Tutorial of Codeforces Round 729 (Div.2)

文章目录

A
- 题目
- C++代码
B
- 题目
- 分析
- C++代码
C
- 题目
- C++代码

A

题目

Example
input

5
2
2 3 4 5
3
2 3 4 5 5 5
1
2 4
1
2 3
4
1 5 3 2 6 7 3 4

output

Yes
No
No
Yes
No

Note
In the first test case, a possible way of splitting the set is (2,3), (4,5).
In the second, third and fifth test case, we can prove that there isn’t any possible way.
In the fourth test case, a possible way of splitting the set is (2,3).

C++代码

#include <iostream>
#include <algorithm>
using namespace std;int main(){int t;cin >> t;while(t--){int n, a, num_odd=0;cin >> n;for(int i=0; i<2*n; i++){cin >> a;if( a%2==1 )num_odd++;}if(num_odd==n)cout << "Yes" <<endl;elsecout << "No" <<endl;}
}

B

题目

There is an infinite set generated as follows:

111 is in this set.
If xxx is in this set, x⋅ax \cdot ax⋅a and x+bx+bx+b both are in this set.

For example, when a=3a=3a=3 and b=6b=6b=6, the five smallest elements of the set are:

111,
333 (111 is in this set, so 1⋅a=31\cdot a=31⋅a=3 is in this set),
777 (111 is in this set, so 1+b=71+b=71+b=7 is in this set),
999 (333 is in this set, so 3⋅a=93\cdot a=93⋅a=9 is in this set),
131313 (777 is in this set, so 7+b=137+b=137+b=13 is in this set).

Given positive integers aaa, bbb, nnn, determine if nnn is in this set.

Input

The input consists of multiple test cases. The first line contains an integer ttt (1≤t≤1051\leq t\leq 10^51≤t≤105) — the number of test cases. The description of the test cases follows.

The only line describing each test case contains three integers nnn, aaa, bbb (1≤n,a,b≤1091\leq n,a,b\leq 10^91≤n,a,b≤109) separated by a single space.

Output

For each test case, print “Yes” if nnn is in this set, and “No” otherwise. You can print each letter in any case.

Example

input

5
24 3 5
10 3 6
2345 1 4
19260817 394 485
19260817 233 264

output

Yes
No
Yes
No
Yes

Note

In the first test case, 242424 is generated as follows:

111 is in this set, so 333 and 666 are in this set;
333 is in this set, so 999 and 888 are in this set;
888 is in this set, so 242424 and 131313 are in this set.

Thus we can see 242424 is in this set.

The five smallest elements of the set in the second test case is described in statements. We can see that 101010 isn’t among them.

分析

当我们对数字+b后再乘a，实际上是加上 a个b，因此每一项SiS_iSi都满足Si=at+k∗bS_i=a^t+k*bSi=at+k∗b 我们可以不断枚举ata^tat，然后判断是否能整除bbb即可

C++代码

#include <iostream>
#include <algorithm>
using namespace std;int main(){int t, n, a, b;cin >> t;while(t--){cin >> n >> a >> b;if(a == 1){ if((n - 1) % b == 0)   cout << "Yes" << endl;else    cout << "No" << endl;continue;}int pow = 1, flag = 0;while(n>=pow){if( (n-pow) % b==0){cout << "Yes" << endl;flag = 1;break;}pow *= a;}if(flag==0)cout << "No" << endl;}return 0;
}

C

题目

Let f(i)f(i)f(i) denote the minimum positive integer xxx such that xxx is not a divisor of iii.

Compute ∑i=1nf(i)\sum_{i=1}^n f(i)∑i=1nf(i) modulo 109+710^9+7109+7. In other words, compute f(1)+f(2)+⋯+f(n)f(1)+f(2)+\dots+f(n)f(1)+f(2)+⋯+f(n) modulo 109+710^9+7109+7.

Input

The first line contains a single integer ttt (1≤t≤1041\leq t\leq 10^41≤t≤104), the number of test cases. Then ttt cases follow.

The only line of each test case contains a single integer nnn (1≤n≤10161\leq n\leq 10^{16}1≤n≤1016).

Output

For each test case, output a single integer ansansans, where ans=∑i=1nf(i)ans=\sum_{i=1}^n f(i)ans=∑i=1nf(i) modulo 109+710^9+7109+7.

Example

input

6
1
2
3
4
10
10000000000000000

output

Note

In the fourth test case n=4n=4n=4, so ans=f(1)+f(2)+f(3)+f(4)ans=f(1)+f(2)+f(3)+f(4)ans=f(1)+f(2)+f(3)+f(4).

111 is a divisor of 111 but 222 isn’t, so 222 is the minimum positive integer that isn’t a divisor of 111. Thus, f(1)=2f(1)=2f(1)=2.
111 and 222 are divisors of 222 but 333 isn’t, so 333 is the minimum positive integer that isn’t a divisor of 222. Thus, f(2)=3f(2)=3f(2)=3.
111 is a divisor of 333 but 222 isn’t, so 222 is the minimum positive integer that isn’t a divisor of 333. Thus, f(3)=2f(3)=2f(3)=2.
111 and 222 are divisors of 444 but 333 isn’t, so 333 is the minimum positive integer that isn’t a divisor of 444. Thus, f(4)=3f(4)=3f(4)=3.

Therefore, ans=f(1)+f(2)+f(3)+f(4)=2+3+2+3=10ans=f(1)+f(2)+f(3)+f(4)=2+3+2+3=10ans=f(1)+f(2)+f(3)+f(4)=2+3+2+3=10.

C++代码

#include <bits/stdc++.h>
#define ll long long
using namespace std;const int mod = 1e9 + 7;ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b) {return a/gcd(a,b)*b;}int main() {int t;cin >> t;while(t--){ll n;cin>>n;ll ans=n*2;ll cur=1;for(int i=3;i<=43;++i){cur=lcm(cur,i-1);ans=(ans+n/cur)%mod;}cout<<ans<<'\n';   }return 0;
}