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Hercy 想要为购买第一辆车存钱。他 每天 都往力扣银行里存钱。
最开始，他在周一的时候存入 1 块钱。从周二到周日，他每天都比前一天多存入 1 块钱。在接下来每一个周一，他都会比 前一个周一 多存入 1 块钱。
给你 n ，请你返回在第 n 天结束的时候他在力扣银行总共存了多少块钱。
示例 1：
输入：n = 4
输出：10
解释：第 4 天后，总额为 1 + 2 + 3 + 4 = 10 。
示例 2：
输入：n = 10
输出：37
解释：第 10 天后，总额为 (1 + 2 + 3 + 4 + 5 + 6 + 7) + (2 + 3 + 4) = 37
。注意到第二个星期一，Hercy 存入 2 块钱。
示例 3：
输入：n = 20
输出：96
解释：第 20 天后，总额为 (1 + 2 + 3 + 4 + 5 + 6 + 7) + (2 + 3 + 4 + 5 + 6 + 7 + 8) + (3
+ 4 + 5 + 6 + 7 + 8) = 96 。
提示：
1 <= n <= 1000
class Solution { public: int totalMoney(int n) { int sum = 0; int
arr[]={1,2,3,4,5,6,7}; if(n<7){ //如果输入的n小于7天则直接遍历返回 for(int i = 0;i<n;i++){ sum
= sum+arr[i]; } return sum; }else{ //如果输入的大于7天 sum = 28; for(int i = 7;i <
n;i++){ arr[i % 7] = arr[i % 7] + 1; sum = sum+arr[i % 7]; } return sum; } } };
