2014 ACM/ICPC Asia Regional Xi'an Online

2014 ACM/ICPC Asia Regional Xi'an Online

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A. Post Robot

把每种单词都kmp跑一遍，顺序输出即可

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <map>
#include <cstring>
#include <cmath>
#include <queue>
#include <set>
#include <vector>
#include <iterator>
#include <string>
#include <deque>
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
#define pb push_back
#define MP make_pair
#define fr first
#define sc second
#define PII pair<int,int>
#define VI vector<int>
typedef long long ll;
typedef unsigned long long ull;

inline int read() {
    char c=getchar();int x=0,f=1;
    while(!isdigit(c)){if(f=='-')f=-1;c=getchar();}
    while(isdigit(c)){x=x*10+c-'0';c=getchar();}
    return x*f;
}
const int N = 1000500;
using namespace std;

int n;
string s,t,s1="Apple",s2="iPhone",s3="iPod",s4="iPad",s5="Sony";
int nxt[N],vis[N];
void kmp_pre(string s) {
    int i,j,m=s.size();
    j=nxt[0]=-1;
    i=0;
    while(i<m){
        while(-1!=j&&s[i]!=s[j])j=nxt[j];
        nxt[++i]=++j;
    }
}
void kmp(string s,string t,int f) {
    int i,j,n=s.size(),m=t.size();
    i=j=0;
    while(i<n){
        while(-1!=j&&s[i]!=t[j])j=nxt[j];
        ++i;++j;
        if(j>=m) {
            //printf("%d ",i);
            vis[i] = f;
            j=nxt[j];
        }
    }
}
int main() {
//    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
    memset(vis,-1,sizeof(vis));
    int f=0;
    while(getline(cin,t)){
        if(f) s+=" ";f=1;
        s+=t;
    }

    kmp_pre(s1); kmp(s,s1,0);
    kmp_pre(s2); kmp(s,s2,0);
    kmp_pre(s3); kmp(s,s3,0);
    kmp_pre(s4); kmp(s,s4,0);
    kmp_pre(s5); kmp(s,s5,1);
    rep(i,1,s.size())
        if(vis[i]==0) puts("MAI MAI MAI!");
        else if(vis[i]==1) puts("SONY DAFA IS GOOD!");
    return 0;
}

E. Game

手推了几组，找规律的。把每一堆的大小都异或起来就星了。

#include <cstdio>
#include <algorithm>
#include <map>
#include <cstring>
#include <cmath>
#include <queue>
#include <set>
#include <vector>
#include <iterator>
#include <string>
#include <deque>
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
#define pb push_back
#define MP make_pair
#define fr first
#define sc second
#define PII pair<int,int>
#define VI vector<int>
typedef long long ll;
typedef unsigned long long ull;

inline int read() {
    char c=getchar();int x=0,f=1;
    while(!isdigit(c)){if(f=='-')f=-1;c=getchar();}
    while(isdigit(c)){x=x*10+c-'0';c=getchar();}
    return x*f;
}
using namespace std;

int n,x;

int main() {
    while(scanf("%d",&n)!=EOF){
        int ans=0;
        rep(i,1,n){
            int x=read();ans^=x;
        }
        if(!ans)puts("Lose");
        else puts("Win");
    }
    return 0;
}

F. Dice

直接从123456开始bfs出到每种状态的距离即可。查询时，替换下标即可。

#include <cstdio>
#include <algorithm>
#include <map>
#include <cstring>
#include <cmath>
#include <iostream>
#include <queue>
#include <set>
#include <vector>
#include <iterator>
#include <string>
#include <deque>
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
#define pb push_back
#define MP make_pair
#define fr first
#define sc second
#define PII pair<int,int>
#define VI vector<int>
typedef long long ll;
typedef unsigned long long ull;
const int N = 654321 + 10;
inline int read() {
    char c=getchar();int x=0,f=1;
    while(!isdigit(c)){if(f=='-')f=-1;c=getchar();}
    while(isdigit(c)){x=x*10+c-'0';c=getchar();}
    return x*f;
}
using namespace std;
int hs(string s) {
    int hs = 0;
    for(int i=0;i<6;++i) hs = hs*10 + (s[i]-'0');
    return hs;
}
string t;
string fhs(int x) {
    t.clear();
    while(x) {
        t += (char)(x%10+'0');
        x/=10;
    }
    reverse(t.begin(),t.end());
    return t;
}
string tx,tx2;
int rt1(int x) {
    tx = fhs(x);
    tx2 = tx;
    tx2[2-1] = tx[3-1];
    tx2[4-1] = tx[2-1];
    tx2[1-1] = tx[4-1];
    tx2[3-1] = tx[1-1];
    return hs(tx2);
}
int rt2(int x) {
    tx = fhs(x);
    tx2 = tx;
    tx2[2-1] = tx[4-1];
    tx2[3-1] = tx[2-1];
    tx2[1-1] = tx[3-1];
    tx2[4-1] = tx[1-1];
    return hs(tx2);
}
int rt3(int x) {
    tx = fhs(x);
    tx2 = tx;
    tx2[2-1] = tx[5-1];
    tx2[6-1] = tx[2-1];
    tx2[1-1] = tx[6-1];
    tx2[5-1] = tx[1-1];
    return hs(tx2);
}
int rt4(int x) {
    tx = fhs(x);
    tx2 = tx;
    tx2[2-1] = tx[6-1];
    tx2[5-1] = tx[2-1];
    tx2[1-1] = tx[5-1];
    tx2[6-1] = tx[1-1];
    return hs(tx2);
}
int dis[N],vis[N];
void bfs(string s) {
    memset(dis,-1,sizeof(dis));
    memset(vis,0,sizeof(vis));
    int u = hs(s);
    dis[u] = 0;
    queue<int> q;
    q.push(u);
    vis[u]=1;
    while(!q.empty()) {
        int u = q.front(); q.pop();
        int tu = u;
        tu = rt1(u);
        if(!vis[tu]) {
            vis[tu] = 1;
            q.push(tu);
            dis[tu] = dis[u] + 1;
        }
        tu = rt2(u);
        if(!vis[tu]) {
            vis[tu] = 1;
            q.push(tu);
            dis[tu] = dis[u] + 1;
        }
        tu = rt3(u);
        if(!vis[tu]) {
            vis[tu] = 1;
            q.push(tu);
            dis[tu] = dis[u] + 1;
        }
        tu = rt4(u);
        if(!vis[tu]) {
            vis[tu] = 1;
            q.push(tu);
            dis[tu] = dis[u] + 1;
        }
    }
}

