一道ACM题，提交好多次了，还是WA，大家看一下怎么回事？
http://acm.pku.edu.cn/JudgeOnline/problem?id=1001
Exponentiation
Time Limit:500MS Memory Limit:10000K
Total Submit:19697 Accepted:4454
Description
Problems involving the computation of exact values of very large magnitude and precision 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 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 in the output. Insignificant trailing zeros must not be printed. Don 't print the decimal 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
Hint
If you don 't know how to determine wheather encounted the end of input:
s is a string and n is an integer
C++
while(cin> > s> > n)
{
...
}
c
while(scanf( "%s%d ",s,&n)==2) //to see if the scanf read in as many items as you want
/*while(scanf(%s%d ",s,&n)!=EOF) //this also work */
{
...
}
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#include <math.h>
#include <string.h>
void compute(char *result,long m);
void print(char *result,int x,int n,int y);
int main(int argc, char *argv[])
{
float m;
int n,x,i,k;
char mm[7],result[125];
while(scanf( "%s%d ",mm,&n)!=EOF)
{
if(n==0)
{
printf( "%d\n ",1);
continue;
}
m=(float)atof(mm);
if(m==0)
{
printf( "%d\n ",0);
continue;
}
k=(int)m;
for(i=0;i <125;i++)
result[i]= '0 ';
result[124]= '1 ';
x=0;
for(i=5;i> =0;i--)
{
if(mm[i]== '. ')
{
x=5-i;
break;
}
}
if(m==k)
x=0;
m=m*pow(10,x);
for(i=0;i <n;i++)
compute(result,(long)m);
print(result,x,n,k);
}
system( "PAUSE ");
return 0;
}
void compute(char *result,long m)
{
int n,y,i;
n=0;
for(i=124;i> =0;i--)
{
y=(result[i]- '0 ')*m+n;
n=y/10;
result[i]=y%10+ '0 ';
}
}
void print(char *result,int x,int n,int y)
{
int i,j,k=124;
for(i=124;i> =124-x*n+1;i--)
{
if(result[i]!= '0 ')
{
k=i;
break;
}
}
if(y)
{
for(i=0;i <=k;i++)
{
if(result[i]!= '0 ')
{
for(j=i;j <=k;j++)
{
printf( "%c ",result[j]);
if((j==125-x*n-1)&&(x!=0))
printf( ". ");
}
printf( "\n ");
break;
}
}
}
else
{
printf( ". ");
for(i=125-n*x;i <=k;i++)
printf( "%c ",result[i]);
printf( "\n ");
}
}
[解决办法]
我是在写出来这个程序后3个月后偶然把程序改过的。
用用特殊的测试用例。比如
10 10
1 1
0.1 25
等
[解决办法]
练细心的:) 这题ac了以后推荐来看看偶们这里的1001
Simple Arithmetics
http://acm.fjnu.edu.cn/show?problem_id=1001
对STL有点感觉后来做作这个~
Searching the Web
http://acm.fjnu.edu.cn/show?problem_id=1015
[解决办法]
Simple Arithmetics
这题（不是楼主说的那题） 偶从写出来到ac也是一个月:)
要自己多想想~
Searching the Web 倒是还好~
