tyvj1022

题目链接:https://www.tyvj.cn/Problem_Show.aspx?id=1022

 1 #include <cstdio>
 2 #include <cstdlib>
 3 #include <cmath>
 4 long long int n, k, r; int a[100];
 5 int main(void) {
 6     scanf("%lld",&n); int j=0; if(!n){printf("0
");return 0;}
 7     while(n) {k=n/(-2); r=n-(-2)*k; if(r==-1) n=k+1,r=1; else n=k; a[j++]=r; }
 8     for(int k=j-1;k>=0;--k)printf("%d",a[k]);printf("
");
 9     return 0;
10 }

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原文地址:https://www.cnblogs.com/liuxueyang/p/3178963.html