A – Darker and Darker
一眼题，直接 BFS 找最长最短路即可。
Continue reading →考虑对仅有的 \(s \leq 3\) 个元素排列顺序然后判就完事了。
// A.cpp
#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 1e5 + 200;
int n, ai[MAX_N], si[MAX_N];
map<int, int> mp, tmp;
int main()
{
scanf("%d", &n);
for (int i = 1, x; i <= n; i++)
scanf("%d", &ai[i]), mp[ai[i]]++;
if (mp.size() <= 3)
{
vector<int> vi;
for (auto x : mp)
vi.push_back(x.first);
bool flag = false;
do
{
int len = vi.size();
tmp.clear();
for (int i = 1; i <= 2; i++)
si[i] = vi[(i - 1) % len];
for (int i = 3; i <= n; i++)
si[i] = si[i - 1] ^ si[i - 2];
for (int i = 1; i <= n; i++)
tmp[si[i]]++;
bool cflag = true;
for (auto x : tmp)
cflag &= mp[x.first] == x.second;
flag |= cflag;
} while (next_permutation(vi.begin(), vi.end()));
puts(flag ? "Yes" : "No");
}
else
puts("No");
return 0;
}
Continue reading → 三角形斜着放即可。
// A.cpp
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll S;
int main()
{
scanf("%lld", &S);
ll y = (S + int(1e9) - 1) / int(1e9), x = y * int(1e9) - S;
printf("%d %d %d %d %lld %lld\n", 0, 0, int(1e9), 1, x, y);
return 0;
}
Continue reading → 这些 3000+ 题目里做过最小清新的。
首先肯定可以想到，分解质因子之后，对于一个 \(p^c\)，把每个元素该质因子的指数拿出来做中位数即为此质因子的贡献。我大概想到这里就断片了。
Continue reading →