Valentine's Day Round（HDU BC比赛）

Ferries Wheel Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 258 Accepted Submission(s): 97 Problem Description The Ferries Wheel is a circle,rounded by many cable cars,and the cars are numbered 1,2,3...K−1,K in order.Every cable car has a unique value and A[i−1]<A[i]<A[i+1](1<i<K). Today,Misaki invites N friends to play in the Ferries Wheel.Every one will enter a cable car. One person will receive a kiss from Misaki,if this person satisfies the following condition: (his/her cable car's value + the left car's value ) % INT_MAX = the right car's value,the 1st car’s left car is the kth car,and the right one is 2nd car,the kth car’s left car is the (k−1)th car,and the right one is the1st car. Please help Misaki to calculate how many kisses she will pay,you can assume that there is no empty cable car when all friends enter their cable cars,and one car has more than one friends is valid. Input There are many test cases. For each case,the first line is a integer N(1<=N<=100) means Misaki has invited N friends,and the second line contains N integers val1,val2,...valN, the val[i] means the ith friend's cable car's value. (0<=val[i]<= INT_MAX). The INT_MAX is 2147483647. Output For each test case, first output Case #X: ,then output the answer, if there are only one cable car, print "-1"; Sample Input 3 1 2 3 5 1 2 3 5 7 6 2 3 1 2 7 5 Sample Output Case #1: 1 Case #2: 2 Case #3: 3 HintIn the third sample, the order of cable cars is {{1},{2}, {3}, {5}, {7}} after they enter cable car,but the 2nd cable car has 2 friends,so the answer is 3. Source Valentine's Day Round Recommend hujie | We have carefully selected several similar problems for you: 5177 5176 5175 5173 5172 Statistic | Submit | Discuss | Note

题意：有N个人去做摩天轮，每个人进一个缆车，问有多少人满足：【(他的缆车的值+左边车的值)%INT_MAX=右边缆车的值】.
解法：暴力，记录每个缆车出现的次数，排序去重，枚举缆车的值，判断是否满足那个等式即可。

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<queue>
#include<vector>
using namespace std;
int aa[111],aaa[111];
struct node{
	int v;
	int id;
}a[111];
bool cmp(node a,node b)
{
	return a.v<b.v;
}
__int64 m[222],ans;
int main()
{
	int n,i,cnt,t;
	t=0;
	while(~scanf("%d",&n))
	{
		t++;
		for(i=1;i<=n;i++)
		{
			scanf("%d",&a[i].v);
			a[i].id=i;
		}
		sort(a+1,a+n+1,cmp);
		int nn=1;
		aa[a[1].id]=1;
		for(i=2;i<=n;i++)
		{
			if(a[i].v!=a[i-1].v)
			{
				nn++;
				aa[a[i].id]=nn;
			}
			else
			{
				aa[a[i].id]=nn;
			}
			
		}
		memset(aaa,0,sizeof(aaa));
		for(i=1;i<=n;i++)
		{
			aaa[aa[i]]++;
		}
		cnt=1;
		m[cnt]=a[1].v;
		for(i=2;i<=n;i++)
		{
			if(a[i].v!=a[i-1].v)
			{
				m[++cnt]=a[i].v;
			}
		}
		if(cnt==1)
		{
			ans=-1;
			printf("Case #%d: %d\n",t,ans);
		}
		else
		{
				ans=0;
		for(i=2;i<=cnt-1;i++)
		{
			if((m[i]+m[i-1])%2147483647==m[i+1])
			{
				ans+=aaa[i];
			}
		}
		if((m[cnt]+m[cnt-1])%2147483647==m[1])
		{
			ans+=aaa[cnt];
		}
		if((m[1]+m[cnt])%2147483647==m[2])
		{
			ans+=aaa[1];
		}
		printf("Case #%d: %I64d\n",t,ans);
		}
		
