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Tokio Marine & Nichido Fire Insurance Programming Contest 2021（AtCoder Regular Contest 122) 題解A Many FormulaeB InsuranceC CalculatorD XOR Game

A Many Formulae 

#include<bits/stdc++.h> 
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
						For(j,m-1) cout<<a[i][j]<<' ';\
						cout<<a[i][m]<<endl; \
						} 
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
	int x=0,f=1; char ch=getchar();
	while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
	while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
	return x*f;
} 
#define MXAN (112345)
int n;
ll dp[112233];
int main()
{
//	freopen("A.in","r",stdin);
//	freopen(".out","w",stdout);
	
	dp[0]=1;
    dp[1]=2;
    Fork(i,2,1e5+10)
        dp[i]=(dp[i-1]+dp[i-2])%F;
    n=read();
    ll ans=0;
    For(i,n) {
    	int x=read();
    	if(i==1) upd(ans,dp[n-1]*x%F);
    	else {
    		upd(ans,sub(mul(dp[max(0,i-2)],dp[n-i]),dp[max(0,i-3)]*dp[max(0,n-i-1)])*x%F);
		}
	}
	cout<<ans<<endl;
	
	return 0;
}

B Insurance 

#include<bits/stdc++.h> 
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
						For(j,m-1) cout<<a[i][j]<<' ';\
						cout<<a[i][m]<<endl; \
						} 
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
	int x=0,f=1; char ch=getchar();
	while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
	while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
	return x*f;
} 
#define MXAN (112345)
int n;
ll a[112233];
double ck(double x) {
    double p=0;
    For(i,n)
        p+=x+a[i]-min(a[i]*1.,2*x);
    return p;
}
int main()
{
//	freopen("A.in","r",stdin);
//	freopen(".out","w",stdout);
	
    n=read();
    For(i,n) a[i]=read();
    double l=0,r=3e9,ans;
    Rep(i,404) {
    	double m=(l+r)/2;
		double l1=l+(r-l)/3,r1=r-(r-l)/3;
		if(ck(l1)>ck(r1)) l=l1;else r=r1;			
	}
    printf("%.20lf\n",ck((l+r)/2)/n);
	return 0;
}

C Calculator

交替進行3，4操作可以構造Fib數列，

如果在其中插入1個1或2操作，得到的會是 F i b x + F i b 1 Fib_x+Fib_1 Fibx​+Fib1​,繼續進行得到 F i b x + k + F i b k − 1 Fib_{x+k}+Fib_{k-1} Fibx+k​+Fibk−1​

将N轉成 F i b a 1 + F i b a 2 + ⋯ + F i b a t ( a 1 ≤ a 2 ≤ . . ≤ a t ) Fib_{a_1}+Fib_{a_2}+\cdots+Fib_{a_t}(a_1\le a_2 \le .. \le a_t) Fiba1​​+Fiba2​​+⋯+Fibat​​(a1​≤a2​≤..≤at​)

按如上方法構造

#include<bits/stdc++.h> 
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
						For(j,m-1) cout<<a[i][j]<<' ';\
						cout<<a[i][m]<<endl; \
						} 
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
	int x=0,f=1; char ch=getchar();
	while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
	while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
	return x*f;
} 
#define MXAN (112345)
int n;
ll dp[112233];
ll a[1123];
ll x=0,y=0;
void work(int t) {
	if(t==1) ++x;
	else if(t==2) ++y;
	else if(t==3) x+=y;
	else y+=x;
}
int main()
{
//	freopen("A.in","r",stdin);
//	freopen(".out","w",stdout);
	
	dp[0]=1;
    dp[1]=1;
    Fork(i,2,130){
        dp[i]=(dp[i-1]+dp[i-2]);
		if(dp[i]>1e18) {
			n=i;break;
		}
	}
    ll p;cin>>p;
    ll N=p;
    int m=0;
    ForD(i,n) {
    	a[i]=p/dp[i];
		p%=dp[i];
		if(!m && a[i]) m=i;
	}
    
    int v[100000],len=0;
    int op=1;
    ForD(i,m) {
    	For(j,a[i]) v[++len]=op;
    	v[++len]=5-op;
    	if(op==1)++op;else --op;
	}
	cout<<len<<endl;
	For(i,len) {
		work(v[i]);
//		cout<<v[i]<<endl;
	}
//	cout<<x<<' '<<y<<endl;
	int h[10];
	h[1]=2,h[2]=1,h[3]=4,h[4]=3;
	if(x==N) {
		For(i,len) {
			cout<<v[i]<<endl;
		}	
	}
	else {
		For(i,len) {
			cout<<h[v[i]]<<endl;
		}	
		
	}
	
	return 0;
}

D XOR Game

不難發現Bob可以湊出所有2n個數分成n對數的所有情況。

考慮把所有數補前導0扔字典樹。

如果字典數左右兩子樹對應的字元串均為偶數，則顯然答案不會出現分别選取兩個子樹中的數湊成一對的情況，遞歸求解

否則答案為左右2子樹湊1對，剩下的分别在子樹内湊。

#include<bits/stdc++.h> 
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
						For(j,m-1) cout<<a[i][j]<<' ';\
						cout<<a[i][m]<<endl; \
						} 
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
	int x=0,f=1; char ch=getchar();
	while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
	while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
	return x*f;
} 
#define MXAN (112345)
int n;
ll a[412345];
#define maxn (400000*31)
struct XorTrie{
    int cnt = 0,dfn = 0;
    int t[maxn][2];
    int L[maxn],R[maxn];

    int s[maxn];
    void _init(){
        cnt = 0;
        dfn = 0;
    }
    void Insert(ll x,int id){
        int rt = 0;s[rt]++;
        for(int k=32;k>=0;k--){
            int op = (x>>k&1ll)?1:0;
            if(!t[rt][op]) t[rt][op] = ++cnt;
            rt = t[rt][op];
            s[rt]++;
			if(!L[rt]) L[rt] = id;
            R[rt] = id;
			
        }
    }
    ll AnswerPos(int rt,int pos,ll x){
        ll ans = 0;
        for(int i=pos;i>=0;i--){
            int op = (x>>i&1ll)?1:0;
            if(t[rt][op]) rt = t[rt][op];
            else{
                rt = t[rt][!op];
                ans += (1ll<<i);
            }
        }return ans;
    }
    ll Divide(int rt,int pos){
        if(t[rt][0]&&t[rt][1]){
            int x = t[rt][0],y = t[rt][1];
            if(s[x]%2==0 && s[y]%2==0) return max(Divide(t[rt][0],pos-1),Divide(t[rt][1],pos-1));
            ll minl = INF;
            for(int k=L[x];k<=R[x];k++) minl = min(minl,AnswerPos(y,pos-1,a[k])+(1ll<<pos));
            return minl;
        }
        else if(t[rt][0]) return Divide(t[rt][0],pos-1);
        else if(t[rt][1]) return Divide(t[rt][1],pos-1);
        return 0;
    }


}S;
int main()
{
//	freopen("A.in","r",stdin);
//	freopen(".out","w",stdout);
	
    n=read()*2;
    For(i,n) cin>>a[i];
	sort(a+1,a+1+n);
	ll p=0;
    for(int i=1;i<=n;i++) S.Insert(a[i],i);	

    printf("%lld\n",S.Divide(0,32));

	return 0;
}
比賽題解 tries Fibonacci
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