玲珑杯”ACM比赛 Round #4 E -- array【DP】

玲珑杯”ACM比赛 Round #4

Start Time：2016-11-05 12:00:00 End Time：2016-11-05 17:00:00 Refresh Time：2016-11-06 21:43:46 Private

E -- array

Time Limit：3s Memory Limit：64MByte

Submissions：465Solved：140

DESCRIPTION

2 array is an array, which looks like:
$$1,2,4,8,16,32,\mathrm{64......}$$
$$a_1=1|\frac{a_{i+1}}{a_i}=2$$
Give you a number array, and your mission is to get the number of subsequences ,which is 2 array, of it.
Note: 2 array is a finite array.

INPUT

There are multiple test cases.The first line is a number T ( $T\le 10$), which means the number of cases.For each case, two integer $n(1\le n\le {10}^5)$.The next line contains $n$ numbers $a_i(1\le a_i\le {10}^9)$

OUTPUT

one line - the number of subsequence which is 2 array.(the answer will $\%{10}^9+7$)

SAMPLE INPUT

241 2 1 241 2 4 4

SAMPLE OUTPUT

54

       

        没想到DP就直接做了，官方题解：注意到${10}^9$范围内的$2$的幂次只有$30$个，所以我们定义$dp\left[30\right]$这样一个dp数组，$dp\left[i\right]$表示以$2^i$为结尾的满足条件的子序列的个数。枚举每一个数来转移，复杂度$O\left(32n\right)$。

AC代码：

#include
#include

typedef long long LL;
const int MOD=1e9+7;

LL p[44],a[44];

int main()
{
	p[1]=1; 
	for(int i=2;i<=33;++i) {
     	p[i]=p[i-1]<<1; //printf("%lld %lld\n",i,p[i]);	
	}
	int T; scanf("%d",&T);
	while(T--) {
		int N; scanf("%d",&N);
		memset(a,0,sizeof(a));
		for(int i=0;i