Codeforces 708 A-E1题解
A. Meximization
这道题问给一些数，如何让前缀的mex之和最大，那么首先，我们要抬mex的话肯定是要把前面都铺垫完的，所以在i位置确定的时候，i + 1自然是越大越好，可以证明i + 1的位置放大至少优于放小，然后其他数搞一搞就行
AC Code
#include <bits/stdc++.h>
using namespace std;
constexpr int limit = (2000000 + 5);//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-9
#define FASTIO ios::sync_with_stdio(false);cin.tie(0),cout.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a, b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].nxt)
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\dabiao.txt", "wt", stdout)
typedef long long ll;
typedef unsigned long long ull;
char buf[1 << 23], *p1 = buf, *p2 = buf, obuf[1 << 23], *O = obuf;
inline ll read() {
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
ll sign = 1, x = 0;
char s = getchar();
while (s > '9' || s < '0') {
if (s == '-')sign = -1;
s = getchar();
}
while (s >= '0' && s <= '9') {
x = (x << 3) + (x << 1) + s - '0';
s = getchar();
}
return x * sign;
#undef getchar
}//快读
void print(ll x) {
if (x / 10) print(x / 10);
*O++ = x % 10 + '0';
}
void write(ll x, char c = 't') {
if (x < 0)putchar('-'), x = -x;
print(x);
if (!isalpha(c))*O++ = c;
fwrite(obuf, O - obuf, 1, stdout);
O = obuf;
}
constexpr ll mod = 1e9 + 7;
ll quick_pow(ll base, ll expo){
ll ans = 1;
while(expo){
if(expo & 1) ans = ans * base % mod;
base = base * base % mod;
expo >>= 1;
}
return ans;
}
ll inv(ll x){
return quick_pow(x, mod - 2);
}
ll C(ll n, ll m){
ll ans = 1;
for(ll i = 1; i <= m; i++){
ans = ans * (n - i + 1) % mod;
ans = ans * inv(i) % mod;
}
return ans;
}
int n,m;
int a[limit];
void solve(){
cin>>n;
map<int, int>mp;
rep(i,1,n) cin>>a[i];
int it = 0;
sort(a + 1, a + 1 + n);
vector<int>v;
rep(i,1,n){
if(!mp[a[i]]) mp[a[i]]++, cout<<a[i]<<" ";
else v.push_back(a[i]);
}
for(auto x : v) cout<<x<<" ";
cout<<endl;
};
int32_t main() {
#ifdef LOCAL
FOPEN;
// FOUT;
#endif
FASTIO
int kase;
cin>>kase;
while (kase--)
invoke(solve);
cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << "s\n";
return 0;
}
B. M-arrays
这道题问的是给一个数列和一个数m，要求选若干组，每组相邻的元素之和可以被m整除，问最少要分多少组
首先我们可以只关心m的余数，然后我们枚举余数i从1-m -1，然后我们可以知道i的数量和m - i的数量，这两个可以凑成符合要求的数字
然后假如有x个i，y个m - i，x < y， 那么x个i最多能带走x + 1个y，然后剩下的y，如果有的话，单独加出来就行
最后要注意特殊处理一下0，显然无论多少个0加起来都是符合要求的，所以 + 1
AC Code
#include <bits/stdc++.h>
using namespace std;
constexpr int limit = (2000000 + 5);//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-9
#define FASTIO ios::sync_with_stdio(false);cin.tie(0),cout.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a, b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].nxt)
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\dabiao.txt", "wt", stdout)
typedef long long ll;
typedef unsigned long long ull;
char buf[1 << 23], *p1 = buf, *p2 = buf, obuf[1 << 23], *O = obuf;
inline ll read() {
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
