UVA 10494 (13.08.02)

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;
}