版权声明：CopyRight@2018 BPM136 https://blog.csdn.net/qq_20139423/article/details/88651970

A

温暖签到

#include <bits/stdc++.h>

using namespace std;

int main() {
int n;
cin >> n;
for (int i = 1; i <= n; ++i) {
int x, ans = 0;
for (int j = 1; j <= 4; ++j) {
cin >> x;
if (x > 0)
++ans;
}
if (ans == 0)
puts("Typically Otaku");
if (ans == 1)
puts("Eye-opener");
if (ans == 2)
puts("Young Traveller");
if (ans == 3)
puts("Excellent Traveller");
if (ans == 4)
puts("Contemporary Xu Xiake");
}
}

B

待补

C

upsolved
对x和y分别排序然后维护左边和右边分别被合并的区间，因为答案只可能出现在角落，所以就可以四个ans来维护(比赛的时候没看到每行每列只有一个这句话写死我了)

#include <bits/stdc++.h>

using namespace std;

using ll = long long;

namespace fastIO{
#define BUF_SIZE 100000
#define OUT_SIZE 100000
//fread->read
bool IOerror=0;
inline char nc(){
static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
if (p1==pend){
p1=buf; pend=buf+fread(buf,1,BUF_SIZE,stdin);
if (pend==p1){IOerror=1;return -1;}
//{printf("IO error!\n");system("pause");for (;;);exit(0);}
}
return *p1++;
}
inline bool blank(char ch){return ch==32||ch==10||ch==13||ch==9;}
inline bool enter(char ch){return ch==10||ch==13;}
inline void read(int &x){
bool sign=0; char ch=nc(); x=0;
for (;blank(ch);ch=nc());
if (IOerror)return;
if (ch==45)sign=1,ch=nc();
for (;ch>=48&&ch<=57;ch=nc())x=x*10+ch-48;
if (sign)x=-x;
}
inline void read(ll &x){
bool sign=0; char ch=nc(); x=0;
for (;blank(ch);ch=nc());
if (IOerror)return;
if (ch==45)sign=1,ch=nc();
for (;ch>=48&&ch<=57;ch=nc())x=x*10+ch-48;
if (sign)x=-x;
}
inline void read(double &x){
bool sign=0; char ch=nc(); x=0;
for (;blank(ch);ch=nc());
if (IOerror)return;
if (ch==45)sign=1,ch=nc();
for (;ch>=48&&ch<=57;ch=nc())x=x*10+ch-48;
if (ch==46){
double tmp=1; ch=nc();
for (;ch>=48&&ch<=57;ch=nc())tmp/=10.0,x+=tmp*(ch-48);
}
if (sign)x=-x;
}
inline void read(char *s){
char ch=nc();
for (;blank(ch);ch=nc());
if (IOerror)return;
for (;!blank(ch)&&!IOerror;ch=nc())*s++=ch;
*s=0;
}<
1aa6f
/span>
inline void readln(char *s) {
char ch=nc();
for (;blank(ch);ch=nc());
if(IOerror)return;
for(;!enter(ch)&&!IOerror;ch=nc())*s++=ch;
*s=0;
}
inline void read(char &c){
for (c=nc();blank(c);c=nc());
if (IOerror){c=-1;return;}
}
#undef OUT_SIZE
#undef BUF_SIZE
}
using fastIO::read;

int const N = 300005;

struct node {
int x, y, id;
};
vector<node> detx, dety;
int ans1, ans2, ans3, ans4;
int lx, rx, ly, ry;
int xmi, xma, ymi, yma;
int dx, dy;
int mark[N];
int n, m;

void setpn(node a, int val) {
if (mark[a.id] == 0) {
mark[a.id] = val;
return;
}
int& t = mark[a.id];
if (t == 1) {
if (val == 3)
++ans1;
if (val == 4)
++ans3;
}
if (t == 2) {
if (val == 3)
++ans2;
if (val == 4)
++ans4;
if (val == 1)
t = val;
}
if (t == 3) {
if (val == 1)
++ans1;
if (val == 2)
++ans2;
}
if (t == 4) {
if (val == 1)
++ans3;
if (val == 2)
++ans4;
if (val == 3)
t = val;
}
}

