【玲珑杯 1050】【DP】array

传送门：http://www.ifrog.cc/acm/problem/1050

思路：注意到10^9范围内的2的幂次只有30个，所以我们定义dp[30]这样一个dp数组，dp[i]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(n)

代码：

#include <iostream>#include <stdio.h> #include <algorithm>#include <string.h>using namespace std;typedef long long ll;const ll mod = 1e9 + 7;int dp[35];int main(){    int T, n;    ll a, ans, sz;    scanf("%d",&T);    while(T--){        memset(dp, 0, sizeof(dp));        scanf("%d", &n);        for(int i = 1 ; i <= n ; i++){            scanf("%lld", &a);            if(a == 1)                dp[0]++;            else{                sz = 0;                while(a % 2 == 0){                    a >>= 1;                    sz++;                }                if(a == 1 && dp[sz - 1] != 0)                    dp[sz] = (dp[sz] + dp[sz - 1]) % mod;            }        }        ans = 0;        for(int i = 0 ; i < 32 ; i++)            ans = (ans + dp[i]) % mod;        printf("%lld\n",ans % mod);    }    return 0;}

描述：

1050 - array

Time Limit：3s Memory Limit：64MByte

Submissions：456Solved：135

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

SOLUTION

“玲珑杯”ACM比赛 Round #4