ll sign = 1, x = 0;
char s = getchar();
while (s > '9' || s < '0') {
if (s == '-')sign = -1;
s = getchar();
}
while (s >= '0' && s <= '9') {
x = (x << 3) + (x << 1) + s - '0';
s = getchar();
}
return x * sign;
#undef getchar
}//快读
void print(ll x) {
if (x / 10) print(x / 10);
*O++ = x % 10 + '0';
}
void write(ll x, char c = 't') {
if (x < 0)putchar('-'), x = -x;
print(x);
if (!isalpha(c))*O++ = c;
fwrite(obuf, O - obuf, 1, stdout);
O = obuf;
}
constexpr ll mod = 1e9 + 7;
ll quick_pow(ll base, ll expo){
ll ans = 1;
while(expo){
if(expo & 1) ans = ans * base % mod;
base = base * base % mod;
expo >>= 1;
}
return ans;
}
ll inv(ll x){
return quick_pow(x, mod - 2);
}
ll C(ll n, ll m){
ll ans = 1;
for(ll i = 1; i <= m; i++){
ans = ans * (n - i + 1) % mod;
ans = ans * inv(i) % mod;
}
return ans;
}
int n,m;
int a[limit];
void solve(){
cin>>n>>m;
map<int, int>mp;
rep(i,1,n)cin >> a[i], a[i] %= m, mp[a[i]]++;
int ans = 0;
rep(i, 1, m / 2){
int fst = mp[i];
int scd = mp[m - i];
if(fst > scd) swap(fst, scd);
if(!fst){
ans += scd;
}else{
scd -= fst + 1;
++ans;
ans += max(0, scd);
}
}
// ans >>= 1;
if(mp[0])ans += 1;
cout << ans << endl;
};
int32_t main() {
#ifdef LOCAL
FOPEN;
// FOUT;
#endif
FASTIO
int kase;
cin>>kase;
while (kase--)
invoke(solve);
cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << "s\n";
return 0;
}
C. k-LCM (easy version) + (hard version)
这题是给一个数字n和一个数字k，要求把数字分解成k个正整数，使得
- a1 + ... + ak == n
- lcm(a) <= n / 2
然后easy version是固定k = 3.
那么这个题稍加观察我们发现，其实就是个奇偶题
对于奇数，我们可以分解成1 ， (n - 1) / 2, (n - 1) / 2
那么偶数有两种情况
- n / 2之后是奇数，那么我们可以从两个n / 2各拆一个1给第一个数字
- n / 2之后是偶数，那么我们可以让前两个数字相等，也就是n / 2 / 2(n / 2)， 然后再加一个n /2
没什么难度
然后这个hard version如果你写了easy version应该问题就不大，我们只需要先凑m - 3个1（因为1不会对lcm造成负面影响），然后n - m + 3就成功转化为了easy version
AC Code
#include <bits/stdc++.h>
using namespace std;
constexpr int limit = (2000000 + 5);//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-9
#define FASTIO ios::sync_with_stdio(false);cin.tie(0),cout.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a, b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].nxt)
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\dabiao.txt", "wt", stdout)
typedef long long ll;
typedef unsigned long long ull;
char buf[1 << 23], *p1 = buf, *p2 = buf, obuf[1 << 23], *O = obuf;
inline ll read() {
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
ll sign = 1, x = 0;
char s = getchar();
while (s > '9' || s < '0') {
if (s == '-')sign = -1;
s = getchar();
}
while (s >= '0' && s <= '9') {
x = (x << 3) + (x << 1) + s - '0';
s = getchar();
}
return x * sign;
#undef getchar
}//快读
void print(ll x) {
if (x / 10) print(x / 10);
*O++ = x % 10 + '0';
}
void write(ll x, char c = 't') {
if (x < 0)putchar('-'), x = -x;
print(x);
if (!isalpha(c))*O++ = c;
fwrite(obuf, O - obuf, 1, stdout);
O = obuf;
}
