PKU OJ的一道题目－－Exponentiation（ID:1001)????????

[hua_zhixing_]

哪位做出来啦？？最后的数据不知道怎么处理，而且我做出来的结果根本没那么大的精度。
Exponentiation
Time Limit: 500MS Memory Limit: 10000K
Total Submissions: 47367 Accepted: 10904

Description

Problems involving the computation of exact values of very large magnitude(['mæɡnitju:d],大小, 积, 量, 长(度)) and precision([pri'siӡən],精确(性); 精密(度)) are common. For example, the computation of the national debt is a taxing experience for many computer systems.

This problem requires that you write a program to compute the exact value of Rn where R is a real number ( 0.0 < R < 99.999 ) and n is an integer such that 0 < n <= 25.

Input

The input will consist of a set of pairs of values for R and n. The R value will occupy columns(['kɔləm]圆柱, 柱壮物,专栏) 1 through 6, and the n value will be in columns 8 and 9.

Output

The output will consist of one line for each line of input giving the exact value of R^n. Leading zeros should be suppressed([sə'pres],止住,不让发表 )in the output. Insignificant([͵insiɡ'nifikənt]无意义的; 无关重要的; 无用的) trailing([treil]痕迹, 足迹,一串, 一系列) zeros must not be printed. Don't print the decimal(['desiməl]以十作基础的, 十进位的, 小数的) point if the result is an integer.
Sample Input

95.123 12
0.4321 20
5.1234 15
6.7592 9
98.999 10
1.0100 12

Sample Output

548815620517731830194541.899025343415715973535967221869852721
.00000005148554641076956121994511276767154838481760200726351203835429763013462401
43992025569.928573701266488041146654993318703707511666295476720493953024
29448126.764121021618164430206909037173276672
90429072743629540498.107596019456651774561044010001
1.126825030131969720661201

[liao05050075]

这题目的输出很麻烦。我改了两次输出才搞定的
下面是我的代码。

#include<stdio.h>
#include<string.h>

int main()
	{
		char s[7];
		int n,lena,decp;
		char b[500],c[500];
lab1:	while(scanf("%s%d",s,&n)>0)
			{
				char a[500]={0};
				if(!strcmp(s,"0.0000")) {printf("0\n");goto lab1;}
				strrev(s);
				int i=0,j=0;
				while(s[i]=='0') i++;
				while(i&&i<(int)strlen(s)+1) 
					s[j++]=s[i++];
				i=0;
				int len=strlen(s);
				while(s[i]!='.') i++;
				decp=i*n;
				for(;i<=len;i++) s[i]=s[i+1];
				len--;
				for(i=0;i<len;i++) {s[i]-=48;a[i]=s[i];}
				lena=len;
				int k=0;
				
				int t;
				while(--n)
					{
						t=0;
						int tmp=0;
						for(j=0;j<500;j++) c[j]=0;
						for(i=0;i<len;i++)
							{	if(!s[i]) goto lab;
								for(j=0;j<500;j++) b[j]=0;
								tmp=0;
								for(j=0;j<lena;j++)
									{
										b[j+t]=s[i]*a[j]+tmp;
										tmp=0;
										if(b[j+t]>9) 
											{
												tmp=b[j+t]/10;
												b[j+t]%=10;
											}
									}
								if(tmp) b[j+t]=tmp;
								for(k=t;k<=t+lena;k++)
									{
										c[k]+=b[k];
										if(c[k]>9)
											{
												c[k+1]+=c[k]/10;
												c[k]%=10;
											}
									}
								while(c[k]>9)
									{
										c[k+1]+=c[k]/10;
										c[k]%=10;
										k++;
									}
lab:							t++;
							}
						j=0;
						while(j<k) {a[j]=c[j];j++;}
						lena=j;
					}

				//output
				while(!a[lena]) lena--;
				if(decp>lena)
					{
						printf(".");
						for(i=decp-lena-1;i>0;i--) printf("0");
						while(lena) 
							printf("%d",a[lena--]);
						printf("%d\n",a[0]);
					}
					else
						{
							for(i=lena;i>=0;i--)
								if(i==decp&&decp!=0) printf("%d.",a[i]);
									else printf("%d",a[i]);
							printf("\n");
						}

			}
		return 0;
	}

[liao05050075]

这题目的输出很麻烦。我改了两次输出才搞定的
下面是我的代码。

#include<stdio.h>
#include<string.h>

int main()
	{
		char s[7];
		int n,lena,decp;
		char b[500],c[500];
lab1:	while(scanf("%s%d",s,&n)>0)
			{
				char a[500]={0};
				if(!strcmp(s,"0.0000")) {printf("0\n");goto lab1;}
				strrev(s);
				int i=0,j=0;
				while(s[i]=='0') i++;
				while(i&&i<(int)strlen(s)+1) 
					s[j++]=s[i++];
				i=0;
				int len=strlen(s);
				while(s[i]!='.') i++;
				decp=i*n;
				for(;i<=len;i++) s[i]=s[i+1];
				len--;
				for(i=0;i<len;i++) {s[i]-=48;a[i]=s[i];}
				lena=len;
				int k=0;
				
				int t;
				while(--n)
					{
						t=0;
						int tmp=0;
						for(j=0;j<500;j++) c[j]=0;
						for(i=0;i<len;i++)
							{	if(!s[i]) goto lab;
								for(j=0;j<500;j++) b[j]=0;
								tmp=0;
								for(j=0;j<lena;j++)
									{
										b[j+t]=s[i]*a[j]+tmp;
										tmp=0;
										if(b[j+t]>9) 
											{
												tmp=b[j+t]/10;
												b[j+t]%=10;
											}
									}
								if(tmp) b[j+t]=tmp;
								for(k=t;k<=t+lena;k++)
									{
										c[k]+=b[k];
										if(c[k]>9)
											{
												c[k+1]+=c[k]/10;
												c[k]%=10;
											}
									}
								while(c[k]>9)
									{
										c[k+1]+=c[k]/10;
										c[k]%=10;
										k++;
									}
lab:							t++;
							}
						j=0;
						while(j<k) {a[j]=c[j];j++;}
						lena=j;
					}

				//output
				while(!a[lena]) lena--;
				if(decp>lena)
					{
						printf(".");
						for(i=decp-lena-1;i>0;i--) printf("0");
						while(lena) 
							printf("%d",a[lena--]);
						printf("%d\n",a[0]);
					}
					else
						{
							for(i=lena;i>=0;i--)
								if(i==decp&&decp!=0) printf("%d.",a[i]);
									else printf("%d",a[i]);
							printf("\n");
						}

			}
		return 0;
	}

[liuyang19890710]

大哥，你写的代码一行注释都麽有，让人情何以堪。
//[Quote=引用 1 楼 liao05050075 的回复:]
这题目的输出很麻烦。我改了两次输出才搞定的
下面是我的代码。

C/C++ code

#include<stdio.h>
#include<string.h>

int main()
{
char s[7];
int n,lena,decp;
char b[500],c[500];
lab1: while(s……
[/Quote]