ZCMU-1037

1037: I want you!

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 179  Solved: 29

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Description

Given an array Ai(0<=i<n), its length is n; We want to find an another array Bi , which is 0 or 1,and its length is n too;

Besides, Ai and Bi meets the following conditions:

If neither A[i]*B[i] nor A[j]*B[j] equals to 0, then A[i]*B[i] < A[j]*B[j];(0<=i<j<n)

Now , we want to know the maximum of ∑Bi(0<=i<n) you can get.

Input

The input consists of multiple test cases。For each test case ,the first line contains one integer n (1<n<=300).Next line contains n integers.

Output

For each case, output the maximum of ∑Bi.

Sample Input

6
1 2 3 4 5 6
4
3 2 3 6

Sample Output

6
3

【解析】

 这道题的话其实就是叫你算最长子序列得到长度是多少，但是0页要算入里面b这个数组里面只有0或者1，其实就是这样的我们就考虑b为1的情况来算就行了，因为他要a[i]*b[i]<a[j]*b[j] (i<j)其实我们这里可以不用说真的给b数组只有1和0我们完全可以用b数组来存放长度，因为只有保证了b为1才能比较，如果a[i]为0则f[i]为0,其实就是找b数组求它的长度是多少的问题。如果a[i]为0那么f[i]也为0，这样的话其实它也是小于后面的。

#include <iostream>
#include <cstdio>
#include<vector>
using namespace std;
int main()
{
int n,i,j,count1=0,sum,num;
while(~scanf("%d",&n))
{
sum=0;
count1=0;
int a[1000]={0};
int b[1000]={0};
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
if(a[i]==0)
{
count1++;
b[i]=0;//如果a[i]为0则给b[i]也设置为0
}
for(j=0;j<i;j+
4000
+)
{
if(a[j]!=0&&a[j]<a[i]&&b[j]>b[i])
b[i]=b[j];//判断下标比i小的元素小于i的如果此刻上升序列把b[j]序列的长度给a[i]之后再继续，
//这里起到一个动态给的过程
}
if(a[i]!=0)
b[i]++;
if(b[i]>sum)
sum=b[i];
}
printf("%d\n",sum+count1);
}
return 0;
}