PTA A1001&A1002
从今天起每天刷1-2题PAT甲级
第一天
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<stdio.h>
#include<stdlib.h>
#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 aN1 N2 aN2... NKaNK
where K is the number of nonzero terms in the polynomial, Niand aNi(i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10，0≤NK<⋯<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 /,pɒlɪ'nəʊmɪəl/ 美 /,pɑlɪ'nomɪəl/
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 /ɪk'spəʊnənt; ek-/ 美 /ɪk'sponənt/
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 /,kəʊɪ'fɪʃ(ə)nt/ 美 /koʊəˈfɪʃənt/
n. [数] 系数；率；协同因素
adj. 合作的；共同作用的
decimal
英 /'desɪm(ə)l/ 美 /'dɛsɪml/
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<stdio.h>
#include<stdlib.h>
#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<stdio.h>
#include<stdlib.h>
#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);
}
}
