sz(a)  int((a).size())
#define pb  push_back
#define FOR(var,from,to) for(int var=(from);var<=(to);++var)
#define rep(var,n)  for(int var=0;var<(n);++var)
#define rep1(var,n)  for(int var=1;var<=(n);++var)
#define repC2(vari,varj,n)  for(int vari=0;vari<(n)-1;++vari)for(int varj=vari+1;varj<(n);++varj)
#define ALL(c)  (c).begin(),(c).end()
#define RALL(c)  (c).rbegin(),(c).rend()
#define tr(i,c)  for(auto i=(c).begin(); i!=(c).end(); ++i)
#define found(s,e)  ((s).find(e)!=(s).end())
#define mset(arr,val)  memset(arr,val,sizeof(arr))
#define mid(x,y) ((x)+((y)-(x))/2)
#define IN(x,a,b) ((a)<=(x)&&(x)<=(b))

#include <vector>
using namespace std;

vector<int> primes;
vector<int> smallest_prime_factor;

int sieve(int nmax){
    primes.clear();
    smallest_prime_factor.assign(nmax+1, 0);

    // ここではあえて愚直な書き方
    for (int n=2; n<=nmax; ++n) {
        if (!smallest_prime_factor[n]) {
            primes.push_back(n);
            for (int kn=n; kn<=nmax; kn+=n) {
                if (!smallest_prime_factor[kn]) {  // ←このチェックをせずに上書きを続けると largest_prime_factor[] が出来上がる
                    smallest_prime_factor[kn] = n;
                }
            }
        }
    }
    return primes.size();
}

// xを素因数分解
vector<int> factorize(int x) {
    vector<int> prime_factors;
    while (x > 1) {
        int s = smallest_prime_factor[x];
        if (!s) break;
        prime_factors.push_back(s);
        x /= s;
    }
    return prime_factors;
}

ll gcd(ll a, ll b) { while(a) swap(a, b%=a); return b; }

const ll M=1000000007LL;

ll ADD(ll x, ll y) { return (x+y) % M; }
ll SUB(ll x, ll y) { return (x-y+M) % M; }
ll MUL(ll x, ll y) { return x*y % M; }
ll POW(ll x, ll e) { ll v=1; for(; e; x=MUL(x,x), e>>=1) if (e&1) v = MUL(v,x); return v; }
ll DIV(ll x, ll y) { assert(y%M!=0); return MUL(x, POW(y, M-2)); }
// ll comb(ll n, ll k) { ll v=1; for(ll i=1; i<=k; i++) v = DIV(MUL(v, n-i+1),i); return v; }


ll solve(int N) {
    // 1-1e3
    map<int,int> st;
    rep1(i,N){
        vi f = factorize(i);
        for (int fi : f){
            ++st[fi];
        }
    }
    // cerr << st << endl;
    ll ans = 1LL;
    for(auto p:st){
        ans = MUL(ans, p.second+1);
    }
    return ans;
}

int main() {
    sieve(1000);
    int N; cin >> N;
    cout << solve(N) << endl;
    return 0;
}

Submission Info

Submission Time
Task C - Factors of Factorial
User naoya_t
Language C++14 (GCC 5.4.1)
Score 300
Code Size 2914 Byte
Status
Exec Time 1 ms
Memory 256 KB

Test Cases

Set Name Score / Max Score Test Cases
Sample 0 / 0 sample_01.txt, sample_02.txt, sample_03.txt
All 300 / 300 sample_01.txt, sample_02.txt, sample_03.txt, subtask_1_certain_01.txt, subtask_1_certain_02.txt, subtask_1_certain_03.txt, subtask_1_certain_04.txt, subtask_1_rand_01.txt, subtask_1_rand_02.txt, subtask_1_rand_03.txt
Case Name Status Exec Time Memory
sample_01.txt 1 ms 256 KB
sample_02.txt 1 ms 256 KB
sample_03.txt 1 ms 256 KB
subtask_1_certain_01.txt 1 ms 256 KB
subtask_1_certain_02.txt 1 ms 256 KB
subtask_1_certain_03.txt 1 ms 256 KB
subtask_1_certain_04.txt 1 ms 256 KB
subtask_1_rand_01.txt 1 ms 256 KB
subtask_1_rand_02.txt 1 ms 256 KB
subtask_1_rand_03.txt 1 ms 256 KB