A. Quintomania
照题意模拟下去即可
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#include <bits/stdc++.h>
using namespace std;
#define ls now<<1
#define rs now<<1|1
#define endl "\n"
#define lowbit(x) ((x)&(-x))
typedef long long ll;
const int N=1e5+7, mod=1e9+7;
int n;
int a[N];
void solve(){
cin>>n;
for(int i=1;i<=n;i++)cin>>a[i];
for(int i=1;i<n;i++){
if(abs(a[i+1]-a[i])!=5&&abs(a[i+1]-a[i])!=7){
cout<<"NO\n";
return;
}
}
cout<<"YES\n";
}
int main(){
ios::sync_with_stdio(0);
cin.tie(0);cout.tie(0);
int t=1;
cin>>t;
while(t--)solve();
return 0;
}
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B. Startup
记录每种牌子的总价值，排序后从大到小取
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#include <bits/stdc++.h>
using namespace std;
#define ls now<<1
#define rs now<<1|1
#define endl "\n"
#define lowbit(x) ((x)&(-x))
typedef long long ll;
#define int long long
const int N=2e5+7, mod=1e9+7;
struct Node{
ll id,val;
bool operator<(const Node &a){return val>a.val;}
};
int n,k;
int b[N],c[N];
map<int,int>vis;
void solve(){
cin>>k>>n;
vis.clear();
for(int i=1;i<=n;i++)cin>>b[i]>>c[i];
for(int i=1;i<=n;i++){
vis[b[i]]+=c[i];
}
vector<Node>v;
for(auto [x,y]:vis){
v.push_back(Node{x,y});
}
sort(v.begin(),v.end());
ll ans=0;
for(int i=0;i<min(k, (int)v.size());i++){
ans+=v[i].val;
}
cout<<ans<<endl;
}
signed main(){
ios::sync_with_stdio(0);
cin.tie(0);cout.tie(0);
int t=1;
cin>>t;
while(t--)solve();
return 0;
}
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C. Anya and 1100
每次修改其实只会对临近的几个位置造成影响，只需要统计附近是否增删了 1100 片段即可
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#include <bits/stdc++.h>
using namespace std;
#define ls now<<1
#define rs now<<1|1
#define endl "\n"
#define lowbit(x) ((x)&(-x))
typedef long long ll;
const int N=1e5+7, mod=1e9+7;
string s;
int q;
void solve(){
cin>>s>>q;
int n=s.size();
s=" "+s;
set<int>st;
for(int i=1;i<=n-3;i++){
if(s.substr(i,4)=="1100")st.insert(i);
}
while(q--){
int x,y;
cin>>x>>y;
if(n<4){
cout<<"NO\n";
continue;
}
s[x]=char('0'+y);
// cout<<s<<endl;
for(int i=max(1, x-4); i<=min(n-3, x+4); i++){
if(s.substr(i,4)=="1100")st.insert(i);
else st.erase(i);
}
if(st.size())cout<<"YES\n";
else cout<<"NO\n";
}
}
int main(){
ios::sync_with_stdio(0);
cin.tie(0);cout.tie(0);
int t=1;
cin>>t;
while(t--)solve();
return 0;
}
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D. I Love 1543
把图一层一层拆分开，对每一层形成的字符串计数即可
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#include <bits/stdc++.h>
using namespace std;
#define ls now<<1
#define rs now<<1|1
#define endl "\n"
#define lowbit(x) ((x)&(-x))
typedef long long ll;
const int N=1e3+7, mod=1e9+7;
int n,m,ans;
char mp[N][N];
bool vis[N][N];
char x;
void getString(int x){
string s="";
for(int i=x;i<=m-x+1;i++){
if(vis[x][i])break;
s+=mp[x][i],vis[x][i]=1;
}
for(int i=x+1;i<=n-x+1;i++){
if(vis[i][m-x+1])break;
s+=mp[i][m-x+1],vis[i][m-x+1]=1;
}
for(int i=m-x;i>x;i--){
if(vis[n-x+1][i])break;
s+=mp[n-x+1][i],vis[n-x+1][i]=1;
}
for(int j=n-x+1;j>x;j--){
