Codeforces Round #697 (Div. 3)
A. Odd Divisor
题意
给你一个n((2leq nleq 10^{14}))问你n是否有奇数因子不包括1但包括它本身。
思路
在质因数分解时只有2是偶数，所以不断对一个数除以2，然后判断最后是不是为1就可以。
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
#define int long long
const int N = 2e5 + 10;
void solve() {
int n; cin >> n;
if (n == 2) {
cout << "NO
";
return ;
}
while(n %2 == 0) {
n /= 2;
}
if (n & 1 && n != 1) cout << "YES
";
else cout << "NO
";
}
signed main() {
ios::sync_with_stdio(0);
cin.tie(0);
int T = 1;
cin >> T;
while (T--) {
solve();
}
}
B. New Year's Number
题意
给你一个数n问你n能否被2020，2021相加求和。
思路
直接判断一个数里有多少个2020，然后如果余数少于20020的话就可以将余数给一些2020构成2021
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
#define int long long
const int N = 2e5 + 10;
void solve() {
int n; cin >> n;
int t = n % 2020;
if (n < 2020) {
cout << "NO
";
return ;
}
if (t <= n / 2020) {
cout << "YES
";
return ;
}
cout << "NO
";
}
signed main() {
ios::sync_with_stdio(0);
cin.tie(0);
int T = 1;
cin >> T;
while (T--) {
solve();
}
}
C. Ball in Berland
题意
有a个男的，b个女的，k个配对，问你两两配对且男不重复女也不重复的可能性有多少种
思路
分别计算某个男生某个女生配对的人数有多少种，然后枚举一种情况计算所有可能性
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
#define int long long
const int N = 2e5 + 10;
int a[N], b[N];
int ca[N], cb[N];
void solve() {
int x, y, k; cin >> x >> y >> k;
for (int i = 1; i <= max(x, y); ++i) {
ca[i] = cb[i] = 0;
}
for (int i = 1; i <= k; ++i) {
cin >> a[i];
++ca[a[i]];
}
for (int i = 1; i <= k; ++i) {
cin >> b[i];
++cb[b[i]];
}
int ans = 0;
for (int i = 1; i <= k; ++i) {
ans = ans + k - (ca[a[i]] + cb[b[i]]) + 1;
}
cout << ans / 2 << endl;
}
signed main() {
ios::sync_with_stdio(0);
cin.tie(0);
int T = 1;
cin >> T;
while (T--) {
solve();
}
}
D. Cleaning the Phone
题意
给你n个数，且每个数都有一个b，问你在清理内存大于等于m的情况下b的和最小
思路
以b值分组，分组后求前缀和，然后二分讨论。
需要特别注意若a组一个都没用或者b组一个都没用的情况
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
//#define int long long
const int N = 2e5 + 10;
struct NODE {
int a, b;
} d[N];
LL s1[N], s2[N];
void solve() {
int n, m;
cin >> n >> m;
LL tot = 0;
int c1 = 0;
vector<int>v1, v2;
for (int i = 1; i <= n; ++i) {
cin >> d[i].a;
tot += d[i].a;
}
for (int i = 1; i <= n; ++i) {
cin >> d[i].b;
if (d[i].b == 1) {
v1.push_back(d[i].a);
} else v2.push_back(d[i].a);
}
if(tot < m) {
cout << -1 << endl;
return ;
}
sort(v1.begin(), v1.end(), greater<int>());
sort(v2.begin(), v2.end(), greater<int>());
for (int i = 0; i < v1.size(); ++i) {
if (i == 0)s1[i] = v1[i];
else s1[i] = s1[i - 1] + v1[i];
}
for (int i = 0; i < v2.size(); ++i) {
if (i == 0)s2[i] = v2[i];
else s2[i] = s2[i - 1] + v2[i];
}
int ans = 1e9;
for (int i = 0; i < v1.size(); ++i) {
if (s1[i] >= m) {
ans = min(ans, i + 1);
break;
}
int idx = lower_bound(s2, s2 + v2.size(), m - s1[i]) - s2;
if (idx == v2.size()) continue;
ans = min(ans, (i + 1) + (idx + 1) * 2);
}
int idx = lower_bound(s2, s2 + v2.size(), m) - s2;
if (idx != v2.size()) {
ans = min(ans, (idx + 1) * 2);
}