void init(vector<node>& t) {
for (int i = 0; i < n; ++i)
mark[i] = 0;
ans1 = ans2 = ans3 = ans4 = 0;
dx = dy = 0;
detx = t;
dety = t;
sort(detx.begin(), detx.end(), [] (node const& a, node const& b) {
return a.x < b.x;
});
sort(dety.begin(), dety.end(), [] (node const& a, node const& b) {
return a.y < b.y;
});
lx = 0, rx = n - 1;
setpn(detx[0], 3);
setpn(detx[n - 1], 4);
xmi = detx[0].x;
xma = detx[n - 1].x;

ly = 0, ry = n - 1;
setpn(dety[0], 1);
setpn(dety[n - 1], 2);
ymi = dety[0].y;
yma = dety[n - 1].y;
}

void moveL(int step) {
dy -= step;
ymi = max(1, ymi - step);
yma = max(1, yma - step);
while (ly < ry && dety[ly + 1].y + dy <= ymi) {
++ly;
if (ly == ry)
return;
setpn(dety[ly], 1);
}
}

void moveR(int step) {
dy += step;
ymi = min(n, ymi + step);
yma = min(n, yma + step);
while (ly < ry && dety[ry - 1].y + dy >= yma) {
--ry;
if (ly == ry)
return;
setpn(dety[ry], 2);
}
}

void moveU(int step) {
dx -= step;
xmi = max(1, xmi - step);
xma = max(1, xma - step);
while (lx < rx && detx[lx + 1].x + dx <= xmi) {
++lx;
if (lx == rx)
return;
setpn(detx[lx], 3);
}
}

void moveD(int step) {
dx += step;
xmi = min(n, xmi + step);
xma = min(n, xma + step);
while (lx < rx && detx[rx - 1].x + dx >= xma) {
--rx;
if (lx == rx)
return;
setpn(detx[rx], 4);
}
}

int query(int const& x, int& mi, int& ma, int d) {
if (x + d <= mi)
return mi;
if (x + d >= ma)
return ma;
return x + d;
}

long long C(int x) {
return (long long) x * (x - 1) / 2;
}

long long QueryAns() {
if (n == 1)
return 0;
if (xmi == xma && ymi == yma)
return C(ans1 + ans2 + ans3 + ans4);
if (xmi == xma)
return C(ans1 + ans3) + C(ans2 + ans4);
if (ymi == yma)
return C(ans1 + ans2) + C(ans3 + ans4);
return C(ans1) + C(ans2) + C(ans3) + C(ans4);
}

int main() {
// freopen("in.txt", "r", stdin);
// freopen("out.txt", "w", stdout);
ios::sync_with_stdio(0);
int T;
read(T);
while (T--) {
read(n);
read(m);
auto a = vector<node>(n);
for (int i = 0; i < n; ++i) {
read(a[i].x);
read(a[i].y);
a[i].id = i;
}
init(a);
/////////////////////
// cerr << "MARK : ";
// for (int i = 0; i < n; ++i)
//     cerr << mark[i] << ' ';
// cerr << '\n';

for (int o = 0; o < m; ++o) {
char op;
int x;
read(op);
if (op != '!')
read(x);
switch(op) {
case 'L':
moveL(x);
break;
case 'R':
moveR(x);
break;
case 'U':
moveU(x);
break;
case 'D':
moveD(x);
break;
case '?': {
int px = query(a[x - 1].x, xmi, xma, dx);
int py = query(a[x - 1].y, ymi, yma, dy);
cout << px << ' ' << py << '\n';
}
break;
default :
cout << QueryAns() << '\n';
}

