2014-03-01 春季PAT 1073-1076解题报告

今天下午的PAT考试状态不理想，回来怒刷了一遍，解题报告如下：

1073. Scientific Notation (20)

基本模拟题，将一长串的科学计数转换为普通的数字表示方式。思路是是数组存储输入，然后找到指数位置，并根据指数的大小和正负对前面的小数进行针对性的处理。

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

class PAT{
public:
    enum { N = , };
    char num[N];
    int finger,epos,length;
    void run();
};

void PAT::run()
{
    scanf("%s", &num);
    length = strlen(num);
    epos = find(num, num + length, 'E') - num;
    sscanf(num + epos + , "%d", &finger);
    if (num[epos + ] == '-') finger = - * finger;
    if(num[]=='-')printf("%c", num[]);
    int i, j;
    if (finger >= )
    {
        printf("%c", num[]);
        for (i = ; i < epos && finger > ; i++,finger--) printf("%c", num[i]);
        if (i < epos) { printf("."); for (; i < epos; i++)printf("%c", num[i]); }
        else if (finger > ) while (finger > ){ printf(""); finger--; }
    }
    else
    {
        printf("0."); finger++;
        while (finger < ) {printf(""); finger++;}
        printf("%c", num[]);
        for (int i = ; i < epos; i++)printf("%c", num[i]);
    }
}
int main()
{
    //freopen("input.txt", "r", stdin);
    PAT *p = new PAT;
    p->run();
    return ;
}

PAT 1074. Reversing Linked List (25)

这个题不难，关键是要细心，首先要看懂题意。题目说的逆转排序是指每隔K个节点就将这k个节点的顺序颠倒过来，而如果最后不足K个节点的话，是不进行颠倒的，我下午考试的时候把这个意思弄错了，以为最后的节点也要进行颠倒，结果当然不对。这个题可以用map来做，不过据我同学说，用map查询会超时，我没有这样试过，不知道究竟如何，我使用的是hash的办法，直接在数组里面存储，速度会快点。

另外，这个题跟前面1052题的链表排序一样，有一个陷阱，就是不是所有给出的节点都属于此链表，需要判断，最后一个测试点是这个陷阱，不过这个点只有1分，姥姥出题的时候还是很人道。不过很多人这个点都没过。

#include <iostream>

using namespace std;

class PAT
{
public:
    enum{N=};
    struct Node
    {
        int addr, data, next;
    };
    Node node[N];
    int hash[N];//addr->index
    int addr[N];//index->addr
    int n,k,head;
    void run();
};

void PAT::run()
{
    scanf("%d%d%d", &head, &n, &k);
    for (int i = ; i < n; i++)
    {
        scanf("%d%d%d", &node[i].addr, &node[i].data, &node[i].next);
        hash[node[i].addr] = i;
    }
    int cur = head,c=;
    while (cur != -)
    {
        addr[c] = node[hash[cur]].addr;
        cur = node[hash[cur]].next;
        c++; //节点计数
    }
    n = c; //排除不在链表中的节点
    for (int i = ; i < n; i = i + k)
    {
        int r = (i + k<= n) ? i + k -  : n - ;
        if (i + k <= n)
        {
            for (int j = r; j > i; j--)
            {
                node[hash[addr[j]]].next = node[hash[addr[j - ]]].addr;
            }
            if (r == n - ) node[hash[addr[i]]].next = -;
            else node[hash[addr[i]]].next = (i +  * k <= n) ? node[hash[addr[i + *k -]]].addr : node[hash[addr[i+k]]].addr;
            if (r!=n-)
                for (int j = r; j >= i; j--) printf("%05d %d %05d\n", node[hash[addr[j]]].addr, node[hash[addr[j]]].data, node[hash[addr[j]]].next);
            else
            {
                for (int j = r; j > i; j--) printf("%05d %d %05d\n", node[hash[addr[j]]].addr, node[hash[addr[j]]].data, node[hash[addr[j]]].next);
                printf("%05d %d -1\n", node[hash[addr[i]]].addr, node[hash[addr[i]]].data);
            }
        }
        else
        {
            for (int j = i; j < n-; j++) printf("%05d %d %05d\n", node[hash[addr[j]]].addr, node[hash[addr[j]]].data, node[hash[addr[j]]].next);
            printf("%05d %d -1\n", node[hash[addr[n-]]].addr, node[hash[addr[n-]]].data);
        }
    }
}

int main()
{
    //freopen("input.txt", "r", stdin);
    PAT *p = new PAT;
    p->run();
    return ;
}

