cf589div2

https://codeforces.com/contest/1228/problem/A

#include<iostream>
#include<cstdio>
 
using namespace std;
 
int pd(int x) {
	int a[10] = {};
	while(x) {
		if(a[x%10]) return 0;
		a[x%10]++;
		x /= 10;
	}
	return 1;
}
 
int main(){
	int l,r; scanf("%d%d", &l, &r);
	for(int i = l; i <= r; i++)
	{
		if(pd(i)) {
			printf("%d\n", i); return 0;
		}
	}
	printf("-1\n");
	return 0;
}
/*

input

121 130

output

123

input

98766 100000

output

-1

*/

　　

https://codeforces.com/contest/1228/problem/B

// solution to a problem such as the meaning of the questions simulation, computational not have to fill in the rest of the grid 2 ^ cnt (cnt little fast with no power will do)
 // But to judge at the time when the output cnt 0 1
 // there when when conditions indicate a conflict there can be no output is 0
 // tips or operational conditions can be a time when the conflict will be a larger value 3 
#include <bits / STDC ++ H.> a using namespace std; // cf589div2b #define _for (I, A, B) for (int I = (A); I <(B); I ++)
 #define _rep (I, A, B) for (int I = (A); I <= (B); ++ I)
 #define LL Long Long
 const LL + 1E9 = MOD . 7 ; int L, R & lt, CNT = 0 ;
 int main () { // ios_base :: sync_with_stdio (0), cin.tie (0), cout.tie (0 ); 
    scanf (
 


"%d%d", &l, &r);
    vector<vector<int> > f(l+5, vector<int>(r+5, 0));
    _rep(i,1,l) {
        int k;  scanf("%d", &k); 
        _rep(j,1,k) f[i][j] |= 1;
        if(k+1 <= r)  f[i][k+1] |= 2;
    }
    _rep(i,1,r)    {
        int k;  scanf("%d", &k); 
        _rep(j,1,k) f[j][i] |= 1;
        if(k+1 <= l)  f[k+1][i] |= 2;
    }
    ll ans = 1;
    _rep(i,1,l) _rep(j,1,r)  if(!f[i][j]) ans *= 2, ans %= mod;
    _rep(i,1,l) _rep(j,1,r)  if(f[i][j] == 3) ans *= 0;
    //if(!f[i][j] && i>=c[i] && j>=d[j]) cnt++;
    //printf("cnt = %d\n", cnt);
    printf("%lld\n", ans);
    return 0;
}

/*
3 4 
0 3 1
0 2 3 0
output
2

1 1
0
1
output
0

19 16
16 16 16 16 15 15 0 5 0 4 9 9 1 4 4 0 8 16 12
6 12 19 15 8 6 19 19 14 6 9 16 10 11 15 4
output          2^47%mod  = 487370169
797922655 == 2^51%mod
*/

 

https://codeforces.com/contest/1228/problem/C

The derivation of the equation it can be seen f (x, 1) * ...... * f (x, n) is the result of the prime factors p1 x * ...... * pk 

1 * ...... * n x remove non-factor of prime factors remaining answer mode is 1e9 + 7, but the actual operation is relatively cumbersome

Product through a series of derivation (magic) can be obtained result is equal to pi ^ (n / pi) (i 1 ...... k) of

! N-th power of 2 x can be divided by x = 0; while (n> 0) {n / = 2, x + = n;} can be calculated from x, interesting (Section 4.4 factorial specific chapter number theory of mathematics related factors derived)

Other prime factor index Seeking Similarly, as follows  

/*
input
10 2
output
2
input
20190929 1605
output
363165664
input
947 987654321987654321
output
593574252
*/
//vector<vector<int> > f(l+5, vector<int>(r+5, 0));
#include<bits/stdc++.h>

using namespace std;
#define ll long long
const ll mod = (int)1e9+7;
ll powmod(ll x, ll a){
	if(a == 0) return 1;
	if(a & 1) return x*powmod(x, a-1)%mod;
	return powmod(x*x%mod, a/2)%mod;
}
int main(){
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	ll x, n;
	cin >> x >> n;
	set<ll>your;
	for(int d = 2; d*d <= x; d++)
		while(x%d == 0)
			dv.insert(d), x /= d;
	if(x > 1) dv.insert(x);
	ll ans = 1;
	for(ll d : dv){
		ll g = 0, r = n;
		while(r > 0) r /= d, g += r;
		ans *= powmod(d, g); ans %= mod;
	}
	return cout << (int)ans << '\n', 0;
}