文章目录

A

思路：就是三个点如果任意两个点的横坐标相同，并且任意两个点的纵坐标相同的话就不行。

#include <cstring>
#include <iostream>
#include <algorithm>
#include <map>
using namespace std;const int N = 100;
int a[N];
map<int, int> b;
map<int, int> c;void solve()
{int flag1 = 1, flag2 = 1;for (int i = 1; i <= 6; i++)cin >> a[i];for (int i = 1; i <= 6; i += 2){b[a[i]]++;if (b[a[i]] == 2)flag1 = 0;}for (int i = 2; i <= 6; i += 2){c[a[i]]++;if (c[a[i]] == 2)flag2 = 0;}if (!flag1 && !flag2){cout << "NO" << endl;}else{cout << "YES" << endl;}
}int main()
{ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);int T;cin >> T;while (T--){b.clear();c.clear();solve();}return 0;
}

B

思路：按题意模拟一遍即可，但是第二个到第n个要先排序。

#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;#define int long long
const int N = 1e6 + 10;
int a[N];void solve()
{int n;cin >> n;int sum = 0;int k;cin >> k;for (int i = 1; i < n; i++){cin >> a[i];}sort(a + 1, a + n);for (int i = 1; i < n; i++){if (k < a[i]){int j = k + a[i];if ((k + a[i]) % 2 == 0){k = (k + a[i]) / 2;a[i] = (k + a[i]) / 2;}else{k = (k + a[i]) / 2 + 1;a[i] = (k + a[i]) / 2;}}}cout << k << endl;
}signed main()
{ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);int T;cin >> T;while (T--){solve();}return 0;
}

C

C题是比较难的一个题
思路：首先我们考虑 a [ 1 ] [ i ] , a [ 2 ] [ i ] , a [ 3 ] [ i ] , . . . , a [ i ] [ i ] a[1][i],a[2][i],a[3][i],...,a[i][i] a[1][i],a[2][i],a[3][i],...,a[i][i]我们可以发现这样它必须是222221111的这种情况（我们先不考虑0的情况）也就是前面是2后面是1的情况。我们定义 f [ L ] [ R ] f[L][R] f[L][R]是1 ~ L-1是2，并且L ~ R是1的串的方案数。然后我们考虑状态转移。状态转移可以分两种情况如果第R和R+1是相同的那么就是 f [ L ] [ R + 1 ] = f [ L ] [ R + 1 ] + f [ L ] [ R ] f[L][R+1] =f[L][R+1]+ f[L][R] f[L][R+1]=f[L][R+1]+f[L][R],如果是不同的 f [ R + 1 ] [ R + 1 ] = f [ R + 1 ] [ R + 1 ] + f [ L ] [ R ] f[R+1][R+1] = f[R+1][R+1]+f[L][R] f[R+1][R+1]=f[R+1][R+1]+f[L][R]，我们在具体实现的时候还要判断一下情况是否成立。借鉴的大佬的思路传送门大佬有图文便于理解。非常建议去康康。

#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;#define int long long
const int N = 1e2 + 10, mod = 998244353;
int a[N][N], dp[N][N];void solve()
{int n;cin >> n;for (int i = 1; i <= n; i++)for (int j = i; j <= n; j++)cin >> a[i][j];dp[1][1] = 2;//一开始没要求，可以存1或者0，所以是两种方案。for (int R = 1; R <= n; R++){for (int L = 1; L <= R; L++){bool flag = true;for (int i = 1; i <= L - 1; i++)if (a[i][R] == 1)flag = false;for (int i = L; i <= R; i++)if (a[i][R] == 2)flag = false;if (!flag)//方案不合理的话就直接赋成0dp[L][R] = 0;(dp[L][R + 1] += dp[L][R]) %= mod;(dp[R + 1][R + 1] += dp[L][R]) %= mod;}}int ans = 0;for (int i = 1; i <= n; i++)(ans += dp[i][n]) % mod;cout << ans % mod << endl;
}signed main()
{ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);int T = 1;while (T--){solve();}return 0;
}

D

思路：佬们直接猜的结论，比赛的时候我并没有猜出来

#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;#define int long long
const int N = 1e6 + 10;
int a[N];void solve()
{int n;string s;int sum1 = 0;int sum0 = 0;cin >> n >> s;for (int i = 0; i < s.size(); i++){if (s[i] == '1')sum1++;elsesum0++;}int left = pow(2, sum1);int right = pow(2, n) - pow(2, sum0) + 1;for (int i = left; i <= right; i++){cout << i << ' ';}cout << endl;
}signed main()
{ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);int T=1;while (T--){solve();}return 0;
}

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Educational Codeforces Round 140 (Rated for Div. 2)A~D题解

文章目录

A

B

C

D

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