}
}
}

D

pypy的小数居然是float，交Python就能过，这里提供相同的C++代码

#include <bits/stdc++.h>

using namespace std;

double const PI = acos(-1.0);

double sqr(double x) {
return x * x;
}

double calc(double a, double b, double r, double th) {
th = th / 180.0 * PI;
double l = sqrt(sqr(a + r) + sqr(b));
double th0 = atan(b / (a + r));
return l * cos(max(th0 - th, 0.0)) - r;
}

int main() {
int T;
cin >> T;
while (T--) {
double a, b, r, th;
cin >> a >> b >> r >> th;
double ans = calc(a, b, r, th);
cout << fixed << setprecision(12) << ans << '\n';
}
}

E

from math import gcd

def primes(mx):
ret = []
for i in range(2, mx):
if all(i % p != 0 for p in ret):
ret.append(i)
return ret

def select(ps, n):
cur, xs = 1, []
for p in ps:
if cur * p > n:
return cur, xs
cur *= p
xs.append(p)

def calc(n, ps):
a, qs = select(ps, n)
b = 1
for q in qs:
b *= q + 1
g = gcd(a, b)
return a // g, b // g

def main():
ps = primes(300)

for _ in range(int(input())):
n = int(input())
a, b = calc(n, ps)
print(f'{a}/{b}')

main()

F

bfs

#include <iostream>
#include <string>
#include <queue>
#include <utility>
#include <algorithm>

using Edges = std::vector<std::vector<int>>;

int bfs(int n, const Edges &es, int ibeg, int iend) {
const int INF = 1000 * 1000 + 10;
auto q = std::queue<int>();
auto dis = std::vector<int>(n, INF);
q.push(ibeg);
dis[ibeg] = 0;
while(!q.empty()) {
int c = q.front();
q.pop();
if(c == iend)
return 1 + dis[c];
for(int to: es[c])
if(dis[to] == INF) {
dis[to] = dis[c] + 1;
q.push(to);
}
}
return -1;
}

int main() {
std::ios::sync_with_stdio(false);

int t;
std::cin >> t;
while(t--) {
int r, c;
std::cin >> r >> c;
std::cin.get(); // Eat space.
int n = r * c, ibeg, iend;

std::vector<std::string> inp;
for(int i = 0; i < 4 * r + 3; ++i) {
std::string s;
std::getline(std::cin, s);
inp.push_back(std::move(s));
}

auto es = Edges(n);

auto adde = [&](int x1, int y1, int x2, int y2) {
es[x1 * c + y1].push_back(x2 * c + y2);
es[x2 * c + y2].push_back(x1 * c + y1);
};

for(int i = 0; i < r; ++i)
for(int j = 0; j < c; ++j) {
int x = (j & 1 ? 4 : 2) + i * 4,
y = 4 + j * 6;

if(inp[x][y] == 'S')
ibeg = i * c + j;
if(inp[x][y] == 'T')
iend = i * c + j;

if(inp[x + 2][y] == ' ')
adde(i, j, i + 1, j);
if(inp[x + 1][y + 3] == ' ') {
if(j & 1)
adde(i, j, i + 1, j + 1);
else
adde(i, j, i, j + 1);
}
if(inp[x - 1][y + 3] == ' ') {
if(j & 1)
adde(i, j, i, j + 1);
else
adde(i, j, i - 1, j + 1);
}
}

int ans = bfs(n, es, ibeg, iend);
std::cout << ans << '\n';

}

return 0;
}

I

贪心取两边即可

#include<iostream>
using namespace std;
long long a[100010];
int main()
{
int t,n;
cin>>t;
while(t--)
{
long long ans=0,sum=0;
cin>>n;
for(int i=1;i<n;i++)cin>>a[i];
for(int i=1;i<n;i++)a[i]+=a[i-1];
for(int i=1;i<=n;i++)
{
if(i&1)ans+=sum;
else sum+=a[n-i/2]-a[i/2-1],ans+=sum;
cout<<ans;
if(n!=i)cout<<" ";
}
cout<<endl;
}
}