A
http://codeforces.com/contest/471/problem/A
解題思路：給你6個數，問是否有至少4個數都相等，沒有的話輸出“Alien”，有的話再看剩下的兩個數，如果相等就輸出"
Elephant"，否則輸出"
Bear"；
#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
#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;
}http://codeforces.com/contest/471/problem/B
解題思路：給你一個序列，問是否有三種不同的方法使它們按非減序排序，顯然只有有2個相等的元素集合數大於等於2時或者有3個相等的元素的集合時纔有解，有解的時候集合內排序一下，我寫的很繁，導致後面的題目沒寫，整場就跪了
#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
#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.dif<b.dif;
else
return a.id<b.id;
}
};
int main()
{
int n,m[2005],cnt,i,j;
bool flag;
Mask arr[2005];
//re;wr;
while(~scanf("%d",&n))
{
flag=false;
cnt=0;
memset(m,0,sizeof(m));
for(i=1;i<=n;i++)
{
scanf("%d",&arr[i].dif);
m[arr[i].dif]++;
arr[i].id=i;
}
for(i=0;i<=2000;i++)
{
if(m[i]>2)
{
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;
//cout<<a<<b<<c<<endl;
for(j=1;j<=n;j++)
{
//cout<<j<<endl;
if(j<i||j>i+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;j<=n;j++)
{
if(j<i||j>i+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(j<i||j>i+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;
//cout<<a<<b<<c<<d<<endl;
//cout<<f1<<f2<<endl;
for(i=1;i<=n;i++)
{
if((i<f1||i>f1+1)&&(i<f2||i>f2+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((i<f1||i>f1+1)&&(i<f2||i>f2+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((i<f1||i>f1+1)&&(i<f2||i>f2+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;
}
}
}http://codeforces.com/contest/471/problem/C
解題思路：首先二分出n個棍子最多可以搭幾層，然後從1開始枚舉，注意到每一層的數目都是一個公差爲3的等差數列，判斷一下這一層能否滿足n的需求即可
#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
#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;
}http://codeforces.com/problemset/problem/471/D
解題思路：做出兩個數列的相鄰兩項的差分數列，KMP判斷短的差分數列在長的差分數列中出現幾次即可，特判n=1和w=1的情況
#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
#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<l1)
{
if(j==-1||a[i]==b[j])
{
i++;j++;
}
else
j=next[j];
if(j==l2)
{
j=next[j];
ans++;
}
}
return ans;
}
int a[Maxn<<1],b[Maxn<<1],ca[Maxn<<1],cb[Maxn<<1];
int main()
{
int m,n;
//re;wr;
while(~scanf("%d%d",&m,&n))
{
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
memset(ca,0,sizeof(ca));
memset(cb,0,sizeof(cb));
for(int i=0;i<m;i++)
{
scanf("%d",a+i);
if(i>=1)
ca[i-1]=a[i]-a[i-1];
}
for(int i=0;i<n;i++)
{
scanf("%d",b+i);
if(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;
}