hdu 2227 Find the nondecreasing subsequences(树状数组)

给出一个长度为n的序列求这个序列中非递减的序列有几个。

1,2,3: {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}.一共有7个。

 1 <= n <= 100000, 0 <= si <= 2^31.数据过大，所以需要离散化。

怎么离散化呢?将值映射到下标。再取原数组可知每个数在什么位置。

再将得到的下标查询，查询后+1，每次都是增加1，再将这个数更新进树状数组。说的不是很清楚，还是看代码吧

/*
Problem ID:
meaning:
Analyzing:
*/
#include <iostream>
#include <algorithm>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<vector>

using namespace std;
typedef struct even{int y1,y2,x;}even;

#define clr(A,k) memset(A,k,sizeof(A))
#define FOR(i,s,t) for(int i=(s); i<(t); i++)
#define LL long long
#define BUG puts("here!!!")
#define print(x) printf("%d\n",x)
#define STOP system("pause")
#define eps 1e-8
#define PI acos(-1.0)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define maxn 100005
#define maxm 1005
#define MM 1<<31
#define MOD  1000000007
#define lowbit(x) x&(-x)
LL gcd(LL a,LL b) {return a?gcd(b%a,a):b;}
int S[maxn],C[maxn],A[maxn];
int n,tot;

void update(int pos,int val){
    while(pos<=n){
        C[pos]+=val;
        if(C[pos]>=MOD) C[pos]%=MOD;
        pos+=lowbit(pos);
    }
}

LL query(int x){
    LL ret=0;
    while(x>0){
        ret+=C[x];
        if(ret>=MOD) ret%=MOD;
        x-=lowbit(x);
    }
    return ret;
}

int Binsearch(int x){
    int l=1,r=tot;
    int m;
    while(l<=r){
        int m=(l+r)>>1;
        if(S[m]==x) return m;
        if(S[m]>x) r=m-1;
        else l=m+1;
    }
}

int main(){
    while(~scanf("%d",&n)){
        clr(C,0);
        for(int i=1;i<=n;i++){
            scanf("%d",&A[i]);
            S[i]=A[i];
        }
        sort(S+1,S+n+1);
        tot=1;
        for(int t=2;t<=n;t++){
            if(S[t]!=S[t-1]){
                S[++tot]=S[t];
            }
        }
        LL ret=0;
        for(int i=1;i<n+1;i++){
            int id=Binsearch(A[i]);
            LL tp=query(id);
            ret+=tp+1;
            if(ret>=MOD) ret%=MOD;
            update(id,tp+1);
        }
        printf("%I64d\n",ret);
    }
return 0;
}