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 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 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 /‘pr?d?kt/ 美 /‘prɑd?kt/

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",&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 N_?2,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 /‘r?d?ks; ‘re?-/ 美 /‘red?ks/

n. 根；[数] 基数

n. (Radix)人名；(法、德、西)拉迪克斯；(英)雷迪克斯

decimal

英 /‘des?m(?)l/ 美 /‘d?s?ml/

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 != ‘‘)

{

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 != ‘‘)
     {
         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!=‘‘)
    {
        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