zafu个人赛 2022.2.16 A B C D E F G总结
A - Three DiceAtCoder - abc202_a
- 参考代码:
#include<bits/stdc++.h>
using namespace std;
#define mp make_pair
#define int long long
#define re register int
#define pb emplace_back
#define lowbit(x) (x&-x)
#define fer(i,a,b) for(re i = a ; i <= b ; i ++)
#define der(i,a,b) for(re i = a ; i >= b ; i --)
#define snow ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
int gcd(int a,int b){return b?gcd(b,a%b):a;}
int lcm(int a,int b){return a*b/gcd(a,b);}
typedef pair<int,int>PII;
typedef pair<int,string>PIS;
signed main(){
int a,b,c;
cin>>a>>b>>c;
int res=21;
cout<<res-a-b-c<<endl;
return 0;
}
B. Lineland Mail(900分)CodeForces - 567A
Tip : greedy、implementation
- 参考代码:
#include<bits/stdc++.h>
using namespace std;
#define mp make_pair
#define int long long
#define re register int
#define pb emplace_back
#define lowbit(x) (x&-x)
#define fer(i,a,b) for(re i = a ; i <= b ; i ++)
#define der(i,a,b) for(re i = a ; i >= b ; i --)
#define snow ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
int gcd(int a,int b){return b?gcd(b,a%b):a;}
int lcm(int a,int b){return a*b/gcd(a,b);}
typedef pair<int,int>PII;
typedef pair<int,string>PIS;
const int N=1e5+10;
int a[N];
signed main(){
int n;
cin>>n;
for(int i=1;i<=n;i++)cin>>a[i];
int l=a[1];
int r=a[n];
for(int i=1;i<=n;i++){
if(i==1)cout<<a[2]-a[1]<<' ';
else if(i==n)cout<<abs(a[i-1]-a[i])<<' ';
else{
cout<<(min(a[i]-a[i-1],a[i+1]-a[i]))<<' ';
}
cout<<max(abs(l-a[i]),r-a[i])<<endl;
}
return 0;
}
B. Code obfuscation(1100分)CodeForces - 765B
Tip : implementation 、string
- 参考代码:
#include<bits/stdc++.h>
using namespace std;
#define mp make_pair
#define int long long
#define re register int
#define pb emplace_back
#define lowbit(x) (x&-x)
#define fer(i,a,b) for(re i = a ; i <= b ; i ++)
#define der(i,a,b) for(re i = a ; i >= b ; i --)
#define snow ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
int gcd(int a,int b){return b?gcd(b,a%b):a;}
int lcm(int a,int b){return a*b/gcd(a,b);}
typedef pair<int,int>PII;
typedef pair<int,string>PIS;
signed main(){
bool ok=true;
char st='a';
string s;
cin>>s;
for(int i=0;i<s.size();i++){
if(s[i]==st){
st++;
}
else if(s[i]<st){
continue;
}
else if(s[i]>st){
ok=false;
break;
}
}
if(ok)cout<<"YES"<<endl;
else cout<<"NO"<<endl;
return 0;
}
D. Restoring Painting(1400分)CodeForces - 675B
Tip : constructive algorithms、math
- 参考代码:
#include<bits/stdc++.h>
using namespace std;
#define mp make_pair
#define int long long
#define re register int
#define pb emplace_back
#define lowbit(x) (x&-x)
#define fer(i,a,b) for(re i = a ; i <= b ; i ++)
#define der(i,a,b) for(re i = a ; i >= b ; i --)
#define snow ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
int gcd(int a,int b){return b?gcd(b,a%b):a;}
int lcm(int a,int b){return a*b/gcd(a,b);}
typedef pair<int,int>PII;
typedef pair<int,string>PIS;
signed main(){
int n,a,b,c,d;
cin>>n>>a>>b>>c>>d;
int res[5];//统计四个角两边和
res[0]=a+b;
res[1]=b+d;
res[2]=c+d;
res[3]=c+a;
sort(res,res+4);
int x=res[3]-res[0];//最小的减到0时,max跟着减最终的值。
x++;
if(x>n){
cout<<0<<endl;
}
else{
int sum=n-x+1;//四个角约束条件下带来的方案数。
cout<<sum*n<<endl;
}
return 0;
}
E. Hard Process(1600分)CodeForces - 660C
Tip:two pointers
- 参考代码:
#include<bits/stdc++.h>
using namespace std;
#define mp make_pair
#define int long long
#define re register int
#define pb emplace_back
#define lowbit(x) (x&-x)
#define fer(i,a,b) for(re i = a ; i <= b ; i ++)
#define der(i,a,b) for(re i = a ; i >= b ; i --)
#define snow ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
