【Analysis】

这题应该贪心和DP都能过；

可以发现，一个数只能变成$-1$−1或$1$1,令$Dp_{i,j}$Dpi,j​表示前i个数有j个变成了$-1$−1,发现转移和j的数量无关，只和奇偶性有关，令$Dp_{i,0}$Dpi,0​表示前i个数用了偶数个$-1$−1,那么方程就是

Dp[0][1] = 1e18;

Dp[i][0] = std::min(Dp[i - 1][0] + D[i][1], Dp[i - 1][1] + D[i][0]);
Dp[i][1] = std::min(Dp[i - 1][0] + D[i][0], Dp[i - 1][1] + D[i][1]);

总复杂度是$O(N)$O(N)

【Code】

#include <cstdio>
#include <set>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <queue>
#include <map>
 
namespace IO
{
    inline char gc()
    {
        static char s[1<<20|1]={0},*p1=s,*p2=s;
        return (p1==p2)&&(p2=(p1=s)+fread(s,1,1<<20,stdin),p1==p2)?EOF:*(p1++);
    }
//	inline char gc() { return getchar(); }
	inline long long read()
	{
		long long ret=0;bool flag=0;char c=gc();
		while ((c<'0')|(c>'9')) flag ^= !(c^'-'),c=gc();
		while ((c>='0')&(c<='9')) ret=(ret<<1)+(ret<<3)+(c^'0'),c=gc();
		return flag?-ret:ret;
 	}
 	char OutputAns[1<<20|1],*OutputCur = OutputAns;
 	inline void output()
 	{
 		OutputCur -= fwrite(OutputAns,1,OutputCur - OutputAns,stdout);
	}
	inline void print(long long ans)
	{
		char s[20]={0};
		if (OutputCur - OutputAns + sprintf(s,"%I64d",ans) >> 20) output();
		OutputCur += sprintf(OutputCur,"%I64d",ans);
	}
	inline void printc(char c)
	{
		if (OutputCur - OutputAns + 1 >> 20) output();
		*(OutputCur++) = c;
	}
}
 
using IO::read;
using IO::print;
using IO::printc;
using IO::output;
 
const int M = 1e5 + 11;
 
long long A[M], Dp[M][2], D[M][2];
 
int n;
 
int main(void)
{
	n = read();
	for (int i = 1;i <= n; ++i)
		A[i] = read();
//	std::sort(A + 1, A + 1 + n);
	for (int i = 1;i <= n; ++i)
		D[i][0] = abs(A[i] + 1), D[i][1] = abs(A[i] - 1);
	Dp[0][1] = 1e18;
	for (int i = 1;i <= n; ++i)
	{
		Dp[i][0] = std::min(Dp[i - 1][0] + D[i][1], Dp[i - 1][1] + D[i][0]);
		Dp[i][1] = std::min(Dp[i - 1][0] + D[i][0], Dp[i - 1][1] + D[i][1]);
	} printf("%I64d\n", Dp[n][0]);
}