Codeforces Round 383

A Arpa’s hard exam and Mehrdad’s naive cheat

求

1378n

个位是几。找规律。。。。

#include <cstdio>#include <cstring>const int d[]={6,8,4,2}; int main(){    int n;    while(~scanf("%d",&n)){        if(n==0)printf("1\n");        else{            printf("%d\n",d[n%4]);        }    }    return 0;}

B Arpa’s obvious problem and Mehrdad’s terrible solution

找有多少对 a[i] xor a[j] == x.
用map记录 x^a[i] 值的个数。因为异或的自反性， x^a[i]^a[i]==x

#include <cstdio>#include <cstring>#include <iostream>#include <map>#define LL long long using namespace std;int n,x;int a;map<int,int> mp;int main(){    scanf("%d%d",&n,&x);    mp.clear();    LL ans=0;    for(int i=1;i<=n;i++){        scanf("%d",&a);        if(mp[x^a]) ans+=mp[x^a];        mp[a]++;    }    cout<<ans<<endl;    return 0;}

C Arpa’s loud Owf and Mehrdad’s evil plan

找f[f[….f[x]…]]=y&&f[f[…f[y]…]]=x，t为嵌套的个数。
要求所有t的最小公倍数？
t为偶数的话可以除以2，然后一起求。

#include <cstdio>#include <cstring>#include <iostream>#include <climits>#define LL long long using namespace std;int a[1010];int len[1010];LL gcd(LL a,LL b){    return b==0?a:gcd(b,a%b);}LL lcm(LL a,LL b){    return a*b/gcd(a,b);}int main(){    int n;    scanf("%d",&n);    for(int i=1;i<=n;i++) scanf("%d",a+i);    for(int i=1;i<=n;i++){        int t=1,x=a[i];        while(1){            x=a[x];            if(x==i){                t++;                if(t%2==0) t=t/2;                len[i]=t;                break;            }            t++;            if(t>n) break;        }    }    LL ans=0;    for(int i=1;i<=n;i++){        if(len[i]==0){            ans=-1;            break;        }        if(ans==0){            ans = len[i];            continue;        }        ans = lcm(ans,len[i]);    }    cout<<ans<<endl;    return 0;}