int gcd(int a,int b){return b?gcd(b,a%b):a;}
int lcm(int a,int b){return a*b/gcd(a,b);}
typedef pair<int,int>PII;
typedef pair<int,string>PIS;
const int N=3e5+10;
int a[N];
signed main(){
int n,k;
cin>>n>>k;
int ma=0;//统计最大1的数量
int ll,rr;//用来记最大连续的左右边界
for(int i=1;i<=n;i++){
cin>>a[i];
}
if(k==0){
int res=0;
for(int i=1;i<=n;i++){
if(a[i]==1){res++;}
else if(a[i]==0){
res=0;
}
ma=max(ma,res);
}
cout<<ma<<endl;
for(int i=1;i<=n;i++)cout<<a[i]<<' ';
cout<<endl;
return 0;
}
int l,r;
l=r=0;
int res=0;
for(int i=1;i<=n;i++){
if(a[i]==1){//如果是1直接右移即可,不需要消耗魔法次数
r++;
}
else if(a[i]==0){
if(res<k){//如果魔法次数还有右移,消耗1个魔法次数
res++;
r++;
}
else if(res>=k){//如果魔法次数不够用了,左边界右移。
while(l<r){
if(a[l+1]==0){
res--;
l++;
}
else if(a[l+1]==1){
l++;
}
if(res<k)break;
}
if(res<k){//如果通过左移左边界得到魔法次数那么可以消耗1魔法次数将当前的0变成1
res++;
r++;
}
}
}
if(r-l>ma){//比较更新
ma=r-l;
ll=l,rr=r;
}
}
cout<<ma<<endl;
for(int i=1;i<=ll;i++)cout<<a[i]<<' ';
for(int i=ll+1;i<=rr;i++)cout<<1<<' ';//区间内都是1,其他老样子输出。
for(int i=rr+1;i<=n;i++)cout<<a[i]<<' ';
cout<<endl;
return 0;
}
F. Restore Permutation(1900分)CodeForces - 1208D
Tip:Binary Indexed Tree、segment tree
- 参考代码(树状数组版本):
#include<bits/stdc++.h>
using namespace std;
#define mp make_pair
#define int long long
#define re register int
#define pb emplace_back
#define lowbit(x) (x&-x)
#define fer(i,a,b) for(re i = a ; i <= b ; i ++)
#define der(i,a,b) for(re i = a ; i >= b ; i --)
#define snow ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
int gcd(int a,int b){return b?gcd(b,a%b):a;}
int lcm(int a,int b){return a*b/gcd(a,b);}
typedef pair<int,int>PII;
typedef pair<int,string>PIS;
const int N=2e5+10;
int tr[N];
int a[N];
int n;
void add(int a,int x){
for(int i=a;i<=n;i+=lowbit(i)){
tr[i]+=x;
}
}
int ask(int x){
int res=0;
for(int i=x;i;i-=lowbit(i)){
res+=tr[i];
}
return res;
}
signed main(){
cin>>n;
vector<int>S;
for(int i=1;i<=n;i++){
cin>>a[i];
add(i,i);//所有数全部存到树中
}
bool ok1=true;//判断1有没有选过
for(int i=n;i>=1;i--){
int l=1;int r=n;
while(l<r){
int mid=l+r+1>>1;
int x=ask(mid);
if(x>a[i])r=mid-1;
else l=mid;
}
if(a[i]==0&&l==1){
if(ok1){
ok1=false;
add(1,-1);
S.push_back(1);
}
else{
add(2,-2);
S.push_back(2);
}
}
else{
l=l+1;
add(l,-l);
S.push_back(l);
}
}
reverse(S.begin(),S.end());//S[0]是尾巴的数,所以得翻一下。
for(auto x:S)cout<<x<<' ';
cout<<endl;
return 0;
}
G. Ryouko’s Memory Note(1800)CodeForces - 433C
Tip:math 、sorting
- 参考代码:
#include<bits/stdc++.h>
using namespace std;
#define mp make_pair
#define int long long
#define re register int
#define pb emplace_back
#define lowbit(x) (x&-x)
#define fer(i,a,b) for(re i = a ; i <= b ; i ++)
#define der(i,a,b) for(re i = a ; i >= b ; i --)
#define snow ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
int gcd(int a,int b){return b?gcd(b,a%b):a;}
int lcm(int a,int b){return a*b/gcd(a,b);}
typedef pair<int,int>PII;
typedef pair<int,string>PIS;
const int N=1e5+10;
int a[N];
vector<int>b[N];//用来存数
signed main(){
int n,m;
cin>>n>>m;
for(int i=1;i<=m;i++)cin>>a[i];
for(int i=1;i<=m;i++){
if(i!=m){
if(a[i]!=a[i+1])
b[a[i]].push_back(a[i+1]);
}
if(i!=1){
if(a[i]!=a[i-1])
b[a[i]].push_back(a[i-1]);
}
}
int sum=0;
for(int i=2;i<=m;i++){
sum+=abs(a[i]-a[i-1]);
}
int ma=0;//存最大节约
for(int i=1;i<=n;i++){
if(b[i].size()==0)continue;//如果没有存的数跳过,不会对答案有贡献。
sort(b[i].begin(),b[i].end());
int y=b[i][(b[i].size()-1)>>1];//中位数
int before,after;
before=after=0;
for(auto x:b[i]){
before+=abs(i-x);
after+=abs(y-x);
}
ma=max(ma,before-after);//更新最大节约
}
cout<<sum-ma<<endl;//最终的最小能量花费
return 0;
}
切了 1900 1900 1900的题还是蛮开心的,虽然 c f cf cf比赛时候可能时间不够,算是一个小进步。