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Good Bye 2022: 2023 is NEAR C

合集 - acm(93)
1.AtCoder Beginner Contest 293(C,D ,E,F)2.Educational Codeforces Round 115 (Rated for Div. 2)（D,E)3.AtCoder Beginner Contest 294(E,F,G)4.AtCoder Beginner Contest 2465.中国石油大学（北京）第三届“骏码杯”程序设计竞赛（同步赛）(D,E,F)6.Codeforces Global Round 16(D,E,F)7.AtCoder Beginner Contest 209（D，E）8.Monoxer Programming Contest 2022（AtCoder Beginner Contest 238）（E，F）9.AtCoder Beginner Contest 285（B,D,E,F)10.AtCoder Beginner Contest 242（D，E)11.AtCoder Beginner Contest 223（D,E,F)12.AtCoder Beginner Contest 207（D,E)13.AtCoder Beginner Contest 247（E,F)14.AtCoder Beginner Contest 226(E,F,G)15.AtCoder Beginner Contest 229(F,G)16.AtCoder Beginner Contest 273(E)17.AtCoder Beginner Contest 286(G)18.AtCoder Beginner Contest 287(C,D,E,F)19.AtCoder Beginner Contest 288(D,E,F)20.AtCoder Beginner Contest 289(E,F)21.AtCoder Beginner Contest 290(D,E)22.AtCoder Beginner Contest 292(E,F,G)23.AtCoder Beginner Contest 298(D,F)24.AtCoder Beginner Contest 299(E,F)25.AtCoder Beginner Contest 300(E,F)26.AtCoder Beginner Contest 302（E，F，G）27.AtCoder Beginner Contest 253(E,F)28.AtCoder Beginner Contest 245(D，E，F)29.AtCoder Beginner Contest 248(D，E，F) 30.AtCoder Beginner Contest 206（Sponsored by Panasonic）（E，F）31.Codeforces Round 875 (Div. 2)(D)32.AtCoder Beginner Contest 178(E,F)33.AtCoder Beginner Contest 307（E,F,G)34.CodeTON Round 5 (Div. 1 + Div. 2, Rated, Prizes!)C35.Educational Codeforces Round 151 (Rated for Div. 2)(C,D)36.AtCoder Beginner Contest 212(E,F)37.牛客小白月赛5138.2022年浙大城市学院新生程序设计竞赛（同步赛）（补题）39.Codeforces Round #770 (Div. 2)B，C40.Codeforces Global Round 24(B,C)41.Codeforces Round #836 (Div. 2）C42.Codeforces Round #840 (Div. 2) C
43.Good Bye 2022: 2023 is NEAR C
44.Codeforces Round #765 (Div. 2)A，B，C45.Codeforces Round #766 (Div. 2)C，D46.Codeforces Round #841 (Div. 2) and Divide by Zero 202247.Codeforces Round #767 (Div. 2)C ,D 48.Codeforces Round #768 (Div. 2)C ,D49. Codeforces Round #769 (Div. 2) B，C50.Educational Codeforces Round 122 (Rated for Div. 2)，C，D51.Educational Codeforces Round 119 (Rated for Div. 2)52.AtCoder Beginner Contest 25853.Codeforces Round #763 (Div. 2)C54.Codeforces Round #843 (Div. 2)（B，C，D，E）55.Educational Codeforces Round 141 (Rated for Div. 2)（B，C，D）56.AtCoder Beginner Contest 275（B，C，D，E，F）57.Codeforces Round #842 (Div. 2)（B，D，E）58.AtCoder Beginner Contest 284（D，E，F）59.The 14th Jilin Provincial Collegiate Programming Contest（补题）60.牛客小白月赛65（C，D，E，F）61.AtCoder Beginner Contest 281（D，E，F）62.Good Bye 2021: 2022 is NEAR D63.Hello 202364.The 15th Jilin Provincial Collegiate Programming Contest(补题)65.Codeforces Round #781 (Div. 2)C 66.Hello 2022（B，D）67.AtCoder Beginner Contest 272（D,E)68.Codeforces Round 751 (Div. 2)（D）69.Codeforces Round 856 (Div. 2)(C,D)70.Codeforces Round 752 (Div. 2)（C，D，E）71.Codeforces Round 855 (Div. 3)（E，F）72.AtCoder Regular Contest 131（A,B,C）73.Educational Codeforces Round 144 (Rated for Div. 2)（A，B，C，D）74.Codeforces Round 853 (Div. 2)（C,D)75.牛客练习赛109（C,D)76.AtCoder Beginner Contest 291（Sponsored by TOYOTA SYSTEMS）（D，E，F）77.Educational Codeforces Round 143 (Rated for Div. 2)（A，C,D)78.Codeforces Round #852 (Div. 2)（C,D)79.Educational Codeforces Round 118 (Rated for Div. 2)（D，E）80.AtCoder Beginner Contest 236（D，E，F）81.Codeforces Round #850 (Div. 2, based on VK Cup 2022 - Final Round)(B,D)82.Codeforces Round #848 (Div. 2)（B,C,D）83.TypeDB Forces 2023 (Div. 1 + Div. 2, Rated, Prizes!) (B,C,D) 84.Codeforces Round #846 (Div. 2)（B，E） 85.Educational Codeforces Round 142 (Rated for Div. 2)（C，D）86.2023牛客寒假算法基础集训营687.Codeforces Round #845 (Div. 2) and ByteRace 2023（A，B，C）88.2023牛客寒假算法基础集训营5 89.2023牛客寒假算法基础集训营3 90.2023牛客寒假算法基础集训营291.2023牛客寒假算法基础集训营192.Educational Codeforces Round 120 (Rated for Div. 2) C，D93.AtCoder Beginner Contest 254（C，D，E，F)
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Good Bye 2022: 2023 is NEAR C

