PTA A1009&A1010
第五天
A1009 Product of Polynomials (25 分)
题目内容
This time, you are supposed to find A×B where A and B are two polynomials.
Input Specification:
Each 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 product 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 up to 1 decimal place.
Sample Input:
2 1 2.4 0 3.2
2 2 1.5 1 0.5
Sample Output:
3 3 3.6 2 6.0 1 1.6
单词
product
英 /'prɒdʌkt/ 美 /'prɑdʌkt/
n. 产品；结果；[数] 乘积；作品
题目分析
多项式相乘，类似于A1002的多项式相加，之前也写过用的开结构体数组的方法，很笨，所以用之前在A1002学到的开一个数组，用下标代表指数，对应元素代表系数的方式来存，不过因为是乘法所以要多开两个数组把数据临时保存一下。代码如下。
具体代码
#include<stdio.h>
#include<stdlib.h>
double p1[1001];
double p2[1001];
double res[2002];
int N, M;
int main(void)
{
scanf("%d", &N);
for (int i = 0; i < N; i++)
{
int e;
double c;
scanf("%d %lf", &e, &c);
p1[e] = c;
}
scanf("%d", &M);
for (int i = 0; i < M; i++)
{
int e;
double c;
scanf("%d %lf", &e, &c);
p2[e] = c;
}
for (int i = 0; i < 1001; i++)
{
if (p1[i] != 0)
for (int j = 0; j < 1001; j++)
if (p2[j] != 0)
res[i + j] += p1[i] * p2[j];
}
int num = 0;
for (int i = 0; i < 2002; i++)
{
if (res[i] != 0)
num++;
}
printf("%d", num);
for (int i = 2001; i >= 0; i--)
{
if (res[i] != 0)
printf(" %d %0.1f", i, res[i]);
}
system("pause");
}
A1010 Radix (25 分)
题目内容
Given a pair of positive integers, for example, 6 and 110, can this equation 6 = 110 be true? The answer is yes, if 6 is a decimal number and 110 is a binary number.
Now for any pair of positive integers N1 and N2, your task is to find the radix of one number while that of the other is given.
Input Specification:
Each input file contains one test case. Each case occupies a line which contains 4 positive integers:
N1 N2 tag radix
Here N1 and N2 each has no more than 10 digits. A digit is less than its radix and is chosen from the set { 0-9, a-z } where 0-9 represent the decimal numbers 0-9, and a-z represent the decimal numbers 10-35. The last number radix is the radix of N1 if tag is 1, or of N2 if tag is 2.
Output Specification:
For each test case, print in one line the radix of the other number so that the equation N1 = N2 is true. If the equation is impossible, print Impossible. If the solution is not unique, output the smallest possible radix.
Sample Input:
6 110 1 10
Sample Output:
2
单词
radix
英 /'rædɪks; 'reɪ-/ 美 /'redɪks/
n. 根；[数] 基数
n. (Radix)人名；(法、德、西)拉迪克斯；(英)雷迪克斯
decimal
英 /'desɪm(ə)l/ 美 /'dɛsɪml/
n. 小数
adj. 小数的；十进位的
题目分析
从9点做到11点多，巨坑，第一次是默认最大基数为36，部分通过，找不到bug很懵，上网查了一会儿看了一些博客才知道，最大的基数不一定是36，可以非常的大，最大值应该是已经算出的值+1，由于数据可以非常的大，所以int的数据大小肯定是不够了，而且用暴力方法遍历查找，必然会超时，这时候就要用二分查找，于是把之前写的全删掉重新写了个二分查找的代码，调试完依然有几个无法通过，于是把自己的代码和别人已经AC的代码对比，发现大佬的代码中在二叉搜索时多了一个判断条件，即算出的值需要判断是否小于0，这是因为如果数字实在太大甚至超过了long long的范围，那么这时我们去另一半继续找基数，我对这里有一点疑惑，如果数据真的特别大，或者基数卡在溢出和mid中间，那么这个方法可能找不到，可是数据又是无限的，所以这个问题深究起来可能是无解的（这么看来这好像是个数学问题，不知道有没有大佬能证明出来这道题到底有没有通解）。最后的最后，大家在开字符数组的时候一定记得要加一。。。。。
对了，关于把字符串转换成数字的问题，我之前一直用的是算出数组长度，减一后用math.h里的pow算出每个进位的值后一个一个加，今天看了大佬的代码，发现了一种更简洁的方法：
int sum=0;
while(\*p != '\0')
{
int n = convert(\*p);
sum = sum \* radix + n;
p++;
}
具体代码
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define MAXSIZE 11
char s1[MAXSIZE];
char s2[MAXSIZE];
int tag;
long long radix;
long long result;
long long int finalradix;
int convert(char a)
{
if (a >= '0'&&a <= '9')
return a - '0';
else
return a - 'a' + 10;
}
long long count_num(char *s, long long radix)
{
char *p = s;
long long sum = 0;
while(*p != '\0')
{
int n = convert(*p);
sum = sum * radix + n;
p++;
}
return sum;
}
long long min_radix(char *s)
{
long long max = 0;
char *p = s;
while(*p!='\0')
{
int f = convert(*p);
if (f > max)
max = f;
p++;
}
return max + 1;
}
long long binary_search(long long result,char *s,long long rmin,long long rmax)
{
while (rmin <= rmax)
{
long long mid = rmin+(rmax-rmin)/2;
long long n = count_num(s, mid);
if (n > result || n < 0 )
rmax = mid - 1;
else if (n < result)
rmin = mid + 1;
else
return mid;
}
return -1;
}
int main(void)
{
char c1[MAXSIZE];
char c2[MAXSIZE];
scanf("%s %s %d %lld", c1, c2, &tag, &radix);
if (tag == 1)
{
strcpy(s1, c1);
strcpy(s2, c2);
}
else
{
strcpy(s2, c1);
strcpy(s1, c2);
}
result = count_num(s1, radix);
long long rmax;
if(result != 0)
rmax = result + 1;
else
rmax = 2;
long long rmin = min_radix(s2);
long long r = binary_search(result, s2, rmin, rmax);
if (r == -1)
printf("Impossible");
else
printf("%lld", r);
system("pause");
}
参考博客
【笨方法学PAT】1010 Radix（25 分）
1010 Radix (25 分)
*甲级PAT 1010 Radix（二分搜索+坑）
1010 Radix （25 分）C++实现-终于AC了
