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HDU 1326 Box of Bricks[解题报告] C++

Box of Bricks

问题描述 :

Little Bob likes playing with his box of bricks. He puts the bricks one upon another and builds stacks of different height. “Look, I’ve built a wall!”, he tells his older sister Alice. “Nah, you should make all stacks the same height. Then you would have a real wall.”, she retorts. After a little con- sideration, Bob sees that she is right. So he sets out to rearrange the bricks, one by one, such that all stacks are the same height afterwards. But since Bob is lazy he wants to do this with the minimum number of bricks moved. Can you help?


The input consists of several data sets. Each set begins with a line containing the number n of stacks Bob has built. The next line contains n numbers, the heights hi of the n stacks. You may assume 1 <= n <= 50 and 1 <= hi <= 100.

The total number of bricks will be divisible by the number of stacks. Thus, it is always possible to rearrange the bricks such that all stacks have the same height.

The input is terminated by a set starting with n = 0. This set should not be processed.


For each set, first print the number of the set, as shown in the sample output. Then print the line “The minimum number of moves is k.”, where k is the minimum number of bricks that have to be moved in order to make all the stacks the same height.

Output a blank line after each set.


5 2 4 1 7 5


Set #1
The minimum number of moves is 5.

#include <stdio.h>
int main()
	int n, hi[51], i, sum, avg, step, count;
	count = 1;
	scanf("%d", &n);
	while (n) 
		sum = 0;
		avg = 0;
		step = 0;
		for (i = 0; i < n; i++) 
			scanf("%d", &hi[i]);
			sum += hi[i];
		avg = sum/n;
		for (i = 0; i < n; i++) 
			if (avg > hi[i]) 
				step += avg - hi[i];
		printf("Set #%d\n", count++);
		printf("The minimum number of moves is %d.\n\n", step);
		scanf("%d", &n);
	return 0;


  1. 算法是程序的灵魂,算法分简单和复杂,如果不搞大数据类,程序员了解一下简单点的算法也是可以的,但是会算法的一定要会编程才行,程序员不一定要会算法,利于自己项目需要的可以简单了解。

  2. 为什么for循环找到的i一定是素数叻,而且约数定理说的是n=p1^a1*p2^a2*p3^a3*…*pk^ak,而你每次取余都用的是原来的m,也就是n