constexpr ll mod = 1e9 + 7;
ll quick_pow(ll base, ll expo){
ll ans = 1;
while(expo){
if(expo & 1) ans = ans * base % mod;
base = base * base % mod;
expo >>= 1;
}
return ans;
}
ll inv(ll x){
return quick_pow(x, mod - 2);
}
ll C(ll n, ll m){
ll ans = 1;
for(ll i = 1; i <= m; i++){
ans = ans * (n - i + 1) % mod;
ans = ans * inv(i) % mod;
}
return ans;
}
int n,m;
int a[limit];
void solve(){
cin>>n>>m;
if(n == m){
rep(i,1,n){
cout<<1<<" ";
}
cout<<endl;
return;
}
if(m > 3){
n -= m - 3;
rep(i,1,m - 3){
cout<<1<<" ";
}
}
if(n & 1){
cout<<1<<" ";
--n;
cout<<n / 2<<" "<<n / 2<<endl;
}else{
if(n % 4 == 0){
cout<<n / 4<<" ";
cout<<(n / 4)<<" "<<(n / 2)<<endl;
}else{
cout<<2<<" "<<(n / 2 - 1)<<" "<<(n / 2 - 1)<<endl;
}
}
};
int32_t main() {
#ifdef LOCAL
FOPEN;
// FOUT;
#endif
FASTIO
int kase;
cin>>kase;
while (kase--)
invoke(solve);
cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << "s\n";
return 0;
}
D. Genius
看了下，感觉题应该是属于dp，但是我没想到什么方法可以快速处理2 ^ i，太大了，所以看道E1容易点先去做了。等会儿补一下。
看了半天似乎不是很懂这个咋搞的有后效性的dp，还有为啥iq是个误导性条件，不懂
AC Code
#include <bits/stdc++.h>
using namespace std;
constexpr int limit = (200000 + 5);//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-9
#define FASTIO ios::sync_with_stdio(false);cin.tie(0),cout.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a, b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].nxt)
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\dabiao.txt", "wt", stdout)
typedef long long ll;
typedef unsigned long long ull;
char buf[1 << 23], *p1 = buf, *p2 = buf, obuf[1 << 23], *O = obuf;
inline ll read() {
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
ll sign = 1, x = 0;
char s = getchar();
while (s > '9' || s < '0') {
if (s == '-')sign = -1;
s = getchar();
}
while (s >= '0' && s <= '9') {
x = (x << 3) + (x << 1) + s - '0';
s = getchar();
}
return x * sign;
#undef getchar
}//快读
void print(ll x) {
if (x / 10) print(x / 10);
*O++ = x % 10 + '0';
}
void write(ll x, char c = 't') {
if (x < 0)putchar('-'), x = -x;
print(x);
if (!isalpha(c))*O++ = c;
fwrite(obuf, O - obuf, 1, stdout);
O = obuf;
}
constexpr ll mod = 1e9 + 7;
ll quick_pow(ll base, ll expo){
ll ans = 1;
while(expo){
if(expo & 1) ans = ans * base % mod;
base = base * base % mod;
expo >>= 1;
}
return ans;
}
ll inv(ll x){
return quick_pow(x, mod - 2);
}
ll C(ll n, ll m){
ll ans = 1;
for(ll i = 1; i <= m; i++){
ans = ans * (n - i + 1) % mod;
ans = ans * inv(i) % mod;
}
return ans;
}
int n,m;
int a[limit];
int b[limit];
int tag[limit];
ll dp[limit];
void solve(){
cin>>n;
rep(i,1,n){
cin>>tag[i];
dp[i] = 0;
}
rep(i,1,n){
cin>>a[i];
}
ll ans = 0;
rep(i,1,n){
per(j,1, i - 1){
if(tag[i] == tag[j])continue;
ll fst = dp[i];
ll snd = dp[j];
ll cost = 1ll * llabs(a[i] - a[j]);
dp[i] = max(dp[i], snd + cost);
dp[j] = max(dp[j], fst + cost);
ans = max(ans, dp[i]);
ans = max(ans, dp[j]);
}
}
