5312 If We Were a Child Again

Problem E: If We Were a Child Again

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 8  Solved: 2
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Description

“Oooooooooooooooh!

If I could do the easy mathematics like my school days!!

I can guarantee, that I’d not make any mistake this time!!”

Says a smart university student!!

But his teacher even smarter – “Ok! I’d assign you such projects in your software lab. Don’t be so sad.”

“Really!!” - the students feels happy. And he feels so happy that he cannot see the smile in his teacher’s face.The first project for the poor student was to make a calculator that can just perform the basic arithmetic operations.

                                                                       

 

But like many other university students he doesn’t like to do any project by himself. He just wants to collect programs from here and there. As you are a friend of him, he asks you to write the program. But, you are also intelligent enough to tackle this kind of people. You agreed to write only the (integer) division and mod (% in C/C++) operations for him.

Input

Input is a sequence of lines. Each line will contain an input number. One or more spaces. A sign (division or mod). Again spaces. And another input number. Both the input numbers are non-negative integer. The first one may be arbitrarily long（less than 100 bit.). The second number n will be in the range (0 < n < 2³¹).

Output

A line for each input, each containing an integer. See the sample input and output. Output should not contain any extra space.

Sample Input

110 / 100
99 % 10
2147483647 / 2147483647
2147483646 % 2147483647

Sample Output

1
9
1
2147483646

HINT

解题思路：

    1、将后面的除数转化unsigned long long类型

    2、知道除数的位数后，先得到相同位数的被除数

    3、进行除法操作，知道被除数到最后一位

    4、判断，若为除，则在循环中每次输出结果，若为取余，则在最后对a取次余并输出

#include <cstdio>#include <cstring>#include <cstdlib>#define Max 550char c_a[Max],c_b[Max];char c;int main () {#ifndef ONLINE_JUDGEfreopen("test_001.in","r",stdin);#endifwhile (~scanf("%s %c %s",c_a, &c, c_b) ) {unsigned long long a=0,b=0,sum=0;int i, n=strlen(c_b), m=strlen(c_a);for (i=0; i<n; i++) {b = b*10 + (c_b[i] - '0');}for (i=0; i<n; i++) {a = a*10 + (c_a[i] - '0');}do {if (c == '/') {printf("%lld",a/b);}a %= b;if (i==m)break;a = a*10 + (c_a[i] - '0');i++;}while (i<=m);//printf("%lld",b);if (c == '%'){a = a % b;printf("%lld",a);}printf("\n");}return 0;}