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贴一个改了N次才提交上的考虑了所有异常输入输出的代码给大家看看,附上几组测试数据

Posted by xiaohui5319 at 2012-05-18 22:28:32 on Problem 1001 and last updated at 2012-05-18 22:32:30
特别需要注意的几组测试数据是:
0.00000 0
1.23456 0
123456. 0
123456. 1
1234.00 1
123.000 2
大家可以试一下这几组异常数据,如果你都跑过了,而且格式完全正确,AC应该就没什么问题了。

至于鄙人的代码,的确写的又臭又长,被这个题折磨的暂时没什么好心情去改了。
#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <cstring>
using namespace std;

char result[200];

void reverse(char * a)
{
	int len=strlen(a);
	for(int i=0; i<len/2; i++)
	{
		int temp=a[i];
		a[i]=a[len-1-i];
		a[len-1-i]=temp;
	}
}

void multiply(char * a, char * b, char * result)//一定要保证输入的a和b指针是不同的
{
	reverse(a);
	reverse(b);
	int a_len, b_len, i, j;
	a_len=strlen(a);
	b_len=strlen(b);
	int * res=new int[a_len+b_len];
	memset(res, 0, (a_len+b_len)*sizeof(int) );
	for(i=0; i<a_len; i++)
		for(j=0; j<b_len; j++)
		{
			res[i+j] += (a[i]-'0')*(b[j]-'0');
		}
		//调整进位
		i=j=0;
		for(i=0; i<a_len+b_len; i++)
		{
			res[i+j+1] += res[i+j]/10;
			res[i+j] = res[i+j]%10;
		}
		//转换为char类型的result
		for(i=0; i<a_len+b_len; i++)
			result[i]=res[i]+'0';
		result[i]='\0';
		reverse(result);
		reverse(a); //我们设定a为输入的R,所以每次乘法颠倒后,应该保证下一次输入的是正序的R
		delete [] res;
}
//函数,求R的n次幂,结果放入result中
void factorial(char * s, char * result, int n)
{
	char temp[200];
	if(n==0)
	{
		result[0]='1';
		result[1]='\0';
	}
	else if(n==1)
	{
		strcpy(result, s);
	}
	else
	{
		strcpy(temp, s);
		for(int i=1; i<=n-1; i++)
		{
			multiply(s, temp, result);
			strcpy(temp, result);
		}
	}
}

bool is_float(char * s, bool & hasDotandIsInt, int & dotIndex)
{
	int len=strlen(s);
	hasDotandIsInt = false;
	for(int i=0; i<len; i++)
	{
		if(s[i]=='.' && i==len-1)
		{
			hasDotandIsInt = true;
			dotIndex = i;
			return false;
		}
		else if(s[i]=='.' && i!=len-1)
		{
			dotIndex = i;
			if(atoi(s+i+1)==0)
			{
				hasDotandIsInt = true;
				return false;
			}
			else
				return true;
		}
	}
	return false;
}

int main()
{
	char s[7];
	int n;
	while(cin>>s>>n)
	{	
		if(atof(s) == 0.000000)
		{
			printf("0\n");
			continue;
		}
		if(n == 0)
		{
			printf("1\n");
			continue;
		}
		bool isFloat = false, hasDotandIsInt = false;
		int dotIndex;
		isFloat = is_float(s, hasDotandIsInt, dotIndex);
		if(hasDotandIsInt)
			strcpy(s+dotIndex, "\0");
		//还需要处理输入,比如95.123转换为95123,然后在输入
		if(strcmp(s,"0")==0)
		{
			printf("0\n");
		}
		else if(isFloat==true)
		{
			char sd[7]; //经过处理,将小数点去掉后的s
			char number[200];
			memset(number, '0', sizeof(char)*200);
			int i, flo_fla, flo_bits, des_flo_len;	
			i=flo_fla=flo_bits=0;
			//首先将最后的0去掉
			i=strlen(s)-1;
			while(s[i]=='0') i--;
			s[i+1]='\0';
			//第二步将小数位去掉,放入sd数组
			i=0;
			while(s[i]!='.') i++;
			flo_fla=i;
			s[flo_fla]='\0';
			strcpy(sd, s);
			strcpy(&sd[flo_fla], &s[flo_fla+1]);
	
			flo_bits=strlen(sd)-flo_fla;
			des_flo_len=n*flo_bits;
	
			factorial(sd, result, n);
			i=0;
			while(result[i]=='0') i++;
			int in_len=strlen(&result[i])-des_flo_len;
			if( in_len<=0 )
			{
				number[0]='.';
				strcpy(&number[ abs(in_len)+1 ], &result[i]);
			}
			else 
			{
				number[in_len]='.';
				strcpy(&number[in_len+1], &result[i+in_len]);
				result[i+in_len]='\0';
				strcpy(number, &result[i]);
				number[in_len]='.';
			}
			cout<<number<<endl;
		}else
		{
			factorial(s, result, n);
			int i=0;
			while(result[i]=='0') i++;
			printf("%s\n", &result[i]);
		}
	}
	return 0;
}

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