hdu-2227-Find the nondecreasing subsequences（DP+离散化+树状数组）

Find the nondecreasing subsequences

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

Total Submission(s) : 2   Accepted Submission(s) : 2

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

31 2 3

 

Sample Output

7

 

题意：

给定序列集合 求有多少个非递减子序列。

思路：每多加一个数的方法数是之前所有小于等于它的方法数之和加一，数据较大所以 所以需要 离散化。然后用树状数组维护。

code：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#define ll long long#define mod 1000000007#define maxn 100005#define lowbit(x) (x&(x^(x-1)))using namespace std;struct node{    int key;///保存原序列的次序，离散化后用来唯一代表这个val值    ll val;///序列的大小};node a[maxn];int c[maxn];int b[maxn];int n;int sum(int p){    ll ans=0;    while(p>0){        ans=(ans+c[p])%mod;        p-=lowbit(p);    }    return ans;}void update(int p,int x){    while(p<=n){        c[p]=(c[p]+x)%mod;        p+=lowbit(p);    }}bool cmp(node a,node b){    return a.val<b.val;}int main(){    while(~scanf("%d",&n)){        for(int i=1;i<=n;i++){            scanf("%lld",&a[i].val);            a[i].key=i;        }        sort(a+1,a+n+1,cmp);        b[ a[1].key ]=1;        int cnt=1;        for(int i=2;i<=n;i++){            if(a[i].val!=a[i-1].val)                cnt++;            b[a[i].key]=cnt;///cnt 用来代表它的大小        }        memset(c,0,sizeof(c));        for(int i=1;i<=n;i++){            int temp=sum(b[i]);            update(b[i],temp+1);        }        printf("%d\n",sum(n));    }    return 0;}