if(vis[j][x])break;
s+=mp[j][x],vis[j][x]=1;
}
int len=s.size();
// cout<<s<<endl;
s=s+s;
for(int i=0;i<len;i++){
if(s.substr(i,4)=="1543")ans++;
}
}
void solve(){
cin>>n>>m;
ans=0;
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
cin>>x;
mp[i][j]=x;
vis[i][j]=0;
}
}
for(int i=1;i<=min(n,m);i++){
if(vis[i][i])break;
getString(i);
}
cout<<ans<<endl;
}
int main(){
ios::sync_with_stdio(0);
cin.tie(0);cout.tie(0);
int t=1;
cin>>t;
while(t--)solve();
return 0;
}
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E. Reverse the Rivers
两个数按位或，只可能变大，那么每个位置都是一个单调递增的序列，易想到通过二分来查找符合的国家
输出符合条件的最小的国家序号即可
赛时没发现 l 初始设置为 0 了，连 wa 数发
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#include <bits/stdc++.h>
using namespace std;
#define ls now<<1
#define rs now<<1|1
#define endl "\n"
#define lowbit(x) ((x)&(-x))
typedef long long ll;
#define int long long
const int N=1e5+7, mod=1e9+7;
const ll inf=1e18;
int n,k,q,m;
void solve(){
cin>>n>>k>>q;
vector<vector<ll>> a(n+10, vector<ll>(k+10));
for(int i=1;i<=n;i++){
for(int j=1;j<=k;j++){
cin>>a[i][j];
a[i][j]|=a[i-1][j];
}
}
vector<vector<int>> s(k + 1);
for(int i=1;i<=k;i++)s[i].push_back(0);
for(int i=1;i<=n;i++){
for(int j=1;j<=k;j++){
s[j].push_back(a[i][j]);
}
}
while(q--){
cin>>m;
int low=1,up=n;
for(int i=1;i<=m;i++){
ll r,c;
char x;
cin>>r>>x>>c;
if(x=='>'){
auto l=upper_bound(s[r].begin(),s[r].end(),c)-s[r].begin();
low = max(low, l);
}
else{
auto l = lower_bound(s[r].begin(), s[r].end(), c) - s[r].begin();
l--;
up = min(up, l);
}
// cout<<low<<" "<<up<<endl;
}
// cout<<low<<" "<<up<<endl;
if(low <= up) cout << low << endl;
else cout << "-1\n";
}
}
signed main(){
ios::sync_with_stdio(0);
cin.tie(0);cout.tie(0);
int t=1;
// cin>>t;
while(t--)solve();
return 0;
}
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F. XORificator 3000
异或前缀和有一个性质
$$ f(x)=\begin{cases}
x, & x\%4=0 \\
1, & x\%4=1 \\
x+1, & x\%4=2 \\
0, & x\%4=3
\end{cases} $$同时题中另一性质的数: $x≡3\mod2^4$
$x=0000011,0001011,0010011,0011011$
可以看出，后几位是始终不变的，而前面的会递增，也可以使用异或前缀和求出
即可以先对前面的位数求异或和，再对后面的位数求异或和
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#include <bits/stdc++.h>
using namespace std;
#define ls now << 1
#define rs now << 1 | 1
#define endl "\n"
#define lowbit(x) ((x) & (-x))
typedef long long ll;
const int N = 1e5 + 7, mod = 1e9 + 7;
ll l,r,i,k;
ll fc(ll n){
if(n%4==0) return n;
if(n%4==1) return 1;
if(n%4==2) return n+1;
return 0;
}
ll fx(ll n,ll i,ll k){
if(!i){
if(!k)return fc(n);
else return 0;
}
ll x=1ll<<i;
if(n<k) return 0;
ll t=(n-k)/x;
ll cnt=t+1;
ll res=fc(t)<<i;
if(cnt%2)res^=k;
return res;
}
void solve() {
cin>>l>>r>>i>>k;
ll s=fc(l-1)^fc(r);
ll tmp=fx(r,i,k)^fx(l-1,i,k);
ll ans=s^tmp;
cout<<ans<<endl;
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
// freopen("..//..//in_out//in.txt", "r", stdin);
// freopen("..//..//in_out//out.txt", "w", stdout);
int t = 1;
cin >> t;
while (t--) solve();
return 0;
}
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