Find the nondecreasing subsequences

最新推荐文章于 2021-09-06 15:08:28 发布

原创最新推荐文章于 2021-09-06 15:08:28 发布 · 495 阅读

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本文介绍了一种算法，用于解决在一个整数序列中寻找所有非递减子序列的问题。通过将输入的数和其索引存入结构体并进行排序，利用树状数组求逆序数的方法来高效计算子序列数量。

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roblem Description

How many nondecreasing subsequences can you find in the sequence S = {s1, s2, s3, ...., sn} ? For example, we assume that S = {1, 2, 3}, and you can find seven nondecreasing subsequences, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}.

Input

The input consists of multiple test cases. Each case begins with a line containing a positive integer n that is the length of the sequence S, the next line contains n integers {s1, s2, s3, ...., sn}, 1 <= n <= 100000, 0 <= si <= 2^31.

Output

For each test case, output one line containing the number of nondecreasing subsequences you can find from the sequence S, the answer should % 1000000007.

Sample Input

3
1 2 3

Sample Output

Author

8600

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保存输入的数和序号在结构体中，然后按数的大小排序。

之后b数组保存排序后的顺序。相当用树状数组求逆序数。

for i=1：n

b[i]

按输入的原来顺序（b[i]保存排序后的顺序)

更新树状数组。统计出比他小的数总共有几个。最后sum[n]为所求。

#include<iostream>
#include<stdio.h>
#include<memory.h>
#include<algorithm>
using namespace std;
#define MAX 100005
#define BASE 1000000007
struct node{
int val,id;
}a[MAX];

bool cmp(node a,node b)
{
return a.val<b.val;
}

int b[MAX],c[MAX],s[MAX],n;

int lowbit(int i)
{
return i&(-i);
}

void update(int i,int x)
{
while(i<=n)
{
s[i]+=x;
if(s[i]>=BASE)
s[i]%=BASE;
i+=lowbit(i);

}
}

int sum(int i)
{
int sum=0;
while(i>0)
{
sum+=s[i];
if(sum>=BASE)
sum%=BASE;
i-=lowbit(i);

}
return sum;

}
int main()
{
int i,res;
while(scanf("%d",&n)!=EOF)
{
memset(b,0,sizeof(b));
memset(s,0,sizeof(s));
for(i=1;i<=n;i++)
{
scanf("%d",&a[i].val);
a[i].id=i;
}
sort(a+1,a+n+1,cmp);
b[a[1].id]=1;
for(i=2;i<=n;i++)
{
if(a[i].val!=a[i-1].val)
b[a[i].id]=i;
else b[a[i].id]=b[a[i-1].id];

}
res=0;
for(i=1;i<=n;i++)
{
c[i]=sum(b[i]);
update(b[i],c[i]+1);
}
printf("%d\n",sum(n));
}
return 0;
}