B. The Number of Products
time limit per test
memory limit per test
input
output
You are given a sequence a1,a2,…,ana1,a2,…,an consisting of nn non-zero integers (i.e. ai≠0ai≠0).
You have to calculate two following values:
- the number of pairs of indices (l,r)(l,r) (l≤r)(l≤r) such that al⋅al+1…ar−1⋅aral⋅al+1…ar−1⋅ar is negative;
- the number of pairs of indices (l,r)(l,r) (l≤r)(l≤r) such that al⋅al+1…ar−1⋅aral⋅al+1…ar−1⋅ar is positive;
Input
The first line contains one integer nn (1≤n≤2⋅105)(1≤n≤2⋅105) — the number of elements in the sequence.
The second line contains nn integers a1,a2,…,ana1,a2,…,an (−109≤ai≤109;ai≠0)(−109≤ai≤109;ai≠0) — the elements of the sequence.
Output
Print two integers — the number of subsegments with negative product and the number of subsegments with positive product, respectively.
Examples
input
Copy
5
5 -3 3 -1 1
output
Copy
8 7
input
Copy
10
4 2 -4 3 1 2 -4 3 2 3
output
Copy
28 27
input
Copy
5
-1 -2 -3 -4 -5
output
Copy
9 6
题意：给你n长度的数组。分别求出任意长度连续区间内乘积为负数和正数的个数。
暴力会超时，int 会炸。需要O(n)才能过。
这需要一个继承的思想。
看这么一组： 1 2 3 令a=0此为正数个数，sum为区间乘积为正数个数。读入1,a++,sum+=a；a=1,sum=1;
读入2,a++,sum+=a；a=2,sum=3;
读入3，a++,sum+=a；a=3,sum=6;
这个sum累加的过程符合求区间个数的过程。
ll n,a=0,b=0,alla=0,allb=0; //a统计当前正数区间数，b统计当前负数区间数。alla为所有区间乘积为正数个数，allb为所有区间乘积为负数个数.
正数的加入不会改变正负号。
比如对于1，2，3。相当于一个继承过程。我们可以把初始看成1,2，？。（只有2个数）“？”处填入的数，并不影响原来只有1,2时的区间数，a++只是多了一个{3}而已。“？”处只是一个可有可无的东西，加入不加入3,。1,2就那么多区间，1，2，3也就那么多区间，两者唯一区别就是多了一个{3}，所以a++即可，当然这是对全正数的处理。
加入一个负数的话，继承之前的正数区间数swap(a,b)，因为负数的加入使正数区间变为负数区间（上段思想），然后有一个{-1}所以再b++即可。同理正数区间继承了负数区间个数，但不需要a++，因为加入了{-1}再乘就不是正数了。
摘自https://blog.csdn.net/Luoriliming/article/details/100883824的一个解题思想：简单来说就是按序遍历，记录他可以组成的正数个数和负数个数，并且在s1,s2中累加，遇到正数时只需将前一个状态继承并++正数个数即可，如果遇到负数，则需调换正数和负数个数，并++负数个数
上代码把：
#include<iostream>
using namespace std;
typedef long long ll;
ll n,a=0,b=0,alla=0,allb=0;
int main()
{
cin>>n;
for(int i=0;i<n;i++)
{
ll x;
cin>>x;
if(x>0)
{
a++;
}
else
{
swap(a,b);
b++;
}
alla+=a;
allb+=b;
}
cout<<allb<<" "<<alla<<endl;
}
C题题解：
C. Everyone is a Winner!
On the well-known testing system MathForces, a draw of nn rating units is arranged. The rating will be distributed according to the following algorithm: if kk participants take part in this event, then the nn rating is evenly distributed between them and rounded to the nearest lower integer, At the end of the drawing, an unused rating may remain — it is not given to any of the participants.
For example, if n=5n=5 and k=3k=3, then each participant will recieve an 11 rating unit, and also 22 rating units will remain unused. If n=5n=5, and k=6k=6, then none of the participants will increase their rating.
Vasya participates in this rating draw but does not have information on the total number of participants in this event. Therefore, he wants to know what different values of the rating increment are possible to get as a result of this draw and asks you for help.
For example, if n=5n=5, then the answer is equal to the sequence 0,1,2,50,1,2,5. Each of the sequence values (and only them) can be obtained as ⌊n/k⌋⌊n/k⌋ for some positive integer kk (where ⌊x⌋⌊x⌋ is the value of xx rounded down): 0=⌊5/7⌋0=⌊5/7⌋, 1=⌊5/5⌋1=⌊5/5⌋, 2=⌊5/2⌋2=⌊5/2⌋, 5=⌊5/1⌋5=⌊5/1⌋.
Write a program that, for a given nn, finds a sequence of all possible rating increments.
Input
The first line contains integer number tt (1≤t≤101≤t≤10) — the number of test cases in the input. Then tt test cases follow.
Each line contains an integer nn (1≤n≤1091≤n≤109) — the total number of the rating units being drawn.
Output
Output the answers for each of tt test cases. Each answer should be contained in two lines.
In the first line print a single integer mm — the number of different rating increment values that Vasya can get.
In the following line print mm integers in ascending order — the values of possible rating increments.
Example
input
Copy
4
5
11
1
3
output
Copy
4
0 1 2 5
6
0 1 2 3 5 11
2
0 1
3
0 1 3
题意：就是给出n，从 k =1 一直分到k=n+1 ，即b/k
比如样例1： n =5
k= 1 2 3 4 5 6
5 2 1 1 1 0
所以输出0 1 2 5
#include<iostream>
using namespace std;
typedef long long ll;
ll n,a=0,b=0,alla=0,allb=0;
int main()
{
cin>>n;
for(int i=0;i<n;i++)
{
ll x;
cin>>x;
if(x>0)
{
a++;
}
else
{
swap(a,b);
b++;
}
alla+=a;
allb+=b;
}
cout<<allb<<" "<<alla<<endl;
}
