整数转化为2进制补码(双指针的一种变形)

#include <iostream>
#include "algorithm" 
#include "cmath"
#include "string"
#include <vector> 
using namespace std;

int main()
{
 int sav[8]={1,2,4,8,16,32,64,128};//2^n事先存储 
 int er[8]={0};//存储二进制01 
 int phigh=7;
 int cur=0;
 int sum=100;
 while(phigh>=0){
 cur+=sav[phigh];
 if(cur>sum){
 cur-=sav[phigh];
 //cout<<"情况1"<<cur<<endl;
 phigh--;//之前忽略了这个    
 //cur=0;
 }
 else if(cur==sum){
     er[phigh]=1;
     //cout<<"情况2"<<cur<<endl;
     break;
 }
 else if(cur<sum){
     er[phigh]=1;
     //cout<<"情况3"<<cur<<endl;
     phigh--;
 }    
     
     
 }
 for(int i=0;i<8;i++){
     cout<<er[i];
 }


}

也可以不用事先存储：（只适合正整数求补码）

#include <iostream>
#include "algorithm" 
#include "cmath"
#include "string"
#include <vector> 
#include "bitset"
using namespace std;

int main()
{
//如果只是输出就不用事先存储了    
 int phigh=7;
 int cur=0;
 int num=0;//位数 
 int sum=100;
 while(phigh>=0){
 cur+=pow(2,phigh);
 if(cur>sum){
 cout<<0;num++;    
 cur-=pow(2,phigh);
 phigh--;//之前忽略了这个    
 }
 else if(cur==sum){
      cout<<1;num++; 
     break;
 }
 else{
     cout<<1;num++;
     phigh--;  
 }
}
 for(int i=1;i<=8-num;i++){//默认8位 
     cout<<0;
 }cout<<endl;
   

}

常规解法：（只适合正整数求补码）

#include <iostream>
#include "algorithm" 
#include "cmath"
#include "string"
#include <vector> 
using namespace std;

int main()
{
int sum=100;
vector<int> res; 
while(sum){
res.push_back(sum%2);
sum=sum/2;        
}
reverse(res.begin(),res.end());
for(vector<int>::iterator it=res.begin();it!=res.end();it++)
cout<<*it; 
}

 库函数：//正负数都可以

#include<iostream>
#include<cstdio>
#include<bitset>
using namespace std;
int main()
{
    bitset<8> t=-125;
    cout<<t;
    return 0;
}