HDU 2227 Find the nondecreasing subsequences dp思想 + 树状数组

http://acm.hdu.edu.cn/showproblem.php?pid=2227

用dp[i]表示以第i个数为结尾的nondecreasing串有多少个。

那么对于每个a[i]

要去找 <= a[i]的数字那些位置，加上他们的dp值即可。

可以用树状数组维护

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;

#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
const int MOD = ;
const int maxn =  + ;
LL a[maxn], b[maxn];
int n;
LL c[maxn];
LL lowbit(LL x) {
    return x & (-x);
}
void UpDate(int pos, LL val) {
    while (pos <= n) {
        c[pos] += val;
        if (c[pos] >= MOD) c[pos] %= MOD;
        pos += lowbit(pos);
    }
}
LL query(int pos) {
    LL ans = ;
    assert(pos >= );
    while (pos) {
        ans += c[pos];
        pos -= lowbit(pos);
    }
    return ans;
}
void work() {
    memset(c, , sizeof c);
    for (int i = ; i <= n; ++i) {
        cin >> a[i];
        b[i] = a[i];
    }
    sort(b + , b +  + n);
    LL ans = ;
    for (int i = ; i <= n; ++i) {
        int pos = lower_bound(b + , b +  + n, a[i]) - b;
        LL tans = query(pos) + ;
        ans += tans;
        if (ans >= MOD) ans %= MOD;
        UpDate(pos, tans);
    }
    cout << ans << endl;
}

int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    IOS;
    while (cin >> n) work();
    return ;
}