HDU 2227 Find the nondecreasing subsequences(DP)，hdunondecreasing

HDU 2227 Find the nondecreasing subsequences(DP)，hdunondecreasing
Problem DescriptionHow 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}. 
InputThe 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. 
OutputFor 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

 題意：問你不降子序列的個數。一看n達到了1e5的層級。就知道得nlogn演算法。然而想到了一個mp的迭代但是每次迭代都得log複雜度太高。所以樹狀數組+離散化搞。
題解：設dp[i]為前i個並且以i結尾的的不降子序列個數。我們知道前面凡是小於等於a[i]的都可以到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;}/**/