点此连接到UVA10494

思路: 采取一种, 边取余边取整的方法, 让这题变的简单许多~

AC代码:

#include<stdio.h>

#include<string.h>

int main() {

	long long mod;

	long long k, tmp;

	int len;

	int ans[10010];

	char num[10010], ch[2];

	while(scanf("%s%s%lld", num, ch, &mod) != EOF) {

		len = strlen(num);

		k = 0;

		tmp = 0;

		memset(ans, 0, sizeof(ans));

		for(int i = 0; i < len; i++) {

			tmp = tmp * 10 + num[i] - '0';

			ans[k++] = tmp / mod; //此步导致ans有前导0, 后续要判断

			tmp = tmp % mod;

		}

		if(ch[0] == '/') {

			int pos;

			for(pos = 0; pos < 10010; pos++)

				if(ans[pos])

					break; //从头判起, 直到第一个不为0的位置

			if(pos == 10010)

				printf("0");

			else

				for(int i = pos; i < k; i++)

					printf("%d", ans[i]);

			printf("\n");

		}

		else

			printf("%lld\n", tmp);

	}

	return 0;

}