HDOJ 5504 GT and sequence (乱搞)

2016-12-02 12:51:49来源:网络收集作者:管理员人点击

GT and sequence
Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2192Accepted Submission(s): 504
Problem Description
You are given a sequence of
N
integers.You should choose some numbers(at least one),and make the product of them as big as possible.It guaranteed that **the absolute value of** any product of the numbers you choose in the initial sequence will not bigger than263−1.Input
In the first line there is a number
T
(test numbers).For each test,in the first line there is a number N,and
in the next line there are N
numbers.1≤T≤1000
1≤N≤62You'd better print the enter in the last line when you hack others.You'd better not print space in the last of each line when you hack others.Output
For each test case,output the answer.Sample Input
1
3
1 2 3
Sample Output
6题意:给出NNN个整数。你要选择至少一个数,使得你选的数的乘积最大。
保证任意选一些数相乘的绝对值都不会大于2^63-1
思路:输入的数包括负数、正数、零,注意负数和0的情况就行了如果负数的个数为,那么除以最大的那个负数就行了
,如果存在0,不能把0乘入总积,到最后在判断情况就行了。
ac代码:
#include
#include
#include
#include
#include
#define fab(a) (a)>0?(a):(-a)
#define LL long long
#define MAXN 100010
#define INF 0x7fffffff
using namespace std;
LL a[65];
LL ans;
int main()
{
int t,i,n;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
if(n==1)
{
scanf("%lld",&a[0]);
printf("%lld/n",a[0]);
continue;
}
LL k=-INF;
LL ans=1;
int cnt=0;
int bz=0;
for(i=1;i<=n;i++)
{
scanf("%lld",&a[i]);
if(a[i]!=0)
ans*=a[i];
if(a[i]<0)
k=max(a[i],k),cnt++;
else if(a[i]>0)
bz=1;
}
if(cnt<=1&&bz==0)
{
printf("0/n");
continue;
}
if(cnt%2)
{
ans=ans/k;
printf("%lld/n",ans);
}
else
printf("%lld/n",ans);
}
return 0;
}提供几组数据:
1000
3
1 1 1
5
1 1 -1 2 2
6
2 -1 -3 -5 1 1
7
0 1 1 1 2 2 1
3
0 -1 1
4
0 -1 -3 4
5
0 0 -3 -1 2
2
0 -1
2
-1 2

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