PAT甲级题目汇总及注意事项

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[PAT A1052] Linked list Sorting

注意事项：

1. 输出格式的控制，比如换行

2. n是一个陷阱，最后链表可能会提前结束，所以需要遍历静态链表确定链表的实际长度。

3. 感慨一句，PAT不亏是ZJU的看门武器，对测试数据的封锁，对思维严密性的要求，都在leetcode之上。

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

const int maxn = 100010;
struct Node
{
    int addr;
    int data;
    int next;
    bool flag;
    Node()
    {
        flag = false;
    }
}node[maxn];


bool cmp(const Node &a, const Node &b)
{
    if(a.flag == false ||  b.flag == false)
        return a.flag > b.flag;
    else
        return a.data < b.data;
}

int main()
{
    int n, begin, addr;
    scanf("%d%d", &n, &begin);

    for(int i = 0; i < n; ++i)
    {
        scanf("%d",&addr);
        scanf("%d%d", &node[addr].data, &node[addr].next);
        node[addr].addr = addr;
    }

    int count = 0;
    for(int p = begin; p != -1; p = node[p].next)
    {
        count += 1;
        node[p].flag = true;
    }

    if(count == 0)
        printf("0 -1");
    else
    {
        sort(node, node + maxn, cmp);
        printf("%d %05d\n", count, node[0].addr);

        for(int i = 0; i < count; ++i)
        {
            if(i != count - 1)
                printf("%05d %d %05d\n", node[i].addr, node[i].data, node[i+1].addr);
            else
                printf("%05d %d -1", node[i].addr, node[i].data);
        }
    }
    return 0;
}

[PAT A1032 ] Sharing

注意事项：

1. = 和== 一定要注意区别

2. 多次使用scanf可能会超时，尽可能的减少使用scanf的次数

3. 注意特殊情况，这就是相当于数学题目中简单情况但是又会忽略的地方，不考虑失分很严重。PAT考察的是细致和耐心。

#include<cstdio>

const int maxn = 100010;
struct Node
{
    char data;
    int next;
    bool flag;
    Node()
    {
        flag = false;
    }
}node[maxn];

int main()
{
    int addr1, addr2, n;
    scanf("%d%d%d", &addr1, &addr2, &n);

    int addr, next;
    char data;
    for(int i = 0; i < n; ++i)
    {
        scanf("%d %c %d", &addr, &data, &next);
        node[addr].data = data;
        node[addr].next = next;
    }

    int p1 = addr1, p2 = addr2;
    while(p1 != -1)
    {
        node[p1].flag = true;
        p1 = node[p1].next;
    }

    while(p2 != -1)
    {
        if(node[p2].flag == true)
            break;
        p2 = node[p2].next;
    }

    if(p2 == -1)
        printf("-1");
    else
        printf("%05d", p2);

    return 0;
}

[PAT A1001 ] A+B Format

注意事项：

1. 利用字符数组存储每一位数，便于处理’,’

2. 数组最高位可能有多余的’,’，索引回退，去掉多余的’,’

3. PAT通式: 现处理特殊情况

#include<cstdio>
const int maxn = 10;
char buf[maxn];

int main()
{
    int a, b, sum;
    bool sign = false;
    int bufIndex = 0;
    int numIndex = 0;
    scanf("%d%d", &a, &b);

    sum = a + b;
    if(sum < 1000 && sum > -1000)
        printf("%d", sum);

    else
    {
        if(sum < 0)
        {
            sign = true;
            sum = 0 - sum;
        }

        while(sum != 0)
        {
            buf[bufIndex++] = sum % 10 + '0';
            sum /= 10;
            numIndex += 1;
            if(numIndex % 3 == 0)
                buf[bufIndex++] = ',';
        }

        if(buf[--bufIndex] == ',')
            bufIndex -= 1;

        if(sign)
            printf("-");

        for(int i = bufIndex; i >= 0; --i)
            printf("%c", buf[i]);
    }
    return 0;
}

