Find the nondecreasing subsequences（数状数组+离散化+dp）

Problem B

Time Limit : 10000/5000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 70   Accepted Submission(s) : 17

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：

给你一串数字，问共有多少个不递减的子序列。

思路：

一看到不递减立马想到了逆序数，一想到逆序数又想到数状数组，题目的难点在于dp思想的运用。这一点是非常不好理解的。就是还要注意求和会溢出，要求余mod。

代码：

#include <iostream>
#include <stdio.h>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define mod 1000000007
typedef long long ll;
using namespace std;
int n;
int c[100010],b[100010],t[100010];
struct node{
int v;
int id;
bool operator<(const node&b)const
{
return v<b.v;
}
}a[100010];
int lowbit(int x)
{
return x&(-x);
}
int sum(int x)
{
int res=0;
while(x>0){
res+=c[x];
if(res>=mod)
res%=mod;
x-=lowbit(x);
}
return res;
}
void add(int i,int v)
{
while(i<=n){
c[i]+=v;
if(c[i]>=mod)
c[i]%=mod;
i+=lowbit(i);
}
}
int main()
{
//ios::sync_with_stdio(false);
int i,j;
while(scanf("%d",&n)!=EOF)
{
memset(b,0,sizeof(b));
memset(c,0,sizeof(c));
for(i=1;i<=n;i++)
{
scanf("%d",&a[i].v);
a[i].id=i;
}
sort(a+1,a+n+1);
b[a[1].id]=1;
for(i=2;i<=n;i++) //离散化
{
if(a[i].v!=a[i-1].v)
b[a[i].id]=i;
else b[a[i].id]=b[a[i-1].id];
}
for(i=1;i<=n;i++) //核心点也是难理解的地方
{
t[i]=sum(b[i]);
add(b[i],t[i]+1);
}
printf("%d\n",sum(n));
}
return 0;
}