题目描述
Every positive number can be presented by the exponential form.For example, 137 = 2^7 + 2^3 + 2^0。 Let’s present a^b by the form a(b).Then 137 is presented by 2(7)+2(3)+2(0). Since 7 = 2^2 + 2 + 2^0 and 3 = 2 + 2^0 , 137 is finally presented by 2(2(2)+2 +2(0))+2(2+2(0))+2(0). Given a positive number n,your task is to present n with the exponential form which only contains the digits 0 and 2.
示例
输入
1315
输出
2(2(2+2(0))+2)+2(2(2+2(0)))+2(2(2)+2(0))+2+2(0)
解法
#include<iostream>
#include<cmath>
using namespace std;
void generate(int j){
int jj = j;
int n=0;
if(jj==1)cout<<"2(0)";
if(jj==2)cout<<"2";
if(jj==3)cout<<"2+2(0)";
if(jj==4)cout<<"2(2)";
if(jj==5)cout<<"2(2)+2(0)";
if(jj==6)cout<<"2(2)+2";
if(jj==7)cout<<"2(2)+2+2(0)";
if(jj>7){
if((jj&(jj-1))==0){
for(int i=0;i<16;i++){
if(pow(2,i)==jj){
cout<<"2(";
generate(i);
cout<<")";
return;
}
}
}
while((jj&(jj-1))!=0){
jj--;
n++;
}
generate(jj);
cout<<"+";
generate(n);
}
}
int main(){
int a;
scanf("%d",&a);
generate(a);
cout<<endl;
return 0;
}