2021-09-19 11:14

# 这段代码编译不通过的原因是什么？为什么不能通过。即可采纳

1455:An Easy Problem

As we known, data stored in the computers is in binary form. The problem we discuss now is about the positive integers and its binary form.

Given a positive integer I, you task is to find out an integer J, which is the minimum integer greater than I, and the number of '1's in whose binary form is the same as that in the binary form of I.

For example, if "78" is given, we can write out its binary form, "1001110". This binary form has 4 '1's. The minimum integer, which is greater than "1001110" and also contains 4 '1's, is "1010011", i.e. "83", so you should output "83".

One integer per line, which is I (1 <= I <= 1000000).

A line containing a number "0" terminates input, and this line need not be processed.

One integer per line, which is J.

1
2
3
4
78
0

2
4
5
8
83

``````
#include <iostream>
#include <cstring>
using namespace std;
int a[21],b[21],len,t,n,m;
void lowtrans(int);
void uptrans(int y[],int l);
int main()
{
while(cin >> n && n)
{
memset(a,0,sizeof(a));    memset(b,0,sizeof(b));
lowtrans(n);
for(int i = 2; i <= len; i++)
{
if(!(a[i]))
{
b[i] = 1;
for(i = 1; i <= t; i++)        b[i] = 0;
break;
}
}
uptrans(b[len + 1],len + 1);
}
return 0;
}
void lowtrans(int x)
{
len = 1;    t = 0;
while(x % 2)
{
x /= 2;
a[len] = x % 2;
if(a[len])    t++;
len++;
}
cout << "lowtrans : ";
for(int i = 1; i <= len; i++)    cout << a[i];
cout << endl;
}
void uptrans(int y[],int l)
{
for(int i = 1; i <= l; i++)
{
if(y[i])
for(int j = i; j > 0; j--)
y[i] *= 2;
b[i] = y[i];
}
cout << "uptrans : ";
for(int i = 1; i <= len; i++)    cout << a[i];
cout << endl;
}
``````
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