文章目录
题目
A. Mezo Playing Zoma
签到题
脑筋急转弯？
#include<bits/stdc++.h>
using namespace std;
int main(){
int n;
string s;
scanf("%d",&n);
cin>>s;
cout<<n+1<<endl;
}
B. Just Eat It!
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll maxn=2e5+10;
const ll INF=0x3f3f3f3f;
ll a[maxn];
ll st[maxn];
ll len[maxn];
int main(){
ll t;
scanf("%lld",&t);
while(t--){
ll n;
ll sum=0;
ll pos=0;
ll l=0;
ll MAX=-INF;
scanf("%lld",&n);
for(ll i=1;i<=n;i++){
scanf("%lld",&a[i]);
sum+=a[i];
}
for(ll i=1;i<=n;i++){
if(st[i-1]>0){
st[i]=st[i-1]+a[i];
len[i]=len[i-1]+1;
}
else{
st[i]=a[i];
len[i]=1;
}
if(st[i]>MAX){
pos=i;
MAX=st[i];
l=len[i];
}
}
if(pos==n&&sum==MAX&&l==n)
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
}
return 0;
}
C. Fadi and LCM
十几行的题用了将近150行？感觉自己像个脑残
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int t;
const ll INF=1e12+10;
ll n,prim[1000];
ll ans;
int tot;
ll mul_mod(ll a,ll b,ll c)
{
ll t=0;
a%=c;
b%=c;
while(b){
if(b&1){
t=(t+a)%c;
}
a<<=1;
a%=c;
b>>=1;
}
return t;
}
ll pow_mod(ll x,ll n,ll mod){
if (n==1) return x%mod;
int bit[90],k=0;
while (n){
bit[k++]=n&1;
n>>=1;
}
ll ret=1;
for (k=k-1;k>=0;k--){
ret=mul_mod(ret,ret,mod);
if (bit[k]==1) ret=mul_mod(ret,x,mod);
}
return ret;
}
bool check(ll a,ll n,ll x,ll t)
{
ll ret=pow_mod(a,x,n);
ll tmp=ret;
for(int i=0;i<t;i++){
ret=mul_mod(ret,ret,n);
if(ret==1&&tmp!=1&&tmp!=n-1) return true;
tmp=ret;
}
if(ret!=1) return true;
return false;
}
bool Miller_rabin(ll n)
{
ll x=n-1,t=0;
while((x&1)==0){
x>>=1;
t++;
}
int ok=1;
if(t>=1&&(x&1)){
for(int i=0;i<20;i++){
ll a=rand()%(n-1)+1;
if(check(a,n,x,t)){
ok=1;break;
}
ok=0;
}
}
if(!ok||n==2) return false;
return true;
}
ll gcd(ll a,ll b){
if (a==0) return 1;
if (a<0) return gcd(-a,b);
while (b){
ll t=a%b; a=b; b=t;
}
return a;
}
ll Pollard_rho(ll x,ll c)
{
ll i=1,k=2;
ll x0=rand()%x,y=x0;
while(1){
i++;
x0=(mul_mod(x0,x0,x)+c)%x;
ll d=gcd(y-x0,x);
if(d>1&&d<x) return d;
if(y==x0) return x;
if(i==k){
y=x0;
k+=k;
}
}
}
void findfac(ll n)
{
if(!Miller_rabin(n)){
prim[tot++]=n;
return;
}
ll p=n;
while(p>=n) p=Pollard_rho(p,rand()%(n-1)+1);
findfac(p);
findfac(n/p);
}
unordered_map<ll,ll>mp;
vector<ll>vec;
int main()
{
srand(time(NULL));
scanf("%lld",&n);
if(n==1||(!Miller_rabin(n))){
printf("1 %lld\n",n);
}
else{
tot=0;
ll ans1=1,ans2=1;
findfac(n);
for(int i=0;i<tot;i++){
if(mp[prim[i]]){
mp[prim[i]]*=prim[i];
}
else{
mp[prim[i]]=prim[i];
}
}
for(auto it=mp.begin();it!=mp.end();it++){
vec.push_back(it->second);
}
ll MAX=INF;
for(int i=0;i<(1<<vec.size());i++){
ll fac1=1,fac2=1;
ll cur=i;
int pos=0;
while(cur){
if(cur&1)
fac1*=vec[pos];
pos++;
cur>>=1;
}
fac2=n/fac1;
if(max(fac1,fac2)<MAX){
MAX=max(fac1,fac2);
ans1=fac1;
ans2=fac2;
}
}
printf("%lld %lld\n",ans1,ans2);
}
}
D. Dr. Evil Underscores
非常经典的01字典树dp
遇到异或之类的位运算可以往二叉树的方向思考
非常经典的一道题目。
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
struct node{
ll num;
node *left,*right;
node(){
num=0;
left=right=NULL;
}
}*root;
void dfs(node* root,ll depth){
if(root->left!=NULL){
dfs(root->left,depth+1);
}
if(root->right!=NULL){
dfs(root->right,depth+1);
}
if(root->right==NULL&&root->left==NULL){
root->num=0;
}
else if(root->right==NULL)
root->num=root->left->num;
else if(root->left==NULL)
root->num=root->right->num;
else
root->num=(1<<(29-depth))+min(root->left->num,root->right->num);
}
int main(){
ll n;
scanf("%lld",&n);
root=new node;
for(ll i=1;i<=n;i++){
ll x;
scanf("%lld",&x);
node* cur=root;
for(ll i=29;i>=0;i--){
if((x>>i)&1){
if(cur->right==NULL){
cur->right=new node;
}
cur=cur->right;
}
else{
if(cur->left==NULL){
cur->left=new node;
}
cur=cur->left;
}
}
}
dfs(root,0);
cout<<root->num<<endl;
return 0;
}
E. Delete a Segment
先求出不删除区间的区间数为ans1，然后算去掉每一个区间造成区间总数变化的量（可能增加也可能减少），然后取最大值ans2
#include<bits/stdc++.h>
using namespace std;
const int maxn=4e5+10;
const int INF=0x3f3f3f3f;
int a[maxn];
int ans[maxn];
int main(){
int t;
scanf("%d",&t);
while(t--){
memset(ans,0,sizeof(ans));
int n;
scanf("%d",&n);
vector<pair<int,int> >vec;
for(int i=1;i<=n;i++){
int l,r;
scanf("%d%d",&l,&r);
vec.push_back(make_pair(l,-i));
vec.push_back(make_pair(r,i));
}
sort(vec.begin(),vec.end());
multiset<int>se;
int ans1=0;
for(int i=0;i<vec.size();i++){
if(vec[i].second<0)
se.insert(-vec[i].second);
else
se.erase(se.find(vec[i].second));
if(se.size()==0)
ans1++;
if(se.size()==1&&vec[i].second>0&&vec[i+1].second<0&&vec[i+1].first>vec[i].first){
ans[*se.begin()]++;
}
if(se.size()==1&&vec[i].second<0&&vec[i+1].second>0)
ans[*se.begin()]--;
}
int ans2=-INF;
for(int i=1;i<=n;i++){
ans2=max(ans2,ans[i]);
}
cout<<ans1+ans2<<endl;
}
}
