Codeforces Round 883 (Div. 3) VP记录
A
很明显找落下来会碰到地的钉子有多少即可
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define haha cout<<"\n"
#define ye cout<<"Yes\n"
#define no cout<<"No\n"
typedef unsigned long long ull;
const int N = 1e6 + 10;
const int M = 10100;
const int mod = 998244353;
const double eps = 1e-8;
double pi = acos(-1);
int dx[10] = {-1, 0, 1, 0};
int dy[10] = {0, 1, 0, -1};
int n, m, k;
int ksm(int x, int y) {
int ans = 1;
while (y) {
if (y & 1)
ans = ans * x;
y >>= 1;
x = x * x;
}
return ans;
}
struct node {
int a, b;
} a[N];
void solve() {
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> a[i].a >> a[i].b;
}
int cnt = 0;
for (int i = 1; i <= n; i++) {
if (a[i].a > a[i].b)
cnt++;
}
cout << cnt << '\n';
}
signed main() {
// ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int t = 1;
// init();
cin >> t;
while (t--)
solve();
return 0;
}
/*
13 5
13
*/
B
暴力遍历
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define haha cout<<"\n"
#define ye cout<<"Yes\n"
#define no cout<<"No\n"
typedef unsigned long long ull;
const int N = 1e6 + 10;
const int M = 10100;
const int mod = 998244353;
const double eps = 1e-8;
double pi = acos(-1);
int dx[10] = {-1, 0, 1, 0};
int dy[10] = {0, 1, 0, -1};
int n, m, k;
int ksm(int x, int y) {
int ans = 1;
while (y) {
if (y & 1)
ans = ans * x;
y >>= 1;
x = x * x;
}
return ans;
}
char a[10][10];
void solve() {
for (int i = 1; i <= 3; i++) {
for (int j = 1; j <= 3; j++) {
cin >> a[i][j];
}
}
for (int i = 1; i <= 3; i++) {
char x = a[i][1];
if (x == '.') continue;
int f = 0;
for (int j = 1; j <= 3; j++) {
if (a[i][j] != x) {
f = 1;
}
}
if (f == 0) {
cout << x << '\n';
return;
}
}
for (int i = 1; i <= 3; i++) {
char x = a[1][i];
if (x == '.') continue;
int f = 0;
for (int j = 1; j <= 3; j++) {
if (a[j][i] != x) {
f = 1;
}
}
if (f == 0) {
cout << x << '\n';
return;
}
}
if (a[1][1] != '.' && a[1][1] == a[2][2] && a[1][1] == a[3][3]) {
cout << a[1][1] << '\n';
return;
}
if (a[1][3] != '.' && a[1][3] == a[2][2] && a[1][3] == a[3][1]) {
cout << a[1][3] << '\n';
return;
}
cout << "DRAW\n";
}
signed main() {
// ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int t = 1;
// init();
cin >> t;
while (t--)
solve();
return 0;
}
/*
13 5
13
*/
C
排序模拟即可，因为罚时计算错误喜提罚时
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define haha cout<<"\n"
#define ye cout<<"Yes\n"
#define no cout<<"No\n"
typedef unsigned long long ull;
const int N = 1e6 + 10;
const int M = 10100;
const int mod = 998244353;
const double eps = 1e-8;
double pi = acos(-1);
int dx[10] = {-1, 0, 1, 0};
int dy[10] = {0, 1, 0, -1};
int n, m, k;
int ksm(int x, int y) {
int ans = 1;
while (y) {
if (y & 1)
ans = ans * x;
y >>= 1;
x = x * x;
}
return ans;
}
char a[10][10];
void solve() {
int h;
cin >> n >> m >> h;
vector<int> v;
for (int i = 1; i <= m; i++) {
int x;
cin >> x;
v.push_back(x);
}
int sum = 0;
sort(v.begin(), v.end());
