1488.Problem_F 八进制模拟
Description
There are two octal integers numbers a and b.You need to calculate the result of a minus b，then output in octal notation.
If the result is negative, you should use the negative sign instead of complement notation.
Input
The first line is an integer T(1 <= T <= 1000), indicating the number of test cases.
For each case, the first and only line contains two octal integers a, b. (0 <= a, b < 2^100).
Output
Print the result contains a list of the fully qualified octal digits，one per line.
Sample Input
176 7
Sample Output
67
Hint
就是给你两个八进制数~让你进行减法，并以八进制的形式输出。思路不难~~但一定要小心各个地方的坑；
顺便提供一组数据（ 1 2000 1）；
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
using namespace std;
int m[1005];
#define mm 1000
int main()
{
int te;
scanf("%d", &te);
string j, k;
char q[1005], p[1005];
int spot;
while (te--)
{
spot = 1;
memset(m, 0, sizeof(m));
memset(q, 0, sizeof(q));
memset(p, 0, sizeof(p));
cin >> j >> k;
int len1 = j.length();
int len2 = k.length();
int cnt = len1 - 1;
for (int s = mm; s > mm - len1; s--)
{
q[s] = j[cnt--];
}
cnt = len2 - 1;
for (int s = mm; s > mm - len2; s--)
{
p[s] = k[cnt--];
}
if (len1 > len2)
{
for (int s = mm; s > mm - len2; s--)
{
int w = q[s] - p[s];
if (w < 0)
{
w += 8;
q[s - 1] = q[s - 1] - 1;
}
m[s] = w;
}
for (int s = mm - len2; s > mm - len1; s--)
{
int t = q[s] - '0';
if (t < 0)
{
t += 8;
q[s - 1]--;
}
m[s] = t;
}
}
else if (len1 == len2)
{
int sp = 1;
for (int s = mm - len1 + 1; s <= mm; s++)
{
if (q[s] > p[s])
{
break;
}
if (q[s] < p[s])
{
sp = 0;
break;
}
}
if (sp)
{
for (int s = mm; s > mm - len2; s--)
{
int w = q[s] - p[s];
if (w < 0)
{
w += 8;
q[s - 1] = q[s - 1] - 1;
}
m[s] = w;
}
}
else
{
// cout << "asdasd" << endl;
spot = 0;
for (int s = mm; s > mm - len2; s--)
{
int w = p[s] - q[s];
if (w < 0)
{
w += 8;
p[s - 1]--;
}
m[s] = w;
}
}
}
else
{
spot = 0;
for (int s = mm; s > mm - len1; s--)
{
int w = p[s] - q[s];
if (w < 0)
{
w += 8;
p[s - 1]--;
}
m[s] = w;
}
for (int s = mm - len1; s > mm - len2; s--)
{
int t = p[s] - '0';
if (t < 0)
{
t += 8;
p[s - 1]--;
}
m[s] = t;
}
}
if (!spot)
{
cout << "-";
}
int sp = 0;
for (int s = 0; s < mm; s++)
{
if (m[s] != 0)
{
sp = 1;
}
if (sp)
cout << m[s];
}
cout << m[mm] << endl;
}
return 0;
}
