版权声明:qq836678589 https://blog.csdn.net/weixin_43924623/article/details/90901609
Bitset
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 32552 Accepted Submission(s): 23859
Problem Description
Give you a number on base ten,you should output it on base two.(0 < n < 1000)
Input
For each case there is a postive number n on base ten, end of file.
Output
For each case output a number on base two.
Sample Input
1
2
3
Sample Output
1
10
11
#include<iostream>
using namespace std;
int main(){
int n,i,a,b[100];
while(cin>>n){
int j=0;
while(n!=0){
a=n%2;
n=n/2;
b[j]=a;
j++;
}
for(int k=j-1;k>=0;k--){
cout<<b[k];
}
cout<<endl;
}
return 0;
}
**总结:**转换成二进制,囤入数组中,然后反转,注意不要越界。