A
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"；
<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
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
<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>
