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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#124979#5874. Mystery Squarezhouhuanyi0 3ms3860kbC++112.6kb2023-07-15 21:14:322023-07-15 21:14:35

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2023-07-15 21:14:35]
  • 评测
  • 测评结果:0
  • 用时:3ms
  • 内存:3860kb
  • [2023-07-15 21:14:32]
  • 提交

answer

#include<iostream>
#include<cstdio>
#include<vector>
#define N 125
#define M 62
using namespace std;
int read()
{
	char c=0;
	int sum=0;
	while (c<'0'||c>'9') c=getchar();
	while ('0'<=c&&c<='9') sum=sum*10+c-'0',c=getchar();
	return sum;
}
int T,n,Base,tong[N+1],length,tong2[N+1],length2;
char c[N+1],cs[N+1];
__int128 res,ans,dst;
long long rst;
vector<long long>p[N+1];
int main()
{
	bool op;
	string s;
	T=read();
	for (int qt=1;qt<=T;++qt)
	{
		cin>>s,n=s.length(),length=length2=0,dst=1;
		for (int i=1;i<=n;++i) c[i]=cs[i]=s[i-1];
		for (int i=n;i>=1;i-=2,dst*=4)
		{
			op=1;
			for (int j=i+1;j<=n;++j)
				if (c[j]=='1')
					op=0;
			if (!op) continue;
			Base=(i+1)>>1,length=length2=0;
			for (int j=i+1;j<=n;++j) cs[j]='0';
			for (int j=1;j<=Base;++j)
				if (c[j]=='?')
					tong[++length]=j;
			for (int j=Base+1;j<=i;++j)
				if (c[j]=='?')
					tong2[++length2]=j;
			if (length<=length2)
			{
				for (int j=0;j<(1<<length);++j)
				{
					for (int k=1;k<=length;++k)
					{
						if (j&(1<<(k-1))) cs[tong[k]]='1';
						else cs[tong[k]]='0';
					}
					res=0;
					for (int k=1;k<=i;++k)
					{
						if (k<=Base) res=(res<<1)|(cs[k]-'0');
						else res=(res<<1)|1;
					}
					rst=0,op=1;
					for (int k=M;k>=0;--k)
						if ((__int128)(rst+(1ll<<k))*(rst+(1ll<<k))<=res)
							rst+=(1ll<<k);
					res=(__int128)(rst)*rst,op=1;
					for (int k=1;k<=n;++k)
						if (cs[k]!='?'&&cs[k]-'0'!=((res>>(n-k))&1))
							op=0;
					if (op) ans=(__int128)(rst)*rst*dst;
				}
			}
			else
			{
				for (int j=0;j<(1<<length2);++j)
				{
					for (int k=1;k<=length2;++k)
					{
						if (j&(1<<(k-1))) cs[tong2[k]]='1';
						else cs[tong2[k]]='0';
					}
					rst=0;
					for (int k=Base+1;k<=i;++k) rst=(rst<<1)|(cs[k]-'0'),p[k].clear();
					p[i].push_back(rst&1);
					for (int k=i-1;k>=Base+1;--k)
						for (int t=0;t<p[k+1].size();++t)
						{
							if ((((__int128)(p[k+1][t])*p[k+1][t])&((1ll<<(n-k+1))-1))==(rst&((1ll<<(n-k+1))-1))) p[k].push_back(p[k+1][t]);
							if ((((__int128)(p[k+1][t]+(1ll<<(n-k)))*(p[k+1][t]+(1ll<<(n-k))))&((1ll<<(n-k+1))-1))==(rst&((1ll<<(n-k+1))-1))) p[k].push_back(p[k+1][t]+(1ll<<(n-k)));
					    }
					for (int k=0;k<p[Base+1].size();++k)
					{
						res=(__int128)(p[Base+1][k])*p[Base+1][k],op=1;
						for (int t=1;t<=n;++t)
							if (cs[t]!='?'&&cs[t]-'0'!=((res>>(n-t))&1))
								op=0;
						if (op) ans=(__int128)(p[Base+1][k])*p[Base+1][k]*dst;
					}
				}
			}
		}
		printf("Case #%d: ",qt);
		for (int i=1;i<=n;++i) printf("%c",(char)(((ans>>(n-i))&1)+'0'));
		puts("");
    }
	return 0;
}

詳細信息

Subtask #1:

score: 0
Wrong Answer

Test #1:

score: 0
Wrong Answer
time: 3ms
memory: 3860kb

input:

25
1???
1
10??110??00??1000??
1??010?0110?1010?0?010?0111011?11100?100010?0??0??1
1??11????00??1?1?0?1??01??110110?11?00100110?00100?0?00
11?1?1???11111?11?1??11110000?00?????00??0?000?000?1
10??000000?0?00000?00000000??0?0000???00??????0000???
101???11??11000?????1??1?1??10??0?0100011?0001?01011001...

output:

Case #1: 1001
Case #2: 1
Case #3: 1011110110000100001
Case #4: 111010001100101000001010111011011100110001000110001
Case #5: 0000111010001100101000001010111011011100110001000110001
Case #6: 1111111111111111111111111000000000000000000000000001
Case #7: 1000000000000000000000000000000000000000000000000...

result:

wrong answer 5th lines differ - expected: 'Case #5: 1101111110000101100101010111011001110010011000010000100', found: 'Case #5: 0000111010001100101000001010111011011100110001000110001'

Subtask #2:

score: 0
Memory Limit Exceeded

Test #2:

score: 0
Memory Limit Exceeded

input:

25
1????????????????????111101010000011100110101111000001011111100110000011000101100000010010110100101000????????????????????
10?11100?000111??11?01010110100??1?100111?001000000??0101?110?0111?011?11?1??00010111?010??100?100??10?010?001001110111110?1
1000100111100100110011010111100001111010?????????...

output:

Case #1: 11100101110101010110111110101000001110011010111100000101111110011000001100010110000001001011010010100001101011000010100001
Case #2: 1011110000001110111001010110100001110011100010000001101010110001111011111110000010111101010100010010101010100100111011111001

result: