关于一些初级ACM竞赛题目的分析和题解（九）

一些逻辑题目，见下题:

Kefa decided to make some money doing business on the Internet for exactly n days. He knows that on the i-th day (1 ≤ i ≤ n) he makes ai money. Kefa loves progress, that's why he wants to know the length of the maximum non-decreasing subsegment in sequence ai. Let us remind you that the subsegment of the sequence is its continuous fragment. A subsegment of numbers is called non-decreasing if all numbers in it follow in the non-decreasing order.

Help Kefa cope with this task!

Input

The first line contains integer n (1 ≤ n ≤ 105).

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109).

Output

Print a single integer — the length of the maximum non-decreasing subsegment of sequence a.

    Examples 
  
      input 
    
6
2 2 1 3 4 1

      output 
    
3

      input 
    
3
2 2 9

      output 
    
3

Note

In the first test the maximum non-decreasing subsegment is the numbers from the third to the fifth one.

In the second test the maximum non-decreasing subsegment is the numbers from the first to the third one.

Kefa决定在互联网上做一些钱，正好n天。他知道在第i天（1≤i≤n），他赚了钱。 Kefa喜欢进步，这就是为什么他想知道序列ai中最大非递减子段的长度。让我们提醒你，序列的子段是它的连续片段。如果数字中的所有数字都是非递减的，则数字的一个子段称为非递减。

在第一个测试中，最大的非递减子段是从第三个到第五个的数字。
在第二个测试中，最大的非递减子段是从第一个到第三个的数字。

using namespace std;
typedef long long ll;
int a,b,c,n,s[100001];
int main()
{
    cin>>n;
    for (int i=0;i<n;i++)
        {
            cin>>s[i];
        }
           for (int i=0;i<n;i++)
            {if(s[i]>=s[i-1])a++;  //找到非递减序列
            else (b=max(a,b),a=1);}// 不断寻找最大非递减序列的长度

    b=max(a,b);
    printf("%d",b);

}

For a positive integer n let's define a function f:

f(n) = - 1 + 2 - 3 + .. + ( - 1)nn

Your task is to calculate f(n) for a given integer n.

Input

The single line contains the positive integer n (1 ≤ n ≤ 1015).

Output

Print f(n) in a single line.

     Examples 
   
       input 
     
4

       output 
     
2

       input 
     
5

       output 
     
-3

Note

f(4) = - 1 + 2 - 3 + 4 = 2

f(5) = - 1 + 2 - 3 + 4 - 5 = - 3

知道题做的时候脑袋傻了，明明是高中知识我还给整的超时了，下面是代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll a,b,c,d,e,n;
int main()
{
    scanf("%lld",&n);
    if(n%2==0)
        printf("%lld",n/2);
    if(n%2!=0)
        printf("%lld",(-1-n)/2);

}

是一个分奇偶的等差数列，分清题意就好了，

关于一些初级ACM竞赛题目的分析和题解（九）

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