Codeforces Round 971 (Div. 4) A-G1
A
b-a
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
mt19937 rnd(time(0));
#define int long long
typedef tuple<int,int,int> tp;
#define x first
#define y second
typedef pair<int,int> pii;
typedef pair<double,double> pdd;
constexpr int N=200010,mod=1e9+7,inf=1e18;
constexpr double pi=3.1415926535897932384626,eps=1e-5;
const ll P=rnd()%mod;
#define all(a) a.begin(),a.end()
#define get_count(x) __builtin_popcount(x)
#define fors(i,a,b) for(int i=a;i<=b;i++)
#define forr(i,a,b) for(int i=b;i>=a;i--)
#define pb(x) push_back(x)
int dx[]={0,1,0,-1,1,1,-1,-1,0};
int dy[]={1,0,-1,1,1,-1,-1,1,0};
int random(int l,int r){
return rand()%r+l;
}
signed main(){
cin.tie(nullptr)->sync_with_stdio(false);
int __=1;
cin >> __;
while(__--){
int a,b;
cin >> a >> b;
int ans=inf;
for(int c=a;c<=b;c++) ans=min(ans,c-a+b-c);
cout << ans << '\n';
}
system("color 04");
return 0;
}
B
模拟，从最后向前遍历找每行的第一个\(#\)记录列数，然后输出
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
mt19937 rnd(time(0));
#define int long long
typedef tuple<int,int,int> tp;
#define x first
#define y second
typedef pair<int,int> pii;
typedef pair<double,double> pdd;
constexpr int N=200010,mod=1e9+7,inf=1e18;
constexpr double pi=3.1415926535897932384626,eps=1e-5;
const ll P=rnd()%mod;
#define all(a) a.begin(),a.end()
#define get_count(x) __builtin_popcount(x)
#define fors(i,a,b) for(int i=a;i<=b;i++)
#define forr(i,a,b) for(int i=b;i>=a;i--)
#define pb(x) push_back(x)
int dx[]={0,1,0,-1,1,1,-1,-1,0};
int dy[]={1,0,-1,1,1,-1,-1,1,0};
int random(int l,int r){
return rand()%r+l;
}
signed main(){
cin.tie(nullptr)->sync_with_stdio(false);
int __=1;
cin >> __;
while(__--){
int n,m=4;
cin >> n;
vector<vector<char>> g(n+1,vector<char>(m+1));
fors(i,1,n){
fors(j,1,m){
cin >> g[i][j];
}
}
vector<int> ans;
forr(i,1,n){
fors(j,1,m){
if(g[i][j]=='#'){
ans.pb(j);
}
}
}
fors(i,0,ans.size()-1) cout << ans[i] << " \n"[i==ans.size()-1];
}
system("color 04");
return 0;
}
C
如果最后一次x没到，y到了，仅需要一次 \((t-1)*2+1\)如果最后一次x到了，y没到，需要俩次 \(t*2\),t为到x或者y的步数最大值
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
mt19937 rnd(time(0));
#define int long long
typedef tuple<int,int,int> tp;
#define x first
#define y second
typedef pair<int,int> pii;
typedef pair<double,double> pdd;
constexpr int N=200010,mod=1e9+7,inf=1e18;
constexpr double pi=3.1415926535897932384626,eps=1e-5;
const ll P=rnd()%mod;
#define all(a) a.begin(),a.end()
#define get_count(x) __builtin_popcount(x)
#define fors(i,a,b) for(int i=a;i<=b;i++)
#define forr(i,a,b) for(int i=b;i>=a;i--)
#define pb(x) push_back(x)
int dx[]={0,1,0,-1,1,1,-1,-1,0};
