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/

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N=210;
int dp[N],a[N],n;
signed main(void) {
	ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
	cin>>n;
	for(int i=1;i<=n;i++){
		cin>>a[i];
	}
	fill(dp,dp+n+1,0x3f3f3f3f3f3f3f3f);dp[1]=0;
	for(int i=1;i<=n;i++){
		if(a[i+1]==a[i]+1) dp[i+1]=min(dp[i+1],dp[i]+1);
		for(int j=1;j<i;j++){
			for(int k=i;k<=n;k++){
				if(a[k]<=a[i-j]+(1ll<<j)) dp[k]=min(dp[k],dp[i]+j+1);
			}
		}
	}
	cout<<(dp[n]==0x3f3f3f3f3f3f3f3f?-1:dp[n])<<'\n';
	return 0;
}

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