UVA1614(贪心）

Time Limit:3000MS

Memory Limit:Unknown

64bit IO Format:%lld & %llu

Description

Most financial institutions had become insolvent during financial crisis and went bankrupt or were bought by larger institutions, usually by banks. By the end of financial crisis of all the financial institutions only two banks still continue to operate. Financial markets had remained closed throughout the crisis and now regulators are gradually opening them. To prevent speculation and to gradually ramp up trading they will initially allow trading in only one financial instrument and the volume of trading will be limited to i contracts for i -th minute of market operation. Two banks had decided to cooperate with the government to kick-start the market operation. The boards of directors had agreed on trading volume for each minute of this first trading session. One bank will be buying a_i contracts ( 1

a_i

i ) during i -th minute ( 1

n ), while the other one will be selling. They do not really care whether to buy or to sell, and the outside observer will only see the volume a_i of contracts traded per minute. However, they do not want to take any extra risk and want to have no position in the contract by the end of the trading session. Thus, if we define b_i = 1 when the first bank is buying and b_i = - 1 when the second one is buying (and the first one is selling), then the requirement for the trading session is that $sum_{{i=1}}^{{n}}$ a_ib_i = 0 . Your lucky team of three still works in the data center (due to the crisis, banks now share the data center and its personnel) and your task is to find such b_i or to report that this is impossible.

Input

The input file contains several test cases, each of them as described below. The first line of the input contains the single integer number n ( 1

100 000 ). The second line of the input contains n integer numbers -- a_i ( 1

a_i

i ).

Output

For each test case, the first line of the output must contain `` Yes'' if the trading session with specified volumes is possible and `` No'' otherwise. In the former option a second line must contain n numbers -- b_i .

Sample Input

Sample Output

No
Yes
1 -1 -1 1

Source

Root :: AOAPC II: Beginning Algorithm Contests (Second Edition) (Rujia Liu) :: Chapter 8. Algorithm Design :: Exercises

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排序之后贪心瞎搞。。。

/*************************************************************************
    > File Name: c.cpp
    > Author: acvcla
    > QQ: 
    > Mail: acvcla@gmail.com 
    > Created Time: 2014年10月11日 星期六 08时42分28秒
 ************************************************************************/
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<cstring>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<cstdlib>
#include<ctime>
#include<set>
#include<math.h>
using namespace std;
typedef long long LL;
const int maxn = 1e5 + 10;
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define pb push_back
int A[maxn],cnt[maxn],f[maxn];
std::vector<int> v;
int main(){
	int n;
	while(~scanf("%d",&n)&&n){
		memset(cnt,0,sizeof(cnt));
		memset(f,0,sizeof f);
		v.clear();
		LL sum=0;
		LL M=0;
		for(int i=1;i<=n;i++){
			scanf("%d",A+i);
			if(A[i]>M)M=A[i];
			sum+=A[i];
			cnt[A[i]]++;
		}
		if(sum&1){
			puts("No");
			continue;
		}
		sum/=2;
		bool ok=false;
		for(int i=M;!ok&&i>=1;i--){
			f[i]=min((LL)cnt[i],sum/i);
			sum-=(LL)f[i]*i;
			if(sum==0){
				ok=true;
				break;
			}
		}
		if(!ok){
			puts("No");
		}else{
			puts("Yes");
			for(int i=1;i<=n;i++){
				if(i==1){
					if(f[A[i]]>0){
						printf("1");
						f[A[i]]--;
					}
					else printf("-1");
				}
				else{
					if(f[A[i]]>0){
						printf(" 1");
						f[A[i]]--;
					}
					else printf(" -1");
				}
			}
			puts("");
		}

	}
	return 0;
}