A Who is better?
The meaning of problems
excrt + Fibonacci Qi Bo Yi
analysis
Java's BigInteger objects defaults to null, can not be directly compared.
Code
import java.math.BigInteger;
import java.util.Scanner;
public class Main {
static int n;
static BigInteger[] a=new BigInteger[25];
static BigInteger[] b=new BigInteger[25];
static BigInteger INF=BigInteger.valueOf((long)1e15);
static BigInteger[] fib=new BigInteger[110];
static int cnt=0;
static void init(){
fib[1]=BigInteger.ONE;
fib[2]=BigInteger.ONE;
cnt=2;
for(int i=3;i<110;i++){
fib[i]=fib[i-1].add(fib[i-2]);
if(fib[i].compareTo(INF)>0){
break;
}
cnt++;
}
}
static boolean check(BigInteger t){
for(int i=1;i<=cnt;i++){
if(fib[i]==null){
return false;
}
if(fib[i].compareTo(t)==0){
return true;
}
}
return false;
}
static BigInteger exgcd(BigInteger a,BigInteger b,BigInteger[] x,BigInteger[] y){
if(a.compareTo(BigInteger.ZERO)==0 && b.compareTo(BigInteger.ZERO)==0){
return new BigInteger("-1");
}
if(b.compareTo(BigInteger.ZERO)==0){
x[0]= BigInteger.ONE;
y[0]=BigInteger.ZERO;
return a;
}
BigInteger d=exgcd(b,a.mod(b),y,x);
y[0]=y[0].subtract(a.divide(b).multiply(x[0]));
return d;
}
static BigInteger excrt(){
BigInteger[] x=new BigInteger[2];
BigInteger[] y=new BigInteger[2];
BigInteger M=b[1];
BigInteger ans=a[1];
for(int i=2;i<=n;i++){
BigInteger ai=M,bi=b[i],ci=(a[i].subtract(ans.mod(bi)).add(bi)).mod(bi);
BigInteger gcd=exgcd(ai,bi,x,y);
BigInteger bg=bi.divide(gcd);
if(ci.mod(gcd).compareTo(BigInteger.ZERO)!=0){
return new BigInteger("-1");
}
x[0]=x[0].multiply(ci.divide(gcd)).mod(bg);
ans=ans.add(x[0].multiply(M));
M=M.multiply(bg);
ans=(ans.mod(M).add(M)).mod(M);
}
return (ans.mod(M).add(M)).mod(M);
}
public static void main(String[] args) {
Scanner cin=new Scanner(System.in);
init();
n=cin.nextInt();
for(int i=1;i<=n;i++){
b[i]=cin.nextBigInteger();
a[i]=cin.nextBigInteger();
}
BigInteger ans=excrt();
if(ans.compareTo(BigInteger.valueOf(-1))==0){
System.out.println("Tankernb!");
}else{
if(check(ans)){
System.out.println("Lbnb!");
}else{
System.out.println("Zgxnb!");
}
}
}
}B so easy
The meaning of problems
Two operations, delete a number, asked a number of follow-up (not necessarily still) a.
analysis
Disjoint-set, initialize each answer queries is itself, the answer is deleted after the \ (the Find (the X-+ 1) \) , to be discrete.
Code
#include <bits/stdc++.h>
using namespace std;
const int N=2e6+50;
struct Orz{
vector<int> a;
void init(){
a.clear();
}
int siz(){
return a.size();
}
void add(int x){
a.push_back(x);
}
void work(){
sort(a.begin(),a.end());
a.erase(unique(a.begin(),a.end()),a.end());
}
int idx(int x){
return lower_bound(a.begin(),a.end(),x)-a.begin()+1;
}
int val(int i){
return a[i-1];
}
}orz;
int n,q,o[N],a[N],p[N];
int find(int x){
return x==p[x]?p[x]:p[x]=find(p[x]);
}
int main(){
scanf("%d%d",&n,&q);
orz.init();
for(int i=1;i<=q;i++){
scanf("%d%d",&o[i],&a[i]);
orz.add(a[i]);
orz.add(a[i]+1);
}
orz.work();
int m=orz.siz();
for(int i=1;i<=m;i++){
p[i]=i;
}
for(int i=1;i<=q;i++){
int k=orz.idx(a[i]);
if(o[i]==1){
if(find(k)==k){
p[k]=find(k+1);
}
}else{
int v=orz.val(find(k));
printf("%d\n",v);
}
}
return 0;
}C Buy Watermelon
Code
#include<bits/stdc++.h>
using namespace std;
int main(){
int n;
scanf("%d",&n);
if(n%2==0 && n!=2)
puts("YES");
else puts("NO");
} D Carneginon
The meaning of problems
A plurality of strings and a string T S, and whether the length of the substring size determination according to an output.
analysis
The complexity of the subject just described is violence KMP.
