Dreaming of Freedom(数论,贪心)

合集 - 刷题集(97)

1.妖梦拼木棒2023-05-232.[USACO1.2]回文平方数 Palindromic Squares2023-05-243.[COCI2012-2013#1] LJUBOMORA2023-05-254.火柴棒等式2023-05-255.Kieszonkowe2023-05-256.[COCI2014-2015#5] TRAKTOR2023-05-267.硬币翻转2023-05-268.[蓝桥杯 2020 省 AB1] 走方格2023-05-269.P2097 资料分发2023-05-2610.[AHOI2005]约数研究2023-05-2611.[NOIP2004 普及组] 火星人2023-05-2612.我焯,原神(这真是个题)2023-05-2613.滑雪 OJ36512023-05-2414.CV的人有福啦!YTU贪心训练2(部分注释)2023-05-2715.砝码称重2023-05-2716.警钟长鸣-----找零钱2023-05-2817.[USACO09OCT]Barn Echoes G2023-05-2818.【CSGRound2】光骓者的荣耀2023-05-2919.初见蓝桥杯2023-05-3020.[蓝桥杯 2022 省 B] 扫雷2023-05-3121.开餐馆2023-05-3122.P9241 [蓝桥杯 2023 省 B] 飞机降落2023-06-0123.P7083 [NWRRC2013] Energy Tycoon2023-06-0324.高精度2023-06-0425.P7954 [COCI2014-2015#6] PAPRIKA2023-06-0426.(输出路径搜索)[USACO06OCT] Cows on Skates G2023-06-0427.畜栏保留问题2023-06-0428.(贪心+搜索+剪枝)P8801 [蓝桥杯 2022 国 B] 最大数字2023-06-0529.蓝桥杯 入门----普及-2023-06-0630.[NOIP2016 提高组] 玩具谜题2023-06-0631.[NOIP2002 提高组] 字串变换(无向图查重)2023-06-0632.[NOIP2000 提高组] 单词接龙2023-06-0733.POJ 硬币2023-06-0734.动态规划学习 1(最长上升子序列问题)2023-06-0735.动态规划学习 2(背包问题及其变形)2023-06-0836. 切割回文2023-06-1037.[NOIP2001 提高组] 数的划分(剪枝)2023-06-1038.字符串转换数字,sscanf和sprintf大法2023-06-1039.[CTSC1997] 选课(树状DP)2023-06-1040. k 倍区间(同余定理,组合数)2023-06-1141.[蓝桥杯 2018 省 AB] 全球变暖2023-06-1142.马的遍历(对bfs的更深一层探讨)2023-06-1143.找零钱2023-06-1144.Pasha Maximizes2023-06-1245.Newspaper Headline2023-06-1246.Balanced Ternary String2023-06-1347.(数论)素数,质数2023-06-1348.(数论) 约数2023-06-1349.Helping the Nature2023-06-1450.(博弈论)Even Number Addicts2023-06-1451.Present2023-06-1552.Mr. Kitayuta's Colorful Graph(二维并查集,弱化版)2023-06-1653.Edgy Trees(dfs,并查集,快速幂,树形结构,红黑树)2023-06-1654.并查集2023-05-2455.再谈 前缀和,差分,离散化2023-06-1656.Best Cow Fences(前缀和+特殊二分)2023-06-1757.二分2023-06-1758.同类型,类背包动态规划,选地dp2023-06-1759.Primes on Interval(欧拉筛+二分+滑动窗口)2023-06-1860.字符串(KMP , Trie树 , STL)2023-06-1861.Codeforces1 #879 div.22023-06-1962.Add Modulo 10(数论,思维,数学,规律)2023-06-2063.期末考试YTU4035: Shmily(数学,等差数列)2023-06-2064.[POI2006] OKR-Periods of Words2023-06-2165.Getting Zero(Bfs)2023-06-2166.快速幂2023-06-21

67.Dreaming of Freedom(数论,贪心)2023-06-23

68.Scoring Subsequences2023-06-2469.Dolce Vita2023-06-2970.Cake Assembly Line2023-06-2971.Constructive Problem2023-06-3072.Forever Winter2023-06-3073.Maximum Sum(前缀和)2023-07-1074.Contrast Value(数学规律)2023-07-1075.Lamps(STL+双端队列)2023-07-1076.Long Long2023-07-1177.Sum in Binary Tree2023-07-1178.Apple Tree(树状搜索,树形DP)2023-07-1179.Strong Password(贪心思想)2023-07-1280.Come Together2023-07-1281.Notepad2023-07-1382.Hamiltonian Wall2023-07-1383.7.17作业2023-07-1684.逛画展(双指针)2023-07-1985.#8852023-07-2086.树状数组2023-07-2087.Subsequence Addition2023-07-2288.7.22作业2023-07-2289.Diverse Substrings2023-07-2490.Tracking Segments(二分,区间前缀)2023-07-2791.最大食物链计数2023-07-2792.线段树12023-07-2293.[TJOI2007] 线段(状态机)2023-07-2994.#8782023-07-3095.测试12023-07-3096.李白打酒2023-07-3097.砍竹子2023-07-30

收起

用nsqrt(n)的时间复杂度就能过

//Dreaming of Freedom:https://codeforces.com/problemset/problem/1826/C

#include <bits/stdc++.h>
//#define int long long
using namespace std;
const int N=1e5+10,mod=1e9+7;
string s;
int n,t,a[N],f[N],res,num,ans,m;
bool vis[N];
int main()
{
    std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    cin>>t;
    while(t--){
        cin>>n>>m;
        if(n>=2&&m>=n){
            cout<<"NO"<<endl;
            continue;
        }
        if(!(n%2)&&m>=2){
            cout<<"NO"<<endl;
            continue;
        }
        if(n>m){
            bool f=false;
            for(int i=2;i<=n/i&&i<=m;++i){
                if(n%i==0){
                    f=true;
                    break;
                }
            }
            if(f){
                cout <<"NO"<<endl;
                continue;
            }
        }
        cout<<"YES"<<endl;
        
    }
    return 0;
}

 

#include <bits/stdc++.h>
//#define int long long
using namespace std;
const int N=1e6+10,mod=1e9+7;
string s;
int n,t,p[N],prime[N],res,num,ans,m;
bool vis[N];
int main()
{
    std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    cin>>t;
    vis[1]=true,prime[1]=1;
    for(int i=2;i<=1000000;i++){
        if(!vis[i]) p[++num]=prime[i]=i;
        for(int j=1;p[j]<=1000000/i;j++){
            vis[p[j]*i]=true,prime[i*p[j]]=p[j];
            if(!(i%p[j])) break;
        }
    }
    while(t--){
        cin>>n>>m;
        if(prime[n]==1||prime[n]>m) cout<<"YES"<<endl;
        else if(prime[n]<=m)cout<<"NO"<<endl;
    }
    return 0;
}

 

__EOF__

本文作者：Sakurajimamai
本文链接：https://www.cnblogs.com/o-Sakurajimamai-o/p/17498803.html
关于博主：评论和私信会在第一时间回复。或者直接私信我。
版权声明：本博客所有文章除特别声明外，均采用 BY-NC-SA 许可协议。转载请注明出处！
声援博主：如果您觉得文章对您有帮助，可以点击文章右下角【推荐】一下。您的鼓励是博主的最大动力！