小红叕战小紫
#include "bits/stdc++.h"
using namespace std;
#define int long long
#define endl '\n'
#define IOS ios::sync_with_stdio(0),cin.tie(0);
#define all(x) x.begin(),x.end()
#define pi pair<int,int>
#define vi vector<int>
#define si set<int>
#define mi map<int,int>
#define mc map<char,int>
#define YES cout<<"Yes"<<endl;
#define NO cout<<"No"<<endl;
#define pb(x) push_back(x);
#define fi first
#define sc second
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return true; } return false; }
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return true; } return false; }
void solve()
{
string s;
cin>>s;
if(s.size()>1){
cout<<"kou";
}
else {
cout<<"yukari";
}
}
signed main()
{
IOS
int t;
t=1;//cin>>t;
while(t--){
solve();
}
}小红的数组移动
#include "bits/stdc++.h"
using namespace std;
#define int long long
#define endl '\n'
#define IOS ios::sync_with_stdio(0),cin.tie(0);
#define all(x) x.begin(),x.end()
#define pi pair<int,int>
#define vi vector<int>
#define si set<int>
#define mi map<int,int>
#define mc map<char,int>
#define YES cout<<"Yes"<<endl;
#define NO cout<<"No"<<endl;
#define pb(x) push_back(x);
#define fi first
#define sc second
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return true; } return false; }
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return true; } return false; }
const int mod=1e9+7;
void solve()
{
int n;
cin>>n;
vi a(n);
for (int i=0;i<n;i++){
cin>>a[i];
}
string s;
cin>>s;
int len=s.size();
int sum=0;
int pos=0;
for (int i=0;i<len;i++){
if(s[i]=='R'){
// cout<<i<<endl;
if(pos!=n-1){
pos++;
sum+=a[pos];
}
else {
sum+=a[pos];
}
}
else {
//cout<<pos<<endl;
if(pos!=0){
pos--;
sum+=a[pos];
}
else {
sum+=a[pos];
}
}
}
cout<<sum%mod;;
}
signed main()
{
IOS
int t;
t=1;//cin>>t;
while(t--){
solve();
}
}小红的素数合并
想要极差尽可能的小，在有偶数个数时，我们可以让首位相乘。
而在奇数时，我们可以空出最后一个数，因为相比空出中间那个数来说的话，最后一个数，不仅最小值变大了，而且最大值变小了。
#include "bits/stdc++.h"
using namespace std;
#define int long long
#define endl '\n'
#define IOS ios::sync_with_stdio(0),cin.tie(0);
#define all(x) x.begin(),x.end()
#define pi pair<int,int>
#define vi vector<int>
#define si set<int>
#define mi map<int,int>
#define mc map<char,int>
#define YES cout<<"Yes"<<endl;
#define NO cout<<"No"<<endl;
#define pb(x) push_back(x);
#define fi first
#define sc second
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return true; } return false; }
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return true; } return false; }
const int mod=1e9+7;
void solve()
{
int n;
cin>>n;
vi a(n);
for (int i=0;i<n;i++){
cin>>a[i];
}
sort(all(a));
int maxn=0,minn=1e18;
if(n%2==0){
for (int i=0;i<n/2;i++){
a[i]*=a[n-i-1];
maxn=max(maxn,a[i]);
minn=min(minn,a[i]);
}
}
else {
for (int i=0;i<n/2;i++){
a[i]*=a[n-i-2];
maxn=max(maxn,a[i]);
minn=min(minn,a[i]);
}
maxn=max(maxn,a[n-1]);
minn=min(minn,a[n-1]);
}
cout<<maxn-minn;
}
signed main()
{
IOS
int t;
t=1;//cin>>t;
while(t--){
solve();
}
}小红的树上删边
树上问题熟悉的遍历树。当发现这个子树的联通块大小为偶数时，便可以切除一条边。
#include "bits/stdc++.h"
using namespace std;
#define int long long
#define endl '\n'
#define IOS ios::sync_with_stdio(0),cin.tie(0);
#define all(x) x.begin(),x.end()
#define pi pair<int,int>
#define vi vector<int>
#define si set<int>
#define mi map<int,int>
#define mc map<char,int>
#define YES cout<<"Yes"<<endl;
#define NO cout<<"No"<<endl;
#define pb(x) push_back(x);
#define fi first
#define sc second
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return true; } return false; }
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return true; } return false; }
const int N = 200010;
int e[N],ne[N],st[N],h[N];
int idx;
void add(int a,int b){
e[idx]=b,ne[idx]=h[a],h[a]=idx++;
}
int ans=0;
int sz[N];
void dfs(int x){
st[x]=1;
sz[x]=1;
for (int i=h[x];~i;i=ne[i]){
int j=e[i];
if(st[j]==0){
dfs(j);
sz[x]+=sz[j];
}
}
if(x!=1 && sz[x]%2==0){
ans++;
}
}
void solve()
{
int n;
cin>>n;
memset(h,-1,sizeof h);
for (int i=1;i<=n-1;i++){
int x,y;
cin>>x>>y;
add(x,y),add(y,x);
}
if(n&1){
cout<<-1<<endl;
return ;
}
dfs(1);
cout<<ans;
}
signed main()
{
IOS
int t;
t=1;//cin>>t;
while(t--){
solve();
}
}小红的子序列求和
使用了组合数的思想。j可以视为第i个数在序列中的位置。除去自身的贡献之外。还有前面j个数和后面j-j-1个数的贡献，并将组合数预处理，字符串的每一个字符的贡献只需要在这次计算就可以了。
#include "bits/stdc++.h"
using namespace std;
#define int long long
#define endl '\n'
#define IOS ios::sync_with_stdio(0),cin.tie(0);
#define all(x) x.begin(),x.end()
#define pi pair<int,int>
#define vi vector<int>
#define si set<int>
#define mi map<int,int>
#define mc map<char,int>
#define YES cout<<"Yes"<<endl;
#define NO cout<<"No"<<endl;
#define pb(x) push_back(x);
#define fi first
#define sc second
#define is insert
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return true; } return false; }
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return true; } return false; }
const int INF =1e18;
const int N = 200010;
const int mod = 1e9+7;
//字符串的子序列有多少个。
int c[1010][1010];
int pow1[1000];
void init(){
for (int i=0;i<=1000;i++){
for (int j=0;j<=i;j++){
if(j==0 || i==j){
c[i][j]=1;
}
else{
c[i][j]=(c[i-1][j-1]+c[i-1][j])%mod;
}
}
}
pow1[0]=1;
for (int i=1;i<=1000;i++){
pow1[i]=pow1[i-1]*10%mod;
}
}
void solve()
{
int n,k;
cin>>n>>k;
string s;
cin>>s;
int ans=0;
for (int i=0;i<n;i++){
for (int j=0;j<k;j++){
int tmp=(s[i]-'0');
ans+=pow1[k-j-1]*tmp%mod*c[i][j]%mod*c[n-i-1][k-1-j]%mod;
ans%=mod;
}
}
cout<<ans;
}
signed main()
{
IOS
init();
int t;
t=1;//cin>>t;
while(t--){
solve();
}
}