Codeforces Round #435 (Div. 2)_codeforces 435 div 2-优快云博客

本文链接：https://blog.youkuaiyun.com/sasuke__/article/details/78052399

本文解析了四道编程题目，包括查找数组元素、构造二分图、生成特定异或结果的序列以及解决交互式搜索问题。涉及数组操作、图论、位运算及交互式算法。

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#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define pill pair<int, int>
#define mst(a, b)	memset(a, b, sizeof a)
#define REP(i, x, n)	for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 1e6 + 10;
const int INF = 1e9 + 10;
bool vis[105];

int main(){
	int n, x;	scanf("%d%d", &n, &x);
	for(int i = 1; i <= n; ++i) {
		int a;	scanf("%d", &a);
		vis[a] = true;
	}
	int cnt = 0;
	if(vis[x])	cnt++;
	for(int i = 0; i < x; ++i) {
		if(!vis[i])	cnt++;
	}
	printf("%d\n", cnt);
	return 0;
}

题意：n个结点n-1条边组成一棵树，现在要把结点分成一个二分图，问最多能加多少条边

思路：直接统计两部分的结点数，求出两部分结点的乘积减去n - 1条边即可

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define pill pair<int, int>
#define mst(a, b)	memset(a, b, sizeof a)
#define REP(i, x, n)	for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 1e6 + 10;
const int INF = 1e9 + 10;
vector<int> G[qq];
LL num[2];
bool vis[qq];

void Dfs(int u, int idx) {
	num[idx]++;
	vis[u] = true;
	for(int i = 0; i < G[u].size(); ++i) {
		int v = G[u][i];
		if(vis[v])	continue;
		Dfs(v, idx ^ 1);
	}
}

int main(){
	int n;	scanf("%d", &n);
	for(int i = 1; i < n; ++i) {
		int u, v;	scanf("%d%d", &u, &v);
		G[u].pb(v), G[v].pb(u);
	}
	Dfs(1, 0);
	printf("%lld\n", num[0] * num[1] - (n - 1));
	return 0;
}

题意：给出n，x，要求输出n个不同的非负数xor的结果是x，如果不存在输出NO，最大数不超过1e6

思路：参考了题解的思路

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define pill pair<int, int>
#define mst(a, b)	memset(a, b, sizeof a)
#define REP(i, x, n)	for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 2e5 + 10;
const int INF = 1e9 + 10;

int main(){
	int n = 100000;
	int maxn = 0;
	int cnt = 0;
	while(maxn > 0) {
		if(n & 1) maxn = max(maxn, cnt);
		n >>= 1;
		cnt++;
	}
	int a = (1 << maxn);
	int b = (1 << (maxn + 1));
	int x;
	scanf("%d%d", &n, &x);
	if(n == 1) {
		printf("YES\n%d\n", x);
	} else if(n == 2) {
		if(x == 0)	puts("NO");
		else	printf("YES\n0 %d\n", x);
	} else {
		puts("YES");
		int ans = 0;
		for(int i = 1; i <= n - 3; ++i) {
			ans ^= i;
			printf("%d ", i);
		}
		if(ans == x) {
			printf("%d %d %d\n", a, b, a + b);
		} else {
			printf("%d %d %d\n", a, a ^ ans ^ x, 0);
		}
	}
	return 0;
}

题意：交互题，答案是一个长度为n的字符串，你可以输出一个字符串，机器会返回给你和答案串的Hamming distance，让你再最多询问15次的前提下求出某个1和0的位置，保证答案串中至少一个1和一个0

思路：首先通过两次询问可以确定答案串的第一个字符是1还是0，假设答案串的前m个字符都是0或者1，如果都是那么我们要找的答案大于m，如果不是，那么可以确定m以内有我们要找的答案

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define pill pair<int, int>
#define mst(a, b)	memset(a, b, sizeof a)
#define REP(i, x, n)	for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 1e6 + 10;
const int INF = 1e9 + 10;
int Query(int n, int t) {
	string str(n, '0');
	for(int i = 0; i < t; ++i) {
		str[i] = '1';
	}
	printf("? %s\n", str.c_str());
	fflush(stdout);
	int ans;	scanf("%d", &ans);
	return ans;
}
int res[2];

int main(){
	int n;	scanf("%d", &n);
	int a = Query(n, 0), b = Query(n, 1);
	int tmp;
	if(a > b) {		// find 0
		res[1] = 1;
		int l = 2, r = n;
		while(l <= r) {
			int m = (l + r) >> 1;
			if(-Query(n, m) + a == m) {
				l = m + 1;
			} else {
				tmp = m;
				r = m - 1;
			}
		}
		res[0] = tmp;
	} else {	//find 1
		res[0] = 1;
		int l = 2, r = n;
		while(l <= r) {
			int m = (l + r) >> 1;
			if(-a + Query(n, m) == m) {
				l = m + 1;
			} else {
				tmp = m;
				r = m - 1;
			}
	//		printf("%d %d %d\n", l, r, tmp);
		}
		res[1] = tmp;
	}
	printf("! %d %d\n", res[0], res[1]);
	return 0;
}