UVa10007 - Count the Trees

最新推荐文章于 2020-08-07 17:20:25 发布

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最新推荐文章于 2020-08-07 17:20:25 发布

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分类专栏： OJ 文章标签： div include null

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本文深入探讨了使用C语言实现数学运算（乘法、除法、减法、加法）和数据处理（整数除法、字符串操作）的详细步骤与技巧。通过实例代码展示了如何高效地进行复杂的数据操作与数学计算，为程序员提供了一套实用的解决方案。

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#include <stdio.h>
#include <string.h>

#define MAX 800
#define N 310

char map[N][MAX] = {"0", "1"};
char buf[MAX], temp[MAX];
char F[MAX], S[MAX], Res[MAX];;

void reverse(char *from, char *to);
void call_mult(char *first, char *second, char *result);
void call_minus(char *first, char *second, char *result);
void call_add(char *first, char *second, char *result);
void call_divide(char *first, int div, char *result);

int main()
{

	int n, i;

	setvbuf(stdout, NULL, _IONBF, 0);

	for (i = 1; i < 300; i++) {
		sprintf(buf, "%d", 4 * i * i + 6 * i + 2);
		call_mult(map[i], buf, temp);
		call_divide(temp, i + 2, map[i + 1]);
	}

	while (scanf("%d", &n) && n != 0) {
		printf("%s\n", map[n]);
	}

	return 0;
}


void reverse(char *from, char *to)
{
	int len = strlen(from), i;

	for (i = 0; i < len; i++) {
		to[i] = from[len - 1 - i];
	}
	to[len] = '\0';
}

void call_mult(char *first, char *second, char *result)
{
	int f_len, s_len, f, s, r, r_len, sum, carry;

	f_len = strlen(first);
	s_len = strlen(second);
	r_len = f_len + s_len;
	r = -1;
	reverse(first, F);
	reverse(second, S);

	for (f = 0; f < r_len; f++)
		Res[f] = '0';
	Res[f] = '\0';

	for (s = 0; s < s_len; s++) {
		for (f = 0, carry = 0; f < f_len; f++) {
			sum = (F[f] - '0') * (S[s] - '0') + (Res[f + s] - '0') + carry;
			Res[f + s] = sum % 10 + '0';
			carry = sum / 10;
			if (f + s > r)
				r = f + s;
		}

		while (carry) {
			sum = (Res[f + s] - '0') + carry;
			Res[f + s] = sum % 10 + '0';
			carry = sum / 10;
			if (f + s > r)
				r = f + s;
			f++;
		}
	}

	for (; r > 0 && Res[r] == '0'; r--);
	Res[r + 1] = '\0';
	reverse(Res, result);
}

void call_minus(char *first, char *second, char *result)
{

	int f_len, s_len, now, borrow, sum;

	f_len = strlen(first);
	s_len = strlen(second);
	reverse(first, F);
	reverse(second, S);

	for (now = 0, borrow = 0; now < f_len && now < s_len; now++) {
		sum = (F[now] - '0') - (S[now] - '0') - borrow;
		if (sum < 0) {
			borrow = 1;
			F[now] = sum + 10 + '0';
		} else {
			borrow = 0;
			F[now] = sum + '0';
		}
	}

	for (; now < f_len && borrow; now++) {
		sum = (F[now] - '0') - borrow;
		if (sum < 0) {
			borrow = 1;
			F[now] = sum + 10 + '0';
		} else {
			borrow = 0;
			F[now] = sum + '0';
		}
	}

	for (now = f_len - 1; now > 0 && F[now] == '0'; now--);
	F[now + 1] = '\0';
	reverse(F, result);

}

void call_add(char *first, char *second, char *result)
{
	int f_len, s_len, now, carry, sum;

	f_len = strlen(first);
	s_len = strlen(second);
	reverse(first, F);
	reverse(second, S);

	for (now = 0, carry = 0; now < f_len && now < s_len; now++) {
		sum = (F[now] - '0') + (S[now] - '0') + carry;
		Res[now] = sum % 10 + '0';
		carry = sum / 10;
	}

	for (; now < f_len; now++) {
		sum = (F[now] - '0') + carry;
		Res[now] = sum % 10 + '0';
		carry = sum / 10;
	}

	for (; now < s_len; now++) {
		sum = (S[now] - '0') + carry;
		Res[now] = sum % 10 + '0';
		carry = sum / 10;
	}

	if (carry)
		Res[now++] = carry + '0';
	Res[now] = '\0';

	reverse(Res, result);
}


void call_divide(char *first, int div, char *result)
{
	int len = strlen(first);
	int remainder = 0, now, sum;

	for (now = 0; now < len; now++) {
		sum = remainder * 10 + (first[now] - '0');
		Res[now] = sum / div + '0';
		remainder = sum % div;
	}
	Res[now] = '\0';

	for (now = 0; Res[now] == '0'; now++);
	strcpy(result, &Res[now]);
	if (strlen(result) == 0)
		strcpy(result, "0");
}