Editorial in english must also be posted as soon as in japanese.

I hope it would be a great Contest

Excited that I will learn something new!

Update: Learned when to use double when not to use doubles and how to use doubles!

2nd and the 3rd question seems easy but is really tricky... i dont know on which cases my code is getting wa...can anyone please help

I'm so sad, it took me less time to D than it took me to do B and I haven't even done C. Everytime, I miss one of B-D. I'm cursed D:

Will a CE submission be calculated into the overall penalty?

I submitted two pieces of codes in the wrong language and will my overall penalty increase by 10 min?

How to solve F?

Knapsack for All Segments vs Knapsack for All Subsets :D

How to solve E!

In the second part, there are essentially two tricks for the problem 1) Sort the array. If the value at index i is 0, then we cannot return -1(no matter what). 2) Multiplying two extremely large numbers even when the data type is long long int will result in negative values, which will result in wrong answers. So, a*b>=c is equivalent to b>=(c/a)(take care of the border cases) Link to my code https://atcoder.jp/contests/abc169/submissions/13810433

Okay, so in the 3rd question, I simply multiplied a and b, and applied the floor function for the product. Link to my code in C++ https://atcoder.jp/contests/abc169/submissions/13814577

I just wonder how to solve B,C...

i was getting TLE Second Question

My solutions to all problems are at https://mirror.codeforces.com/blog/entry/78195.

Can someone please tell why this fails for C?

n = input().split()
i1 = int(n[0])
i2 = int(float(n[1]) * 100.0)
i = i1 * i2
print(i // 100)

In C, why is math.trunc(a*b) in python giving WA,but in CPP, this is accepted-

ll a;
long double b;
cin>>a>>b;
ll ans;
ans=a*b;
cout<<ans;

nvm, got it

can anybody find mistake in my D solution.

#include<bits/stdc++.h>
using namespace std; 
bool ispower(int x,set<int> s)
{
	int f=0;
	int p=1;
	while(x%2==0)
	{
		p*=2;
		x=x/2;
		f=1;
	}
	if(x==1 && s.find(p)==s.end())return true;
	if(f==1)return false;
	int t=sqrt(x);
	for(int i=3;i<=t;i+=2)
	{
		int p=1;
		while(x%i==0)
		{
			f=1;
			x=x/i;
			p*=i;
		}
		if(f==1 && x==1 && s.find(p)==s.end())return true;
		if(f==1)return false;
	}
	return false;
}
bool isprime(int x)
{
	for(int i=2;i<=sqrt(x);i++)if(x%i==0)return false;
	return true;
}
int main()
{
	
	long long x;
	cin>>x;
	long long ans=0;
	int r=sqrt(x);
	set<int> s;
	bool seive[r+1];
	for(int i=2;i<=r;i++)seive[i]=true;
	for(int i=2;i<=r;i++)
	{
		if(seive[i])
		{
			for(int j=i;j<=r;j*=i)
			{
				if(x%j==0)
				{
					x=x/j;
					ans++;
					s.insert(j);
				}
			}
			for(int j=2*i;j<=r;j+=i)
			{
				seive[j]=false;
			}
		}
	}
	if(x>=2 && s.find(x)==s.end() && (ispower(x,s) || isprime(x)))ans++;
	cout<<ans;
	return 0;
}

Any care to explain why is this wrong for problem C ?

#define int long long 

void solve()
{
    int a,b;
    double c;
    
    cin >> a;
    cin >> c;
    
    b = (c*100);
    
    int res = (a*b);
    res/=100;
    
    cout << res;   
}

It gives WA for random_12.txt

Edit : I always define int as long long in my template.

In the fourth question, I simply used the Sieve of Eratosthenes to get the prime numbers less than 1e6, after that I iterate from 2 to the largest prime less than 1e6, and simply code out for the number of ways the number can be broken down! Just be careful when we multiply two numbers of digits 5 and 6! Link to my code- https://atcoder.jp/contests/abc169/submissions/13855795

Why this fails for problem B: https://atcoder.jp/contests/abc169/submissions/13804989

How to solve c? I multiplied a*b and took floor(a*b). What's wrong? submission

problem B solution : https://atcoder.jp/contests/abc169/submissions/13824612

Please someone help me out with Problem C ....... My Submission

Whats wrong wit C

import java.math.BigDecimal; import java.util.Scanner;

public class Main {

public static void main(String args[])
{
    Scanner sc=new Scanner(System.in);
    long a=sc.nextLong();
    long b=(long)(sc.nextDouble()*100);
    long ans=(a*b)/100;
    System.out.println(ans);