1075. PAT Judge （25）

这个题是一个常规的排序题，排序的时候仔细点。有一个要注意的地方是，如果提交了没有编译通过，submit里面显示的分数是-1，但是最终输出要输出为0。

#include <iostream>
#include <algorithm>
using namespace std;

struct User
{
    int id, total, score[], rank, solved;
    bool onlist;
};

class PAT
{
public:
    enum{N=};
    User user[N];
    int fullscore[N];
    int n, k, m;
    void run();
    void printscore(const User &u);
};

bool cmp(const User &u1, const User &u2)
{
    if (u1.total != u2.total) return u1.total > u2.total;
    else if (u1.solved != u2.solved) return u1.solved > u2.solved;
    else return u1.id < u2.id;
}

void PAT::printscore(const User &u)
{
    for (int i = ; i <= k; i++)
    {
        if (u.score[i] == -) printf(" -");
        else if (u.score[i] == -) printf("");
        else printf(" %d", u.score[i]);
    }
}

void PAT::run()
{
    scanf("%d%d%d", &n, &k, &m);
    for (int i = ; i<n; i++)
    {
        user[i].id = i+;
        user[i].total = ;
        user[i].solved = ;
        user[i].onlist = false;
        for (int j = ; j <= k; j++)
            user[i].score[j] = -;
    }
    for (int i = ; i <= k; i++) scanf("%d", &fullscore[i]);
    int tid, tpid, tscore;
    while (m-- > )
    {
        scanf("%d%d%d", &tid, &tpid, &tscore);
        if (user[tid-].score[tpid] < tscore) user[tid-].score[tpid] = tscore;
    }
    for (int i = ; i < n; i++)
    {
        for (int j = ; j <= k; j++)
        {
            if (user[i].score[j] >= )
            {
                user[i].total += user[i].score[j]; user[i].onlist = true;
            }
            if (user[i].score[j] == fullscore[j]) user[i].solved++;
        }
    }
    sort(user, user + n, cmp);
    user[].rank = ;
    for (int i = ; i < n; i++)
    {
        if (user[i].total == user[i - ].total) user[i].rank = user[i - ].rank;
        else user[i].rank = i+;
    }
    for (int i = ; i < n; i++)
    {
        if (user[i].onlist)
        {
            printf("%d %05d %d", user[i].rank, user[i].id, user[i].total);
            printscore(user[i]);
            printf("\n");
        }
        else break;
    }
}

int main()
{
    //freopen("input.txt", "r", stdin);
    PAT *p = new PAT;
    p->run();
    return ;
}

1076. Forwards on Weibo (30)

这个题直接用BFS搜索遍历即可。

#include <iostream>
#include <vector>

using namespace std;

class PAT
{
public:
    enum{N=};
    vector<int> fans[N];
    vector<int> curfans;
    int vis[N];
    int n, level, k;
    void run();
    void bfs(int);
    int maxforward;
};

void PAT::run()
{
    scanf("%d%d", &n, &level);
    int tnum,tfol;
    for (int i = ; i <= n; i++)
    {
        scanf("%d", &tnum);
        while (tnum-- > )
        {
            scanf("%d", &tfol);
            fans[tfol].push_back(i);
        }
    }
    scanf("%d", &k);
    int qnum;
    while (k-- > )
    {
        scanf("%d", &qnum);
        fill(vis, vis + N, );
        curfans.clear();
        maxforward = ;
        vis[qnum] = ;
        int fansize = fans[qnum].size();
        for (int i = ; i < fansize; i++)
        {
            curfans.push_back(fans[qnum][i]);
            vis[fans[qnum][i]] = ;
        }
        bfs(level);
        printf("%d\n", maxforward);
    }

}

void PAT::bfs(int l)
{
    if (l == ) return;
    if (curfans.empty()) return;
    int fansize = curfans.size();

    maxforward += fansize;
    vector<int> tmp;
    for (int i = ; i < fansize; i++)
    {
        int fansize2 = fans[curfans[i]].size();
        vector<int> &cur = fans[curfans[i]];
        for (int j = ; j < fansize2; j++)
        {
            if (vis[cur[j]] == ) {
                tmp.push_back(cur[j]); vis[cur[j]] = ;
            }
        }
    }
    curfans.clear();
    curfans.resize(tmp.size());
    copy(tmp.begin(), tmp.end(), curfans.begin());
    bfs(l-);
}
int main()
{
    //freopen("input.txt", "r", stdin);
    PAT *p = new PAT;
    p->run();
    return ;
}