PTA A1001&A1002

Posted 2022-10-27 z-y-k

从今天起每天刷1-2题PAT甲级

A1001 A+B Format (20 分)

题目内容

Calculate a+b and output the sum in standard format -- that is, the digits must be separated into groups of three by commas (unless there are less than four digits).

Input Specification:

Each input file contains one test case. Each case contains a pair of integers a and b where ?10⁶≤a,b≤10⁶. The numbers are separated by a space.

Output Specification:

For each test case, you should output the sum of a and b in one line. The sum must be written in the standard format.

Sample Input:

-1000000 9

Sample Output:

-999,991

单词

commas

n. [语] 逗号（comma的复数）

题目分析

输入两个数，计算结果后以标准格式输出，标准格式为每隔三个数字加一个逗号（这里插一句，个人感觉这里说的不太严谨，因为并没有具体说每隔三个数字是从末尾开始还是从开头开始，我一开始以为是从开头算，如2000就是，200,0，但是这题实际上是要从结尾算，2000应该是2,000，而且输入输出样例也无法看出到底是从开头算还是结尾算，导致我一直AC不了很难受）。
如果题目没有像我一样误读的话，这题应该有两种方法，一种是处理数字，结合取余取整运算符慢慢输出，我第一次做时用的是这种方法。第二种是处理字符串，用sprintf把数字转化为字符串，然后从末尾每隔三个插入一个‘,‘，类似于顺序表插入，要分正负情况，因为负数时第一个符号是-。

具体代码

#include
#include
#define MAXSIZE 20
int main(void)

    int a, b;
    scanf("%d %d", &a, &b);
    char C[MAXSIZE];
    sprintf(C, "%d", a + b);
    int M = 0;
    while (C[M] != ‘\0‘)M++;
    int N = M;
    if (a + b > 0)
    
        for (int n = 1; (M - 3 * n) > 0; n++)
        
            for (int m = N - 1; m >= M - 3 * n; m--)
                C[m + 1] = C[m];
            C[M - 3 * n] = ‘,‘; N++;
        
    
    else
    
        for (int n = 1; (M - 3 * n) > 1; n++)
        
            for (int m = N - 1; m >= M - 3 * n; m--)
                C[m + 1] = C[m];
            C[M - 3 * n] = ‘,‘; N++;
        
    
    C[N] = ‘\0‘;
    printf("%s", C);
    system("pause");

1002 A+B for Polynomials (25 分)

题目内容

Input Specification:

ach input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial:
K N₁ a_N1 N₂ a_N2... N_Ka_NK
where K is the number of nonzero terms in the polynomial, N_iand a_Ni(i=1,2,?,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10，0≤N_K<?<N₂<N₁≤1000.

Output Specification:

For each test case you should output the sum of A and B in one line, with the same format as the input. Notice that there must be NO extra space at the end of each line. Please be accurate to 1 decimal place.

Sample Input:

2 1 2.4 0 3.2
2 2 1.5 1 0.5

Sample Output:

3 2 1.5 1 2.9 0 3.2

单词

polynomial

英 /,p?l?‘n??m??l/ 美 /,pɑl?‘nom??l/

n. 多项式；多词拉丁学名；表示任何多项数之和
adj. 多项的，多词的；多项式的

exponent

英 /?k‘sp??n?nt; ek-/ 美 /?k‘spon?nt/

n. [数] 指数；典型；说明者，说明物

n. 倡导者，鼓吹者，代表者，拥护者
adj. 说明的

coefficient

英 /,k???‘f??(?)nt/ 美 /ko???f???nt/

n. [数] 系数；率；协同因素
adj. 合作的；共同作用的

decimal

英 /‘des?m(?)l/ 美 /‘d?s?ml/

n. 小数
adj. 小数的；十进位的

题目分析

多项式相加，由于指数范围不大，所以可以构造一个数组，用下标代表指数，对应元素为系数，这样的话可以把代码简化非常多。
一开始我用的是一种略笨的方法，做了三个结构体数组，并且默认其指数是以递减的方式输入的，所以一直有两个测试点过不了，现在把错误代码也一并贴出。

具体代码

正确

#include
#include
#define MAXSIZE 1001
int n;
double m[MAXSIZE];
int k=0;

int main(void)

    scanf("%d", &n);
    for (int i = 0; i < n; i++)
    
        int e;
        double c;
        scanf("%d %lf", &e, &c);
        m[e] += c;
    
    scanf("%d", &n);
    for (int i = 0; i < n; i++)
    
        int e;
        double c;
        scanf("%d %lf", &e, &c);
        m[e] += c;
    
    for (int i = 0; i < MAXSIZE; i++)
        if (m[i] != 0)
          k++;
    printf("%d", k);
    for (int i = MAXSIZE; i >= 0; i--)
    
        if (m[i] != 0)
        
            printf(" %d %.1f",i,m[i]);
        
    

错误

#include
#include
#define maxsize 10

struct a 
    int e;
    float c;
;

struct a A[maxsize];
struct a B[maxsize];
struct a C[maxsize];

int M, N, K ;

int main(void)

    int M, N;
    scanf("%d", &M);
    for (int i = 0; i < M; i++)
        scanf("%d %f", &A[i].e, &A[i].c);
    scanf("%d", &N);
    for (int i = 0; i < N; i++)
        scanf("%d %f", &B[i].e, &B[i].c);
    int p = 0, q = 0;
    while (p != M && q != N)
    
        if (A[p].e > B[q].e)
        
            C[K].c = A[p].c;
            C[K].e = A[p].e;
            K++;
            p++;
        
        else if (A[p].e == B[q].e)
        
            C[K].c = A[p].c + B[q].c;
            C[K].e = A[p].e;
            K++;
            p++;
            q++;
        
        else if (A[p].e < B[q].e)
        
            C[K].c = B[p].c;
            C[K].e = B[p].e;
            K++;
            q++;
        
    
    if(p==M)
        while (q != N)
        
            C[K].c = B[p].c;
            C[K].e = B[p].e;
            K++;
            q++;
        
    else if(q==N)
        while (p != M)
        
            C[K].c = A[p].c;
            C[K].e = A[p].e;
            K++;
            p++;
        
    printf("%d ", K);
    for (int i = 0; i < K; i++)
    
        if (i == K - 1)
            printf("%d %.1f", C[i].e, C[i].c);
        else
            printf("%d %.1f ", C[i].e, C[i].c);