cout<<ans<<endl;
};
int32_t main() {
#ifdef LOCAL
FOPEN;
// FOUT;
#endif
FASTIO
// get_prime();
int kase;
cin>>kase;
while (kase--)
invoke(solve);
cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << "s\n";
return 0;
}
E1.Square-Free Division (easy version)
这个题要把数组分成连续的段，使得段中任意两个元素乘积都不是完全平方数，问最少分段。
这个题肯定是贪心没得跑。
说到完全平方数，就是说所有的质因子数量都是偶数，然后我们分解一下，把奇数分子单列出来，然后排序后给凑成一个字符串，如果有相同的，那么说明已经出现的段内和当前的数字互补，能够形成完全平方数，需要分段。
然后string tle了，我然后试了试整数乘法，可以，但是还是会t。那么预处理这个值呢？1e7的还没处理完就tle了，我真的会谢。那就加个map存fact的结果吧，然后优化了一大圈，请出来了平板电视里的gp_hash_table才过，艰难。
AC Code
#include <bits/stdc++.h>
using namespace std;
constexpr int limit = (2000000 + 5);//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-9
#define FASTIO ios::sync_with_stdio(false);cin.tie(0),cout.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a, b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].nxt)
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\dabiao.txt", "wt", stdout)
typedef long long ll;
typedef unsigned long long ull;
char buf[1 << 23], *p1 = buf, *p2 = buf, obuf[1 << 23], *O = obuf;
inline ll read() {
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
ll sign = 1, x = 0;
char s = getchar();
while (s > '9' || s < '0') {
if (s == '-')sign = -1;
s = getchar();
}
while (s >= '0' && s <= '9') {
x = (x << 3) + (x << 1) + s - '0';
s = getchar();
}
return x * sign;
#undef getchar
}//快读
void print(ll x) {
if (x / 10) print(x / 10);
*O++ = x % 10 + '0';
}
void write(ll x, char c = 't') {
if (x < 0)putchar('-'), x = -x;
print(x);
if (!isalpha(c))*O++ = c;
fwrite(obuf, O - obuf, 1, stdout);
O = obuf;
}
constexpr ll mod = 1e9 + 7;
ll quick_pow(ll base, ll expo){
ll ans = 1;
while(expo){
if(expo & 1) ans = ans * base % mod;
base = base * base % mod;
expo >>= 1;
}
return ans;
}
ll inv(ll x){
return quick_pow(x, mod - 2);
}
ll C(ll n, ll m){
ll ans = 1;
for(ll i = 1; i <= m; i++){
ans = ans * (n - i + 1) % mod;
ans = ans * inv(i) % mod;
}
return ans;
}
int n,m;
int a[limit];
#include <bits/extc++.h>
using namespace __gnu_pbds;
gp_hash_table<ll, ll>dp;
ll fact(ll x){
ll xx = x;
if(dp.find(x) != dp.end()) return dp[x];
map<ll, int>mp;
for(ll i = 2; i * i <= x; i++){
ll& px = x;
if(px % i == 0){
while(px % i == 0){
px /= i;
mp[i]++;
}
}
}
if(x > 1) mp[x]++;
ll ans = 1;
for(auto && [k, v] : mp){
if(v & 1)ans *= k;
}
return dp[xx] = ans;
}
ll b[limit];
void solve(){
cin>>n>>m;
rep(i,1,n){
cin>>a[i];
b[i] = fact(a[i]);
}
unordered_set<ll>st;
int ans = 1;
rep(i,1,n){
if(st.count(b[i])){
++ans;
st.clear();
st.insert(b[i]);
}else{
st.insert(b[i]);
}
}
cout<<ans<<endl;
};
int32_t main() {
#ifdef LOCAL
FOPEN;
// FOUT;
#endif
FASTIO
// get_prime();
int kase;
cin>>kase;
while (kase--)
invoke(solve);
cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << "s\n";
return 0;
}
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