A. Odd Divisor
题意
询问一个数 (n) 是否包含除 1 外的质因子
思路
不停除 2
/*
* @Author: zhl
* @LastEditTime: 2021-01-25 23:02:11
*/
/* Author: wxhgkk
* Time: 2021-01-25 23:00:02
**/
#include<bits/stdc++.h>
//#define int long long
using namespace std;
typedef long long ll;
const int N = 3e5 + 100;
int T, k, m, A[N];
ll n;
string s;
void init() {
cin >> n;
}
void solve() {
while(n > 1){
if (n & 1ll) {
cout << "YES" << endl;
return;
}
n >>= 1ll;
}
cout << "NO" << endl;
}
signed main() {
//ios::sync_with_stdio(0);cin.tie(0);
cin >> T;
while (T--) {
init();
solve();
}
}
B. New Year's Number
题意
给 (n) ，判断 (n) 能否表示为若干个 2020 与 2021 之和
思路
暴力枚举
/*
* @Author: zhl
* @LastEditTime: 2021-01-25 23:05:20
*/
/* Author: wxhgkk
* Time: 2021-01-25 23:00:02
**/
#include<bits/stdc++.h>
//#define int long long
using namespace std;
typedef long long ll;
const int N = 3e5 + 100;
int T, n, k, m, A[N];
string s;
void init() {
cin >> n;
}
void solve() {
for(int i = 0;i <= n;i += 2020){
if((n - i) % 2021 == 0){
cout << "YES" << endl;
return;
}
}
cout << "NO" << endl;
}
signed main() {
//ios::sync_with_stdio(0);cin.tie(0);
cin >> T;
while (T--) {
init();
solve();
}
}
C. Ball in Berland
题意
给一个二分图，选两条不同的边，问有多少种选择使得选中的四个点各不相同
思路
(C_k^2 - sum_v{C_{cnt_v} ^2})
总方案减去非法方案，枚举重合的点
/*
* @Author: zhl
* @LastEditTime: 2021-01-25 23:13:11
*/
/* Author: wxhgkk
* Time: 2021-01-25 23:00:02
**/
#include<bits/stdc++.h>
//#define int long long
using namespace std;
typedef long long ll;
const int N = 3e5 + 100;
int T;
int A[N], B[N];
int a, b, k;
string s;
void init() {
cin >> a >> b >> k;
for (int i = 0;i <= a;i++)A[i] = 0;
for (int i = 0;i <= b;i++)B[i] = 0;
for(int i = 1;i <= k;i++){
int x;cin >> x;A[x]++;
}
for(int i = 1;i <= k;i++){
int x;cin >> x;B[x]++;
}
}
void solve() {
ll ans = 1ll * k * (k - 1) / 2ll;
for(int i = 1;i <= a;i++){
ans -= 1ll * A[i] * (A[i] - 1ll) / 2ll;
}
for(int i = 1;i <= b;i++){
ans -= 1ll * B[i] * (B[i] - 1ll) / 2ll;
}
cout << ans << endl;
}
signed main() {
//ios::sync_with_stdio(0);cin.tie(0);
cin >> T;
while (T--) {
init();
solve();
}
}
D. Cleaning the Phone
题意
思路
枚举 2 的次数，二分 1 的次数
/*
* @Author: zhl
* @LastEditTime: 2021-01-26 00:56:27
*/
#include<bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
typedef long long ll;
int T, n;
ll m;
ll A[N], cnta;
ll B[N], cntb;
ll tmp[N];
int main() {
cin >> T;
while (T--) {
cin >> n >> m;
for (int i = 1; i <= n; i++)cin >> tmp[i];
cnta = cntb = 0;
for (int i = 1; i <= n; i++) {
int x; cin >> x;
if (x == 1)A[++cnta] = tmp[i];
else B[++cntb] = tmp[i];
}
sort(A + 1, A + 1 + cnta, greater<ll>());
sort(B + 1, B + 1 + cntb, greater<ll>());
for (int i = 1; i <= cnta; i++)A[i] += A[i - 1];
for (int i = 1; i <= cntb; i++)B[i] += B[i - 1];
if (A[cnta] + B[cntb] < m)puts("-1");
else {
ll ans = cnta + 2 * cntb;
for (int i = 0; i <= cntb; i++) {
ll now = B[i];
if (A[cnta] < m - now)continue;
int pos = lower_bound(A, A + 1 + cnta, m - now) - A;
ans = min(ans, 2ll * i + pos);
}
printf("%lld
", ans);
}
}
}
E. Advertising Agency
思路