Code
#include <bits/stdc++.h>
using namespace std;
const int N=1e6+50;
int n;
char s[N],t[N];
int nex[N];
void getNext(char s[],int n){
int i=0,j=-1;
nex[0]=-1;
while(i<n){
if(j==-1 || s[i]==s[j]){
nex[++i]=++j;
}else{
j=nex[j];
}
}
}
int kmp(char s[],int n,char p[],int m){
int i=0,j=0;
getNext(p,m);
while(i<n && j<m){
if(j==-1 || s[i]==p[j]){
i++;
j++;
}else{
j=nex[j];
}
if(j==m){
return i-j+1;
}
}
return -1;
}
int main(){
scanf("%s",t);
int tl=strlen(t);
scanf("%d",&n);
for(int i=1;i<=n;i++){
scanf("%s",s);
int sl=strlen(s);
if(tl>sl){
int p=kmp(t,tl,s,sl);
if(p!=-1){
printf("my child!\n");
}else{
printf("oh, child!\n");
}
}else if(tl<sl){
int p=kmp(s,sl,t,tl);
if(p!=-1){
printf("my teacher!\n");
}else{
printf("senior!\n");
}
}else if(tl==sl){
if(strcmp(t,s)==0){
printf("jntm!\n");
}else{
printf("friend!\n");
}
}
}
return 0;
}E XKC's basketball team
The meaning of problems
For each \ (A [I] \) , the right side than the query \ (a [i] + m \) large maximum distance.
analysis
Not discrete, not the Chairman of the tree, without weights tree line. . .
Only need to build a normal maximum segment tree maintenance can, try to go right sub-tree query.
Code
#include <bits/stdc++.h>
using namespace std;
#define ls i<<1
#define rs i<<1|1
#define mid (l+r)/2
const int N=5e5+50;
int n,m,a[N],ans[N];
int mx[N*4];
void pushup(int i){
mx[i]=max(mx[ls],mx[rs]);
}
void build(int i,int l,int r){
mx[i]=0;
if(l==r){
return;
}
build(ls,l,mid);
build(rs,mid+1,r);
}
void insert(int i,int l,int r,int p){
if(l==r && l==p){
mx[i]=a[p];
return;
}
if(p<=mid){
insert(ls,l,mid,p);
}else{
insert(rs,mid+1,r,p);
}
pushup(i);
}
int query(int i,int l,int r,int k){
if(l==r){
if(mx[i]>=k){
return l;
}else{
return -1;
}
}
if(mx[rs]>=k){
return query(rs,mid+1,r,k);
}else{
if(mx[ls]>=k){
return query(ls,l,mid,k);
}else{
return -1;
}
}
}
void debug(int i,int l,int r){
printf("%d %d %d\n",l,r,mx[i]);
if(l==r){
return;
}
debug(ls,l,mid);
debug(rs,mid+1,r);
}
int main(){
// freopen("in.txt","r",stdin);
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++){
scanf("%d",&a[i]);
}
build(1,1,n);
for(int i=n;i>=1;i--){
int t=query(1,1,n,a[i]+m);
if(t==-1){
ans[i]=t;
}else{
ans[i]=t-i-1;
}
insert(1,1,n,i);
//
}
for(int i=1;i<=n;i++){
printf("%d%c",ans[i],i==n?'\n':' ');
}
return 0;
}G Colorful String
The meaning of problems
Substring defined value for the number of different characters, find all palindromic substring of the value of the sum.
analysis
Palindrome tree template, insert the time by another palindrome substring transferred from, the newly added character to determine whether a palindrome substring then appear, if not, the value of +1.
