解題區►解題區 - UVa

UVa 997 - Show the Sequence

contents

  1. 1. Input
  2. 2. Output
  3. 3. Sample Input
  4. 4. Sample Output
  5. 5. Solution

The problem of finding the next term of a given sequence of numbers is usually proposed in QI tests. We want to generate the N terms of a sequence from a given codification of the sequence.

Let S = (Si)i $\scriptstyle \in$$\scriptstyle \mathbb {N}$ denote a sequence of real numbers whose i -order term is Si . We codify a constant sequence with the following operator:

S = [ n] meaning that Si = n $\displaystyle \forall$i$\displaystyle \in$$\displaystyle \mathbb {N}$,

where n$ \in$$ \mathbb {Z}$ . We also define the following operators on a given sequence of numbers S = (Si)i $\scriptstyle \in$$\scriptstyle \mathbb {N}$ :

V = [ m + S ] meaning that

$$Vi = \displaystyle \cases{m & , $i=1$ \cr V_{i-1}+ S_{i-1} & , $i > 1$ \cr};$$

V = [ m * S ] meaning that

$$Vi = \displaystyle \cases{m \ast S_{1} & , $i=1$ \cr V_{i-1} \ast S_i & , $i > 1$ \cr};$$

where m$ \in$$ \mathbb {N}$ . For example we have the following codifications:

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[2 + [1]] = 2, 3, 4, 5, 6 ... 	
[1 + [2 + [1]]] = 1, 3, 6, 10, 15, 21, 28, 36 ...
[2 * [1 + [2 + [1]]]] = 2, 6, 36, 360, 5400, 113400 ...
[2 * [5 + [- 2]]] = 10, 30, 30, -30, 90, -450, 3150 ...

Given a codification, the problem is to write the first N terms of the sequence.

Input

The input file contains several test cases. For each of them, the program input is a single line containing the codification, without any space, followed by an integer N(2$ \le$N$ \le$50) .

Output

For each test case, the program output is a single line containing the list of first N terms of the sequence.

Sample Input

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[2+[1]] 3 
[2*[5+[-2]]] 7

Sample Output

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2 3 4 
10 30 30 -30 90 -450 3150

Solution

遞迴建造

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#include <stdio.h> 
#include <string.h>
long long a[128];
void parsing(char s[]) {
	int sign = 1, i, m = 0;
	char pos;
	for(i = 1; s[i]; i++) {
		if(s[i] == '-')	sign = -1;
		else if(s[i] >= '0' && s[i] <= '9') 
			m = m * 10 + s[i] - '0';
		else {
			pos = s[i];
			break;
		}
	}
	m = sign * m;
	if(pos == '+') {
		parsing(s + i + 1);
		long long d = a[0];
		a[0] = m;
		for(int i = 1; i < 50; i++) {
			long long t = a[i];
			a[i] = a[i - 1] + d, d = t;
		}
	} else if(pos == '*') {
		parsing(s + i + 1);
		a[0] *= m;
		for(int i = 1; i < 50; i++)
			a[i] = a[i] * a[i-1];
	} else {
		for(int i = 0; i < 50; i++)
			a[i] = m;
	}
}
int main() {
	char s[1024];
	int n;
	while(scanf("%s %d", s, &n) == 2) {
		memset(a, 0, sizeof(a));
		parsing(s);
		for(int i = 0; i < n; i++)
			printf("%lld%c", a[i], i == n - 1 ? '\n' : ' ');
	}
	return 0;
}
/*
[2+[1]] 3
[2*[5+[-2]]] 7
*/