int dy[]={1,0,-1,1,1,-1,-1,1,0};
int random(int l,int r){
return rand()%r+l;
}
signed main(){
cin.tie(nullptr)->sync_with_stdio(false);
int __=1;
cin >> __;
while(__--){
int x,y,k;
cin >> x >> y >> k;
int a=(x+k-1)/k,b=(y+k-1)/k;
int t=max(a,b);
if(a>b){
cout << (t-1)*2+1 << '\n';
}
else cout << t*2 << '\n';
}
system("color 04");
return 0;
}
D
\(0<=x<=n,0<=y<=1\),可以枚举x的位置，组成直角三角形的点可以有，俩个点在同一列，其余各点都能组成直角三角形，一个点上面或者下面左右俩侧皆有点也可以组成直角三角形。
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
mt19937 rnd(time(0));
#define int long long
typedef tuple<int,int,int> tp;
#define x first
#define y second
typedef pair<int,int> pii;
typedef pair<double,double> pdd;
constexpr int N=200010,mod=1e9+7,inf=1e18;
constexpr double pi=3.1415926535897932384626,eps=1e-5;
const ll P=rnd()%mod;
#define all(a) a.begin(),a.end()
#define get_count(x) __builtin_popcount(x)
#define fors(i,a,b) for(int i=a;i<=b;i++)
#define forr(i,a,b) for(int i=b;i>=a;i--)
#define pb(x) push_back(x)
int dx[]={0,1,0,-1,1,1,-1,-1,0};
int dy[]={1,0,-1,1,1,-1,-1,1,0};
int random(int l,int r){
return rand()%r+l;
}
signed main(){
cin.tie(nullptr)->sync_with_stdio(false);
int __=1;
cin >> __;
while(__--){
int n;
cin >> n;
vector<vector<int>> a(2,vector<int>(n+1));
fors(i,1,n){
int x,y;
cin >> x >> y;
a[y][x]=1;
}
int ans=0;
fors(i,0,n){
if(a[0][i]&&a[1][i]){
ans+=n-2;
}
if(i>=1&&i<=n-1){
if(a[0][i]&&a[1][i+1]&&a[1][i-1]) ans++;
if(a[1][i]&&a[0][i+1]&&a[0][i-1]) ans++;
}
}
cout << ans << '\n';
}
system("color 04");
return 0;
}
E
找到一个位置i分割俩侧，使得俩侧差绝对值最小，有俩种做法，一种是二元一次方程找根，一种是凸凹函数用三分。
二元一次方程
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
mt19937 rnd(time(0));
#define int long long
typedef tuple<int,int,int> tp;
#define x first
#define y second
typedef pair<int,int> pii;
typedef pair<double,double> pdd;
constexpr int N=200010,mod=1e9+7,inf=1e18;
constexpr double pi=3.1415926535897932384626,eps=1e-5;
const ll P=rnd()%mod;
#define all(a) a.begin(),a.end()
#define get_count(x) __builtin_popcount(x)
#define fors(i,a,b) for(int i=a;i<=b;i++)
#define forr(i,a,b) for(int i=b;i>=a;i--)
#define pb(x) push_back(x)
int dx[]={0,1,0,-1,1,1,-1,-1,0};
int dy[]={1,0,-1,1,1,-1,-1,1,0};
int random(int l,int r){
return rand()%r+l;
}
signed main(){
cin.tie(nullptr)->sync_with_stdio(false);
int __=1;
cin >> __;
while(__--){
int n,k;
cin >> n >> k;
int a=2,b=2*(2*k-1),c=-n*(2*k+n-1);
// cout << a << ' ' << b << ' ' << c << '\n';
vector<int> r(2);
auto get=[&](int a,int b,int c)->vector<int>{
double t=(double)b*b-(double)4ll*a*c;
vector<int> ans;
if(t<0) return ans;
int ts=sqrt(t);
// cout << ts << '\n';
ans.pb((ts-b)/(2*a));
ans.pb((-b-ts)/(2*a));
return ans;