int cnt = 0;
int y = 0;
for (int i = 0; i < v.size(); i++) {
int t = v[i];
if (y + t <= h) {
sum = sum + y + t;
cnt++;
y += t;
}
}
v.clear();
// cout << sum << ' ' << cnt << '\n';
int ans = 1;
for (int i = 2; i <= n; i++) {
vector<int> v;
int sum1 = 0;
int cnt1 = 0, y = 0;
for (int j = 1; j <= m; j++) {
int x;
cin >> x;
v.push_back(x);
}
sort(v.begin(), v.end());
for (int i = 0; i < v.size(); i++) {
if (y + v[i] <= h) {
sum1 = sum1 + y + v[i];
cnt1++;
y += v[i];
}
}
// cout << cnt1 << ' ' << sum1 << '\n';
if (cnt1 > cnt) ans++;
else if (cnt1 == cnt) {
if (sum1 < sum) ans++;
}
v.clear();
}
cout << ans << '\n';
}
signed main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int t = 1;
// init();
cin >> t;
while (t--)
solve();
return 0;
}
/*
13 5
13
*/
计算几何题，关键在于计算底的长度，我用到了相似三角形，算面积的时候应该计算\(总面积-被遮盖面积\)更加好写
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define haha cout<<"\n"
#define ye cout<<"Yes\n"
#define no cout<<"No\n"
typedef unsigned long long ull;
const int N = 1e6 + 10;
const int M = 10100;
const int mod = 998244353;
const double eps = 1e-8;
double pi = acos(-1);
int dx[10] = {-1, 0, 1, 0};
int dy[10] = {0, 1, 0, -1};
int n, m, k;
int ksm(int x, int y) {
int ans = 1;
while (y) {
if (y & 1)
ans = ans * x;
y >>= 1;
x = x * x;
}
return ans;
}
double a[N];
void solve() {
double m, h;
cin >> n >> m >> h;
for (int i = 1; i <= n; i++) {
cin >> a[i];
}
double sum = 0;
//(a[i] + h - a[i + 1]) * m / h
double a1 = m * h * 0.5;
for (int i = 1; i < n; i++) {
if (a[i + 1] - a[i] < h) {
double s1 = ( ( a[i] + h - a[i + 1] ) * m / h) * 0.5 * (a[i] + h - a[i + 1]);
// cout << s1 << '\n';
sum += s1;
}
}
double ans = a1 * n;
// printf("%.8lf\n", ans);
// printf("%.8lf\n", sum);
printf("%.8lf\n", ans - sum);
}
signed main() {
// ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int t = 1;
// init();
cin >> t;
while (t--)
solve();
return 0;
}
/*
13 5
13
*/
E1
题目比较难懂，实际需要预处理和一些树论的知识将可能的点全部求出即可。题意：是否存在n个结点的满k叉树。
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define haha cout<<"\n"
#define ye cout<<"Yes\n"
#define no cout<<"No\n"
typedef unsigned long long ull;
const int N = 1e6 + 10;
const int M = 10100;
const int mod = 998244353;
const double eps = 1e-8;
double pi = acos(-1);
int dx[10] = {-1, 0, 1, 0};
int dy[10] = {0, 1, 0, -1};
int n, m, k;
int ksm(int x, int y) {
int ans = 1;
while (y) {
if (y & 1)
ans = ans * x;
y >>= 1;
x = x * x;
}
return ans;
}
double a[N];
map<int, bool> st;
void init() {
int sum;
for (int i = 2; i <= 1000; i++) {
sum = 1;
for (int j = 1; j <= 20; j++) {
int kk = 1;
for (int k = 1; k <= j; k++) kk *= i;
sum += kk;
if (sum >= 1e7) break;
if (j >= 2) st[sum] = 1;
}
}
}
void solve() {
cin >> n;
if (n <= 5) {
no;
return;
}
if (st[n] == 1)
ye;
else
no;
}
signed main() {
// ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int t = 1;
init();
cin >> t;
while (t--)
solve();
return 0;
}
/*
13 5
13
*/
vp名次还可以。
