CF#269DIV2A,B,C,D_html/css_懂视

CF#269DIV2A,B,C,D_html/css

2020-11-27 15:55:51 责编:小采

http://codeforces.com/contest/471/problem/A

解题思路：给你6个数，问是否有至少4个数都相等，没有的话输出“Alien”，有的话再看剩下的两个数，如果相等就输出"

Elephant"，否则

输出"

Bear"；

#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #define Maxn 100005#define Maxm 1000005#define lowbit(x) x&(-x)#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define PI acos(-1.0)#define make_pair MP#define LL long long #define Inf (1LL<<62)#define inf 0x3f3f3f3f#define re freopen("in.txt","r",stdin)#define wr freopen("out.txt","w",stdout)using namespace std;int main(){	int a[6],flag;	//re;wr;	while(~scanf("%d%d%d%d%d%d",&a[0],&a[1],&a[2],&a[3],&a[4],&a[5]))	{	flag=0;	sort(a,a+6);	if(a[0]==a[1]&&a[2]==a[1]&&a[3]==a[2])	flag=1;	if(a[1]==a[2]&&a[2]==a[3]&&a[3]==a[4])	flag=2;	if(a[2]==a[3]&&a[3]==a[4]&&a[4]==a[5])	flag=3;	if(flag==0)	{	puts("Alien");	continue;	}	if(flag==1)	{	if(a[4]==a[5])	{	puts("Elephant");	continue;	}	else	{	puts("Bear");	continue;	}	}	else if(flag==2)	{	if(a[0]==a[5])	{	puts("Elephant");	continue;	}	else	{	puts("Bear");	continue;	}	}	else	{	if(a[0]==a[1])	{	puts("Elephant");	continue;	}	else	{	puts("Bear");	continue;	}	}	}	return 0;}
B

http://codeforces.com/contest/471/problem/B

解题思路：给你一个序列，问是否有三种不同的方法使它们按非减序排序，显然只有有2个相等的元素集合数大于等于2时或者有3个相等的元素的集合时才有解，有解的时候集合内排序一下，我写的很繁，导致后面的题目没写，整场就跪了

#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #define Maxn 100005#define Maxm 1000005#define lowbit(x) x&(-x)#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define PI acos(-1.0)#define make_pair MP#define LL long long #define Inf (1LL<<62)#define inf 0x3f3f3f3f#define re freopen("in.txt","r",stdin)#define wr freopen("out.txt","w",stdout)using namespace std;struct Mask{	int dif;	int id;	friend bool operator <(Mask a,Mask b)	{	if(a.dif!=b.dif)	return a.dif2)	{	flag=true;	break;	}	if(m[i]==2)	cnt++;	}	if(flag||cnt>=2)	{	puts("YES");	sort(arr+1,arr+1+n);	if(flag)	{	for(i=1;i<=n;i++)	{	if(m[arr[i].dif]>2)	break;	}	int a=arr[i].id;	int b=arr[i+1].id;	int c=arr[i+2].id;	//couti+2)	{	printf("%d%c",arr[j].id,j==n?'\n':' ');	}	else if(i==n-2)	{	printf("%d %d %d\n",b,a,c);	j+=2;	}	else	{	printf("%d %d %d ",b,a,c);	j+=2;	}	}	for(j=1;j<=n;j++)	{	if(ji+2)	{	printf("%d%c",arr[j].id,j==n?'\n':' ');	}	else if(i==n-2)	{	printf("%d %d %d\n",c,b,a);	j+=2;	}	else	{	printf("%d %d %d ",c,b,a);	j+=2;	}	}	}	else	{	int f1,f2;	for(i=1;i<=n;i++)	if(m[arr[i].dif]==2)	{	f1=i;	break;	}	for(i=i+2;i<=n;i++)	if(m[arr[i].dif]==2)	{	f2=i;	break;	}	int a=arr[f1].id;	int b=arr[f1+1].id;	int c=arr[f2].id;	int d=arr[f2+1].id;	//coutf2+1))	printf("%d%c",arr[i].id,i==n?'\n':' ');	else if(i>=f1&&i<=f1+1)	{	printf("%d %d ",a,b);	i+=1;	}	else if(i>=f2&&i<=f2+1)	{	if(f2==n-1)	printf("%d %d\n",c,d);	else	printf("%d %d ",c,d);	i+=1;	}	}	for(i=1;i<=n;i++)	{	if((if1+1)&&(if2+1))	printf("%d%c",arr[i].id,i==n?'\n':' ');	else if(i>=f1&&i<=f1+1)	{	printf("%d %d ",a,b);	i+=1;	}	else if(i>=f2&&i<=f2+1)	{	if(f2==n-1)	printf("%d %d\n",d,c);	else	printf("%d %d ",d,c);	i+=1;	}	}	for(i=1;i<=n;i++)	{	if((if1+1)&&(if2+1))	printf("%d%c",arr[i].id,i==n?'\n':' ');	else if(i>=f1&&i<=f1+1)	{	printf("%d %d ",b,a);	i+=1;	}	else if(i>=f2&&i<=f2+1)	{	if(f2==n-1)	printf("%d %d\n",c,d);	else	printf("%d %d ",c,d);	i+=1;	}	}	}	}	else	{	puts("NO");	continue;	}	}}
C

