Round 80 (Rated for Div.2)

Deadline CodeForces - 1288A

题目

Adilbek was assigned to a special project. For Adilbek it means that he has n days to run a special program and provide its results. But there is a problem: the program needs to run for d days to calculate the results.

Fortunately, Adilbek can optimize the program. If he spends x (x is a non-negative integer) days optimizing the program, he will make the program run in ⌈dx+1⌉ days (⌈a⌉ is the ceiling function: ⌈2.4⌉=3, ⌈2⌉=2). The program cannot be run and optimized simultaneously, so the total number of days he will spend is equal to x+⌈dx+1⌉.

Will Adilbek be able to provide the generated results in no more than n days?

Input
The first line contains a single integer T (1≤T≤50) — the number of test cases.

The next T lines contain test cases – one per line. Each line contains two integers n and d (1≤n≤109, 1≤d≤109) — the number of days before the deadline and the number of days the program runs.

Output
Print T answers — one per test case. For each test case print YES (case insensitive) if Adilbek can fit in n days or NO (case insensitive) otherwise.

Example
Input
3
1 1
4 5
5 11
Output
YES
YES
NO
Note
In the first test case, Adilbek decides not to optimize the program at all, since d≤n.

In the second test case, Adilbek can spend 1 day optimizing the program and it will run ⌈52⌉=3 days. In total, he will spend 4 days and will fit in the limit.

In the third test case, it’s impossible to fit in the limit. For example, if Adilbek will optimize the program 2 days, it’ll still work ⌈112+1⌉=4 days.

题意

思路

根据基本不等式可以求出

所以x在等于根号d-1的时候取得最小值，如果不放心可以从1枚举到sqrt(d) 这样一定可以找到最小值，如果最小值满足条件直接输出YES即可

代码

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

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin>>t;
    while(t--)
    {
        double n,d;
        cin>>n>>d;
        if(d<=n)
        cout<<"YES\n";
        else
        {
            int flag = 0;
            ll now = sqrt(d);
            for(int i=1;i<=now;i++)
            {
                double a = i;
                double b = d/(a+1);
                // cout<<a<<" "<<b<<" "<<endl;
                if(b-(ll)b>0)
                {
                    b++;
                    if((ll)b+a<=n)
                    {
                        flag=1;
                        break;
                    }
                }
                else
                {
                    if((ll)b+a<=n)
                    {
                        flag=1;
                        break;
                    }
                }
            }
            if(flag)
            cout<<"YES\n";
            else
            cout<<"NO\n";
        }
    }
}

Yet Another Meme Problem CodeForces - 1288B

题目

Try guessing the statement from this picture http://tiny.cc/ogyoiz.

You are given two integers A and B, calculate the number of pairs (a,b) such that 1≤a≤A, 1≤b≤B, and the equation a⋅b+a+b=conc(a,b) is true; conc(a,b) is the concatenation of a and b (for example, conc(12,23)=1223, conc(100,11)=10011). a and b should not contain leading zeroes.

Input
The first line contains t (1≤t≤100) — the number of test cases.

Each test case contains two integers A and B (1≤A,B≤109).

Output
Print one integer — the number of pairs (a,b) such that 1≤a≤A, 1≤b≤B, and the equation a⋅b+a+b=conc(a,b) is true.

Example
Input
3
1 11
4 2
191 31415926
Output
1
0
1337

题意

给定两个数AB 让你在[1-A]和[1-B]中任意选择两个数，只要他们的a*b+a+b=ab（两数拼起来）满足条件，就算一种结果，问最多可以选出来多少种结果。

思路

当时我是根据样例模拟了一下1 1 到1 9发现只有9可以，然后2 9 3 9… 发现好像是因为前面的乘完以后正好少了a这么多个1 而后面正好加上了a，而b最后一位是9正好满足拼起来是ab，找到规律直接写就完了。需要注意的是999这种情况需要特判一下

代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        ll a,b,sum=0;
        cin>>a>>b;
        int flag=1;
        while(b)
        {
            if(b%10!=9)
            flag=0;
            b/=10;
            sum++;
        }
        if(flag)
        cout<<sum*a<<"\n";
        else
        cout<<(sum-1)*a<<"\n";
    }
}