你最愿意做的哪件事，才是你的天赋所在

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牛客多校10题解报告

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B

题目链接

B

思路

观察到S(n)=S(n-2)S(n-3)S(n-2),按照位置和长度进行搜索，因为只有1e12长度，先前缀处理出这个长度，就可以搜索了。

代码实现

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#include<bits/stdc++.h>
using namespace std;
#define ll long long
const ll md = 1e12+20;
ll f[501];
string s[5];
int n;
ll k;
void dfs(ll pos,int len,int num)
{
	if(num<=3)
	{
	cout<<s[num].substr(pos,len);
	return;
	}
	if(num>56)
	{
		dfs(pos,len,56);
		return ;
	}
	ll len1=f[num-2],len2=f[num-2]+f[num-3];
	if(pos+len<len1)
	{  
		dfs(pos,len,num-2);
	}
	else if(pos<=len1&&pos+len>=len1)
	{
		if(pos+len>=len2)
		{
			dfs(pos,10,num-2);
		    dfs(0,pos-len1+len,num-3);
		    dfs(0,pos-len2+len,num-2);
		}
		else
		{
		dfs(pos,10,num-2);
		dfs(0,pos-len1+len,num-3);
	    }
	}
	else if(pos+len<=len2)
	{
		dfs(pos-len1,len,num-3);
	}
	else if(pos<len2&&pos+len>=len2)
	{
		dfs(pos-len1,10,num-3);
		dfs(0,pos-len2+len,num-2);
	}
	else if(pos>=len2)
	{
		dfs(pos-len2,len,num-2);
	}
}
int main()
{
	f[1]=6;
	f[2]=7;
	bool ok =false;
	for(int i=3;i<=500;i++)
	{
		f[i]=(f[i-1]+f[i-2])%md;
	}
	for(int i=56;i<=500;i++)f[i]+=md;
	int n;
	scanf("%d",&n);
	s[1]="COFFEE";
	s[2]="CHICKEN";
	s[3]="COFFEECHICKEN";
	int a[34];
	/*for(int i=1;i<=500;i++)
	{   printf("i:%d ",i);
		dfs(i-1,10,499);
		printf("\n");
	}*/
	for(int i=0;i<n;i++)
	{
		ll x,y;
		cin>>x>>k;
		dfs(k-1,10,x);
		printf("\n");
	}
 }

D

题目链接

D

思路

是一道EXCRT的板子题，但是需要用__int128来维护，打比赛的时候过了%75样例，代码不规范的原因。
具体的思路在这篇博客中说的很清楚了
中国剩余定理

代码实现

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#include<bits/stdc++.h>
using namespace std;
#define ll __int128
long long AK[105],AC[105];
long long m;
int n;
ll exgcd(ll a,ll b,ll &x,ll &y)
{
	if(b==0){x=1;y=0;return a;}
	ll gcd = exgcd(b,a%b,x,y);
	ll temp = x;
	x=y;
	y=temp-a/b*y;
	return gcd;
}
ll excrt()
{
	ll a=AK[0],b=AC[0];
	ll M=a,sum=b;
	int ok=0;
	int ok1=0;
	for(int i=1;i<n;i++)
	{
		ll x,y;
		a=AK[i],b=AC[i];
		b=((b-sum)%a+a)%a;
		ll gcd = exgcd(M,a,x,y);
		if(b%gcd)ok=-1;
		x=x*(b/gcd)%(a/gcd); 
		x=(x+a)%(a/gcd);
		sum+=x*M;
		M=M/gcd*a;
		if(sum>m)
		{
			ok1=1;
		}
	}
	if(ok)puts("he was definitely lying");
	else if(ok1==1||sum>m) puts("he was probably lying");
	else return printf("%lld\n",(long long)sum);
}
int main()
{
	scanf("%d %lld",&n,&m);
	for(int i=0;i<n;i++)scanf("%lld %lld",&AK[i],&AC[i]);
	excrt();
}

E

题目链接

E

思路

用分治的思想，第一象限关于主对角线对称，第二象限关于副对角线对称，然后2*2的可以轻易求出来。
得到一个点以及大小后，不断地往回推即可

代码实现

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using namespace std;
#define ll long long
ll a[1025][1025];
ll b[2][2]={{1,4},{2,3}};
const int maxn = 1e6+10;
struct node
{
	ll x, y;
	int cp;
}arr[maxn];
int check(ll x,ll y,int num)
{
	ll k = 1<<num;
	if(x<=k/2&&y<=k/2)return 1;
	else if(x>=k/2&&y<=k/2)return 3;
	else if(x<=k/2&&y>=k/2)return 2;
	else return 4;
}
ll solve(ll x,ll y,int k)
{
	ll num = 1<<k;
	ll half = num/2;
	if(k==1)return b[--x][--y];
	int area = check(x,y,k);
	if(area==1)
	{
	return 	solve(y,x,k-1);
	}
	else if(area==2)
	{
		y-=half;
	return	half*half*3+solve(half-y+1,half-x+1,k-1);
	}
	else if(area==3)
	{
		return half*half*1+solve(x-half,y,k-1);
	}
	else if(area==4)
	{
		return half*half*2+solve(x-half,y-half,k-1);
	}
}
int cmp(node s,node t)
{
	return s.cp<t.cp;
}
int main()
{
	int n,k;
	scanf("%d %d",&n,&k);
	ll x,y;
	for(int i=0;i<n;i++)
	{
		scanf("%lld %lld",&arr[i].x,&arr[i].y);
		arr[i].cp=solve1(arr[i].x,arr[i].y,k);
	}
	sort(arr,arr+n,cmp);
	for(int i=0;i<n;i++)
	{
		printf("%lld %lld\n",arr[i].x,arr[i].y);
	}
}
-------------你最愿意做的哪件事才是你的天赋所在-------------