2024牛客寒假算法基础集训营3
A 智乃与瞩目狸猫、幸运水母、月宫龙虾
题干
题解：
就看两个字符串的第一个字母是否相同
#include
using namespace std;
#define int long long
const int N =1e6+10;
const int INF=0x3f3f3f3f;
typedef pair pII;
#define x first
#define y second
typedef long long ll;
#define pb push_back
const int mod=1e9+7;
#define endl '\n'
const double eps=1e-5;
int ddx[] = {1,0,0,-1};
int ddy[] = {0,1,-1,0};
void solve(){
string s1,s2;
cin>>s1>>s2;
if(s1[0]==s2[0] or s1[0]-32==s2[0] or s1[0]+32==s2[0]) cout<<"Yes"<>T;
for(int i=1;i<=T;i++){
solve();
}
return 0;
} L：智乃的36倍数(easy version)
题干：
题解：
#include
using namespace std;
#define int long long
const int N =1e6+10;
const int INF=0x3f3f3f3f;
typedef pair pII;
#define x first
#define y second
typedef long long ll;
#define pb push_back
const int mod=1e9+7;
#define endl '\n'
const double eps=1e-5;
int ddx[] = {1,0,0,-1};
int ddy[] = {0,1,-1,0};
void solve(){
int n;
cin>>n;
vectora(n);
for(int i=0;i>a[i];
int cnt=0,tmp=0;
for(int i=0;i>T;
for(int i=1;i<=T;i++){
solve();
}
return 0;
} B ：智乃的数字手串
题干：
题解：
诈骗题，n为奇数就位清楚姐姐，反之为智乃。
#include
using namespace std;
#define int long long
const int N =1e6+10;
const int INF=0x3f3f3f3f;
typedef pair pII;
#define x first
#define y second
typedef long long ll;
#define pb push_back
const int mod=1e9+7;
#define endl '\n'
const double eps=1e-5;
int ddx[] = {1,0,0,-1};
int ddy[] = {0,1,-1,0};
void solve(){
int n;
cin>>n;
vectora(n);
for(int i=0;i>a[i];
if(n==1){
cout<<"qcjj"<>T;
for(int i=1;i<=T;i++){
solve();
}
return 0;
} D：chino's bubble sort and maximum subarray sum(easy version)
题干：
题解：
#include
using namespace std;
#define int long long
const int N =1e6+10;
const int INF=0x3f3f3f3f;
typedef pair pII;
#define x first
#define y second
typedef long long ll;
#define pb push_back
const int mod=1e9+7;
#define endl '\n'
const double eps=1e-5;
int ddx[] = {1,0,0,-1};
int ddy[] = {0,1,-1,0};
void solve(){
int n,k;
cin>>n>>k;
vectora(n);
for(int i=0;i>a[i];
int res=-1e9,tmp=0;
vectorb;
if(k==1){
for(int i=0;i>T;
for(int i=1;i<=T;i++){
solve();
}
return 0;
} G：智乃的比较函数(easy version)
题干：
题解：
#include
using namespace std;
#define int long long
const int N =1e6+10;
const int INF=0x3f3f3f3f;
typedef pair pII;
#define x first
#define y second
typedef long long ll;
#define pb push_back
const int mod=1e9+7;
#define endl '\n'
const double eps=1e-5;
int ddx[] = {1,0,0,-1};
int ddy[] = {0,1,-1,0};
void solve(){
int n;
cin>>n;
if(n==1){
int x,y,z;
cin>>x>>y>>z;
if(x==y and z==1) cout<<"No"<a(4,-1),b(4);
bool f=1;
for(int i=1;i<=2;i++){
int x,y,z;
cin>>x>>y>>b[i];
if(x==y and b[i]==1) f=0;
if(a[x]!=-1 and a[y]!=-1){
if(b[i]==0){
if(b[i-1]==0) a[x]=a[y];
if(a[x]=a[y])f=0;
}
}else{
if(!b[i])a[x]=2,a[y]=1;
else a[x]=1,a[y]=2;
}
}
if(f)cout<<"Yes\n";
else cout<<"No\n";
}
}
signed main(){
ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
int T= 1;
cin>>T;
for(int i=1;i<=T;i++){
solve();
}
return 0;
} H 智乃的比较函数(normal version)
题干：
题解：
#include
using namespace std;
#define int long long
const int N =1e6+10;
const int INF=0x3f3f3f3f;
typedef pair pII;
#define x first
#define y second
typedef long long ll;
#define pb push_back
#define P push
const int mod=1e9+7;
#define endl '\n'
const double eps=1e-5;
int ddx[]={1,1,0,-1};
int ddy[]={0,1,1,1};
int dp[1001];
vector > v;
int check(){
for(auto [x,y,z]:v){
if(z==1){
if(dp[x]=dp[y])continue;
else return 0;
}
}
return 1;
}
void solve(){
v={};
int n;
cin>>n;
int a,b,c;
while(n--){
cin>>a>>b>>c;
v.push_back({a,b,c});
}
int ans=0;
for(int i=1;i<=3;i++)
for(int j=1;j<=3;j++)
for(int k=1;k<=3;k++){
dp[1]=i; dp[2]=j; dp[3]=k;
ans|=check();
}
if(ans)cout<<"Yes\n";
else cout<<"No\n";
}
signed main(){
ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
int T= 1;
cin>>T;
for(int i=1;i<=T;i++){
solve();
}
return 0;
}