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1103.Distribute Candies to People

1103.Distribute Candies to People

难度:Easy

排排坐,分糖果。 我们买了一些糖果 candies,打算把它们分给排好队的 n = num_people 个小朋友。 给第一个小朋友 1 颗糖果,第二个小朋友 2 颗,依此类推,直到给最后一个小朋友 n 颗糖果。 然后,我们再回到队伍的起点,给第一个小朋友 n + 1 颗糖果,第二个小朋友 n + 2 颗,依此类推,直到给最后一个小朋友 2 * n 颗糖果。 重复上述过程(每次都比上一次多给出一颗糖果,当到达队伍终点后再次从队伍起点开始),直到我们分完所有的糖果。注意,就算我们手中的剩下糖果数不够(不比前一次发出的糖果多),这些糖果也会全部发给当前的小朋友。 返回一个长度为 num_people、元素之和为 candies 的数组,以表示糖果的最终分发情况(即 ans[i] 表示第 i 个小朋友分到的糖果数)。

示例 1:

输入:candies = 7, num_people = 4
输出:[1,2,3,1]
解释:
第一次,ans[0] += 1,数组变为 [1,0,0,0]。
第二次,ans[1] += 2,数组变为 [1,2,0,0]。
第三次,ans[2] += 3,数组变为 [1,2,3,0]。
第四次,ans[3] += 1(因为此时只剩下 1 颗糖果),最终数组变为 [1,2,3,1]。
示例 2:

输入:candies = 10, num_people = 3
输出:[5,2,3]
解释:
第一次,ans[0] += 1,数组变为 [1,0,0]。
第二次,ans[1] += 2,数组变为 [1,2,0]。
第三次,ans[2] += 3,数组变为 [1,2,3]。
第四次,ans[0] += 4,最终数组变为 [5,2,3]。
 

提示:

1 <= candies <= 10^9
1 <= num_people <= 1000

每轮糖果呈等差数列,设第一轮等差数列和为S,接下来一轮每个人多n颗糖果,总数为S+n*n。每经过一轮,总糖果数都多了n*n,以此类推,这又是一个等差数列。 因此初步想法是,对于给定的糖果数,假设已经经过了M次完整的循环,那么可以根据总糖果数求出M。剩下的糖果数,从M+1次开始分糖果,当满足当前people需要的所有糖果时,分其所需要的;当不满足时,全给他,终止循环。 代码如下: 求M的公式最后可以化简为一个一元二次方程,其中a,b如代码中所示,c就是-candies。 sum表示前M次完整循环,一共用了多少糖果。 接下来一个循环,求出前M次分糖果,每个人分得的堂果蔬,又是一个等差数列求和。 r表示残留糖果数。 最后一个for循环,从r里分,tmp表示当前people需要的糖果数,当剩余糖果数充足时,分tmp,不充足时,全给。

class Solution {
public:
    vector<int> distributeCandies(int candies, int num_people) {
        double a=num_people*num_people/2.0;
        double b=num_people/2.0;
        int count=(sqrt(b*b+4*a*candies) -b)/(2.0*a);
        vector<int >res(num_people);
        int sum=(count-1)*count/2*num_people*num_people +count*(num_people+1)*num_people/2;
            for(size_t i=0; i<num_people;i++)
            res[i]+=num_people*(count-1)*count/2 + count*(i+1);
        int r=candies-sum;
        for(size_t i=0;i<num_people;i++)
        {
         int tmp=count*num_people+i+1;
            if(tmp<r)
            {res[i]+=tmp;
             r-=tmp;
        }
            else
            {res[i]+=r;
            break;}
        }

        
        cout<<"a: "<< a << "b: " << b<< "count: "<<count<<"sum: "<<sum<<endl;
        return res;
    }
};

执行用时 :4ms, 在所有 C++ 提交中击败了91.38%的用户 内存消耗 :8.6MB, 在所有 C++ 提交中击败了100%的用户

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