LeetCode 1498. 满足条件的子序列数目（排序+二分查找+快速幂）

示例 1：
输入：nums = [3,5,6,7], target = 9
输出：4
解释：有 4 个子序列满足该条件。
[3] -> 最小元素 + 最大元素 <= target (3 + 3 <= 9)
[3,5] -> (3 + 5 <= 9)
[3,5,6] -> (3 + 6 <= 9)
[3,6] -> (3 + 6 <= 9)

示例 2：
输入：nums = [3,3,6,8], target = 10
输出：6
解释：有 6 个子序列满足该条件。（nums 中可以有重复数字）
[3] , [3] , [3,3], [3,6] , [3,6] , [3,3,6]

示例 3：
输入：nums = [2,3,3,4,6,7], target = 12
输出：61
解释：共有 63 个非空子序列，其中 2 个不满足条件（[6,7], [7]）
有效序列总数为（63 - 2 = 61）

示例 4：
输入：nums = [5,2,4,1,7,6,8], target = 16
输出：127
解释：所有非空子序列都满足条件 (2^7 - 1) = 127
 
提示：
1 <= nums.length <= 10^5
1 <= nums[i] <= 10^6
1 <= target <= 10^6

class Solution {	//C++
    int mod = 1e9+7;
public:
    int numSubseq(vector<int>& nums, int target) {
        sort(nums.begin(),nums.end());
        int i = 0, j;
        unsigned long long count = 0;
        for(i = 0; i < nums.size(); ++i)
        {
            if(nums[i] > target/2+1)
                break;
            j = bs(nums,target-nums[i]);
            if(j != -1 && j >= i)
                count = (count+mypow(j-i))%mod;
        }
        return count;
    }
    
    int bs(vector<int>& a, int t)
    {
        int i = 0, j = a.size()-1, mid;
        while(i <=j)
        {
            mid = (i+j)/2;
            if(a[mid] > t)
                j = mid-1;
            else
            {
                if(mid==a.size()-1 || a[mid+1] > t)
                    return mid;
                else
                    i = mid+1;
            }
        }
        return -1;
    }
    int mypow(int n)
    {
        long long s = 1, p = 2;
        while(n)
        {
            if(n&1)
                s *= p, s %= mod;
            p *= p;
            p %= mod;
            n /= 2;
        }
        return s;
    }
};

class Solution:# py3
    def numSubseq(self, nums: List[int], target: int) -> int:
        mod = int(1e9+7)
        nums.sort()
        def bs(t):
            i,j = 0, len(nums)-1
            while i <= j:
                mid = (i+j)>>1
                if nums[mid] > t:
                    j = mid-1
                else:
                    if mid==len(nums)-1 or nums[mid+1] > t:
                        return mid
                    else:
                        i = mid+1
            return -1
        def mypow(n):
            s, p = 1, 2
            while n:
                if n&1:
                    s *= p
                    s %= mod
                p *= p
                p %= mod
                n //= 2
            return s
        count = 0
        for i in range(len(nums)):
            if nums[i] > target//2+1:
                break;
            j = bs(target-nums[i])
            if j != -1 and j >= i:
                count = (count + mypow(j-i))%mod
        return count