AtCoder Beginner Contest 059 C - Sequence(贪心)

C - Sequence

Time limit : 2sec / Memory limit : 256MB

Score : 300 points

Problem Statement

You are given an integer sequence of length N.
The i-th term in the sequence is ai.
In one operation, you can select a term and either increment or decrement it by one.
At least how many operations are necessary to satisfy the following conditions?
For every i (1≤i≤n),
the sum of the terms from the 1-st through i-th
term is not zero.
For every i (1≤i≤n−1),
the sign of the sum of the terms from the 1-st through i-th
term, is different from the sign of the sum of the terms from the 1-st through(i+1)-th
term.

Constraints

2≤n≤105
|ai|≤109
Each ai is
an integer.

Input

Input is given from Standard Input in the following format:

n
a1 a2 … an

Output

Print the minimum necessary count of operations.

Sample Input 1

Copy

4
1 -3 1 0

Sample Output 1

Copy

4

For example, the given sequence can be transformed into 1,−2,2,−2 by
four operations. The sums of the first one, two, three and four terms are 1,−1,1 and −1,
respectively, which satisfy the conditions.

Sample Input 2

Copy

5
3 -6 4 -5 7

Sample Output 2

Copy

0

The given sequence already satisfies the conditions.

Sample Input 3

Copy

6
-1 4 3 2 -5 4

Sample Output 3

Copy

8

题目大意:给n个数字组成的数组，可以对任何数字进行删减添加，每次删减添加1为一次操作步数，问满足以下两个条件下，最小操作次数

1.要求1-i个数字和不为0，1<=i<=n

2.要求1-(i-1)的和与i的正负相反，也就是说，当前i-1个数字为正数时候，要求加上第i个数字后，之后为负数

解题思路:这个就是当某个数不符合条件的时候，尽可能的使得和为-1或者1，这样才能让操作次数尽可能的小，考虑到这种情况，1到n只可能是正负正负正负，或者负正负正负正两种情况，同时判断，取个最小值，同时要注意的是，当第一个数是0的时候的处理，WA了无数发才发现对第一个数为0的设定不符合情况就尴尬了

#include<iostream>
#include<cstdio>
#include<stdio.h>
#include<cstring>
#include<cstdio>
#include<climits>
#include<cmath>
#include<vector>
#include <bitset>
#include<algorithm>
#include <queue>
#include<map>
#include<stack>
using namespace std;

long long int ans1, ans2, sum1, sum2;
int n, i;
long long int a[100005];
int main()
{
cin >> n;
for (i = 1; i <= n; i++)
{
cin >> a[i];
}
if (a[1] >0)
{
sum1 = a[1];
ans2= a[1] + 1;
sum2 = -1;
}
else if (a[1] == 0)
{
sum1 = 1;
ans1 = ans2 = 1;
sum2 = -1;
}
else
{
sum1 = 1;
ans1 = abs(a[1]) + 1;
sum2 = a[1];
}
for (i = 2; i <= n; i++)
{
if (sum1 > 0)
{
if (a[i] + sum1 >= 0)
{
ans1 += a[i] + sum1 + 1;
sum1 = -1;
}
else
{
sum1 += a[i];
}
}
else
{
if (a[i] + sum1 <= 0)
{
ans1 += abs(sum1 + a[i]) + 1;
sum1 = 1;
}
else
{
sum1 += a[i];
}
}
}
if (sum1 == 0)
{
ans1++;
}
for (i = 2; i <= n; i++)
{
if (sum2 > 0)
{
if (a[i] + sum2 >= 0)
{
ans2 += a[i] + sum2 + 1;
sum2 = -1;
}
else
{
sum2 += a[i];
}
}
else
{
if (a[i] + sum2 <= 0)
{
ans2 += abs(sum2 + a[i]) + 1;
sum2 = 1;
}
else
sum2 += a[i];
}
}
if (sum2 == 0)
{
ans2++;
}
cout << min(ans1, ans2) << endl;
}