POJ 1555: 复杂模拟

——复杂模拟
原题传送门

Description

Given the coefficients of a polynomial from degree 8 down to 0, you are to format the polynomial in a readable format with unnecessary characters removed.
For instance, given the coefficients 0, 0, 0, 1, 22, -333, 0, 1, and -1, you should generate an output line which displays x^5 + 22x^4 - 333x^3 + x - 1.
The formatting rules which must be adhered to are as follows:

1. Terms must appear in decreasing order of degree.
2. Exponents should appear after a caret “^”.
3. The constant term appears as only the constant.
4. Only terms with nonzero coefficients should appear, unless all terms have zero coefficients in which case the constant term should appear.
5. The only spaces should be a single space on either side of the binary + and - operators.
6. If the leading term is positive then no sign should precede it; a negative leading term should be preceded by a minus sign, as in -7x^2 + 30x + 66.
7. Negated terms should appear as a subtracted unnegated term (with the exception of a negative leading term which should appear as described above). That is, rather than x^2 + -3x, the output should be x^2 - 3x.
8. The constants 1 and -1 should appear only as the constant term. That is, rather than -1x^3 + 1x^2 + 3x^1 - 1, the output should appear as-x^3 + x^2 + 3x - 1.

Data

Input
The input will contain one or more lines of coefficients delimited by one or more spaces.
There are nine coefficients per line, each coefficient being an integer with a magnitude of less than 1000.
Output
The output should contain the formatted polynomials, one per line.

	Sample Input
	0 0 0 1 22 -333 0 1 -1
	0 0 0 0 0 0 -55 5 0
	Sample Output
	x^5 + 22x^4 - 333x^3 + x - 1
	-55x^2 + 5x

思路

这是一道模拟题.
通常为了增大难度，出题者会设置各种各样的限制条件，来增大代码量.( 不理解的可以去做一下 魔方 )
本题也不例外.
本来，不加任何限制的多项式，可以写成这样:

for (int i=8;i>=1;i--)
	printf("+%dx^%d",a[i],i);

其中a[i]是第 i 项的系数.
题目这样繁杂的性质，让我们无从下手.

所谓复杂模拟，其实是指代码量.

对于这样的题目，我们的办法是:
忍了! 不忍还能怎么办
毕竟这是模拟题，一般而言找不到捷径.
所以可以一个条件一个条件的写代码.
当然，如果是赶时间，不宜做这类题目 ?
但是换句话说，这类题目也是对代码静态查错的很好训练.
比如说，加上8个条件，需要静态查错的代码就变成了:

Code

//已经AC,不必查错
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int a[15]; 
bool output(int c,int d)
{
	if (c==0) return false; 
	else if (c==-1) printf(" - x^%d",d);
	else if (c==1) printf(" + x^%d",d); 
	else if (c>1) printf(" + %dx^%d",c,d);
	else if (c<-1) printf(" - %dx^%d",-c,d);
	return true; 
}
int main()
{
	while (scanf("%d",&a[8])!=EOF)
	{
		bool all_zero=true;
		for (int i=7;i>=0;i--) scanf("%d",&a[i]);
		for (int i=1;i<=8;i++)
			if(a[i]!=0) all_zero=false;
		if (all_zero)
		{
			printf("%d\n",a[0]);
			continue;
		}
		int k=9,h=9;
		while (a[k]==0) k--,h--; 
		if (h==1)
		{
			if (a[1]==-1) printf("-x");
			else if (a[1]==1) printf("x"); 
			else if (a[1]>1) printf("%dx",a[1]);
			else if (a[1]<-1) printf("%dx",a[1]); 
			if (a[0]!=0)
			{
				if (a[0]>0) printf(" + %d",a[0]);
				else printf(" - %d",-a[0]);
			}
			puts("");
			continue; 
		} 
		if (a[k]==-1) printf("-x^%d",k);
		else if (a[k]==1) printf("x^%d",k); 
		else if (a[k]>1) printf("%dx^%d",a[k],k);
		else if (a[k]<-1) printf("-%dx^%d",-a[k],k);
		for (int i=k-1;i>=2;i--)
			output(a[i],i); 
		if (a[1]==-1) printf(" - x");
		else if (a[1]==1) printf(" + x"); 
		else if (a[1]>1) printf(" + %dx",a[1]);
		else if (a[1]<-1) printf(" - %dx",-a[1]); 
		if (a[0]!=0)
		{
			if (a[0]>0) printf(" + %d",a[0]);
			else printf(" - %d",-a[0]);
		}
		puts("");
	}
	return 0;
}

感谢奆老关注 qwq ?