“玲珑杯”ACM比赛 Round #4 E题

E -- array

Time Limit：3s Memory Limit：64MByte

Submissions：443Solved：130

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

2

4

1 2 1 2

4

1 2 4 4

SAMPLE OUTPUT

5

4

SOLUTION

“玲珑杯”ACM比赛 Round #4

记录一下1,2,4-------各自的个数---

代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define LL long longint lp;LL a[60],b[100010];LL dp[60],mod;void ss(){    lp=0;    LL op=1;    while (op<10000000000)    {        a[lp++]=op;        op*=2;    }    return ;}void slove(){    int n;scanf("%d",&n);    for (int i=1;i<=n;i++)        scanf("%lld",&b[i]);    memset(dp,0,sizeof(dp));    LL s1,ans=0;    for (int i=1;i<=n;i++)    {        if (upper_bound(a,a+lp,b[i])-lower_bound(a,a+lp,b[i]))        {            int k=lower_bound(a,a+lp,b[i])-a;            if (k==0)            {                s1=1;            }            else                s1=dp[k-1];            ans=(ans+s1)%mod;            dp[k]=(dp[k]+s1)%mod;        }    }    printf("%lld\n",ans);}int main(){    mod=1000000007;    ss();    int t;scanf("%d",&t);    while (t--)        slove();    return 0;}