CF #269 DIV2 A,B,C,D_html/css_WEB-ITnose

A

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

Problem-solving idea: Give you 6 numbers and ask if at least 4 numbers are equal. If not, output "Alien". If yes, look at the remaining two numbers. If they are equal, output "

Elephant"，否则输出"

Bear"；

<pre name="code" class="n">#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <set>#include <map>#include <list>#include <queue>#include <stack>#include <deque>#include <vector>#include <bitset>#include <cmath>#include <utility>#define Maxn 100005#define Maxm 1000005#define lowbit(x) x&(-x)#define lson l,m,rtB</utility></cmath></bitset></vector></deque></stack></queue></list></map></set></algorithm></string></cstring></cstdio></iostream>

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

<pre name="code" class="n">#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <set>#include <map>#include <list>#include <queue>#include <stack>#include <deque>#include <vector>#include <bitset>#include <cmath>#include <utility>#define Maxn 100005#define Maxm 1000005#define lowbit(x) x&(-x)#define lson l,m,rt2)			{				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;i2)						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",a,b,c);						j+=2;					}					else					{						printf("%d %d %d ",a,b,c);						j+=2;					}				}				for(j=1;ji+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;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;if1+1)&&(i<f2>f2+1))						printf("%d%c",arr[i].id,i==n?'\n':' ');					else if(i>=f1&&i=f2&&if1+1)&&(i<f2>f2+1))						printf("%d%c",arr[i].id,i==n?'\n':' ');					else if(i>=f1&&i=f2&&if1+1)&&(i<f2>f2+1))						printf("%d%c",arr[i].id,i==n?'\n':' ');					else if(i>=f1&&i=f2&&iC </f2></f2></f2></utility></cmath></bitset></vector></deque></stack></queue></list></map></set></algorithm></string></cstring></cstdio></iostream>

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

Problem-solving idea: first divide n sticks into two to build up to how many layers, and then start enumerating from 1. Note that the number of each layer is one For an arithmetic sequence with a tolerance of 3, just judge whether this layer can meet the needs of n

<pre name="code" class="n">#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <set>#include <map>#include <list>#include <queue>#include <stack>#include <deque>#include <vector>#include <bitset>#include <cmath>#include <utility>#define Maxn 100005#define Maxm 1000005#define lowbit(x) x&(-x)#define lson l,m,rt=l)		{			LL m=(l+r)>>1;			if(cal(m)<strong>D</strong></utility></cmath></bitset></vector></deque></stack></queue></list></map></set></algorithm></string></cstring></cstdio></iostream>

<strong>http://codeforces.com/problemset/problem/471/D</strong>

<strong>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</strong>

<strong></strong><pre name="code" class="n">#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <set>#include <map>#include <list>#include <queue>#include <stack>#include <deque>#include <vector>#include <bitset>#include <cmath>#include <utility>#define Maxn 100005#define Maxm 1000005#define lowbit(x) x&(-x)#define lson l,m,rt=1)				ca[i-1]=a[i]-a[i-1];		}		for(int i=0;i<n scanf if>=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;}</n></utility></cmath></bitset></vector></deque></stack></queue></list></map></set></algorithm></string></cstring></cstdio></iostream>