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Problem Solution

1. Convert the input string into an integer and store it in number, multiplying the number each time.

2. Inversely store the input string in the integer array a [] to store the results after multiplying the number each time.

3. The specific operation process is to find the n-th power of number, each operation multiplied by the number of each digit in a [].

4. Use point to calculate the number of decimal digits, output a [] in reverse order, and deal with the number of decimal digits.

Code

#include <stdio.h>
#include <string.h>
const int N = 200;
 
int main() {
	char num[6]; //Store input strings 
	long number; //Converts the input string to an integer without decimal points 
	int n,point,count;  
	int a[N]; //Store operation results
	while (scanf("%s%d", num, &n) == 2) {
		number = 0; point = -1; count = 0;
		
		//Input Settings 
		int note1,note2; //Remove the redundant 0 before and after the string 
		for (int i = 0; i <= 5; i++) {
			if (num[i] != '0') {
				note1 = i;
				break;
			} 		
		}	
		for (int i = 5; i >= 0; i--) {
			if (num[i] != '0') {
				note2 = i;
				break;
			} 		
		}
		for (int i = note1; i <= note2; i++) {
			if (num[i] != '.') {
				number = number*10 + (num[i]-48);
			}
		}
		for (int i = note2; i >= note1; i--) {
			if (num[i] != '.') {
				a[count++] = num[i]-48;		
			}
		}
		for (int i = note1; i <= note2; i++) {
			if (num[i] == '.') {
				point = (note2-i)*n;
				break;
			}
		}
 
		//operation 
		long s,d;
		for (int k = 1; k < n; k++) {
			d = 0;
			for (int i = 0; i < count; i++) {
				s = a[i]*number+d;	
				a[i] = s%10;
				d = s/10;
			}
			while (d != 0) {
				a[count++] = d % 10;
				d /= 10;
			}
		}
		
		//output setting	
		bool flag = true;
		if (count < point) {  
			flag = false;
			printf(".");
			while (point != count) {
				point--;
				printf("0");
			}
		}
		for (int i = count-1; i >= 0; i--) {
			if (flag == true && i == point-1) printf(".");
			printf("%d", a[i]);
		}
		printf("\n");
	}
	return 0;
}