cout << ans << endl;
}
signed main() {
ios::sync_with_stdio(0);
cin.tie(0);
int T = 1;
cin >> T;
while (T--) {
solve();
}
}
E. Advertising Agency
题意
给你n个数让你找k个数，然后其和最大，问你有多少种可能
思路
显然是把最大的k个数相加，然后看和k个数中最小的数相同的有几个，用了几个。(C_{共计}^{用了})
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 2e5 + 10;
const int mod = 1e9 + 7;
int qmi(int a, int k) {
int res = 1;
while(k) {
if(k & 1) res = (LL)res * a % mod;
k >>= 1;
a = (LL)a * a % mod;
}
return res;
}
int fact[N], infact[N];
void init(int n) {
fact[0] = infact[0] = 1;
for(int i = 1; i <= n; i ++) {
fact[i] = (LL)fact[i - 1] * i % mod;
infact[i] = qmi(fact[i], mod - 2);
}
}
int C(int a, int b) {
if (b > a) return 0;
return (LL)fact[a] * infact[a - b] % mod * infact[b] % mod;
}
int v[1010], a[1010];
void solve() {
int n, k; cin >> n >> k;
for (int i = 1; i <= n; ++i) {
v[i] = 0;
}
for (int i = 1; i <= n; ++i) {
cin >> a[i];
}
sort(a + 1,a + 1 +n);
if (n == k) {
cout << 1 << endl;
return ;
}
LL tot = 0;
LL c1 = 0;
LL c2 = 0;
for (int i = n - k + 1; i <= n; ++i) {
if (a[i] == a[n - k + 1]) ++ c2;
}
for (int i = 1; i <= n - k; ++i) {
if (a[i] == a[n - k + 1])++c1;
}
cout << (C(c1 + c2, c2)) << endl;
}
signed main() {
ios::sync_with_stdio(0);
cin.tie(0);
init(1000);
int T = 1;
cin >> T;
while (T--) {
solve();
}
}
F. Unusual Matrix
题意
给你两个矩阵a,b问你在规则下将a能不能变成b。
规则是每次只能对一行或者一列异或1
思路
如果a,b的相同位置不相同则这个位置需要异或奇数次，如果相同需要异或偶数次，偶数次相当于不异或。
我们先将一行或者一列变为与b相同，然后确定了这一行或一列的异或次数且不能再去改变，然后去枚举其他的情况。
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 2e5 + 10;
int a[1005][1005], b[1005][1005];
void solve() {
int n; scanf("%d", &n);
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; ++j) {
scanf("%1d", &a[i][j]);
}
}
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; ++j) {
scanf("%1d", &b[i][j]);
}
}
for (int i = 1; i <= n; ++i) {
if (a[i][1] != b[i][1]) {
for (int j = 1 ; j <= n; ++j) {
a[i][j] ^= 1;
}
}
}
bool ok = true;
for (int i = 2; i <= n; ++i) {
if (a[1][i] != b[1][i]) {
for (int j = 1; j <= n; ++j) {
if (a[j][i] == b[j][i]) ok = false;
}
} else {
for (int j = 1; j <= n; ++j) {
if (a[j][i] != b[j][i]) ok = false;
}
}
}
if (ok) cout << "YES
";
else cout << "NO
";
}
signed main() {
ios::sync_with_stdio(0);
cin.tie(0);
int T = 1;
scanf("%d", &T);
while (T--) {
solve();
}
}
G. Strange Beauty
题意
给你n个数，问你最少要删除多少个才能使这个数组漂亮
这个数组漂亮：任意两个数都可以作为除数或被除数
思路
挑选倍数,(dp[i])表示最大值为i的情况下这个数列有多少个数
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 2e5 + 10;
int dp[N], num[N];
void solve() {
int n; cin >> n;
memset(num, 0, sizeof num);
memset(dp, 0, sizeof dp);
for (int i = 1; i <= n; ++i) {
int t; cin >> t;
++num[t];
}
int ans = -1;
for (int i = 1; i < N; ++i) {
dp[i] += num[i];
for (int j = 2 * i; j < N; j += i) {
dp[j] = max(dp[i], dp[j]);
}
ans = max(ans, dp[i]);
}
cout << n - ans << endl;
}
signed main() {
ios::sync_with_stdio(0);
cin.tie(0);
int T = 1;
cin >> T;
while (T--) {
solve();
}
}