B
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
int n,k,cnt = 0;
cin >> n >> k;
while(n) {
n /= k;
cnt ++;
}
cout << cnt ;
return 0;
}
C
暴力
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
const int N = 200;
int a[N];
int main() {
ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
int n,ans = 0x3f3f3f3f;
cin >> n;
for(int i = 0;i < n ; ++i) {
cin >> a[i];
}
for(int i = 1;i <= 100; ++i) {
int t = 0;
for(int j = 0;j < n; ++j) {
t += (a[j] - i) * (a[j] - i);
}
ans = min(ans,t);
}
cout << ans << endl;
return 0;
}
整数三分
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
typedef double db;
typedef long long LL;
const db EPS = 1e-9;
const int N = 2e5 + 100,INF = 1 << 31 - 1;
int a[N],n;
int f(int mid) {
int res = 0;
for(int i = 0;i < n; ++i) {
res += (a[i] - mid) * (a[i] - mid);
}
return res;
}
int main() {
ios::sync_with_stdio(0),cout.tie(0),cin.tie(0);
int lans,rans;
cin >> n;
for(int i = 0;i < n; ++i) cin >> a[i];
int l = 1,r = 100;
while(l < r) {
int lmid = l + (r - l) / 3;
int rmid = r - (r - l) / 3;
lans = f(lmid),rans = f(rmid);
if(lans <= rans) r = rmid - 1;
else l = lmid + 1;
}
cout << min(lans,rans) << endl;
return 0;
}
D
组合数取模,快速幂求逆元
\(ans = 2^n - 1 - C_{n}^{a} - C_{n}^{b}\)
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
const int mod = 1e9 + 7,N = 2e5 + 10;
typedef long long LL;
int qmi(int a,int b) {
int res = 1 % mod;
while(b) {
if(b & 1) res = (LL)res * a % mod;
a = (LL)a * a % mod;
b >>= 1;
}
return res;
}
void addmod(int &x,int y) {
x += y;
if(x >= mod) x -= mod;
if(x < 0) x += mod;
}
int main() {
ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
int n,a,b,ans,s = 1;
cin >> n >> a >> b;
ans = qmi(2,n);
addmod(ans, -1);
for(int i = 1;i <= b; ++i) {
s = (LL)s * (n - i + 1) % mod;
s = (LL)s * qmi(i,mod - 2) % mod;
if(i == a) addmod(ans,-s);
if(i == b) addmod(ans,-s);
}
cout << ans << endl;
return 0;
}