大家好，又见面了，我是全栈君，今天给大家准备了Idea注册码。
Problem 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
7
题意：问你不降子序列的个数。
一看n达到了1e5的级别。就知道得nlogn算法。
然而想到了一个mp的迭代可是每次迭代都得log复杂度太高。所以树状数组+离散化搞。
题解：设dp[i]为前i个而且以i结尾的的不降子序列个数。
我们知道前面凡是小于等于a[i]的都能够到dp[i],所以dp[i]+=dp[j](a[j]<=a[i]&&j<i).
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<string>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<bitset>
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
typedef long long LL;
typedef pair<int,int>pil;
const int INF = 0x3f3f3f3f;
const int MOD=1e9+7;
const int maxn=1e5+10;
int a[maxn],n,b[maxn+100];
LL dp[maxn],c[maxn+100];
int lowbit(int x)
{
return x&(-x);
}
void update(int x,LL val)
{
while(x<maxn)
{
c[x]=(c[x]+val)%MOD;
x+=lowbit(x);
}
}
LL query(int x)
{
LL sum=0;
while(x>0)
{
sum=(sum+c[x])%MOD;
x-=lowbit(x);
}
return sum;
}
int main()
{
while(~scanf("%d",&n))
{
int cnt=1;
CLEAR(c,0);
REPF(i,1,n)
{
scanf("%d",&a[i]);
b[cnt++]=a[i];
}
sort(b+1,b+cnt);
cnt=unique(b+1,b+cnt)-(b+1);
for(int i=1;i<=n;i++)
dp[i]=1;
LL ans=0;
for(int i=1;i<=n;i++)
{
int x=lower_bound(b+1,b+1+cnt,a[i])-b;
dp[i]=(dp[i]+query(x))%MOD;
update(x,dp[i]);
ans=(ans+dp[i])%MOD;
}
printf("%I64d\n",ans);
}
return 0;
}
/*
*/
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