很显然从大到小贪心，看最小的数有多少个就可以
/*
* @Author: zhl
* @LastEditTime: 2021-01-25 23:41:59
*/
/* Author: wxhgkk
* Time: 2021-01-25 23:00:02
**/
#include<bits/stdc++.h>
//#define int long long
using namespace std;
typedef long long ll;
const int N = 3e5 + 100, mod = 1e9 + 7;
int T, n, k, m, A[N];
string s;
ll fac[N], invfac[N], inv[N];
void _init() {
fac[0] = invfac[0] = 1;
fac[1] = invfac[1] = 1;
inv[1] = 1;
for (int i = 2;i < N;i++) {
fac[i] = fac[i - 1] * i % mod;
inv[i] = (mod - mod / i) * inv[mod % i] % mod;
invfac[i] = invfac[i - 1] * inv[i] % mod;
}
}
int C(int n, int m) {
return 1ll * fac[n] * invfac[n - m] % mod * invfac[m] % mod;
}
map<int, int>cnt;
void init() {
cin >> n >> k;cnt.clear();
for (int i = 1;i <= n;i++)cin >> A[i], cnt[A[i]]++;
}
void solve() {
sort(A + 1, A + 1 + n, greater<int>());
int now = 0;int pre = -1;
for (int i = 1;i <= k;i++) {
if (A[i] == pre) {
now++;
}
else {
now = 1;
}
pre = A[i];
}
cout << C(cnt[A[k]], now) << endl;
}
signed main() {
//ios::sync_with_stdio(0);cin.tie(0);
cin >> T;_init();
while (T--) {
init();
solve();
}
}
F. Unusual Matrix
思路
A， B异或一下，其实就是看这个矩阵能否能由初始全 0 的矩阵得到
很显然对于每一列或每一行，只有两种状态，判断就可以了
/*
* @Author: zhl
* @LastEditTime: 2021-01-26 00:16:28
*/
/* Author: wxhgkk
* Time: 2021-01-25 23:00:02
**/
#include<bits/stdc++.h>
//#define int long long
using namespace std;
typedef long long ll;
const int N = 1e3 + 100;
int T, n;
int A[N][N], B[N][N];
string s;
void init() {
cin >> n;
for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++)scanf("%01d", &A[i][j]);
//for (int i = 1;i <= 2 * n;i++)fa[i] = i;
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
scanf("%01d", &B[i][j]);; B[i][j] ^= A[i][j];
}
}
}
void solve() {
for (int i = 1; i <= n; i++) {
int ans = B[i][1] ^ B[1][1];
for (int j = 1; j <= n; j++) {
if (B[i][j] ^ B[1][j] != ans) {
cout << "NO" << endl;
return;
}
}
}
cout << "YES" << endl;
}
signed main() {
//ios::sync_with_stdio(0);cin.tie(0);
cin >> T;
while (T--) {
init();
solve();
}
}
G. Strange Beauty
思路
发现对于一个序列，只要当前项 f[i] 是前一项 f[i-1] 的倍数就满足题意
所以直接对所有因数转移就可以，因为 (A[i] le 2e5)
处理出 2e5 的所有因数就可以
(n(dfrac 1 1 + dfrac 1 2 + ... + dfrac 1 n)) 括号里是调和级数，是 (logn) 级别，
复杂度 (O(Tnlogn))
/*
* @Author: zhl
* @LastEditTime: 2021-01-26 00:45:24
*/
/* Author: wxhgkk
* Time: 2021-01-25 23:00:02
**/
#include<bits/stdc++.h>
//#define int long long
using namespace std;
typedef long long ll;
const int N = 3e5 + 100;
int T, n, k, m, A[N];
int f[N];
string s;
vector<int>G[N];
void __init() {
for (int i = 1; i <= 200000; i++) {
for (int j = 1; j * i <= 200000; j++) {
G[j * i].push_back(i);
}
}
}
void init() {
cin >> n;
for (int i = 1; i <= n; i++)cin >> A[i];
sort(A + 1, A + 1 + n);
}
void solve() {
memset(f, 0, sizeof f);
int ans = 0;
for (int i = 1; i <= n; i++) {
int val = A[i];
int res = 0;
for (int v : G[val])res = max(res, f[v] + 1);
ans = max(ans, f[val] = res);
}
cout << n - ans << endl;
}
signed main() {
ios::sync_with_stdio(0);cin.tie(0);
cin >> T; __init();
while (T--) {
init();
solve();
}
}