poj 1001 Exponentiation

Exponentiation

Time Limit: 500MS	Memory Limit: 10000K
Total Submissions: 109780	Accepted: 26662

Description

Problemsinvolving the computation of exact values of very large magnitude and precisionare common. For example, the computation of the national debt is a taxingexperience 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 inputwill consist of a set of pairs of values for R and n. The R value will occupycolumns 1 through 6, and the n value will be in columns 8 and 9.

Output

The outputwill 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 zerosmust 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 youdon't know how to determine wheather encounted the end of input: 
s is a string and n is an integer 

 

求幂

Time Limit: 500MS	Memory Limit: 10000K
Total Submissions: 109780	Accepted: 26662

Description

涉及高精度计算的问题是很常见的. 比如说，计算国债，就是一项费力的工程
这个问题要求你写个程序来计算Rⁿ ，其中R是个实数( 0.0 <R < 99.999 )，n是个整形数（0 < n<= 25）

Input

输入将包含几组数据，R和n。R占6列，n占第8,9列

Output

输出将对每一组数据计算出一个R^n 。开头和尾巴的零应该省略，如果是整形数，不应该有小数点存在。

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 youdon't know how to determine wheather encounted the end of input: 
s is a string and n is an integer 

 

这代码重新写的看起来好看点依然很破，求高速算法…….

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
 
void Exponen(char *s, int n, char *res);
void Output(char *res, int point);
void init(char *s, int *point);
void multi(char *res, char*s);
 
int main()
{
   char s[6];
    char res[555];
   int n;
 
   freopen("in.txt", "r", stdin);
   while(~scanf("%s %d", s, &n)){
       //puts("o");
       Exponen(s,n,res);
    }
 
   return 0;
}
 
void init(char *s, int *point){
   int k = 5, p = 0, i;
   char s1[6];
   while((*(s+k))=='0')  {k--;}//消除小数点后多余的0
   for (i = k;i >= 0; i--){//记录小数点位置，并将处理后的数给一个新字符串。
       if (s[i] == '.'){
           *point = i;
           (*point) = k - (*point);
           //printf("%d\n", *point);
       }else{
           s1[p++] = s[i];
        }
    }
   s1[p] = 0;
   p--;
   while(*(s1+p)=='0') {*(s1+p) = 0;p--;}//消除最前面的多余的0
   strcpy(s, s1);//拷贝回s
}
 
void multi(char *res, char*s){
   int i, j;
   int p, q, k, cnt;
   char temp[555]={0};
    p= 0;q = 0;
   while(p != strlen(s)){
       k = p;
       cnt = 0;
       while(q != strlen(res)){
           *(temp + k) += (*(res + q)-'0') * (*(s + p)-'0') + cnt;
           //printf("\n");
           //printf("here = %d, k = %d, pro = %d, next = %d\n", *(temp +k), k, *(res + q)-'0', *(s + p)-'0');
           cnt = *(temp + k) / 10;
           *(temp + k) %= 10;
           k ++;
           q ++;
       }
       *(temp + k) += cnt;//最后一个进位要加上
       q = 0;
       p ++;
    }
 
   //printf("\n");
   for (i = 0;i <= k;i ++){
        temp[i]+='0';
    }
   if (*(temp + k) != '0')
       temp[k+1] = 0;
   else temp[k] = 0;
   strcpy(res, temp);
}
 
void Exponen(char *s, int n, char *res){
   int i, point=-1, pointt;
   init(s, &point);//初始化字符串，方便运算
 
   strcpy(res, s);//res作为主串
   //printf("%s\n", res);
 
   pointt = point;
   //printf("%d\n", pointt);
   for (i = 0;i < n -1;i ++){
       point += pointt;
    }
   //printf("point = %d\n", point);
   for (i = 0; i < n - 1;i ++){
       multi(res,s);
       //printf("res=%s\n", res);
    }
   Output(res, point);
   //printf("%s\n", res);
}
 
void Output(char *res, int point){
   int len = strlen(res);
   int p, ps;
   if (len >= point){//长度大于
       //point = len - point;
       ps = len;p = len-1;
       while((ps != -1) && (p != -1)){
           if (ps == point){
                putchar('.');ps--;
           }else {
                printf("%c",*(res+p));p--;ps--;
           }
       }
   }else{
       putchar('.');
       p = point;
       while(p != 0){
           if (p > len){
                putchar('0');p--;
           }else{
                p--;
                printf("%c",*(res+p));
           }
       }
 
    }
   putchar('\n');
}