Codeforces Round #159 (Div. 2) D sum

Codeforces Round #159 (Div. 2) D sum

 

 

D. Sum

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasya has found a piece of paper with an array written on it. The array consists of n integers a₁, a₂, ..., a_n. Vasya noticed that the following condition holds for the array a_i ≤ a_i + 1 ≤ 2·a_i for any positive integer i (i < n).

Vasya wants to add either a "+" or a "-" before each number of array. Thus, Vasya will get an expression consisting of n summands. The value of the resulting expression is the sum of all its elements. The task is to add signs "+" and "-" before each number so that the value of expression s meets the limits 0 ≤ s ≤ a₁. Print a sequence of signs "+" and "-", satisfying the given limits. It is guaranteed that the solution for the problem exists.

Input

The first line contains integer n (1 ≤ n ≤ 10⁵) — the size of the array. The second line contains space-separated integers a₁, a₂, ..., a_n(0 ≤ a_i ≤ 10⁹) — the original array.

It is guaranteed that the condition a_i ≤ a_i + 1 ≤ 2·a_i fulfills for any positive integer i (i < n).

Output

In a single line print the sequence of n characters "+" and "-", where the i-th character is the sign that is placed in front of number a_i. The value of the resulting expression s must fit into the limits 0 ≤ s ≤ a₁. If there are multiple solutions, you are allowed to print any of them.

Sample test(s)

input

4
1 2 3 5

output

+++-

input

3
3 3 5

output

++-

 

 題意：給出一組遞增的數a1，a2,a3.....an（a(i - 1) <= a(i) <= 2 * a (i - 1)），在數之間加上‘+’或‘-’運算符算出一個sum，要求0 <=

 

 sum <= a1;（sum開始爲0；）輸出其中一組符合的運算符串（答案有可能不唯一，輸出其中一組可以了且一定有解）；

 

 

 

這題一開始不會做。。。想得太複雜了。。。。看別人代碼才知道純數學推導出來

 

~~~~~經驗不足，慚愧慚愧。。。。。

 

 

 

推導：

設數據a1,a2,a3......a(n - 2),a(n - 1),a(n);

 

因爲數據間滿足a(i - 1) <= a(i) <= 2 * a (i - 1);

 

所以a(n) 是最大的；

 

所以從a(n)開始減；

 

因爲a(n - 1) <= a(n) <= 2 * a (n - 1);

 

所以設k = a(n) - a (n - 1)；

 

0 <= k <= a(n - 1);

 

因爲 a(n - 2) <= a(n - 1) <= 2 * a (n - 2);且k > 0

 

 0 <= k <= 2 * a(n - 2);

 

所以 k = k - a(n - 2);

 

所以 -a(n - 2) <= k <= a(n - 2);

 

所以 0 <= abs (k) <= a(n - 2);

 

所以當n > 2 時 abs（k）一定在0到 a(n - 2)之間；

 

當K < 0 時， -k就大於0了；

 

在判斷n == 2,1也符合；

 

所以做法很簡單；

 

代碼：

 

 

#include <stdio.h>
__int64 num[100100];
char ch[100100];
int main ()
{
	int n,i,j,k;
	scanf ("%d",&n);
	for (i = 0 ; i < n ; i ++)
		scanf ("%I64d",&num[i]);
	k = num[n - 1];//k 就是 sum； 
	ch[n - 1] = '+';
	for (i = n - 2 ; i >= 0 ; i --)
	{
		if (k >= 0)
		{
			ch[i] = '-';
			k -= num[i];
		}
		else
		{
			ch[i] = '+';
			k += num[i];
		}
	}
	if (k < 0) //當K 小於0 所有符合都要相反，k就邊正了 
	{
		for (i = 0 ; i < n ; i ++)
			printf ("%c",ch[i] == '+' ? '-' : '+');
		printf ("\n");	
	}	
	else
		printf ("%s\n",ch);
}