UVa 10494 If We Were a Child Again

本題連結
想法：
　　與直式除法一樣，一邊作商數，一邊作餘數，直到被除數的個位數為止。

	#include <cstdio>
	using namespace std;
	char num[10001];
	long long int b,R;
	char oper[2];
	int main()
	{
	//freopen ("input.txt","rt",stdin);
	while (scanf("%s%s%lld",num,oper,&b) != EOF){
	int Q[10001],pos = 0;
	int i;
	for (i = 0, R = 0; num[i]; i++){
	R = R * 10 + (num[i] - '0');
	Q[pos++] = R / b; // ans前面幾個數字可能為0，等等輸出要從非0開始
	R = R % b;
	}
	if (oper[0] == '/'){
	for (i = 0; i < pos && Q[i] == 0; i++)
	;
	if (i == pos) printf("0"); // 如果被除數比除數小的情況
	else
	for (; i < pos; i++) printf ("%d",Q[i]);
	printf("\n");
	}
	else
	printf("%lld\n",R);
	}
	return 0;
	}

view raw [UVa] 10494 If We Were a Child Again.cpp hosted with ❤ by GitHub