先说C

题目链接：http://codeforces.com/problemset/problem/471/C

题目意思：有 n 张卡，问能做成多少种不同楼层（floor）的 house。注意这 n 张卡都要用光。每层 floor 都由一定的 room 构成，每两个相邻 room 规定要有一个公共的ceiling。规定从上到下看，每层 floor 的 room 的数量呈递增的形式排布。

这样的东西一般就是看图，先自己从小数開始推算找规律 能够发现第i层须要(3*i+2)个，那么前i层总的最少须要就是等差数列求和得(3*i+1)*i/2。  由于不能剩余。那么n-(3*i+1)*i/2必须能被3整除。那么从1到(3*i+1)*i/2<=n遍历一下即可 n=10^12  所以O(1e6)时间还是够的

//#pragma comment(linker, "/STACK:102400000,102400000")
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <iostream>
#include <iomanip>
#include <cmath>
#include <map>
#include <set>
#include <queue>
using namespace std;#define ls(rt) rt*2
#define rs(rt) rt*2+1
#define ll long long
#define ull unsigned long long
#define rep(i,s,e) for(int i=s;i<e;i++)
#define repe(i,s,e) for(int i=s;i<=e;i++)
#define CL(a,b) memset(a,b,sizeof(a))
#define IN(s) freopen(s,"r",stdin)
#define OUT(s) freopen(s,"w",stdout)
const ll ll_INF = ((ull)(-1))>>1;
const double EPS = 1e-8;
const double pi = acos(-1.0);
const int INF = 100000000;int main()
{ll n;while(~scanf("%I64d",&n)){ll ans=0,tmp;for(ll i=1;(tmp=(3*i+1)*i/2)<=n;i++){if( ( n-tmp )%3 == 0)ans++;}printf("%I64d\n",ans);}return 0;
}

B,  首先能不能出现三种以上的排列，写代码时用了类似离散化的写法，能的话  就随便改两个数的次序就能凑够3种

//#pragma comment(linker, "/STACK:102400000,102400000")
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <iostream>
#include <iomanip>
#include <cmath>
#include <map>
#include <set>
#include <queue>
using namespace std;#define ls(rt) rt*2
#define rs(rt) rt*2+1
#define ll long long
#define ull unsigned long long
#define rep(i,s,e) for(int i=s;i<e;i++)
#define repe(i,s,e) for(int i=s;i<=e;i++)
#define CL(a,b) memset(a,b,sizeof(a))
#define IN(s) freopen(s,"r",stdin)
#define OUT(s) freopen(s,"w",stdout)
const ll ll_INF = ((ull)(-1))>>1;
const double EPS = 1e-8;
const double pi = acos(-1.0);
const int INF = 100000000;const int MAXN = 2000+20;
struct Node{int id,v;
}p[MAXN];
int n;
bool cmp(Node a, Node b)
{return a.v<b.v;
}void print()
{printf("%d",p[1].id);for(int i=2;i<=n;i++)printf(" %d",p[i].id);putchar('\n');
}void SW(Node &a, Node &b)
{Node tmp;tmp=a;a=b;b=tmp;
}int main()
{while(~scanf("%d",&n)){for(int i=1;i<=n;i++){scanf("%d",&p[i].v);p[i].id=i;}sort(p+1,p+n+1,cmp);p[n+1].v=-1;int tmp=1,pre=p[1].v;ll cnt=1;int flag=0;for(int i=2;i<=n+1;i++)if(pre!=p[i].v){if(tmp>=3){flag=1;break;}cnt*=tmp;if(cnt>=3){flag=1;break;}tmp=1;pre=p[i].v;}else{tmp++;}if(!flag)puts("NO");else{puts("YES");print();int pr=p[1].v,tmp=1;int cnt=1;for(int i=2;i<=n+1;i++)///if(pr!=p[i].v){if(tmp==2){SW(p[i-1],p[i-2]);if(cnt<3)print(),cnt++;;}if(tmp >= 3){SW(p[i-1],p[i-2]);if(cnt<3) print(),cnt++;//cnt++;SW(p[i-1],p[i-3]);if(cnt<3)print(),cnt++;//cnt++;}if(cnt >=3)break;tmp=1;pr=p[i].v;}else{tmp++;}}}return 0;
}

A  水  只是由于flag少写了一个 WA了一次

//#pragma comment(linker, "/STACK:102400000,102400000")
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <iostream>
#include <iomanip>
#include <cmath>
#include <map>
#include <set>
#include <queue>
using namespace std;#define ls(rt) rt*2
#define rs(rt) rt*2+1
#define ll long long
#define ull unsigned long long
#define rep(i,s,e) for(int i=s;i<e;i++)
#define repe(i,s,e) for(int i=s;i<=e;i++)
#define CL(a,b) memset(a,b,sizeof(a))
#define IN(s) freopen(s,"r",stdin)
#define OUT(s) freopen(s,"w",stdout)
const ll ll_INF = ((ull)(-1))>>1;
const double EPS = 1e-8;
const double pi = acos(-1.0);
const int INF = 100000000;int len[10],vis[12];int main()
{while(~scanf("%d%d%d%d%d%d",len,len+1,len+2,len+3,len+4,len+5)){int ans=0;CL(vis,0);sort(len,len+6);int cnt=0,flag=0;for(int i=0;i<6;i++){vis[len[i]]++;if(vis[len[i]] == 4)cnt=i,flag=1;}if(vis[len[cnt]] == 5){puts("Bear");continue;}if(vis[len[cnt]] == 6){puts("Elephant");continue;}int last=-1,ff=0;for(int i=0;i<6;i++)if(i!=cnt){if(vis[len[i]] == 2)ff=1;if(last==-1){len[0]=len[i];last=1;}else  len[5]=len[i];}if(ff && flag){puts("Elephant");continue;}if(len[0]!=len[5] && flag){puts("Bear");continue;}if(len[0]==len[5] && flag){puts("Elephant");continue;}puts("Alien");}return 0;
}