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02-2. 一元多项式求导 (PAT)

时间:2015-02-03 22:47:32      阅读:214      评论:0      收藏:0      [点我收藏+]

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设计函数求一元多项式的导数。

输入格式:以指数递降方式输入多项式非零项系数和指数(绝对值均为不超过1000的整数)。数字间以空格分隔。

输出格式:以与输入相同的格式输出导数多项式非零项的系数和指数。数字间以空格分隔,但结尾不能有多余空格。注意“零多项式”的指数和系数都是0,但是表示为“0 0”。

输入样例:

3 4 -5 2 6 1 -2 0

输出样例:

12 3 -10 1 6 0
 
 
#include <iostream>
#include <string>
#include <stdlib.h>
using namespace std;

typedef struct node{
    int coef;
    int expon;
    struct node *next;
} *PtrPolyNode, PolyNode;

PtrPolyNode createList();
void deriPoly( PtrPolyNode headNode );
void printPoly( PtrPolyNode headNode );

int main()
{
    PtrPolyNode headNode;
    headNode = createList();
    deriPoly( headNode );
    printPoly( headNode );
    return 0;
}

PtrPolyNode createList()
{
    PtrPolyNode headNode = NULL, currentNode = NULL, prevNode = NULL;
    do
    {
        currentNode = (PtrPolyNode)malloc( sizeof( PolyNode ) );
        cin >> currentNode->coef >> currentNode->expon;
        currentNode->next = NULL;
        if ( headNode == NULL )
        {
            headNode = currentNode;
            prevNode = currentNode;
            continue;
        }
        prevNode->next = currentNode;
        prevNode = currentNode;
    }
    while ( cin.get() != \n );
    return headNode;
}

void deriPoly( PtrPolyNode headNode )
{
    PtrPolyNode prev, cur;
    cur = prev = headNode;
    if ( headNode->next == NULL && headNode->expon == 0 )    //point
    {
        headNode->coef = 0;
        return;
    }
    while( cur != NULL )
    {
        if ( cur->expon == 0 )    //point
        {
            prev->next = NULL;
            free( cur );
            break;
        }
        cur->coef = cur->coef * cur->expon;
        cur->expon = cur->expon - 1;
        prev = cur;
        cur = cur->next;
    }
    return;
}

void printPoly( PtrPolyNode headNode )
{
    PtrPolyNode ptr;
    ptr = headNode;
    while ( ptr != NULL )
    {
        if ( ptr->next == NULL )
        {
            cout << ptr->coef <<   << ptr->expon; 
        }
        else
        {
            cout << ptr->coef <<   << ptr->expon <<  ;
        }
        ptr = ptr->next;
    }
    return;
}

02-2. 一元多项式求导 (PAT)

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原文地址:http://www.cnblogs.com/liangchao/p/4271176.html

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