[C language brush leetcode] 1744 Can you eat your favorite candy on your favorite day (m)

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I'll give you a subscript from 0 Starting array of positive integers  candiesCount , among  candiesCount[i]  It means that you have no  i  The number of candies . And give you a two-dimensional array  queries , among  queries[i] = [favoriteTypei, favoriteDayi, dailyCapi] .

You play a game according to the following rules ：

You start with  0  One day I started eating candy .
Are you eating all   The first i - 1  Before candy , You can't   Eat any one of them i  Candy like .
Before eating all the candy , You have to... Every day At least   eat One   candy .
Please build a Boolean array  answer , To give queries The corresponding answer to each item in . This array satisfies ：

answer.length == queries.length .answer[i] yes queries[i] The answer .
answer[i] by  true  Is the condition of ： Eat every day No more than dailyCapi  On the premise of a candy , You can do it at  favoriteDayi  In the first day  favoriteTypei  Candy like ; otherwise answer[i]  by false .
Be careful , As long as you satisfy the above 3 The second rule of the rules , You can eat different types of candy on the same day .

Please return the resulting array  answer .

Example 1：

Input ：candiesCount = [7,4,5,3,8], queries = [[0,2,2],[4,2,4],[2,13,1000000000]]
Output ：[true,false,true]
Tips ：
1- In the 0 God eat 2 Candy ( type 0）, The first 1 God eat 2 Candy （ type 0）, The first 2 God, you can eat the type 0 Of candy .
2- The most you eat every day is 4 Candy . Even if the 0 God eat 4 Candy （ type 0）, The first 1 God eat 4 Candy （ type 0 And type 1）, You can't be in the second 2 God eat type 4 Of candy . In other words , You can't eat every day 4 Under the limit of one candy, in the second 2 In the first day 4 Candy like .
3- If you eat... Every day 1 Candy , You can do it at 13 God eat type 2 Of candy .
Example 2：

Input ：candiesCount = [5,2,6,4,1], queries = [[3,1,2],[4,10,3],[3,10,100],[4,100,30],[1,3,1]]
Output ：[false,true,true,false,false]
 

Tips ：

1 <= candiesCount.length <= 105
1 <= candiesCount[i] <= 105
1 <= queries.length <= 105
queries[i].length == 3
0 <= favoriteTypei < candiesCount.length
0 <= favoriteDayi <= 109
1 <= dailyCapi <= 109

source ： Power button （LeetCode）
link ：https://leetcode.cn/problems/can-you-eat-your-favorite-candy-on-your-favorite-day
Copyright belongs to the network . For commercial reprint, please contact the official authority , Non-commercial reprint please indicate the source .

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One is to eat as much as possible , Edible range , One is to eat at a minimum , Edible range , Compare and judge the conditions between the two ranges and the meaning of the topic . The focus is on boundary treatment .

bool* canEat(int* candiesCount, int candiesCountSize, int** queries, int queriesSize, int* queriesColSize, int* returnSize){
    int i;
    bool *retarr = NULL;
    int idx = 0;
    long long presum[candiesCountSize + 1];

    retarr = (bool *)malloc(sizeof(bool) * queriesSize);
    *returnSize = queriesSize;
    memset(presum, 0, sizeof(presum));

    presum[0] = 0;
    for (i = 0; i < candiesCountSize; i++) {
        presum[i + 1] = presum[i] + candiesCount[i]; 
    }

    for (i = 0; i < queriesSize; i++) {
        long long min = (queries[i][1]) * 1; //  Take one in front every day ,
        long long max = (queries[i][1] + 1) * ((long long)queries[i][2]); //  Eat the most every day , Including today 
        long long left = presum[queries[i][0]]; //  The maximum amount of the last candy type 
        long long right = presum[queries[i][0] + 1]; //  The maximum amount of candy you want 

        //  If you eat as much as you can every day, you will only finish the last kind of candy , Or eat at least the current type of candy in the previous days 
        if ((max <= left) || (min >= right)) {
            retarr[idx] = false;
        } else {
            retarr[idx] = true;
        }
        idx++;
    }
    return retarr;
}