POJ 1001-Exponentiation（NYOJ 155 求高精度幂):大数问题

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Exponentiation

Time Limit: 500MS		Memory Limit: 10000K
Total Submissions: 120472		Accepted: 29411

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

大数问题,而且需要注意的地方比较多,以前做的了,有些细节可能想不全,想到哪些再不上吧

1.整数部分,头部的0可以去掉,小数部分,尾部的0可以去掉

2.运算选忽略小数点,当做整数运算,最后再移动小数点的位置

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int result[1001];
char a[11];
int top1, top2;

void fun(int num)
{
int i, j;
for(i = 0; i < top1; i++)
{
result[i] = result[i] * num;
}
for(i = 0; i < top1; i++)
{
result[i + 1] += result[i] /10;
result[i] = result[i] % 10;
if(i + 1 == top1 && result[i + 1] != 0)
top1 ++;
}
}

int main()
{
int flag;
char str[11];
int num;
int n;
int i;
int len;
double number;
while(~scanf("%s", str))
{
scanf("%d", &num);
flag = -1;
number = atof(str);
sprintf(str, "%.10g", number);
len = strlen(str);
top1 = 1;
top2 = 0;
result[0] = 1;
for(i = 0; str[i]; i++)
{
if(str[i] =='.')
flag = i;
else
{
a[top2++] = str[i];
}
}
if(flag != -1)
{
flag = len - (flag + 1);
flag *= num;
}
n = atoi(a);
for(i = 0; i < num; i ++)
{
fun(n);
}
if(top1 - 1 < flag)
printf(".");
for(i = flag - top1; i> 0; i --)
printf("0");
for(i = top1 - 1; i >= 0; i--)
{
if((i + 1) == flag)
printf(".");
printf("%d", result[i]);
}
printf("\n");
memset(a, 0, sizeof(a));
memset(result, 0, sizeof(result));

}
return 0;
}