对于想了解CodeforcesRound#435(Div.2)C.MahmoudandEhabandthexor(异或的性质)的读者，本文将提供新的信息，我们将详细介绍异或结果，并且为您提供关于B类-
对于想了解Codeforces Round #435 (Div. 2)C. Mahmoud and Ehab and the xor(异或的性质)的读者，本文将提供新的信息，我们将详细介绍异或结果，并且为您提供关于B类-Codeforces Round #535 (Div. 3)C. Nice Garland、CF 959 E. Mahmoud and Ehab and the xor-MST、CF959E Mahmoud and Ehab and the xor-MST 思维、Codeforces #396 (Div. 2) D. Mahmoud and a Dictionary （并查集+map的有价值信息。
本文目录一览：- Codeforces Round #435 (Div. 2)C. Mahmoud and Ehab and the xor(异或的性质)（异或结果）
- B类-Codeforces Round #535 (Div. 3)C. Nice Garland
- CF 959 E. Mahmoud and Ehab and the xor-MST
- CF959E Mahmoud and Ehab and the xor-MST 思维
- Codeforces #396 (Div. 2) D. Mahmoud and a Dictionary （并查集+map
Codeforces Round #435 (Div. 2)C. Mahmoud and Ehab and the xor(异或的性质)（异或结果）
总结
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B类-Codeforces Round #535 (Div. 3)C. Nice Garland
Codeforces Round #535 (Div. 3)C. Nice Garland
题意：
由‘R‘,‘G‘ and ‘B‘ 三个字母组成的一个字符串，每两个相同的字母需要相差3，找出最小需要交换次数。
分析：
这个字符串的长度大于等于3的时候，一定是RBG这三个字符的某一个排列的循环。
RBG一共最多有6种排列方式{"RGB","RBG","BGR","BRG","GRB","GBR"};
所以直接暴力循环6次即可。
代码：
#include<iostream> using namespace std; string dir[6]={"RGB","RBG","BGR","BRG","GRB","GBR"}; int main(){ int n; cin>>n; string s; cin>>s; int minn=100000000; int flag=0; //cout<<dir[5][1]; for(int j=0;j<6;j++){ int count=0; for(int i=0;i<n;i+=3){ if(s[i]!=dir[j][0]) count++; if(i+1 >= n) break; else if(s[(i+1)]!=dir[j][1]) count++; if(i+2 >= n) break; else if(s[(i+2)]!=dir[j][2]) count++; } if(count<minn){ minn=count; flag=j; } } cout<<minn<<endl; int i; for(i=0;i + 3 <n;i+=3){ cout<<dir[flag]; } int j = 0; for(i;i < n;i++) cout << dir[flag][j++]; //if(n%3==) return 0; }//比赛结束了几分钟才改好，emmmmmmm，笨死啦。
CF 959 E. Mahmoud and Ehab and the xor-MST
E. Mahmoud and Ehab and the xor-MST
https://codeforces.com/contest/959/problem/E
分析：
每个点x应该和x ^ lowbit(x)连边，那么现在就是求$\sum_{i=1}^{n}lowbit(i)$，然后打表找规律。
代码：
1 #include<cstdio>
2 #include<algorithm>
3 #include<cstring>
4 #include<iostream>
5 #include<cmath>
6 #include<cctype>
7 #include<set>
8 #include<queue>
9 #include<vector>
10 #include<map>
11 using namespace std;
12 typedef long long LL;
13
14 inline int read() {
15 int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch==''-'')f=-1;
16 for(;isdigit(ch);ch=getchar())x=x*10+ch-''0'';return x*f;
17 }
18
19 int main() {
20 LL n, res = 0;
21 cin >> n; n--;
22 for (LL x = 1; x <= n; x <<= 1)
23 res += ((n - x) / (x + x) + 1) * x;
24 cout << res;
25 return 0;
26 }
CF959E Mahmoud and Ehab and the xor-MST 思维
Ehab is interested in the bitwise-xor operation and the special graphs. Mahmoud gave him a problem that combines both. He has a complete graph consisting of n vertices numbered from 0 to n?-?1. For all 0?≤?u?<?v?<?n,vertex u and vertex v are connected with an undirected edge that has weight
You can read about complete graphs in https://en.wikipedia.org/wiki/Complete_graph
You can read about the minimum spanning tree in https://en.wikipedia.org/wiki/Minimum_spanning_tree
The weight of the minimum spanning tree is the sum of the weights on the edges included in it.
The only line contains an integer n (2?≤?n?≤?1012),the number of vertices in the graph.
The only line contains an integer x,the weight of the graph‘s minimum spanning tree.
4
4
In the first sample:
题意翻译
n个点的完全图标号(0-n-1),i和j连边权值为i^j,求MST的值
不妨先手算几项，可以发现每一位上的贡献为当前n 的一半；
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize("O3")
using namespace std;
#define maxn 300005
#define inf 0x3f3f3f3f
#define INF 9999999999
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int,int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
inline ll rd() {
ll x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == ‘-‘) f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a,ll b) {
return b == 0 ? a : gcd(b,a%b);
}
ll sqr(ll x) { return x * x; }
/*ll ans;
ll exgcd(ll a,ll b,ll &x,ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b,a%b,x,y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/
ll qpow(ll a,ll c) {
ll ans = 1;
a = a % c;
while (b) {
if (b % 2)ans = ans * a%c;
b /= 2; a = a * a%c;
}
return ans;
}
int main()
{
//ios::sync_with_stdio(0);
ll n; rdllt(n);
ll ans = 0;
ll tmp = 1;
while (n>1) {
ans += tmp * (n >> 1); tmp <<= 1; n -= (n >> 1);
// cout << n<<‘ ‘<<ans << endl;
}
cout << ans << endl;
return 0;
}
Codeforces #396 (Div. 2) D. Mahmoud and a Dictionary （并查集+map
总结
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