Code
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=3e5+50;
struct PT{
int next[N][26];
int vis[N][26];
int val[N];
int fail[N];
int cnt[N];
int num[N];
int len[N];
int S[N];
int last;
int id[N];
int n;
int p;
int newnode(int l){
for(int i=0;i<26;i++){
next[p][i]=0;
}
cnt[p]=0;
num[p]=0;
len[p]=l;
return p++;
}
void init(){
p=0;
newnode(0);
newnode(-1);
last=0;
n=0;
S[n]=-1;
fail[0]=1;
}
int getFail(int x){
while(S[n-len[x]-1]!=S[n]){
x=fail[x];
}
return x;
}
void add(int c){
c-='a';
S[++n]=c;
int cur=getFail(last);
if(!next[cur][c]){
int now=newnode(len[cur]+2);
for(int i=0;i<26;i++){
vis[now][i]=vis[cur][i];
}
val[now]=val[cur];
vis[now][c]=1;
if(!vis[cur][c]){
val[now]++;
}
fail[now]=next[getFail(fail[cur])][c];
num[now]=num[fail[now]]+1;
next[cur][c]=now;
}
last=next[cur][c];
cnt[last]++;
id[last]=n;
}
void count(){
for(int i=p-1;i>=0;i--){
cnt[fail[i]]+=cnt[i];
}
}
ll solve(){
ll ans=0;
for(int i=2;i<p;i++){
ans+=1ll*val[i]*cnt[i];
}
return ans;
}
}ac;
char s[N];
int main(){
// freopen("in.txt","r",stdin);
scanf("%s",&s);
int len=strlen(s);
ac.init();
for(int i=0;i<len;i++){
ac.add(s[i]);
}
ac.count();
ll ans=ac.solve();
printf("%lld\n",ans);
}J Random Access Iterator
Code
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN=1e6+50;
const int mod=1e9+7;
ll dp[MAXN];
vector<int>v[MAXN];
int d[MAXN];
int ans=0;
ll q_pow(ll a,ll k)
{
ll ans=1;
while(k)
{
if(k&1)
{
ans*=a;
ans%=mod;
}
a*=a;
a%=mod;
k>>=1;
}
return ans;
}
void dfs(int now,int pre,int x)
{
ans=max(ans,x);
d[now]=x;
int l=v[now].size();
for(int i=0;i<l;i++)
{
int to=v[now][i];
if(to==pre)continue;
dfs(to,now,x+1);
}
}
void solve(int now,int pre)
{
int l=v[now].size();
if(l==1 && now!=1)
{
dp[now]=(d[now]==ans);
return ;
}
ll ans=0;
for(int i=0;i<l;i++)
{
int to=v[now][i];
if(to==pre)continue;
solve(to,now);
ans+=dp[to];
ans%=mod;
}
if(now!=1)l--;
ans=ans*q_pow(l,mod-2)%mod;
ans=(1-ans+mod)%mod;
ans=q_pow(ans,l);
ans=(1-ans+mod)%mod;
dp[now]=ans;
}
int main()
{
ios::sync_with_stdio(false);
int n;
cin>>n;
for(int i=1;i<n;i++)
{
int x,y;
cin>>x>>y;
v[x].push_back(y);
v[y].push_back(x);
}
dfs(1,0,1);
solve(1,0);
cout<<dp[1]<<endl;
// cout<<(1-bas%mod+mod)%mod;
}K Center
The meaning of problems
To more than one point, asked to increase the minimum number of points, so that these points have a center point.
analysis
Enumerate any two points, find their center point, the center point of each deposit with a map corresponding to the number of points, the final choice of the most.
Code
#include<bits/stdc++.h>
using namespace std;
const int MAXN=1005;
map<pair<int,int> ,int>mp;
pair<int,int>a[MAXN];
int ans=0;
int main()
{
ios::sync_with_stdio(false);
int n;
cin>>n;
for(int i=1;i<=n;i++)
{
cin>>a[i].first>>a[i].second;
a[i].first*=2;
a[i].second*=2;
mp[a[i]]++;
ans=max(ans,mp[a[i]]);
}
for(int i=1;i<=n;i++)
{
for(int j=i+1;j<=n;j++)
{
int X=a[i].first+a[j].first,Y=a[i].second+a[j].second;
X/=2,Y/=2;
pair<int,int>p=make_pair(X,Y);
mp[p]+=2;
ans=max(ans,mp[p]);
}
}
cout<<n-ans<<endl;
}M Longest subsequence
The meaning of problems
To a predetermined sequence two longest string S and T, S satisfies the requirements lexicographically strictly greater than T.
Code
#include<bits/stdc++.h>
using namespace std;
const int MAXN=1e6+50;
const int inf=-1e6;
int ans=-1;
char s[MAXN],t[MAXN];
int cnt[MAXN][30];
int n,m;
int solve(int i,int j)
{
if(j>m && i<=n) return n-i+1;
if(i>n)return inf;
int x=t[j]-'a';
int p1=cnt[i][x];
// cout<<i<<' '<<j<<' '<<p1<<endl;
if(p1==0) return inf;
if(s[p1]>t[j])return n-p1+1;
else if(s[p1]==t[j])
{
int y=x+1;
int p2=cnt[i][y];
// cout<<i<<' '<<j<<' '<<p1<<' '<<p2<<"????"<<endl;
if(p2==0) return solve(p1+1,j+1)+1;
else return max(n-p2+1,solve(p1+1,j+1)+1);
}
}
int main()
{
ios::sync_with_stdio(false);
cin>>n>>m;
cin>>s+1>>t+1;
for(int i=n;i>=1;i--)
{
int x=s[i]-'a';
for(int j=x;j>=0;j--)
{
cnt[i][j]=i;
}
for(int j=x+1;j<26;j++)
{
cnt[i][j]=cnt[i+1][j];
}
}
int ans=solve(1,1);
if(ans<=0)
cout<<-1<<endl;
else cout<<ans<<endl;
}