MiYu原创, 转帖请注明 : 转载自 ______________白白の屋

题目地址:

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

题目描述:

Find the nondecreasing subsequences

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 211 Accepted Submission(s): 80

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

题目分析:

整整一天的时间, 终于是把这个题A了 ,HDU 第一, 纪念下.....................

HUT-MiYu

375MS

1824K

2358B

G++

2010-08-27 09:44:32

但也没有什么值得高兴的, 题目的思路是小A 的, 0rz....................... 现在还是没有理解树状数组是怎么解这一题的, 它的思路是怎样的,

一点也没明白, 刚开始做的时候, 还以为输入数据已经是按非递减排序了, 所以直接用公式 2^n - 1, WA ...

..... 问过小A 才发现, 输入数据是随机的. 这时就要像找上升子串一样, 找到它所有的上升子串. 这里用到了树状数组, 继续理解-ing......

先发代码 :

MiYu原创, 转帖请注明 : 转载自 ______________白白の屋

http://www.cnblog.com/MiYu

Author By : MiYu

Test : 1

Program : 2227

#include <iostream>

#include <algorithm>

using namespace std;

#define lowbit(x) (x&(-x))

int num[100005];

int numcopy[100005];

int hash[100005];

int com[100005];

int nCount = 1;

void add ( int x,int k )

{

while ( x <= nCount )

{

com[x] += k;

if ( com[x] >= 1000000007 ) com[x] %= 1000000007;

x += lowbit(x);

}

int sum ( int x )

{

int s = 0;

while ( x > 0 )

{

s += com[x];

if ( s >= 1000000007 ) s %= 1000000007;

x -= lowbit(x);

}

return s %= 1000000007;

}

int cmp (const void *a, const void *b)

{

return *((int*)a) - *((int*)b);

}

int sfind ( int x )

{

int *p = (int *)bsearch ( &x,hash+1,nCount+1,sizeof ( int ),cmp );

return p - hash;

}

int find(int num){

int top=1,bottom=nCount,mid=(top+bottom)/2,ans=mid;

while(num!=hash[ans]){

if(hash[mid]<=num){

top=(ans=mid)+1;

}else{

bottom=mid-1;

}

mid=(top+bottom)/2;

}

return ans;

}

inline bool scan_d(int &num) //整数输入

{

char in;bool IsN=false;

in=getchar();

if(in==EOF) return false;

while(in!='-'&&(in<'0'||in>'9')) in=getchar();

if(in=='-'){ IsN=true;num=0;}

else num=in-'0';

while(in=getchar(),in>='0'&&in<='9'){

num*=10,num+=in-'0';

}

if(IsN) num=-num;

return true;

}

int main ()

{

int N;

while ( scan_d ( N ) )

{

for ( int i = 0; i != N; ++ i )

scan_d ( num[i] ),numcopy[i] = num[i];

sort ( num, num + N );

memset ( com,0,sizeof (com) );

nCount = 1;

hash[1] = num[0];

for ( int i = 1; i < N; ++ i )

{

if ( num[i] != num[i-1] )

hash[++nCount] = num[i];

}

for ( int i = 0; i < N; ++ i )

{

int pos = find ( numcopy[i] );

int res = sum ( pos ) + 1;

add ( pos,res );

}

cout << sum ( nCount ) << endl;

}

return 0;

}

转载于:https://www.cnblogs.com/MiYu/archive/2010/08/27/1809842.html

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