	}
	return 0;
}

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
using namespace std;
#define mst(A,k) memset(A,k,sizeof(A))
typedef long long ll;
const ll MOD = 2147483647;
const double eps = 1e-8;
const int INF = 100000000;
const int MAX_N =100005;
const int MAX_M = 100005;
int main()
{
	int n,cnt,cas=0;
	ll a[105],k;
	map<ll,int>mp;
	while(scanf("%d",&n)!=EOF){
		mp.clear();
		cnt=0;
		cas++;
		for(int i=0;i<n;i++)
		{
			scanf("%I64d",&k);
			a[cnt]=k;
			mp[k]++;
			if(mp[k]==1)
			{
				cnt++;
			}
		}
		sort(a,a+cnt);
		int ans = 0;
		for(int i=0;i<cnt;i++){
			if((a[i]+a[(i+1)%cnt])%MOD ==a[(i+2)%cnt])
			{
				ans+=mp[a[(i+1)%cnt]];
			}
		}
		if(cnt==1)
		{
			ans=-1;
		}
		cout<<"Case #"<<cas<<": "<<ans<<"\n";
	}
	return 0;
}

3
5
15

Case #1:
1
2
Case #2:
1
4
Case #3:
3
10 12 14

HintIn the third sample, gcd(15,10)=5 and (15 xor 10)=5, gcd(15,12)=3 and (15 xor 12)=3,gcd(15,14)=1 and (15 xor 14)=1

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
using namespace std;
#define mst(A,k) memset(A,k,sizeof(A))
typedef long long ll;
ll n,ans[10000],p[10000];
int m,c;
ll gcd(ll a,ll b){
	if(a<b) swap(a,b);
	ll c = a%b;
	while(c>0)
	{
		a=b;
		b=c;
		c=a%b;
	}
	return b;
}
int main()
{
	int t=0;
	while(~scanf("%I64d",&n)){
		m=c=0;
		for(ll i=1;i*i<=n;i++)
		{
			if(n%i==0){
				p[m++]=i;
				if(n/i!=i){
					p[m++]=n/i;
				}
			}
		}
		sort(p,p+m);
		for(int i=0;i<m;i++)
		{
			ll x=n^p[i];
			if(0<x&&x<=n&&gcd(x,n)==p[i]){
				ans[c++]=x;
			}
		}
		printf("Case #%d:\n%d\n",++t,c);
		sort(ans,ans+c);
		for(int i=0;i<c;i++)
		{
			if(i) putchar(' ');
			printf("%I64d",ans[i]);
		}
		puts("");
	}
	return 0;
}

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<map>
#include<vector>
using namespace std;
__int64 gcd(__int64 a,__int64 b)
{
	if(a<b)
	{
		swap(a,b);
	}
	__int64 c = a%b;
	while(c>0)
	{
		a=b;
		b=c;
		c=a%b;
	}
	return b;
}

int main()
{
	__int64 n,m,k,k2,ans[100011];
	int t=0,cnt;
	while(~scanf("%I64d",&n))
	{
		t++;
		cnt=0;
		for(k=1;k*k<=n;k++)
		{
			if(n%k==0)
			{
				m=n^k;
				if(m>=1&&m<=n&&gcd(n,m)==k)
				{
					ans[++cnt]=m;
				} 
				k2=n/k;
				if(k!=k2)
				{
					m=n^k2;
					if(m>1&&m<=n&&gcd(n,m)==k2)
					{
						ans[++cnt]=m;
					}
				}
			}
		}
		sort(ans+1,ans+cnt+1);
		 printf("Case #%d:\n%d\n", t, cnt);
		if(cnt>0)
		{
			printf("%I64d",ans[1]);
			for(int i=2;i<=cnt;i++)
			{
				printf(" %I64d",ans[i]);
			}
			//printf("\n");<strong>//注意格式，写在这里会PE！！！因为题目输出要求是3行，即使cnt==0也要输出空行！！！故只能写在if语句外！！！</strong>
			
		}
		printf("\n");
	}
	return 0;
}