题目描述
Given an integer N, find the base −2 representation of N.
Here, S is the base −2 representation of N when the following are all satisfied:
S is a string consisting of 0 and 1.
Unless S= 0, the initial character of S is 1.
Let S=SkSk−1…S0, then S0×(−2)0+S1×(−2)1+…+Sk×(−2)k=N.
It can be proved that, for any integer M, the base −2 representation of M is uniquely determined.
Constraints
·Every value in input is integer.
·−109≤N≤109
输入
Input is given from Standard Input in the following format:
N
输出
Print the base −2 representation of N.
样例输入
-9
样例输出
1011
提示
As (−2)0+(−2)1+(−2)3=1+(−2)+(−8)=−9, 1011 is the base −2 representation of −9.
首先说明这个题就是 进制 转换，思路基本和之前的进制转换的思路差不多，就是对-2进行取余
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math") #pragma GCC optimize("Ofast") #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native") #pragma comment(linker, "/stack:200000000") #pragma GCC optimize (2) #pragma G++ optimize (2) #include <bits/stdc++.h> #include <algorithm> #include <map> #include <queue> #include <set> #include <stack> #include <string> #include <vector> using namespace std; #define wuyt main typedef long long ll; #define HEAP(...) priority_queue<__VA_ARGS__ > #define heap(...) priority_queue<__VA_ARGS__,vector<__VA_ARGS__ >,greater<__VA_ARGS__ > > template<class T> inline T min(T &x,const T &y){return x>y?y:x;} template<class T> inline T max(T &x,const T &y){return x<y?y:x;} //#define getchar()(p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++) //char buf[(1 << 21) + 1], *p1 = buf, *p2 = buf; ll read(){ll c = getchar(),Nig = 1,x = 0;while(!isdigit(c) && c!='-')c = getchar(); if(c == '-')Nig = -1,c = getchar(); while(isdigit(c))x = ((x<<1) + (x<<3)) + (c^'0'),c = getchar(); return Nig*x;} #define read read() const ll inf = 1e15; const int maxn = 2e5 + 7; const int mod = 1e9 + 7; #define start int wuyt() #define end return 0 #define N 1005 int num[maxn]; ///int sum[maxn]; map<ll,ll>mp; int n; ll m; start{ ll n=read; int cnt=0; do{ cnt++; num[cnt]=abs(n%2); n=(n-num[cnt])/(-2); } while(n!=0); for(int i=cnt;i>0;i--) printf("%d",num[i]); putchar('\n'); end; }
