51nod 1376 最长递增子序列的数量
数组A包含N个整数（可能包含相同的值）。设S为A的子序列且S中的元素是递增的，则S为A的递增子序列。如果S的长度是所有递增子序列中最长的，则称S为A的最长递增子序列（LIS）。A的LIS可能有很多个。例如A为：{1 3 2 0 4}，1 3 4，1 2 4均为A的LIS。给出数组A，求A的LIS有多少个。由于数量很大，输出Mod 1000000007的结果即可。相同的数字在不同的位置，算作不同的，例如 {1 1 2} 答案为2。
Input
第1行：1个数N，表示数组的长度。(1 <= N <= 50000)
第2 - N + 1行：每行1个数A[i]，表示数组的元素(0 <= A[i] <= 10^9)
Output
输出最长递增子序列的数量Mod 1000000007。
Input示例
5
1
3
2
0
4
Output示例
2
/*
51nod 1376 最长递增子序列的数量 problem:
给你一个数组,求其中最长递增子序列有多少个 solve:
对于第i个数a[i]而言,它需要知道已经出现的[1,a[i]-1]中最长递增子序列的长度以及数量. 所以可以利用线段树来维护,
然后利用长度和数量来更新a[i] hhh-2016/09/03-16:41:1
*/
#pragma comment(linker,"/STACK:124000000,124000000")
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <vector>
#include <math.h>
#include <queue>
#include <set>
#include <map>
#define lson i<<1
#define rson i<<1|1
#define ll long long
#define clr(a,b) memset(a,b,sizeof(a))
#define scanfi(a) scanf("%d",&a)
#define scanfs(a) scanf("%s",a)
#define scanfl(a) scanf("%I64d",&a)
#define scanfd(a) scanf("%lf",&a)
#define key_val ch[ch[root][1]][0]
#define eps 1e-7
#define inf 0x3f3f3f3f3f3f3f3f
using namespace std;
const ll mod = 1000000007;
const int maxn = 50010;
const double PI = acos(-1.0);
int a[maxn],t[maxn];
struct node
{
int l,r;
ll Max,num;
int mid;
} tree[maxn << 2]; void push_up(int i)
{
if(tree[lson].Max == tree[rson].Max)
{
tree[i].Max = tree[lson].Max;
tree[i].num = tree[lson].num + tree[rson].num;
tree[i].num %= mod;
}
else if(tree[lson].Max > tree[rson].Max)
{
tree[i].Max = tree[lson].Max;
tree[i].num = tree[lson].num;
}
else
{
tree[i].Max = tree[rson].Max;
tree[i].num = tree[rson].num;
}
} void build(int i,int l,int r)
{
tree[i].l = l,tree[i].r = r;
tree[i].Max = tree[i].num = 0;
tree[i].mid = (l+r) >> 1;
if(l == r)
return ;
build(lson,l,tree[i].mid);
build(rson,tree[i].mid + 1,r);
push_up(i);
} void push_down(int i)
{ } void update(int i,int k,ll len,ll many)
{
if(tree[i].l == tree[i].r && tree[i].l == k)
{
if(tree[i].Max < len) tree[i].Max = len,tree[i].num = many;
else if(tree[i].Max == len) tree[i].num += many;
tree[i].num %= mod;
return ;
}
int mid = tree[i].mid;
if(k <= mid)
update(lson,k,len,many);
else update(rson,k,len,many);
push_up(i);
}
ll tans,tnum;
void query(int i,int l,int r)
{
if(l > r)
{
tans = 0,tnum = 1;
return ;
}
if(tree[i].l >= l && tree[i].r <= r)
{
if(tans == -1)
tans = tree[i].Max,tnum = tree[i].num;
else
{
if(tans == tree[i].Max) tnum += tree[i].num;
else if(tans < tree[i].Max)
{
tans = tree[i].Max,tnum = tree[i].num;
}
tnum %= mod;
}
return ;
}
push_down(i);
int mid = tree[i].mid;
if(l <= mid)
query(lson,l,r);
if(r > mid)
query(rson,l,r);
return;
} int cnt = 0;
int main()
{
int n;
while(scanfi(n) != EOF)
{
for(int i = 0; i< n;i++){
scanfi(a[i]);
t[i] = a[i];
}
sort(a,a+n);
cnt = 0;
for(int i = 1;i < n;i++)
{
if(a[i] != a[cnt])
a[++cnt] = a[i];
}
// for(int i = 0;i <= cnt;i++)
// printf("%d\n",a[i]);
build(1,0,cnt);
ll tMax = 0,ans = 0;
int id = lower_bound(a,a+cnt+1,t[0]) -a;
update(1,id,1,1);
tMax = 1,ans = 1;
for(int i = 1;i < n;i++)
{
id = lower_bound(a,a+cnt+1,t[i]) -a;
// cout << id <<" " ;
tans = -1,tnum = 0;
query(1,0,id-1);
if(!tans && !tnum)
tnum = 1;
if(tMax < tans+ 1) tMax = tans+1,ans = tnum;
else if(tMax == tans+1) ans += tnum;
ans %= mod;
update(1,id,tans + 1,tnum);
}
// cout <<endl;
printf("%I64d\n",ans);
}
return 0;
}