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3483. 2的幂次方

⭐️难度：中等
⭐️类型：分治

📖题目：题目链接

📚题解：

#define_CRT_SECURE_NO_WARNINGS#include<stdio.h>#include<string>#include<vector>usingnamespacestd;stringGet2sExponet(intn){if(n==0){return"0";}vector<int>exp;for(inti=15;i>=0;--i){if((n&(1<<i))!=0){exp.push_back(i);}}// n = 2^(exp[0]) + 2^(exp[1]) + ... + 2^(exp[size-1])string res="";for(inti=0;i<exp.size();++i){if(i!=0){res+="+";}if(exp[i]==1){res+="2";}else{res+="2("+Get2sExponet(exp[i])+")";}}returnres;}intmain(){intn;while(scanf("%d",&n)!=EOF){printf("%s\n",Get2sExponet(n).c_str());}return0;}