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Appearance Analysis
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#include <bits/stdc++.h>
using namespace std;
int r, c;
char s[200][200];
set<string> st;
int main(void)
{
scanf("%d%d", &r, &c);
for (int i = 1; i <= r; i++)
scanf("%s", s[i] + 1);
int r2, c2;
int ans = 0;
for (int i = 2; s[2][i] != '#'; i++)
c2 = i - 1;
for (int i = 2; s[i][2] != '#'; i++)
r2 = i - 1;
//cout<<c2<<" "<<r2<<endl;
for (int i = 2; i <= r; i += r2 + 1)
{
for (int j = 2; j <= c; j += c2 + 1)
{
ans++;
string a = "", b = "", c = "", d = "";
for (int ii = 1; ii <= r2; ii++)
{
for (int jj = 1; jj <= c2; jj++)
{
a.push_back(s[i + ii - 1][j + jj - 1]);
}
}
for (int ii = 1; ii <= c2; ii++)
{
for (int jj = 1; jj <= r2; jj++)
{
b.push_back(s[i + r2 - jj][j + ii - 1]);
}
}
for (int ii = 1; ii <= r2; ii++)
{
for (int jj = 1; jj <= c2; jj++)
{
c.push_back(s[i + r2 - ii][j + c2 - jj]);
}
}
for (int ii = 1; ii <= c2; ii++)
{
for (int jj = 1; jj <= r2; jj++)
{
d.push_back(s[i + jj - 1][j + c2 - ii]);
}
}
if (st.count(a) || st.count(b) || st.count(c) || st.count(d))
{
ans--;
}
else
st.insert(a), st.insert(b), st.insert(c), st.insert(d);
//cout<<a<<endl<<b<<endl<<c<<endl<<d<<endl;
}
}
printf("%d\n", ans);
}
Convex Contour
计算几何鲨我！
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#include <bits/stdc++.h>
#define ld long double
using namespace std;
int n;
char s[100];
int main(void)
{
const ld sq3 = sqrt(3);
const ld pi = 3.1415926535898;
scanf("%d%s", &n, s + 1);
int loc1 = 0, loc2 = 0;
// if(n==1&&s[1]=='T'){printf("3.000000000\n");return 0;}
for (int i = 1; i <= n; i++)
{
if (s[i] != 'T')
{
loc1 = i;
break;
}
}
for (int i = n; i >= 1; i--)
{
if (s[i] != 'T')
{
loc2 = i;
break;
}
}
// cout<<loc1<<" "<<loc2<<endl;
if (loc1 != 0 && loc2 != 0)
{
ld ans = 0;
ans += (loc2 - loc1);
if (s[loc1] == 'S')
{
ans += 0.5;
if (loc1 == 1)
{
ans += 1;
}
else
{
ans += sqrt((loc1 - 1.5) * (loc1 - 1.5) + (1 - sq3 / 2) * (1 - sq3 / 2));
ans += 1;
}
}
else if (s[loc1] == 'C')
{
if (loc1 != 1)
{
ld dis = (loc1 - 1) * (loc1 - 1) + (sq3 / 2 - 0.5) * (sq3 / 2 - 0.5);
ans += sqrt(dis - 0.25);
ans += (pi / 2 - asin((sq3 / 2 - 0.5) / sqrt(dis)) - acos(0.5 / sqrt(dis))) / 2;
ans += 1;
}
else
{
ans += pi / 2;
}
}
if (s[loc2] == 'S')
{
ans += 0.5;
if (loc2 == n)
ans += 1;
else
{
ans += sqrt((n - loc2 - 0.5) * (n - loc2 - 0.5) + (1 - sq3 / 2) * (1 - sq3 / 2));
ans += 1;
}
}
else if (s[loc2] == 'C')
{
if (loc2 != n)
{
ld dis = (n - loc2) * (n - loc2) + (sq3 / 2 - 0.5) * (sq3 / 2 - 0.5);
// cout<<dis<<" "<<ans<<endl;
ans += sqrt(dis - 0.25);
// cout<<ans<<endl;
ans += (pi / 2 - acos((n - loc2) / sqrt(dis)) - acos(0.5 / sqrt(dis))) / 2;
// cout<<pi/2<<" "<<acos((n-loc2)/sqrt(dis))<<" "<<acos(0.5/sqrt(dis))<<" "<<ans<<endl;
ans += 1;
}
else
{
ans += pi / 2;
}
}
ans += n;
if (s[1] == 'C')
ans -= 0.5;
if (s[n] == 'C')
ans -= 0.5;
printf("%.10Lf\n", ans);
}
else
{
printf("%.10Lf\n", (ld)2 * n + 1);
}
}
Free Figurines
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#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 9;
int n, ans, p[N], q[N];
void open(int u)
{
if (!u)
return;
++ans;
while (p[u])
{
++ans;
int pre = u;
u = p[u];
p[pre] = 0;
}
}
int main()
{
scanf("%d", &n);
for (int i = 1; i <= n; ++i)
scanf("%d", &p[i]);
for (int i = 1; i <= n; ++i)
{
scanf("%d", &q[i]);
if (p[i] != q[i])
open(p[i]), open(q[i]);
}
printf("%d", ans);
}
Key Knocking
三个三个的考虑，但每组实际上考虑的 4 个元素。
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#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 9;
char s[N];
int main()
{
scanf("%s", s);
int len = strlen(s);
s[len] = '0';
vector<int> ans;
for (int i = len - 3; i >= 0; i -= 3)
{
string t(s + i, s + i + 4);
if (t[t.size() - 1] == '1')
for (int i = 0; i < t.size(); ++i)
t[i] ^= 1;
#define ANS(i) (ans.push_back(i), s[i] ^= 1, s[(i) + 1] ^= 1)
if (t == "0000")
ANS(i + 1);
else if (t == "0010")
;
else if (t == "0100")
;
else if (t == "1000")
ANS(i);
else if (t == "0110")
;
else if (t == "1100")
ANS(i + 1);
else if (t == "1010")
;
else if (t == "1110")
ANS(i);
}
cout << ans.size() << '\n';
for (int i = 0; i < ans.size(); ++i)
cout << ans[i] + 1 << ' ';
}