Codeforces Round #696 (Div. 2)
A. Puzzle From the Future
贪心
#include <bits/stdc++.h>
//#pragma GCC optimize(2)
#define int long long
using namespace std;
//const int mod = 998244353;
const int mod = 1e9 + 7;
const int maxn = 2e5 + 10;
void solve() {
int n;
string b;
cin>>n>>b;
int now;
if(b[0]=='0')
now=1;
else now=2;
cout<<1;
for (int i = 1; i <n; ++i) {
if(b[i]=='1'){
if(now!=2){
cout<<1,now=2;
} else cout<<0,now=1;
} else{
if(now==1){
cout<<0,now=0;
} else cout<<1,now=1;
}
}
cout<<"\n";
}
signed main() {
// ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int _ = 1;
cin >> _;
while (_--) {
solve();
}
return 0;
}
B. Different Divisors
素数筛，1，a,b,a*b
#include <bits/stdc++.h>
//#pragma GCC optimize(2)
#define int long long
using namespace std;
//const int mod = 998244353;
const int mod = 1e9 + 7;
const int maxn = 2e5 + 10;
#define MAXN 1008611
char prime[MAXN];
void getPrime()
{
memset(prime,1,sizeof(prime)); //默认全为素数
prime[0]=prime[1]=0; //0和1不是素数
for(int i=2;i<MAXN;i++)
{
if(prime[i]==1) //如果i是素数
{
for(int j=2;j*i<MAXN;j++)
{
prime[i*j]=0; //将所有i的倍数标记为非素数
}
}
}
}
vector<int > v;
void solve() {
int d;
cin>>d;
int i=lower_bound(v.begin(),v.end(),1+d)-v.begin();
int j=lower_bound(v.begin(),v.end(),v[i]+d)-v.begin();
cout<<v[i]*v[j]<<'\n';
}
signed main() {
// ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int _ = 1;
getPrime();
for (int i = 0; i < 1008611; ++i) {
if(prime[i])
v.push_back(i);
}
cin >> _;
while (_--) {
solve();
}
return 0;
}
//4 1 5 9 6 12 16
C. Array Destruction
枚举谁和最大数一起删除
#include <bits/stdc++.h>
#pragma GCC optimize(2)
//#define int long long
using namespace std;
//const int mod = 998244353;
const int mod = 1e9 + 7;
const int maxn = 2e5 + 10;
int a[maxn];
vector<int > check(int n,int sum){
multiset<int> me(a,a+2*n);
vector<int >res;
for (int i = 0; i < n ; ++i) {
auto it1=--me.end();
int y=sum-*it1;
res.push_back(*it1);
me.erase(it1);
auto it2=me.find(y);
if (it2==me.end()) return {
};
res.push_back(*it2);
me.erase(it2);
sum-=y;
}
return res;
}
void solve() {
int n;
cin>>n;
n*=2;
for (int i = 0; i < n; ++i) cin>>a[i];
sort(a,a+n);
for (int i = 0; i < n - 1; ++i) {
int sum=a[i]+a[n-1];
vector<int > res=check(n/2,sum);
if (res.size()){
cout<<"YES\n"<<sum<<"\n";
for (int j = 0; j < n; j+=2)
cout<<res[j]<<" "<<res[j+1]<<"\n";
return;
}
}
cout<<"NO\n";
}
signed main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int _ = 1;
cin >> _;
while (_--) {
solve();
}
return 0;
}
D. Cleaning
根据题意从俩头都来一遍遍历消除，检查是否有相等的时候，或者考虑互换i和i+1，检查是否满足
#include <bits/stdc++.h>
#pragma GCC optimize(2)
#define int long long
using namespace std;
//const int mod = 998244353;
const int inf=1e18;
const int mod = 1e9 + 7;
const int maxn = 2e5 + 10;
int a[maxn];
int b[maxn],c[maxn];
void solve() {
memset(b,0,sizeof(b));
memset(c,0,sizeof(c));
int n;
cin>>n;
for (int i = 1; i <=n; ++i)
cin>>a[i];
for (int i = 1; i <=n; ++i)
if (a[i]>=b[i-1]) b[i]=a[i]-b[i-1];
else b[i]=inf+1;
for (int i = n; i >0; --i)
if (a[i]>=c[i+1]) c[i]=a[i]-c[i+1];
else c[i]=inf-1;
int f1=0;
for (int i = 1; i <n; ++i) {
if (b[i]==c[i+1]){
f1=1;
break;
}else{
if (a[i]>=c[i+2]&&a[i+1]>=b[i-1]){
int t1=a[i]-c[i+2];
int t2=a[i+1]-b[i-1];
if (t1==t2) {
f1=1;
break;
}
}
}
}
cout<<(f1?"YES":"NO")<<"\n";
}
signed main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int _ = 1;
cin >> _;
while (_--) {
solve();
}
return 0;
}