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Link：http://acm.hdu.edu.cn/showproblem.php?pid=5174
Ferries WheelTime 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 the 1st 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
Sample Output
Source
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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。
AC代码1（直接模拟+离散化）：
#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;
}
代码2（用map容器记录数据）：
#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;
}Link：http://acm.hdu.edu.cn/showproblem.php?pid=5175
Misaki's Kiss againTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 301 Accepted Submission(s): 81
Problem Description
After the Ferries Wheel, many friends hope to receive the Misaki's kiss again,so Misaki numbers them
1,2...N−1,N
,if someone's number is M and satisfied the
GCD(N,M)
equals to
N
XOR
M
,he will be kissed again.
Please help Misaki to find all M(1<=M<=N) . Note that: GCD(a,b) means the greatest common divisor of a and b . A XOR B means A exclusive or B
Input
There are multiple test cases.
For each testcase, contains a integets N(0<N<=1010)
Output
For each test case,
first line output Case #X:, second line output k means the number of friends will get a kiss. third line contains k number mean the friends' number, sort them in ascending and separated by a space between two numbers
Sample Input
Sample Output
Source
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AC代码：
#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;
}
BC#30 1002 解题思路
题意即求解出 M ，如果通过遍历所有的 m 来求解，算法复杂度是 O(n)，会 TLE．
枚举n的约数，将n的约数看成最大公约数，看是否存在0<m<n满足题意，也就是倒过来推 。
N^M=K，这里的M和K是一一对应的，唯一的M对应唯一的K，唯一的K对应唯一的M，而且异或运算满足逆运算，即N^K=M
#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;
}
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