};
r=get(a,b,c);
auto sum=[&](int l,int r)->int{
return (r-l+1)*(l+r)/2;
};
int ans=sum(k,k+n-1);
// cout << r[0] << ' ' << r[1] << '\n';
for(int i=0;i<2;i++){
for(int j=r[i]-10;j<=r[i]+10;j++){
if(j<1||j>n) continue;
int x1=sum(k,k+j-1);
int x2=sum(k+j,k+n-1);
ans=min(ans,abs(x1-x2));
}
}
cout << ans << '\n';
}
system("color 04");
return 0;
}
三分
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
mt19937 rnd(time(0));
#define int long long
typedef tuple<int,int,int> tp;
#define x first
#define y second
typedef pair<int,int> pii;
typedef pair<double,double> pdd;
constexpr int N=200010,mod=1e9+7,inf=1e18;
constexpr double pi=3.1415926535897932384626,eps=1e-5;
const ll P=rnd()%mod;
#define all(a) a.begin(),a.end()
#define get_count(x) __builtin_popcount(x)
#define fors(i,a,b) for(int i=a;i<=b;i++)
#define forr(i,a,b) for(int i=b;i>=a;i--)
#define pb(x) push_back(x)
int dx[]={0,1,0,-1,1,1,-1,-1,0};
int dy[]={1,0,-1,1,1,-1,-1,1,0};
int random(int l,int r){
return rand()%r+l;
}
signed main(){
cin.tie(nullptr)->sync_with_stdio(false);
int __=1;
cin >> __;
while(__--){
int n,k;
cin >> n >> k;
auto get=[&](int l,int r)->int{
return (r-l+1)*(l+r)/2;
};
auto check=[&](int i)->int{
return abs(get(k,k+i-1)-get(k+i,k+n-1));
};
int l=1,r=n;
while(r-l>10){
int m1=(2*l+r)/3;
int m2=(l+2*r)/3;
if(check(m1)<check(m2)) r=m2;
else l=m1;
}
int ans=check(l);
fors(i,l,r) ans=min(ans,check(i));
cout << ans << '\n';
}
system("color 04");
return 0;
}
F
一个小技巧破环成链前缀和，可以分为三部分来计算，完整的数组加上左右俩侧，sum+[l,L]+[R,r],算出[l,L]和[R,r]属于第几次偏移量进行前缀和即可
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
mt19937 rnd(time(0));
#define int long long
typedef tuple<int,int,int> tp;
#define x first
#define y second
typedef pair<int,int> pii;
typedef pair<double,double> pdd;
constexpr int N=200010,mod=1e9+7,inf=1e18;
constexpr double pi=3.1415926535897932384626,eps=1e-5;
const ll P=rnd()%mod;
#define all(a) a.begin(),a.end()
#define get_count(x) __builtin_popcount(x)
#define fors(i,a,b) for(int i=a;i<=b;i++)
#define forr(i,a,b) for(int i=b;i>=a;i--)
#define pb(x) push_back(x)
int dx[]={0,1,0,-1,1,1,-1,-1,0};
int dy[]={1,0,-1,1,1,-1,-1,1,0};
int random(int l,int r){
return rand()%r+l;
}
signed main(){
cin.tie(nullptr)->sync_with_stdio(false);
int __=1;
cin >> __;
while(__--){
int n,q,sum=0;
cin >> n >> q;
// 开一个俩倍长度的前缀和
vector<int> a(2*n+1),pre(2*n+1);
fors(i,1,n) cin >> a[i],sum+=a[i],a[i+n]=a[i];
fors(i,1,2*n) pre[i]=pre[i-1]+a[i];
while(q--){
int l,r;
cin >> l >> r;
int L=(l%n==0?l:l+(n-l%n));
L=min(L,r);
int R=(r%n==0?r:r-r%n+1);
R=max(R,l);
// cout << l << ' ' << L << ' ' << R << ' ' << r << '\n';
int len=r-l+1;
