算法

POJ2183总结

Bovine Math Geniuses

题目来源

Description

Farmer John loves to help the cows further their mathematical skills. He has promised them Hay-flavored ice cream if they can solve various mathematical problems.

He said to Bessie, “Choose a six digit integer, and tell me what it is. Then extract the middle four digits. Square them and discard digits at the top until you have another number six digits or shorter. Tell me the result.”

Bessie, a mathematical genius in disguise, chose the six digit number 655554. “Moo: 6 5 5 5 5 4”, she said. She then extracted the middle four digits: 5555 and squared them: 30858025. She kept only the bottom six digits: 858025. “Moo: 8 5 8 0 2 5”, she replied to FJ.

FJ nodded wisely, acknowledging Bessie’s prowess in arithmetic. “Now keep doing that until you encounter a number that repeats a number already seen,” he requested.

Bessie decided she’d better create a table to keep everything straight:


Middle Middle Shrunk to
Num 4 digits square 6 or fewer
655554 5555 30858025 858025
858025 5802 33663204 663204
663204 6320 39942400 942400
942400 4240 17977600 977600
977600 7760 60217600 217600 <-+
217600 1760 3097600 97600 |
97600 9760 95257600 257600 |
257600 5760 33177600 177600 |
177600 7760 60217600 217600 --+

Bessie showed her table to FJ who smiled and produced a big dish of delicious hay ice cream. “That’s right, Bessie,” he praised. “The chain repeats in a loop of four numbers, of which the first encountered was 217600. The loop was detected after nine iterations.”

Help the other cows win ice cream treats. Given a six digit number, calculate the total number of iterations to detect a loop, the first looping number encountered, and also the length of the loop.

FJ wondered if Bessie knew all the tricks. He had made a table to help her, but she never asked:


Middle Middle Shrunk to
Num 4 digits square 6 or fewer
200023 0002 4 4
4 0 0 0
0 0 0 0 [a self-loop]

whose results would be: three iterations to detect a loop, looping on 0, and a length of loop equal to 1.

Remember: Your program can use no more than 16MB of memory.

Input

* Line 1: A single six digit integer that is the start of the sequence testing.

Output

* Line 1: Three space-separated integers: the first number of a loop, the length of the loop, and the minimum number of iterations to detect the loop.

Sample Input

655554

Sample Output

217600 4 9

我的代码

设置标记位记录计算得到某一个数的步骤数,初始化为-1,如果在某一步中发现步骤数不等于-1,说明之前已经遇到过这个数了,直接输出这个数、循环长度(两次的步骤数之差)和当前步骤数。

#include<stdio.h>
#include<string.h>
int flag[1000010];
int main()
{
	int integer,step=0;
	memset(flag,-1,sizeof(flag));
	scanf("%d",&integer);
	while (true){
		if (flag[integer]==-1){
			flag[integer]=step;
			integer=(integer/10)%10000;
			integer=integer*integer;
			integer=integer%1000000;
		}else{
			printf("%d %d %d\n",integer,step-flag[integer],step);
			break;
		}
		step++;
	}
	return 0;
}

反思

这题要求的是六位数的循环,我一开始写的是四位数的,显然是错的。