Codeforces Round #770 (Div. 2)B，C
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，C
40.Codeforces Global Round 24(B,C)41.Codeforces Round #836 (Div. 2）C42.Codeforces Round #840 (Div. 2) C43.Good Bye 2022: 2023 is NEAR C44.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) Codeforces Round #770 (Div. 2)B，C
还是惨绝人寰的只做了一个题，ε=(´ο｀*)))唉
B
这一道题大意是是首先有一个d，然后有n个操作，从1到n，每一次我们都需要选择让d=d+a[i]还是d=d^a[i]，有两个人，Alice最初的d=x,Bob最初的d=x+3,然后我们的答案是谁可以在进行n次操作后把d变成y，就输出那个人的名字
首先我们可以看到x,和x+3的区别是什么？
x和x+3的奇偶性一定不同，而无论是+a[i],还是^a[i]对于d的奇偶性的变化都是一样的，那么在我眼里，不管是加法还是异或，d的奇偶性变化只和数值有关，而一开始x和x+3的奇偶性就不同，而且题目也明确告知一定有一个人可以变成y,那么我们只需要直接判断即可
#include <iostream>
#include <stdlib.h>
using namespace std;
const int maxn=2e5+10;
#define int long long
int t,n,x,y;
void solve()
{
cin>>n>>x>>y;
for (int i=1;i<=n;i++)
{
int tmp;
cin>>tmp;
x=(x%2+tmp%2)%2;
}
if (x%2==y%2)
{
cout<<"Alice\n";
}
else
{
cout<<"Bob\n";
}
return ;
}
signed main ()
{
cin>>t;
while (t--)
{
solve();
}
system ("pause");
return 0;
}
C
这一题大意是有nk件物品，每一件物品的价值在1到nk的范围里，并且每两件物品的价值一定不一样，有n个货架，每个货架里有k件物品，但是我们需要知道怎样摆放物品才可以让i个货架的任意一段区间里的物品的平均值为整数
要使任意一段区间里的平均值为整数，只有这一个货架里的所有价值的奇偶性相同
所以我们只要把奇数和奇数放在一起，偶数和偶数放在一起即可
详见代码
#include <iostream>
#include <vector>
using namespace std;
#define int long long
int n,k,t;
void solve()
{
cin>>n>>k;
// cout<<'\n';
vector<int>ans[3];
int tot=n*k;
if (tot/2%k==0&&(tot+1)/2%k==0)
{
cout<<"YES\n";
// cout<<tot/2<<" "<<(tot+1)/2<<'\n';
for (int i=2;i<=tot;)
{
for (int j=1;j<=k;j++)
{
cout<<i<<" ";
i+=2;
}
cout<<'\n';
}
for (int i=1;i<=tot;)
{
for (int j=1;j<=k;j++)
{
cout<<i<<" ";
i+=2;
}
cout<<'\n';
}
}
else
{
cout<<"NO\n";
}
//cout<<'\n';
return ;
}
signed main ()
{
cin>>t;
while (t--)
{
solve();
}
system ("pause");
return 0;
}
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