if(l%n==1) L=-1;
if(r%n==0) R=-1;
if(L!=-1){
len-=(L-l+1);
}
if(R!=-1){
len-=(r-R+1);
}
len=max(0ll,len);
int k=len/n;
// cout << "len=" << len << '\n';
// cout << l << ' ' << L << ' ' << R << ' ' << r << '\n';
int ans=k*sum;
// cout << "First:ans " << ans << '\n';
int i=(l+n-1)/n-1,j=(r+n-1)/n-1;
int res=0;
int pl=(l%n==0?n:l%n);
int pr=(r%n==0?n:r%n);
if(L!=-1){
int pl=(l%n==0?n:l%n);
int pr=(L%n==0?n:L%n);
// cout << "L:" << pl << ' ' << pr << '\n';
res+=pre[pr+i]-pre[pl+i-1];
}
// cout << "L " << res << '\n';
if(R!=-1){
int pl=(R%n==0?n:R%n);
int pr=(r%n==0?n:r%n);
// cout << "R:" << pl << ' ' << pr << '\n';
res+=pre[pr+j]-pre[pl+j-1];
}
// cout << "R " << res << '\n';
ans+=res;
if(l==R&&L==r) ans-=res/2;
cout << ans << '\n';
}
}
system("color 04");
return 0;
}
G1
给一个定长的子数组[l,r]求使得[l,r]为连续子数组的最小操作数。有一个小技巧，让\(ai=ai-i\) 就可以转换成区间众数问题，可以使用滑动窗口或者莫队来解决。
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
mt19937 rnd(time(0));
#define int long long
typedef tuple<int,int,int> tp;
#define x first
#define y second
typedef pair<int,int> pii;
typedef pair<double,double> pdd;
constexpr int N=200010,mod=1e9+7,inf=1e18;
constexpr double pi=3.1415926535897932384626,eps=1e-5;
const ll P=rnd()%mod;
#define all(a) a.begin(),a.end()
#define get_count(x) __builtin_popcount(x)
#define fors(i,a,b) for(int i=a;i<=b;i++)
#define forr(i,a,b) for(int i=b;i>=a;i--)
#define pb(x) push_back(x)
int dx[]={0,1,0,-1,1,1,-1,-1,0};
int dy[]={1,0,-1,1,1,-1,-1,1,0};
int random(int l,int r){
return rand()%r+l;
}
struct node{
int l,r,id;
}e[N];
signed main(){
cin.tie(nullptr)->sync_with_stdio(false);
int __=1;
cin >> __;
while(__--){
int n,k,q;
cin >> n >> k >> q;
vector<int> a(n+1),alls;
map<int,int> cnt,p;
fors(i,1,n) cin >> a[i],a[i]-=i,alls.pb(a[i]);
sort(all(alls));
alls.erase(unique(all(alls)),alls.end());
fors(i,1,n) a[i]=lower_bound(all(alls),a[i])-alls.begin()+1;
fors(i,1,q){
int l,r;
cin >> l >> r;
e[i]={l,r,i};
}
int len=sqrt(n),res=0;
auto get=[&](int i)->int{
return i/len;
};
auto cmp=[&](const node& a,const node& b)->bool{
int i=get(a.l),j=get(b.l);
if(i!=j) return i<j;
if(i&1) return a.r<b.r;
return a.r>b.r;
};
auto add=[&](int x)->void{
cnt[++p[x]]++;
res=max(res,p[x]);
};
auto del=[&](int x)->void{
if(cnt[p[x]]==1&&p[x]==res) res--;
cnt[p[x]--]--;
};
sort(e+1,e+q+1,cmp);
vector<int> ans(q+1);
for(int t=1,i=1,j=0;t<=q;t++){
auto [l,r,id]=e[t];
while(i<l) del(a[i++]);
while(i>l) add(a[--i]);
while(j<r) add(a[++j]);
while(j>r) del(a[j--]);
ans[id]=res;
}
fors(i,1,q) cout << max(0ll,k-ans[i]) << '\n';
// 区间众数
}
system("color 04");
return 0;
}
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