第一道题真是没理解，wa了好多次

还好后来知道了题意就好做了，我现在在这里就补一下c，之前想了一会，还是不明白，后来想到了一些，容我讲一讲

C

这道题是给你n个数，问我们可以找到一个x，让每一个a[i]+x，然后让这些数两两gcd=1

而这里最关键的消息就是我不明白的那个点

这里用到一个知识点就是如果存在一个x ,每一个数对x取余，众所周知，对x取余一定有0到x-1这些数，只要每余数出现的次数都大于2，那么就一定不存在一个数可以让两两gcd=1

我理解的就是，假如这一个x是5，那么余数为0,1,2,3,4,的数量都大于等于2，不管加哪一个数，都可以让任意两个数gcd不等于1

如加4，那么余数为1的这些数的gcd是5，如果是2，那么余数为3的这些数gcd是5，如果是10，（10%5=0),那么那些余数是0的那些数gcd是5，任意数%5一定是0到4（设为t)，那么对于那n个数对5取余的余数加上t为5（凑满5）的数量一定会大于等于2，那么一定会存在一对数gcd不为1,故此时一定是不存在一个数满足题目要求的

刚好这道题最大的数量是100（n<=100),那么我们只需要例举2到50的情况，大于50一定不存在让0到x-1的余数的数量大于2

详细的做法见代码

#include <iostream>
#include <vector>
#include <map>
using namespace std;
const int maxn=110;
#define int long long 
int a[maxn];
int n;
void solve()
{
    cin>>n;
    map<int,int>vis;
    bool yes=true;
    for (int i=1;i<=n;i++)
    {
        cin>>a[i];
        if (vis[a[i]]) yes=false;
        vis[a[i]]++;
    }
    if (!yes)
    {
        cout<<"NO\n";
        return ;
    }
    for (int i=2;i<=50;i++)
    {
        vis.clear();
        bool ans=false;
        for (int j=1;j<=n;j++)
        {
            vis[a[j]%i]++;
        }
        for (int j=0;j<i;j++)
        {
            if (vis[j]<2)
            {
                ans=true;
                break;
            }
        }
        if (!ans)
        {
            cout<<"NO\n";
            return ;
        }
    }
    cout<<"YES\n";
    return ;
}
signed main ()
{
    int t;
    cin>>t;
    while (t--)
    {
        solve();
    }
    system ("pause");
    return 0;

}
posted @   righting  阅读(53)  评论(0)  编辑  收藏  举报
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