这题的核心思路就是用题目的提示来判断能不能填充1 直接来看代码(此代码为不要三个一的目前最短代码 但不是最快的)

#include<cstdio>
char s[101];int n,m,a,i;
main(){freopen("three.in","r",stdin);freopen("three.out","w",stdout);scanf("%d%d",&n,&m);for(i=0;i<n&&a<m;++i)if(i<2||!(s[i-1]=='1'&&s[i-2]=='1'))s[i]='1',a++;else s[i]='0';while(i<n)s[i++]='0';printf("%s",s);}

https://theresmoregame.g8hh.com/

命令提示符炸服:ping -t -a -l 65500 -f 10.131.7.177