Problem B. Useful Algorithm

题意:交互题，有个分数的分子是x，分母是y，他们都大于等于1且不大于1000，每次询问会告诉你你猜的数是大还是小，请你用不超过20次的询问来才出来x和y，

思路：x和y一共可以组成不超过1e6组方案，并且可以按照大小去排序，然后就可以根据这个来二分出答案。

#include using namespace std;
#define int long long
const int N = 1e6+100;
const int up=1e3;
struct node
{int p,q;bool operator<(const node &a)const{return p*a.q,int>mp;
signed main()
{cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);int n=0;for(int i=1;i<=up;i++)for(int j=1;j<=up;j++){int p=j,q=i;int g=__gcd(p,q);p/=g;q/=g;if(!mp[{p,q}]){n++;b[n].p=p;b[n].q=q;}mp[{p,q}]=1;}sort(b+1,b+n+1);int l=1,r=n;for(int i=1;i<=20;i++){int mid=l+r>>1;cout<<"compare "<>ch;if(ch=='>') r=mid-1;else if(ch=='<') l=mid+1;else break;}return 0;
}

A. AquaMoon and Two Arrays
边角情况注意

其他的倒是没什么

#include using namespace std;
#define int long long
const int N = 1e5+100;
int a[N];
int b[N];
signed main()
{cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);int t;for(cin>>t;t;t--){int n;cin>>n;int suma=0;int sumb=0;for(int i=1;i<=n;i++){cin>>a[i];suma+=a[i];}int cnt=0;for(int i=1;i<=n;i++){cin>>b[i];sumb+=b[i];cnt+=abs(b[i]-a[i]);}if(suma!=sumb){cout<<-1<cout<<-1<cout<<-1<>v;for(int i=1;i<=n;i++){if(a[i]>b[i]){for(int j=i+1;j<=n;j++){while(a[j]b[i]){v.push_back(make_pair(i,j));a[j]++;a[i]--;}}}if(a[i]for(int j=i+1;j<=n;j++){while(a[j]>b[j]&&a[i]v.push_back(make_pair(j,i));a[i]++;a[j]--;}}}}cout<

#include
#include
#include
#include
#include
#include
#include
# include

#include using namespace std;const int N = 1e6+100;
int a[N];
int main()
{cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);int t;for(cin>>t;t;t--){int x,t,k,n,d;cin>>x>>t>>k>>n>>d;for(int i=1;i<=n;i++){cin>>a[i];a[i+n]=a[i]=a[i]<=d?-1:1;}bool falg=false;int len=0;int xx=x;for(int i=1;i<=n*2;i++){x+=a[i];if(x<=k){len++;if(len>=t){falg=true;}}else{len=0;}}if(xfalg=true;}else if(x==xx&&(len==n*2)&&t>=n*2)falg=true;if(falg){cout<<"YES"<cout<<"NO"<

#include
using namespace std;
int main() {int t;cin >> t;while (t--) {int a;cin >> a;a--;int b = a / 2;while (b == a - b || __gcd(b, a - b) > 1) b--;cout << a - b << " " << b << " " << 1 << endl;}
}

#include
using namespace std;
#define int long long
#define endl '\n'
const int N = 1e5+100;
int a[N];
const int mod = 1e9+7;
signed main()
{cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);int t;for(cin>>t;t;t--){int n;int ans=0;cin>>n;if(n&1ll){ans=n/2;}else{int temp=n/2;if(temp&1){ans=n/2-2;}else{ans=n/2-1;}}if(ans>n/2||anscout<<-1<cout<