http://codeforces.com/contest/471/problem/C

解题思路：首先二分出n个棍子最多可以搭几层，然后从1开始枚举，注意到每一层的数目都是一个公差为3的等差数列，判断一下这一层能否满足n的需求即可

#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #define Maxn 100005#define Maxm 1000005#define lowbit(x) x&(-x)#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define PI acos(-1.0)#define make_pair MP#define LL long long #define Inf (1LL<<62)#define inf 0x3f3f3f3f#define re freopen("in.txt","r",stdin)#define wr freopen("out.txt","w",stdout)using namespace std;LL cal(LL n){	return (3*n+1)*n/2;}LL calc(LL a,LL k,LL n){	return a+(n-1)*k;}int main(){	LL n,ans;	//re;wr;	while(~scanf("%I64d",&n))	{	ans=0;	LL l=0,r=(Maxm<<1),lv;	while(r>=l)	{	LL m=(l+r)>>1;	if(cal(m)<=n)	{	lv=m;	l=m+1;	}	else	r=m-1;	}	for(LL i=1;i<=lv;i++)	{	LL a=cal(i);	if((n-a)%3==0)	ans++;	}	printf("%d\n",ans);	}	return 0;}
D

http://codeforces.com/problemset/problem/471/D

解题思路：做出两个数列的相邻两项的差分数列，KMP判断短的差分数列在长的差分数列中出现几次即可，特判n=1和w=1的情况

#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #define Maxn 100005#define Maxm 1000005#define lowbit(x) x&(-x)#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define PI acos(-1.0)#define make_pair MP#define LL long long #define Inf (1LL<<62)#define inf 0x3f3f3f3f#define re freopen("in.txt","r",stdin)#define wr freopen("out.txt","w",stdout)using namespace std;int next[Maxn<<1];void get_next(int *arr,int len){	int j=0,k=-1;	next[0]=-1;	while(j<=len)	{	if(k==-1||arr[j]==arr[k])	{	j++;k++;	next[j]=k;	}	else	k=next[k];	}}int kmp(int *a,int *b,int l1,int l2){	get_next(a,l1);	get_next(b,l2);	int i=0,j=0,ans=0;	while(i=1)	ca[i-1]=a[i]-a[i-1];	}	for(int i=0;i=1)	cb[i-1]=b[i]-b[i-1];	}	if(m==1||n==1)	{	printf("%d\n",max(m,n));	continue;	}	printf("%d\n",m>=n?kmp(ca,cb,m-1,n-1):kmp(cb,ca,n-1,m-1));	}	return 0;}

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