A - Three Strings
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
int t;
cin>>t;
while(t--){
string a,b,c;
cin>>a>>b>>c;
bool flag=true;
for(int i=0;i<a.size();i++){
if(c[i]!=a[i]&&c[i]!=b[i]){
flag=false;
break;
}
}
if(flag)
puts("YES");
else
puts("NO");
}
}
B - Motarack’s Birthday
By JLU_GLHF#, contest: Codeforces Round #619 (Div. 2), problem: (B) Motarack's Birthday, Accepted, #, Copy
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll INF=1e18;
const ll maxn=1e5+10;
ll a[maxn];
ll ABS(ll x){
return x<0?-x:x;
}
int main(){
ll t;
cin>>t;
while(t--){
ll n;
cin>>n;
ll MAX=0,MIN=INF;
for(ll i=1;i<=n;i++){
scanf("%lld",&a[i]);
}
for(ll i=1;i<=n;i++){
if(a[i]!=-1){
if((i>1&&a[i-1]==-1)||(i<n&&a[i+1]==-1))
MAX=max(MAX,a[i]),MIN=min(MIN,a[i]);
}
}
ll k;
if(MAX==0&&MIN==INF)
k=0;
else
k=MIN+(MAX-MIN+1)/2;
for(int i=1;i<=n;i++){
if(a[i]==-1)
a[i]=k;
}
ll ans=0;
for(int i=2;i<=n;i++){
ans=max(ABS(a[i]-a[i-1]),ans);
}
cout<<ans<<' '<<k<<endl;
}
}
C - Ayoub’s function
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
ll t;
cin>>t;
while(t--){
ll n,m;
cin>>n>>m;
ll SIZE=(n+1)/(m+1)-1;
ll a=(m+1)*(SIZE+2)-n-1;
ll b=m+1-a;
ll ans=n*(n+1)/2-a*SIZE*(SIZE+1)/2-b*(SIZE+1)*(SIZE+2)/2;
cout<<ans<<endl;
}
}
D - Time to Run
#include<bits/stdc++.h>
using namespace std;
vector<pair<string ,int>>vec,ans;
int main(){
int n,m,k;
cin>>n>>m>>k;
for(int i=1;i<n;i++){
if(m-1){
vec.push_back(make_pair("R",m-1));
vec.push_back(make_pair("L",m-1));
}
vec.push_back(make_pair("D",1));
}
if(m-1){
vec.push_back(make_pair("R",m-1));
}
for(int i=0;i<m-1;i++){
if(n-1){
vec.push_back(make_pair("U",n-1));
vec.push_back(make_pair("D",n-1));
}
vec.push_back(make_pair("L",1));
}
if(n-1){
vec.push_back(make_pair("U",n-1));
}
for(int i=0;i<vec.size();i++){
if(k>=vec[i].second){
k-=vec[i].second;
ans.push_back(vec[i]);
}
else{
if(k){
ans.push_back(make_pair(vec[i].first,k));
k=0;
}
break;
}
}
if(k){
puts("NO");
}
else{
puts("YES");
cout<<ans.size()<<endl;
for(int i=0;i<ans.size();i++){
cout<<ans[i].second<<' '<<ans[i].first<<endl;
}
}
}
E - Nanosoft
很好的一个板子二维st板子，查找矩阵中的最大最小值，预处理O(nmlogn),查询O(1)。
#include<bits/stdc++.h>
using namespace std;
const int maxn = 505;
const int maxm = 10;
string s[maxn];
int R[4][maxn][maxn];
int n,m,q;
int val[maxn][maxn],f[maxm][maxm][maxn][maxn],lg[maxn];;
void Build_2D_Sparse_Table(int n, int m){
int i, j, k1, k2;
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
f[0][0][i+1][j+1]=val[i][j];
}
}
for(i = 2; i < maxn; i++)
lg[i] = 1 + lg[i/2];
for(i = 1; i <= n; i++)
for(k2 = 1; (1 << k2) <= m; k2++)
for(j = 1; j <= m - (1 << k2) + 1; j++)
f[0][k2][i][j] = max(f[0][k2 - 1][i][j], f[0][k2 - 1][i][j + (1 << (k2 - 1))]);
for(k1 = 1; (1 << k1) <= n; k1++)
for(i = 1; i <= n - (1 << k1) + 1; i++)
for(k2 = 0; (1 << k2) <= m; k2++)
for(j = 1; j <= m - (1 << k2) + 1; j++)
f[k1][k2][i][j] = max(f[k1 - 1][k2][i][j], f[k1 - 1][k2][i + (1 << (k1 - 1))][j]);
}
int Query(int x1, int y1, int x2, int y2){
int k1 = lg[x2 - x1 + 1], k2 = lg[y2 - y1 + 1];
x2 = x2 - (1 << k1) + 1;
y2 = y2 - (1 << k2) + 1;
return max(max(f[k1][k2][x1][y1],f[k1][k2][x1][y2]),max(f[k1][k2][x2][y1],f[k1][k2][x2][y2]));
}
void color(char c,int fx,int fy,int k){
int stx=0,sty=0;
if(fx==-1)stx=n-1;
if(fy==-1)sty=m-1;
while(stx<n&&stx>=0){
while(sty<m&&sty>=0){
if(s[stx][sty]==c){
R[k][stx][sty]=1;
if(stx-fx>=0&&stx-fx<n&&sty-fy>=0&&sty-fy<m){
if(s[stx-fx][sty]==c&&s[stx][sty-fy]==c&&s[stx-fx][sty-fy]==c){
R[k][stx][sty]=min(R[k][stx-fx][sty],
min(R[k][stx][sty-fy],R[k][stx-fx][sty-fy]))+1;
}
}
}
sty+=fy;
}
sty=0;if(fy==-1)sty=m-1;
stx+=fx;
}
}
bool check(int r1,int c1,int r2,int c2,int mid){
if(mid==0)return true;
if(mid==min(r2-r1,c2-c1)+1)return false;
r1=r1+mid-1;
c1=c1+mid-1;
r2=r2-mid;
c2=c2-mid;
if(r1>r2||c1>c2)return false;
return Query(r1,c1,r2,c2)>=mid;
}
int main(){
cin>>n>>m>>q;
for(int i=0;i<n;i++){
cin>>s[i];
}
color('R',1,1,0);
color('Y',-1,1,1);
color('G',1,-1,2);
color('B',-1,-1,3);
for(int i=0;i<n-1;i++){
for(int j=0;j<m-1;j++){
if(s[i][j]=='R'&&s[i+1][j]=='Y'&&s[i][j+1]=='G'&&s[i+1][j+1]=='B'){
val[i][j]=min(R[0][i][j],min(R[1][i+1][j],min(R[2][i][j+1],R[3][i+1][j+1])));
}
}
}
Build_2D_Sparse_Table(n,m);
for(int i=0;i<q;i++){
int r1,c1,r2,c2;
scanf("%d%d%d%d",&r1,&c1,&r2,&c2);
int l=0,r=min(r2-r1,c2-c1)+1;
while(r-l>1){
int mid=(l+r)/2;
if(check(r1,c1,r2,c2,mid)){
l=mid;
}else{
r=mid;
}
}
printf("%d\n",l*l*4);
}
}