Codeforces Round #676 (Div. 2) A - D个人题解（E题待补）

1421A. XORwice

题目链接：Click Here

cpp

// Author : RioTian
// Time : 20/10/18
#include <bits/stdc++.h>
#define ms(a, b) memset(a, b, sizeof a)
using namespace std;
typedef long long ll;
ll n, m, _;
void solve() {cin >> n >> m;cout << (n ^ m) << endl;
}
int main() {// freopen("in.txt", "r", stdin);ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);cin >> _;while (_--) solve();
}

1421B. Putting Bricks in the Wall

题目链接：Click Here

如果选择S的邻居为1，我们就可以F使为0的邻居，就没有办法从S到F，但这在最坏的情况下需要4个开关，这是不够的。幸运的是，为了减少到2个开关，只需反过来考虑，使相邻S的正方形变为0，使相邻的正方形变为F 1。必须存在最多具有两个开关的两个正方形的解，并且您不会从S到F，因为被迫选择1(或0)，并且无法通过与F相反的邻居。

cpp

// Author : RioTian
// Time : 20/10/18
#include <bits/stdc++.h>
#define ms(a, b) memset(a, b, sizeof a)
using namespace std;
typedef long long ll;
ll n, m, _;
char x[210][210];
void solve() {cin >> n;for (int i = 1; i <= n; i++) cin >> x[i] + 1;int a = x[1][2] - '0', b = x[2][1] - '0', c = x[n][n - 1] - '0',d = x[n - 1][n] - '0';if (a == b && c == d && a != c)cout << 0 << endl;else if (a != b && c != d)cout << 2 << endl<< 1 << " " << 2 << endl<< (a == c ? n - 1 : n) << " " << (a == c ? n : n - 1) << endl;else if (a != b)cout << 1 << endl<< (a == c ? 1 : 2) << ' ' << (a == c ? 2 : 1) << endl;else if (c != d)cout << 1 << endl<< (a == c ? n : n - 1) << ' ' << (a == c ? n - 1 : n) << endl;elsecout << 2 << endl << 1 << ' ' << 2 << endl << 2 << " " << 1 << endl;
}
int main() {// freopen("in.txt", "r", stdin);ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);cin >> _;while (_--) solve();
}

python

#python
import sys
input = sys.stdin.readline
I = lambda : list(map(int,input().split()))t,=I()
for _ in range(t):n,=I()l=[input().strip() for i in range(n)]an=[]le=[0,0,1,1,1];p=[1,1,0,0,0]rq=[l[0][1],l[1][0],l[1][1],l[2][0],l[0][2]]pos=[[1,2],[2,1],[2,2],[3,1],[1,3]]ct=cp=0;a1=[]for i in range(5):if le[i]!=int(rq[i]):ct+=1a1.append(pos[i])for i in range(5):if p[i]!=int(rq[i]):cp+=1an.append(pos[i])if ct<=cp:an=a1print(len(an))for i in an:print(*i)

1421C. Palindromifier

题目链接：Click Here

这道题写的挺懵的，一开始都没理解题意

被dalao提醒以后然后发现是构造题，根据题目说的两种方案：

对于任何字符串都可以先 $L ,i = 2 ，然后，然后R，i = 2 $ + R,i=7R,i=7。一定能构造需所需的字符串

cpp

// Author : RioTian
// Time : 20/10/18
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
string s;
int main() {// freopen("in.txt","r",stdin);ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);cin >> s;cout << "3\nL 2\nR 2\n";cout << "R " << 2 * s.size() - 1 << endl;
}

python

print('3\nL 2\nR 2\nR',len(input())*2-1)

1421D. Hexagons (补)

题目链接：Click Here

cpp

// Author : RioTian
// Time : 20/10/18
#include <bits/stdc++.h>
using namespace std;
#define ll long long
int inf = 1e9 + 7;
ll x, y, c1, c2, c3, c4, c5, c6;
ll check(ll ad) {ll ans = 0;if (ad > 0)ans += ad * c1;elseans += -ad * c4;if (x - ad > 0)ans += (x - ad) * c6;elseans += (ad - x) * c3;if (y - ad > 0)ans += (y - ad) * c2;elseans += (ad - y) * c5;return ans;
}
void solve() {cin >> x >> y >> c1 >> c2 >> c3 >> c4 >> c5 >> c6;ll l = -inf, r = inf;while (l < r) {ll mid = l + r >> 1;if (check(mid) < check(mid + 1))r = mid;elsel = mid + 1;}cout << min({check(l), check(l + 1), check(l - 1)}) << '\n';
}
int main() {int t;cin >> t;while (t--) solve();
}

1421E. Swedish Heroes (补)

题目链接：Click Here

鞋子
美容
母婴
箱包
53货源网
特产
货源关键词大全
学院
女鞋
情侣装
内衣
男鞋
女鞋
童鞋

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Codeforces Round #676 (Div. 2) A - D个人题解（E题待补）

1421A. XORwice

1421B. Putting Bricks in the Wall

1421C. Palindromifier

1421D. Hexagons (补)

1421E. Swedish Heroes (补)

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