D. Buying Jewels

合集 - 构造(16)

1.C. Ehab and Path-etic MEXs2024-03-172.Increase Subarray Sums2024-04-103.F. Alex's whims2024-04-124.C. Torn Lucky Ticket2024-04-275.E. Cells Arrangement2024-05-046.俄罗斯方块2024-05-147.B. Omkar and Heavenly Tree2024-05-168.C. Colorful Grid2024-06-169.C. Lexicographically Largest2024-06-1710.A. Bitwise Operation Wizard2024-06-1711.B. Neutral Tonality2024-06-1712.M. 渚千夏的串2024-06-1813.D. Nene and the Mex Operator2024-07-1114.C. Nezzar and Symmetric Array2024-07-16

15.D. Buying Jewels2024-07-19

16.G. A/B Matrix2024-07-20

收起

原题链接

题解

构造题，先想特殊情况再验证

易得当 $n<k$ 时不成立

当 $n=k$ 时输出 1

否则，第一个柜台卖的价格 $p_1\ge 2$

且最多能卖 $\lfloor \frac{n}{p_1}\rfloor +n\mathrm{mod}{p}_1$ 个，且最大值出现在 $p_1==2$ 时

（感性的理解）

code

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

void solve()
{
    ll n,k;

    cin>>n>>k;

    if(n==k)
    {
        cout<<"yes\n1\n"<<1<<'\n';
    }
    else if(2LL*(n-k+1LL)<=n)
    {
        cout<<"no\n";
    }
    else
    {
        cout<<"yes\n2\n";
        cout<<n-k+1LL<<' '<<1<<'\n';
    }
}
int main()
{
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    int t=1;
    cin>>t;
    while(t--) solve();
    return 0;
}