51NOD 1376 最长递增子序列的数量 dp+BIT

题目链接：点这里！！

题意：中文题

题解：

　　f[i]:以第i个数结尾的LIS的长度,和该长度的LIS数量 
　　转移的话,显然f[i].first=max{f[j].first}+1,j<i且a[j]<a[i] 
　　f[i].second=∑jf[j].second,f[j].first=f[i].first−1 
　　我们可以用BIT来优化这个转移,维护≤i的最大长度和对应数量

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
const long long INF = 1e18+1LL;
const double pi = acos(-1.0);
const int N = 5e5+10, M = 1e3+20,inf = 2e9,mod = 1e9+7;



pii C[N];
int n,a[N],b[N];
int mx = 0;
void update(int x,pii ret) {
    for(int i = x; i <= n; i += i & (-i)) {
       if(ret.first > C[i].first) {
            C[i] = ret;
        }
        else if(ret.first == C[i].first){
            C[i].second += ret.second;
            C[i].second %= mod;
        }
    }
}
pii ask(int x) {
    pii  ret = MP(0,1);
    for(int i = x; i; i -= i & (-i))
    {
        if(C[i].first > ret.first) {
            ret = MP(C[i].first,C[i].second);
        }
        else if(C[i].first == ret.first) {
            ret.second += C[i].second;
            ret.second %= mod;
        }
    }
    return ret;
}
int main() {

    scanf("%d",&n);
    for(int i = 1; i <= n; ++i) {
        scanf("%d",&a[i]);
        b[i] = a[i];
    }
    sort(b+1,b+n+1);
    int c = unique(b+1,b+n+1) - b - 1;
    for(int i = 1; i <= n; ++i) {
        a[i] = lower_bound(b+1,b+c+1,a[i]) - b;
    }
    int ans = 0;
    for(int i = 1; i <= n; ++i) {
        pii ret = ask(a[i]-1);
        ret.first++;
        if(ret.first > mx) {
            mx = ret.first;
            ans = ret.second;
        }
        else if(mx == ret.first) {
            ans = (ans + ret.second) % mod;
        }
        update(a[i],ret);
    }
    cout<<ans<<endl;
    return 0;
}