}

I usually get one hit on both Codeforces and Atcoder, but this time seems horrible xD

Whats wrong in D

import java.util.Scanner;

public class Main {

public static void main(String args[])
{
    Scanner sc=new Scanner(System.in);
   long n=sc.nextLong();
   long copyn=n;
   long x=(long)Math.sqrt(n);
   long copyx=x;
   long ans=0;
   for(int i=2;i<=x;i++)
   {
       int count=1;
       while(n%((Math.pow(i,count)))==0)
       {

           n= (long) (n/Math.pow(i,count));
           count++;
           ans++;
       }

       while(n%(Math.pow(i,count))==0)
       {
           n= (long) (n/Math.pow(i,count));
       }
   }

   if(copyn==1)
   {
       System.out.println(0);
   }
   else

   if(n>x)
   {
       System.out.println(1+ans);
   }
   else
   {
       System.out.println(ans);
   }

}

}

What is the problem with this solution of C?Can someone help?

include<bits/stdc++.h>

using namespace std; typedef long long ll;

define mod 1000000007

int main() { ll a; float b1; cin>>a>>b1;
b1=b1*100; ll b=b1; ll s=a*b; s=s/100; printf("%lld\n",s); }

https://atcoder.jp/contests/abc169/submissions/13900202 [WA] https://atcoder.jp/contests/abc169/submissions/13900438 [AC]

Both are identical solutions for problem B of AtCoder Beginner Contest 169. Does anyone knows why this happen?

I think I got a beautiful solution for F: the answer is just the coefficient of x^s in polynominal (2 + x^a1)(2 + x^a2)...(2 + x^an) here '^' is for power. we can just ignore higher order coefficient than x^s.

https://atcoder.jp/contests/abc169/submissions/13847100

please ignore unremoved code for problem D, E.

In B:
For the statement: if(inp[i] <= 1e18/ans) I got WA. But when i wrote : if(inp[i] <= 1000000000000000000/ans) I got AC. Is there any differences between 1e18 and 1000000000000000000?? Otherwise what is the reason??

In editorial there is a typo in problem D English version.

Wrong:$$$O(N)$$$

Correct:$$$O(sqrt(N))$$$

Can AnyBody please tell how can i reduce the submission time of my solution to the problem F?

https://atcoder.jp/contests/abc169/submissions/13920879

I have seen people getting AC in less than 100ms while mine is giving AC in 1000 ms.

thanks in advance. :).

In the editorial of problem F it says:

dp[i][j] = When the choices for the first i options is already determined, the number of combination such that the sum of ak for each k that the first option was chosen is equal to j

What does it even mean?

Can anyone help me debug my solution? I get TLE on 6-7 test cases, no idea why. Thank you!

#include <iostream>
#include <vector>
using namespace std;

int solve(long long n) {
    if (n == 1) {
        return 0;
    }
    vector<int> f;
    for (int i = 2; i * i <= n; i++) {
        while (n % i == 0) {
            f.push_back(i);
            n /= i;
        }
    }
    if (n > 1) f.push_back(n);

    int ans = 0;
    int i = 0;
    while (i < f.size()) {
        int j = i + 1;
        while (j < f.size() && f.at(j) == f.at(i)) j++;
        int count = j - i;
        i = j;
        int sum = 0;
        for (int x = 1; count >= sum + x; x++, ans++) sum += x;
    }
    return ans;
}

int main() {
    long long n;
    cin >> n;
    cout << solve(n) << endl;
}

what is my wrong with my code in problem-c? https://atcoder.jp/contests/abc169/submissions/13926562

For problem D, Maybe this sounds stupid, But can someone please explain why is it optimal to compute for each prime and its powers first, then similarly with next prime and its powers, and so on instead of going from 2 onwards ?

Can anyone provide any proof for problem-E for both odd and even case.

If anyone is still stuck at D here is a Solution

I am Facing problems in the C . No matter what I submit , I am shown wrong answer in 6/22 cases . I am not understanding where is the error or fault exists in that code .. If anyone knows , please give me the solution . My code is given below ::::

include<stdio.h>

int main() { long long int a,pro1; double b,pro=1.0; scanf("%lld %lf",&a,&b); b=b*100; pro=(a*b); pro1=(long long int)(pro/100); printf("%lld",pro1); return 0; }

Can someone help me where I'm going wrong in problem D, I'm getting hand_22 as the only test case as WA. Here is my code https://atcoder.jp/contests/abc169/submissions/14102528 Thanks