玲瓏杯”ACM比賽 Round #4 E -- array【DP】

玲瓏杯”ACM比賽 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

       

        沒想到DP就直接做了，官方題解：注意到${10}^9$範圍內的$2$的冪次只有$30$個，所以我們定義$dp\left[30\right]$這樣一個dp數組，$dp\left[i\right]$表示以$2^i$爲結尾的滿足條件的子序列的個數。枚舉每一個數來轉移，複雜度$O\left(32n\right)$。

AC代碼：

#include<cstdio>
#include<cstring>

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<N;++i) {
			int tem=0; 
			scanf("%d",&tem);
			if(tem==1) {
				a[1]=(a[1]+1)%MOD; 
				continue;
			}
			for(int j=2;j<33;++j) {
				if(tem==p[j]) {
					//printf("11111111\n");
					a[j]=(a[j]%MOD+a[j-1]%MOD)%MOD;
					break;
				}
				else if(tem<p[j] ) break;
			}
		}
		LL ans=0;
		for(int i=1;i<=32;++i)  ans=(ans+a[i])%MOD;
		printf("%lld\n",ans); 
	}
	return 0;
}