結果

問題 No.2365 Present of good number
コンテスト
ユーザー deuteridayo
提出日時 2023-06-30 22:02:43
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 236 ms / 2,000 ms
コード長 3,907 bytes
コンパイル時間 4,455 ms
コンパイル使用メモリ 272,440 KB
最終ジャッジ日時 2025-02-15 03:55:16
ジャッジサーバーID
（参考情報） 		judge1 / judge4
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 2
other AC * 39
権限があれば一括ダウンロードができます

ソースコード

diff # 

#include<bits/stdc++.h>
#include<atcoder/all>
using namespace std;
using namespace atcoder;
using lint = long long;
using ulint = unsigned long long;
#define endl '\n'
int const INF = 1<<30;
lint const INF64 = 1LL<<61;
lint const mod = 1e9+7;
using mint = modint1000000007;

lint ceilDiv(lint x, lint y){if(x >= 0){return (x+y-1)/y;}else{return x/y;}}
lint floorDiv(lint x, lint y){if(x >= 0){return x/y;}else{return (x-y+1)/y;}}
lint Sqrt(lint x){lint upper = 1e9;lint lower = 0;while(upper - lower > 0){lint mid = (1+upper + lower)/2;if(mid * mid > x){upper = mid-1;}else{lower = mid;}}return upper;}
lint gcd(lint a,lint b){if(a<b)swap(a,b);if(a%b==0)return b;else return gcd(b,a%b);}
lint lcm(lint a,lint b){return (a / gcd(a,b)) * b;}
lint chmin(vector<lint>&v){lint ans = INF64;for(lint i:v){ans = min(ans, i);}return ans;}
lint chmax(vector<lint>&v){lint ans = -INF64;for(lint i:v){ans = max(ans, i);}return ans;}
double dist(double x1, double y1, double x2, double y2){return sqrt(pow(x1-x2, 2) + pow(y1-y2,2));}
string toString(lint n){string ans = "";if(n == 0){ans += "0";}else{while(n > 0){int a = n%10;char b = '0' + a;string c = "";c += b;n /= 10;ans = c + ans;}}return ans;}
string toString(lint n, lint k){string ans = toString(n);string tmp = "";while(ans.length() + tmp.length() < k){tmp += "0";}return tmp + ans;}
vector<lint>prime;void makePrime(lint n){prime.push_back(2);for(lint i=3;i<=n;i+=2){bool chk = true;for(lint j=0;j<prime.size() && prime[j]*prime[j] <= i;j++){if(i % prime[j]==0){chk=false;break;}}if(chk)prime.push_back(i);}}
lint Kai[20000001]; bool firstCallnCr = true; 
lint ncrmodp(lint n,lint r,lint p){ if(firstCallnCr){ Kai[0] = 1; for(int i=1;i<=20000000;i++){ Kai[i] = Kai[i-1] * i; Kai[i] %= mod;} firstCallnCr = false;} if(n<0)return 0;
if(n < r)return 0;if(n==0)return 1;lint ans = Kai[n];lint tmp = (Kai[r] * Kai[n-r]) % p;for(lint i=1;i<=p-2;i*=2){if(i & p-2){ans *= tmp;ans %= p;}tmp *= tmp;tmp %= p;}return ans;}
#define rep(i, n) for(int i = 0; i < n; i++)
#define repp(i, x, y) for(int i = x; i < y; i++)
#define vec vector
#define pb push_back
#define se second
#define fi first
#define all(x) x.begin(),x.end()

struct edge{
    int to;
};
using graph = vector<vector<edge>>;

unordered_map<lint, map<lint, lint>> mp;
unordered_map<lint, bool> went;



int main(){
    makePrime(1000000);
    lint n,k;
    cin >> n >> k;
    for(int i = 1; i <= n+20; i++) {
        lint N = i;
        for(auto p: prime) {
            if(p*p > N)break;
            while(N % p == 0) {
                mp[i][p]++;
                N /= p;
            }
        }
        if(N != 1) {
            mp[i][N]++;
        }
    }
    map<lint, lint>cnt;
    cnt = mp[n];
    cerr << "SMILE";
    rep(i, k) {
        cerr << i << " ";
        map<lint, lint>ncnt;
        for(auto p: cnt) {
            for(auto q: mp[p.first + 1]) {
                ncnt[q.first] += (q.second * p.second);
            }
        }
        cnt = ncnt;
        if((ncnt.size() == 1 && ncnt[2] != 0)) {
            cnt.clear();
            lint rest = k - i - 1;
            int t = 2;
            if(rest % 2 == 1) t++;
            
            cnt[t] = (ncnt[2] * pow_mod(2, rest/2, mod-1)) % (mod-1);
            break;
        } else if(ncnt.size() == 2 && ncnt[2] != 0 && ncnt[3] != 0) {
            cnt.clear();
            lint rest = k - i - 1;
            if(rest % 2 == 1) {
                cnt[3] = (ncnt[2] * pow_mod(2, rest/2, mod-1)) % (mod-1);
                cnt[2] = (ncnt[3] * pow_mod(2, rest/2 + 1, mod-1))  % (mod-1);
            } else {
                cnt[2] = (ncnt[2] * pow_mod(2, rest/2, mod-1))  % (mod-1);
                cnt[3] = (ncnt[3] * pow_mod(2, rest/2, mod-1)) % (mod-1);
            }
            
            break;
        }
    }
    mint ans = 1;
    for(auto p: cnt){
        ans *= pow_mod(p.first, p.second, mod);
    }
    cout << ans.val() << endl;
}
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