Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Mr. Cooper, the CEO of CozyWalk Co., receives a report of daily sales every day since the company has been established. Starting from the second day since its establishment, on receiving the report, he compares it with each of the previous reports in order to calculate the number of previous days whose sales amounts are less than or equal to it. After obtaining the number of such days, he writes it in a list.

This problem can be stated more formally as follows. Let A = (a₁, a₂,..., a_n) denote the list of daily sales amounts. And let B = (b₁, b₂,..., b_n-1) be another integer list maintained by Mr. Cooper, each value representing the number of such previous days. On the i-th day ( 2in), he calculates b_i-1, the number of a_k's such that a_ka_i ( 1k < i).

For example, suppose that A = (20, 43, 57, 43, 20). For the fourth day's sales amount, a₄ = 43, the number of previous days whose sales amounts are less than or equal to it is 2 since a₁a₄, a₂a₄, and a₃ > a₄. Therefore, b₃ = 2. Similarly, b₁, b₂, and b₄ can be obtained and it results in B = (1, 2, 2, 1).

Given an array of size n for the list of daily sales amounts, write a program that prints the sum of the n - 1 integers in the list B.

Input 

Your program is to read the input from standard input. The input consists of T test cases. The number of test cases T is given in the first line of the input. Each test case starts with a line containing an integer n ( 2n1, 000), which represents the size of the list A . In the following line, n integers are given, each represents the daily sales amounts a_i ( 1a_i5, 000 and 1in) for the test case.

Output 

Your program is to write to standard output. For each test case, print the sum of the n - 1 integers in the list B which is obtained from the list A.

The following shows sample input and output for two test cases.

Sample Input 

2
5
38 111 102 111 177
8
276 284 103 439 452 276 452 398

Sample Output 

9
20

---

模擬, 使用插入排序法

#include <stdio.h>
#include <stdlib.h>
int main() {
	int T, n, A[1001];
	int i, j;
	scanf("%d", &T);
	while(T--) {
		scanf("%d", &n);
		int ans = 0, tmp;
		for(i = 0; i < n; i++) {
			scanf("%d", &tmp);
			for(j = i-1; j >= 0; j--)
				if(tmp < A[j])
					A[j+1] = A[j];
				else
					break;
			A[j+1] = tmp;
			ans += j+1;
		}
		printf("%d\n", ans);
	}
    return 0;
}