HDU_2227_FindTheNondecreasingSubsequences

Find the nondecreasing subsequences

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1688 Accepted Submission(s): 612



Problem Description

How 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}.

Input

The 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.

Output

For 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

3
1 2 3

Sample Output

7

Author

8600

Recommend

lcy

水题

坐标离散化一下

然后每个点的和应该是这个点之前的子串个数

因此每次add加的操作是加的之前所有满足条件子串的个数+1

1 2 3 4

如读到1 则前面没有 就0加1

读到2 则前面的有1 1+1=2;

读到3 则前面的有 1 2 1+2+1=4;

读到4 则前面的有 1 2 3 1+2+4+1=8;

因此总共有1+2+4+8=15

注意下数据量还有取模

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;

typedef long long LL;
const int M=1e5+5;
const int MO=1e9+7;
LL tree[M];

struct NODE
{
int v,p;
}node[M];

bool cmp1(NODE a,NODE b)
{
if(a.v==b.v)
return a.p<b.p;
return a.v<b.v;     //因为输入没有保证数字不重复，防止排序和离散造成原本可以形成的序列被否定
}

bool cmp2(NODE a,NODE b)
{
return a.p<b.p;
}

inline int lowbit(int x)
{
return x&(-x);
}

LL getsum(int x)
{
LL s=0;
for(int i=x;i;i-=lowbit(i))
s=(s+tree[i])%MO;
return s;
}

void add(int x,LL v)
{
for(int i=x;i<M;i+=lowbit(i))
tree[i]=(tree[i]+v)%MO;
}

int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
memset(tree,0,sizeof(tree));
for(int i=0;i<n;i++)
{
scanf("%d",&node[i].v);
node[i].p=i;
}
sort(node,node+n,cmp1);
for(int i=0;i<n;i++)
node[i].v=i+1;
sort(node,node+n,cmp2);
for(int i=0;i<n;i++)
add(node[i].v,getsum(node[i].v)+1); //加的是之前满足条件数字情况的和
printf("%I64d\n",getsum(M-1));
}
return 0;
}