牛客等级之题N2
牛客等级之题N2（8.3场）
等级之题N2
设大半圆的半径为\(R\)，左小半圆半径为\(r\)
\(r = kR(0<k<1)\)
\(\pi R^2 - \pi k^2R^2 - \pi (R-kR)^2 = 2s\)
\(R^2(k-k^2)=\frac{s}{\pi}\)
\(k = \frac{1}{2}\)时\(R^2\)最小
#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1);
int main() {
double s;
scanf("%lf",&s);
printf("%.3lf\n",2*sqrt(s/pi));
return 0;
}
拯救单身狗
#include <bits/stdc++.h>
using namespace std;
int main() {
int t; cin >> t;
while (t--) {
unordered_map<int,int> mp;
bool flag = true;
int num;
while (cin >> num && num) {
mp[num]++;
}
for (unordered_map<int,int>::iterator it = mp.begin(); it != mp.end(); ++it) {
if (it->second & 1) {
flag = false;
break;
}
}
if (flag) cout << "Yes" << endl;
else cout << "No" << endl;
}
return 0;
}
做题
#include <bits/stdc++.h>
using namespace std;
int main() {
int t; cin >> t;
while (t--) {
unordered_map<int,int> mp;
bool flag = true;
int num;
while (cin >> num && num) {
mp[num]++;
}
for (unordered_map<int,int>::iterator it = mp.begin(); it != mp.end(); ++it) {
if (it->second & 1) {
flag = false;
break;
}
}
if (flag) cout << "Yes" << endl;
else cout << "No" << endl;
}
return 0;
}
Rinne Loves Study
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5+5;
int row[maxn], col[maxn];
int main() {
int n, m, T;
cin >> n >> m >> T;
for (int i = 1; i <= T; ++i) {
int opt; cin >> opt;
if (opt == 1) {
int x; cin >> x;
row[x] = i;
}
else {
int y; cin >> y;
col[y] = i;
}
}
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= m; ++j) {
if (j == 1) cout << max(row[i],col[j]);
else cout << " " << max(row[i],col[j]);
}
cout << endl;
}
return 0;
}
小sun的假期
#include <bits/stdc++.h>
using namespace std;
struct node {
int l, r;
};
vector<node> ve;
bool cmp(const node x, const node y) {
return x.l < y.l;
}
int main() {
int n, m; cin >> n >> m;
while (m--) {
int l, r; cin >> l >> r;
ve.push_back(node{l,r});
}
sort(ve.begin(),ve.end(),cmp);
int pos = 0, ans = 0, pre = 1;
while (pos < ve.size()) {
int l = ve[pos].l, r = ve[pos].r;
pos++;
while (pos < ve.size() && l <= ve[pos].l && ve[pos].l <= r) {
r = max(r,ve[pos].r);
pos++;
}
ans = max(ans,l-pre);
pre = r+1;
}
ans = max(ans,n+1-pre);
cout << ans << endl;
return 0;
}
数数
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int p = 998244353, maxn = 1e7+5;
int mul[maxn];
ll qpow(ll a, ll b) {
ll res = 1;
while (b) {
if (b&1) res = res*a%p;
a = a*a%p;
b >>= 1;
}
return res%p;
}
int main() {
mul[0] = 1;
for (int i = 1; i <= 10000000; ++i) {
mul[i] = ll(mul[i-1])*i%p;
}
int t; cin >> t;
while (t--) {
ll n; cin >> n;
ll sum = n*(n+1)/2 % p;
cout << sum*sum%p << " " << qpow(mul[n],n+n) << endl;
}
return 0;
}
小w的a+b问题
#include <bits/stdc++.h>
using namespace std;
int main() {
int c; cin >> c;
if (c == -1) cout << "No solution" << endl;
else {
int a = 2147483647;
cout << a << " " << c-a << endl;
}
return 0;
}
斐波那契
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll p = 1e9+7;
struct mat {
ll a[2][2];
mat() {
memset(a,0,sizeof(a));
}
};
mat mul(mat A, mat B) {
mat C;
for (int i = 0; i < 2; ++i) {
for (int j = 0; j < 2; ++j) {
for (int k = 0; k < 2; ++k) {
C.a[i][j] = (C.a[i][j] + A.a[i][k]*B.a[k][j]%p)%p;
}
}
}
return C;
}
mat qpow(mat A, ll b) {
mat res;
for (int i = 0; i < 2; ++i) res.a[i][i] = 1;
while (b) {
if (b & 1) res = mul(res,A);
A = mul(A,A);
b >>= 1;
}
return res;
}
int main() {
ll n; cin >> n;
mat A;
A.a[0][0] = 1, A.a[0][1] = 1;
A.a[1][0] = 1, A.a[1][1] = 0;
A = qpow(A,n-1);
mat B;
B.a[0][0] = 1, B.a[0][1] = 0;
B.a[1][0] = 1, B.a[1][1] = 0;
B = mul(A, B);
cout << B.a[0][0]*B.a[1][0]%p << endl;
return 0;
}
Audio
#include <bits/stdc++.h>
using namespace std;
double x[4], y[4];
int main() {
for (int i = 1; i <= 3; ++i) {
scanf("%lf%lf",&x[i],&y[i]);
}
double k1 = (x[2]-x[1])/(y[1]-y[2]);
double k2 = (x[3]-x[1])/(y[1]-y[3]);
double b1 = (y[1]+y[2])/2 - (x[1]+x[2])/2*k1;
double b2 = (y[1]+y[3])/2 - (x[1]+x[3])/2*k2;
x[0] = (b2-b1)/(k1-k2);
y[0] = k1*x[0] + b1;
printf("%.3lf %.3lf\n",x[0],y[0]);
return 0;
}
牛妹和01串
#include <bits/stdc++.h>
using namespace std;
int main() {
string s; cin >> s;
int a = 0, b = 0, m = 0;
for (int i = 0; i < s.size(); ++i) {
if (s[i] == '0') a++;
else b++;
if (a && b) {
m++;
a = b = 0;
}
}
cout << m << endl;
return 0;
}
爵士
#include <bits/stdc++.h>
using namespace std;
int main() {
int t; cin >> t;
while (t--) {
int n; cin >> n; getchar();
int num = 0;
for (int i = 1; i <= n; ++i) {
string s; getline(cin,s);
if (s.find('2') != -1) num++;
}
printf("%.4f\n",double(num)/n);
}
return 0;
}
