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Codeforces Round #769 (Div. 2) B，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) 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 ,D
49. Codeforces Round #769 (Div. 2) B，C
50.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)
收起

Codeforces Round #769 (Div. 2) B，C

B

B

这道题我在vp的时候一直没有想出来，一直不知道到底怎么写，只是想到和幂有关，wa了一发，后来看了大佬的题解，真是觉得自己太菜了，自愧不如

这里我还有一个误区，就是在vp的时候我一直想着把最大的n-1和1放在一起，其实是不对的，因为那时我觉得答案的max一定是n-2(我觉得n-1和1异或一定会减一，我也不知道我当时怎么想出来的，赛后自己仔细想了想，好像n-1和1异或不一定=n-2,只有当n-1是奇数的时候异或1才会减一，当时我应该是傻了吧)

#include <iostream>
using namespace std;
int n,t;
void solve()
{
    cin>>n;
    int mid=1;//mid是二进制中只有一个1且此时1的位置是最高位，把mid和0放在一起，那么这个一定是就是这个数，而其他的都和相邻的进行异或，那么这样
    //他们的值一定不会很大，不会超过mid，对于x,和x+1，只有其中x+1是1000式二进制的时候，异或值才是x+x+1（两个值相加），
    //而x+x+1一定不会等于mid，最大只能等于mid-1,其他的异或值不一定等于x+x+1哦
    for (int i=1;i<=n;i<<=1)
    {
        if ((i<<1)>n-1)
        {
            mid=i;
            break;
        }
    }
    for (int i=1;i<mid;i++)
    {
        cout<<i<<" ";
    }
    cout<<0<<" ";
    for (int i=mid;i<n;i++)
    {
        cout<<i<<" ";
    }
    cout<<'\n';
    return ;
}
int main ()
{
    cin>>t;
    while (t--)
    {
        solve();
    }
    system ("pause");
    return 0;
}

C

C

vp的时候看这道题就只有十多分钟了，脑子还是不清醒，以为总共只有两种操作方法

1，a+x(x次加法)，然后在或运算，最后等于b

2,a先异或，然后再对b进行加法（因为或运算最后的值一定会大于等于max(a,b))

其实还有

3.b+x(先对b进行x次加法)，然后再或

4,直接取差值

然后求取最小操作数

#include <iostream>
using namespace std;
int a,b,n;
void solve()
{
    cin>>a>>b;
    if (a>=b)
    {
        cout<<a-b<<'\n';
        return ;
    }
    //cout<<(a|b)<<'\n';
    int ans1=(a|b)-b;
    ans1++;
    //cout<<ans1<<" ";
    int ta=a,tb=b;
    int ans=0;
    while ((ta|tb)!=tb)
    {
        ta++;
        ans++;
    }
    ans1=min(ans1,ans+1);
    int tmpb=b*2+100;
    ta=a,tb=b;
    ans=0;
    while ((ta|tb)!=tb&&tb<tmpb)
    {
        tb++;
        ans++;
    }
    ans1=min(ans1,ans+1);
    ans1=min(ans1,b-a);
    cout<<ans1<<'\n';
    return ;
}
int main ()
{
    int t;
    cin>>t;
    while (t--)
    {
        solve();
    }
    system ("pause");
    return 0;
}
posted @   righting  阅读(22)  评论(0)  编辑  收藏  举报
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