大数四则运算（高精度）

UVA 424 Integer Inquiry

题意：求给出的几组数字之和

#include 
#include 

char s[105];
int num[105];
int t[10000] = {0};

int main() {
	int i, j, l, flag;
	/*freopen("D:\\in.txt", "r", stdin);*/
	while (scanf("%s", s) != EOF) {
		l = strlen(s);
		memset(num, 0, sizeof(num));
		for (i = 0; i < l; i++)
			num[l - i - 1] = s[i] - '0';

		for (i = 0; i < l; i++)
			t[i] += num[i];

		for (i = 0; i < 10000; i++) {
			t[i+1] += (t[i] / 10);
			t[i] %= 10;
		}
	}
	flag = 0;
	for (i = 10000 - 1; i>=0; i--) {
		if (flag)
			printf("%d", t[i]);
		else if (t[i]) {
			printf("%d", t[i]);
			flag = 1;
		}
	}
	printf("\n");
	return 0;
}

UVA 10106 Product

题意：求出所给的每两组数字的乘积

#include
#include

char n1[300], n2[300];
int x[300], y[300], mul[100000];

int main() {
	int i, j, c;
	/*freopen("D:\\in.txt", "r", stdin);*/
	while (scanf("%s%s", n1, n2) != EOF) {
		int l1 = strlen(n1);
		int l2 = strlen(n2);
		memset(x, 0, sizeof(x));
		memset(y, 0, sizeof(y));
		memset(mul, 0, sizeof(mul));

		for (i = 0; i < l1; i++) 
			x[l1 - 1 - i] = n1[i] - '0';

		for (i = 0; i < l2; i++) 
			y[l2 - 1 - i] = n2[i] - '0';


		for (i = 0; i < l1; i++) {
			c = i;
			for (j = 0; j < l2; j++) {
				mul[c] += x[i] * y[j];
				mul[c + 1] += mul[c] / 10;
				mul[c] %= 10;
				c++;
			}
		}

		int flag = 0;
		for (i = 1000 - 1; i >= 0; i--) {
			if (flag)
				printf("%d", mul[i]);
			else if (mul[i]) {
				printf("%d", mul[i]);
				flag = 1;
			}
		}
		if (flag == 0)
			printf("0");
		printf("\n");
	}
	return 0;
}

UVA 465 Overflow

题意：判断所给的两个加数以及和是否超过了int型所能存储的范围

#include 
#include 
#include 

int main() {
	int max = pow(2.0, 31) - 1;
	char n1[310], n2[310];
	char sign;
	double a, b;
	/*freopen("D:\\in.txt", "r", stdin);*/
	while (scanf("%s %c %s", n1, &sign, n2) != EOF) {
		printf("%s %c %s\n", n1, sign, n2);
		a = atof(n1);
		b = atof(n2);
		if (a > max) printf("first number too big\n");
		if (b > max) printf("second number too big\n");
		if (sign == '+' && a + b > max) printf("result too big\n");
		else if (sign == '*' && a * b > max) printf("result too big\n");
	}
	return 0;
}

UVA 748 Exponentiation

题意：求出所给实数的次方

#include 
#include 

char d[10];
int mul[30000], num[10], m[30000];

int main() {
	int i, j, k, n, t, temp;
	/*freopen("D:\\in.txt", "r", stdin);*/
	while (scanf("%s%d", d, &n) != EOF) {
		memset(mul, 0, sizeof(mul));
		memset(num, 0, sizeof(num));
		k = 5;
		for (i = 0; i < 6; i++) {
			if (d[i] == '.')
				temp = (5 - i) * n;
			else num[--k] = d[i] - '0';
		}

		for (i = 0; i < 5; i++)
			mul[i] = num[i];

		for (i = 1; i < n; i++) {
			memset(m, 0, sizeof(m));
			for (j = 0; j < 5; j++) {
				int l = j;
				for (k = 0; k < 29999; k++) {
					m[l] += mul[k] * num[j];
					m[l + 1] += m[l] / 10;
					m[l] %= 10;
					l++;
				}
			}
			for (j = 0; j < 29999; j++)
				mul[j] = m[j];
		}

		for (i = 29999; i >= 0; i--)
		if (mul[i] || i == temp) break;
		for (j = 0; j <= i; j++)
		if (mul[j]) break;
		for (t = i; t >= j; t--) {
			if (temp == t && t == i) {
				if (mul[t]) printf("%d.", mul[t]);
				else printf(".");
				continue;
			}
			printf("%d", mul[t]);
			if (temp == t && t != i) printf(".");
		}
		printf("\n");
	}
	return 0;
}

UVA 10494 If We Were a Child Again

题意：求商

#include 
#include 
#define MAX 10000

long int quo[MAX] = {0};
char d[MAX] = {0};

int main() {
	long int i, l, count, temp, t;
	int n, flag;
	char sign;
	/*freopen("D:\\in.txt", "r", stdin);*/
	while (scanf("%s %c %d", d, &sign, &n) != EOF) {
		temp = flag = count = 0;
		for (i = 0; i < strlen(d); i++) {
			temp = temp * 10 + d[i] - '0';
			t = temp / n;
			if (t != 0 || t == 0 && flag == 1) {
				quo[count++] = t;
				flag = 1;
			}
			temp %= n;
		}
		if (sign == '%') printf("%d\n", temp);
		else {
			if (count == 0) printf("0");
			else for (i = 0; i < count; i++)
				printf("%d", quo[i]);
			printf("\n");
		}
		memset(quo, 0, sizeof(quo));
		memset(d, 0, sizeof(d));
	}
	return 0;
}