[PAT A1025 ] PAT Ranking

注意事项：

思路简单，但是处理起来还是比较麻烦的，PAT题目有时候在输入输出的格式上花的时间比算法本身还要多……

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;

const int maxn = 100010;
struct Student
{
    char registration_number[20];     //考号
    int score;                        //分数
    int location_number;              //考点
    int local_rank;                   //本考场排名
}student[maxn];

bool cmp(const Student &a, const Student &b)
{
    return (a.score != b.score) ? a.score > b.score : \
            strcmp(a.registration_number, b.registration_number) < 0;
}

int main()
{
    int n;      //考点数
    scanf("%d", &n);

    int i = 0;      //遍历所有的考点
    int total = 0;  //考生总人数
    int studentIndex = 0;     //填充结构体数组
    while(i < n)
    {
        int k;    //本考点的考生数
        scanf("%d", &k);
        total += k;
        for(int j = 0; j < k; ++j)
        {
            scanf("%s %d", student[studentIndex].registration_number,\
                            &student[studentIndex].score);
            student[studentIndex].location_number = i + 1;
            studentIndex += 1;
        }
        sort(student + studentIndex - k, student + studentIndex, cmp);
        //第一名排第一
        student[studentIndex - k].local_rank = 1;
        //注意并列的排名
        for(int j = studentIndex - k + 1; j < studentIndex; ++j)
        {
            if(student[j].score == student[j - 1].score)
                student[j].local_rank = student[j - 1].local_rank;
            else
                student[j].local_rank = j + 1 - (studentIndex - k);
        }
        i += 1;
    }

    //关注总排名
    sort(student, student + studentIndex, cmp);

    //输出信息
    printf("%d\n", total);
    //输出第一名
    int final_rank = 1;
    printf("%s %d %d %d\n", student[0].registration_number,final_rank,\
            student[0].location_number, student[0].local_rank);

    for(int i = 1; i < total; ++i)
    {
        if(student[i].score == student[i - 1].score)
            printf("%s %d %d %d\n", student[i].registration_number,final_rank,\
                student[i].location_number, student[i].local_rank);
        else
        {
            final_rank = i + 1;
            printf("%s %d %d %d\n", student[i].registration_number,final_rank,\
                student[i].location_number, student[i].local_rank);
        }

    }

    return 0;
}

[PAT A1028 ] List Sorting

注意事项：

Presentation Error，这是格式错误，出现这个错误最可能的原因是，在输出结果的后面，多了或少了没什么意义的空格，tab，换行符等等。所以，请先认真检查程序的输出结果是否与标准完！全！一！致！OJ平台对格式的检查可以说是非！常！严！格！

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int maxn = 100010;
struct Student
{
    int ID;
    char name[10];
    int score;
}student[maxn];

bool cmp1(const Student &a, const Student &b)
{
    return a.ID < b.ID;
}

bool cmp2(const Student &a, const Student &b)
{
    if(strcmp(a.name, b.name) != 0)
        return strcmp(a.name, b.name) < 0;
    else
        return a.ID < b.ID;
}

bool cmp3(const Student &a, const Student &b)
{
    if(a.score != b.score)
        return a.score < b.score;
    else
        return a.ID < b.ID;
}
int main()
{
    int n, c; //n 总人数    c待排序的列
    scanf("%d%d", &n, &c);

    for(int i = 0; i  < n; ++i)
    {
        scanf("%d %s %d", &student[i].ID, student[i].name, &student[i].score);
    }

    if(c == 1)
        sort(student, student + n, cmp1);
    else if(c == 2)
        sort(student, student + n, cmp2);
    else
        sort(student, student + n, cmp3);


    //输出信息
    for(int i = 0; i < n; ++i)
    {
        printf("%06d %s %d\n", student[i].ID, student[i].name, student[i].score);
    }

    return 0;
}

[PAT A1012] Reversible Primes

注意事项：

1. 这道题本身不难，但是花了很长时间，主要是对素数的判断考虑情况不全：判断一个数是否是素数，注意两点：（1）对1进行特判，1不是素数。（2）循环终止条件是i * i <= num, = 不能少。

#include<cstdio>
#include<cmath>

bool isPrime(int num)
{
    if(1 == num)
        return false;
    for(int i = 2; i * i <= num; ++i)
    {
        if(num % i == 0)
            return false;
    }
    return true;
}


int reverseWithRadix(int num, int d)
{
    int i = 0;
    int ans = 0;
    while(num != 0)
    {
        ans = ans * d + num % d;
        num /= d;
    }
    return ans;
}


int main()
{
    int n, d;   //n是待判断的数，d是基底
    int i = 0;
    while(scanf("%d%d", &n, &d), n > 0)
    {
        if(isPrime(n) && isPrime(reverseWithRadix(n, d)))
            printf("Yes\n");
        else
            printf("No\n");
    }
    return 0;
}

[PAT A1023 ]Have Fun with Numbers

注意事项：

1. 注意读入数据时，数据高位在数组的低位索引

2. 字符数组转换成int只要-‘0’

