HDU 6222 Heron and His Triangle 找规律打表 大数减法 大数乘法

HDU 6222 Heron and His Triangle

A triangle is a Heron’s triangle if it satisfies that the side lengths of it are consecutive integers t−1, t, t+ 1 and thatits area is an integer. Now, for given n you need to find a Heron’s triangle associated with the smallest t bigger
than or equal to n.

Input
The input contains multiple test cases. The first line of a multiple input is an integer T (1 ≤ T ≤ 30000) followedby T lines. Each line contains an integer N (1 ≤ N ≤ 10^30).

Output
For each test case, output the smallest t in a line. If the Heron’s triangle required does not exist, output -1.

Sample Input
4
1
2
3
4

Sample Output
4
4
4
4

简要题意

每给出一个数n, 求一个大于等于n的数x,
要求: x-1, x, x+1可构成一个三角形

题目分析

首先暴力枚举出较小的结果
得到4,14,52,194…一系列数字
由此可以得到规律a[n] == a[n-1]*4 - a[n-2]
由于该数列数字较大, 可知涉及大数乘法和大数减法
打表后将输入的n与表中数字对比即可

代码

#include
#include
#include

#define MAXT 200
#define MAXN 50
#define CON 4

char reserve[MAXT][MAXN];

int lenfin;

void BigNumOpe( char* small, char* big, int m)
{
	int i, j, k;
	int lens, lenb, lensum, lent;
	char result[MAXN] = {0};
	char temp[MAXN] = {0};
	char t[MAXN] = {0};
	
	lenb = strlen(big);
	lensum = lenb + 1;
	lens = strlen(small);
	

	for( i=0; i<lenb; i++)
	{
		temp[i] = (big[lenb-i-1]-'0') * 4;
	}
	for( i=0; i<lensum; i++)
	{
		temp[i+1] += temp[i] / 10;
		temp[i] %= 10;
	}
//	lent = strlen(temp);			错误, 出现0 会自动减少 
	if( temp[lensum-1]>0)
	{
		lent = lensum;
	}
	else
	{
		lent = lensum-1;
	}
	
	
	
	for( i=0; i<lens; i++)
	{
		t[i] = small[lens-i-1] - '0';
	}
	
	for( i=0; i<lent; i++)
	{
		if( temp[i]< t[i])
		{
			temp[i] += 10;
			temp[i+1] --;
		}
		result[i] = temp[i] - t[i];
	}
	
	for( i=0; i<lent; i++)
	{
		reserve[m][i] = result[lent-i-1] + '0';
	}
	reserve[m][i] = '\0';
	lenfin = strlen(reserve[m]);
//	puts(reserve[m]);
}

int compare(char *x, char *y)
{
	int lenx = strlen(x);
	int leny = strlen(y);
	int k = -1, i;
	if( lenx>leny)
	{
		k = 1;
	}
	else if(lenx < leny)
	{
		k = -1;
	}
	else if( lenx == leny)
	{
		for( i=0; i<lenx; i++)
		{
			if( x[i] - y[i] > 0)
			{
				k = 1;
				break;
			}
			else if( x[i] - y[i] < 0)
			{
				k = -1;
				break;
			}
		}
	}
	return k;
}

int main(void)
{
	int i, j, k, t;
	char x[MAXN];
	strcpy(reserve[0], "4");
	strcpy(reserve[1], "14");
	for( i=2; lenfin<32; i++)
	{
		BigNumOpe( reserve[i-2], reserve[i-1], i);
	}
		
	scanf("%d", &t);
	while(t--)
	{
		memset(x, 0, sizeof(x));
		scanf("%s", x);
		for( i=0; ;i++)
		{
			k = compare(x, reserve[i]);
			if( k<=0)
			{
				break;
			}			
		}
		puts(reserve[i]);
	}
	return 0;
}

收获与反思

1. 程序中自己写了一个compare()函数用来比较两数大小, 而使用strcmp()函数时会出错, 暂时不知道原因
2. 对大数减法和大数乘法有了初步了解
3. 程序中的大数减法部分有偷懒欠缺的地方, 因为数与数之间相差较大, 不会存在最高位是0的情况, 所以程序中没有对此进行特判, 之后应该将此部分补全
4. 第一次用打表的方法ac题目, 不失为一个做有确定答案题目的好方法