int n,a[10],b[10];
map<int,int> c;
int main() {
    string s = "123456";
    bfs(s);
    while(scanf("%d%d%d%d%d%d",&a[1],&a[2],&a[3],&a[4],&a[5],&a[6])!=EOF) {
        scanf("%d%d%d%d%d%d",&b[1],&b[2],&b[3],&b[4],&b[5],&b[6]);
        c.clear();
        rep(i,1,6) c[a[i]] = i;
        int hs = 0;
        rep(i,1,6)hs=hs*10+c[b[i]];
        printf("%d\n",dis[hs]);
    }
    return 0;
}

H. Number Sequence

一开始想起一道题，觉得要从高位到低位分治贪心。然后发现过的人有点多啊。就开始找规律。。。于是顺利浪费了大量时间。。。首先发现最大值就是\(n*(n+1)\) ，而每一个值都可以用一些\(2^i-1\)组合成，于是我们从大到小贪心的把每个数都异或成\(2^i-1\)中的尽可能大的值即可

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <map>
#include <cstring>
#include <cmath>
#include <queue>
#include <set>
#include <vector>
#include <iterator>
#include <string>
#include <deque>
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
#define pb push_back
#define MP make_pair
#define fr first
#define sc second
#define PII pair<int,int>
#define VI vector<int>
typedef long long ll;
typedef unsigned long long ull;
const int N = 1e6 +100000;
inline int read() {
    char c=getchar();int x=0,f=1;
    while(!isdigit(c)){if(f=='-')f=-1;c=getchar();}
    while(isdigit(c)){x=x*10+c-'0';c=getchar();}
    return x*f;
}
using namespace std;
ll n;
ll a[N],b[N];
int vis[N];
void solve() {
    memset(vis,0,sizeof(vis));
    for(int i=n;i>=0;--i) {
        int e = 0;
        for(ll j=20;j>=0;--j) {
            e=(1<<j)-1;
            if(!vis[(e^i)]&&(e^i)<=n) {
                vis[(e^i)]=1;
                b[i] = (e^i);
                break;
            }
        }
    }
}
int main() {
    while(scanf("%lld",&n)!=EOF){
        rep(i,0,n)a[i]=read();
        solve();int f=0;
        printf("%lld\n",(n+1ll)*n);
        rep(i,0,n) {
            if(f)printf(" ");f=1;
            printf("%lld",b[a[i]]);
        }puts("");
    }
    return 0;
}

I. 233 Matrix

发现n非常小，m很大，于是想到矩阵快速幂，推一下就ok了

#include <cstdio>
#include <algorithm>
#include <map>
#include <cstring>
#include <cmath>
#include <queue>
#include <set>
#include <vector>
#include <iterator>
#include <string>
#include <deque>
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
#define pb push_back
#define MP make_pair
#define fr first
#define sc second
#define PII pair<int,int>
#define VI vector<int>
#define rg register
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 10000007;
inline int read() {
    char c=getchar();int x=0,f=1;
    while(!isdigit(c)){if(f=='-')f=-1;c=getchar();}
    while(isdigit(c)){x=x*10+c-'0';c=getchar();}
    return x*f;
}
using namespace std;
ll ans[20][20],c[20][20],d[20][20];
inline void mul(ll a[][20], ll b[][20], int n){
    for(rg int i=1;i<=n;++i)
        for(int j=1;j<=n;++j)ans[i][j]=0,c[i][j]=a[i][j],d[i][j]=b[i][j];
    for(rg int i=1;i<=n;++i)
    for(rg int k=1;k<=n;++k)if(c[i][k])
    for(rg int j=1;j<=n;++j)if(d[k][j]){
        ans[i][j] = (ans[i][j] + (c[i][k]*d[k][j]))%mod;
    }
    for(rg int i=1;i<=n;++i)
        for(rg int j=1;j<=n;++j)a[i][j]=ans[i][j];
}
ll E[20][20],A[20][20];
inline void mx_pow(ll A[][20],ll b,int n) {
     while(b) {
        if(b&1) mul(E,A,n);
        mul(A,A,n);
        b>>=1LL;
     }
}

int n;
ll m,a[20];
int main() {
    while(scanf("%d%lld",&n,&m)!=EOF) {
        memset(A,0,sizeof(A));
        memset(E,0,sizeof(E));
        rep(i,1,n) a[i] = read();
        a[n+1] = 233; a[n+2] = 1;
        rep(i,1,n+2)E[i][i] = 1;
        rep(i,1,n)rep(j,1,i) A[i][j] = 1;
        rep(i,1,n)A[i][n+1]=1;
        A[n+1][n+1] = 10; A[n+1][n+2] = 3;
        A[n+2][n+2]=1;

        mx_pow(A,m,n+2);

        ll res = 0;
        rep(i,1,n+2) res = (res + (E[n][i]*a[i])%mod)%mod;
        printf("%lld\n",res);
    }
    return 0;
}