#include<cstdio>
#include<cstring>
const int maxn = 11;
int check[maxn];

int *multi(char *num, int len)
{
    int *result, carry = 0;
    int resLen = (num[0]  >= '5') ? (len + 1) : len;
    result = new int[resLen];
    for(int i = len - 1; i >= 0; --i)
    {
        int tmp =  2 * (num[i] - '0') + carry;
        result[i] = tmp % 10;
        carry = tmp / 10;
    }
    if(carry != 0)
        result[len] = carry;
    return result;
}

int main()
{
    //输入大整数
    char num[20];
    scanf("%s", num);
    int len = strlen(num);

    //解析大整数
    for(int i = 0; i < len; ++i)
        check[num[i] - '0'] += 1;

    //double 大整数
    int *result= multi(num, len);


    int resLen = (num[0]  >= '5') ? (len + 1) : len;
    //判断double大整数
    if(num[0] >= '5')
        printf("No\n");
    else
    {
        for(int i = 0; i < len; ++i)
            if(--check[result[i]] == -1)
            {
                printf("No\n");
                goto logo;
            }
        printf("Yes\n");
    }
    logo:
        if(resLen > len)
          printf("%d", result[len]);
        for(int i = 0; i < len; ++i)
          printf("%d", result[i]);

    return 0;
}

[PAT A1015 ]The Best Rank

注意事项：

1. 这个题目直接爆破，竟然没有超时！！！

2. 不要轻易粘贴代码，节省的时间远远比不上Debug的时间。

3. emmm, 写代码要认真……

#include<cstdio>
#include<algorithm>
#include<map>
#include<string>
#include<cmath>
using namespace std;

const int maxn = 2010;
typedef struct Student
{
    char ID[8];
    int C, math, English;
    int average;
    int rank1, rank2, rank3, rank4;
}Student;


bool cmp1(const Student &a, const Student &b)
{
    return a.C > b.C;
}

bool cmp2(const Student &a, const Student &b)
{
    return a.math > b.math;

}

bool cmp3(const Student &a, const Student &b)
{
    return a.English > b.English;
}

bool cmp4(const Student &a, const Student &b)
{
    return a.average > b.average;

}


void print(const Student &a)
{
    int m = maxn;
    char ch;

    if(a.rank3 < m)
    {
        m = a.rank3;
        ch = 'E';
    }
    if(a.rank2 <= m)
    {
        m = a.rank2;
        ch = 'M';
    }
    if(a.rank1 <= m)
    {
        m = a.rank1;
        ch = 'C';
    }
    if(a.rank4 <= m)
    {
        m = a.rank4;
        ch = 'A';
    }
    printf("%d %c\n", m, ch);
}

int main()
{
    Student student[maxn];
    int n, m; // n 学生总人数   m 查询的总人数
    scanf("%d %d", &n ,&m);
    for(int i = 0; i < n; ++i)
    {
        scanf("%s %d %d %d", student[i].ID, &student[i].C, &student[i].math,\
                  &student[i].English);
        student[i].average = (int)round((student[i].C + student[i].math + student[i].English)\
                                    / 3.0);
    }


    sort(student, student + n, cmp1);
    student[0].rank1 = 1;
    for(int i = 1; i < n; ++i)
    {
        if(student[i].C == student[i-1].C)
            student[i].rank1 = student[i-1].rank1;
        else
            student[i].rank1 = i + 1;
    }

    sort(student, student+n, cmp2);
    student[0].rank2 = 1;
    for(int i = 1; i < n; ++i)
    {
        if(student[i].math == student[i-1].math)
            student[i].rank2 = student[i-1].rank2;
        else
            student[i].rank2 = i + 1;
    }

    sort(student, student+n, cmp3);
    student[0].rank3 = 1;
    for(int i = 1; i < n; ++i)
    {
        if(student[i].English == student[i-1].English)
            student[i].rank3 = student[i-1].rank3;
        else
            student[i].rank3 = i + 1;
    }

    sort(student, student+n, cmp4);
    student[0].rank4 = 1;
    for(int i = 1; i < n; ++i)
    {
        if(student[i].average == student[i-1].average)
            student[i].rank4 = student[i-1].rank4;
        else
            student[i].rank4 = i + 1;
    }

    map<string, Student> mp;
    for(int i = 0; i < n; ++i)
    {
        string ID = student[i].ID;
        mp[ID] = student[i];
    }


    char str[8];
    Student tmp;
    for(int i = 0; i < m; ++i)
    {
        scanf("%s", str);
        if(mp.count(str) > 0)
        {
            tmp = mp[str];
            print(tmp);
        }
        else
            printf("N/A\n");

    }

    return 0;

}

…….未完待续