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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#668388#9302. Caesar Cipher4eyebird#TL 440ms6132kbC++172.0kb2024-10-23 14:11:092024-10-23 14:11:09

Judging History

你现在查看的是最新测评结果

  • [2024-11-04 17:10:09]
  • hack成功,自动添加数据
  • (/hack/1110)
  • [2024-10-23 14:11:09]
  • 评测
  • 测评结果:TL
  • 用时:440ms
  • 内存:6132kb
  • [2024-10-23 14:11:09]
  • 提交

answer

#include <algorithm>
#include <bitset>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <map>
#include <string>
#include <vector>
using namespace std;
typedef long long ll;

inline char gtc()
{
    static constexpr size_t bufferlen = 5e6;
    static char BB[bufferlen], *S = BB, *T = BB;
    return S == T && (T = (S = BB) + fread(BB, 1, bufferlen, stdin), S == T) ? EOF : *S++;
}

inline int read()
{
    int x = 0, f = 1;
    char ch = gtc();
    while (ch < '0' || ch > '9')
    {
        if (ch == '-')
            f = -1;
        ch = gtc();
    }
    while (ch >= '0' && ch <= '9')
        x = (x << 3) + (x << 1) + (ch ^ 48), ch = gtc();
    return x * f;
}

constexpr int NB = 500000;

typedef bitset<NB> bt;
bt a[16];
bt x1, x2;

signed main()
{
    int n = read(), m = read();
    x1.reset();
    x1.set(0);
    for (int i = 0; i < n; i++)
    {
        int x = read(), j = 0;
        while (x > 0)
        {
            if (x & 1)
                a[j] |= x1;
            j++;
            x >>= 1;
        }
        x1 <<= 1;
    }
    while (m--)
    {
        int o = read();
        if (o == 1)
        {
            int l = read(), r = read();
            x1.set();
            x1 >>= (NB - (r - l + 1));
            x1 <<= (l - 1);
            int i = 0;
            while (x1.any() && i < 16)
            {
                x2 = x1;
                x1 &= a[i];
                a[i] ^= x2;
                i++;
            }
        }
        else if (o == 2)
        {
            int l1 = read(), l2 = read(), len = read();
            bool ok = true;
            for (int i = 0; i < 16; i++)
            {
                if (((a[i] >> (l1 - 1)) << (NB - len)) != ((a[i] >> (l2 - 1)) << (NB - len)))
                {
                    ok = false;
                    break;
                }
            }
            if (ok)
                printf("yes\n");
            else
                printf("no\n");
        }
    }

    return 0;
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 2ms
memory: 4276kb

input:

5 6
1 2 1 2 1
2 1 2 2
2 1 3 3
1 1 1
1 3 5
2 1 2 4
2 1 2 2

output:

no
yes
no
yes

result:

ok 4 token(s): yes count is 2, no count is 2

Test #2:

score: 0
Accepted
time: 0ms
memory: 5124kb

input:

3 3
0 65535 65535
2 1 2 2
1 2 3
2 1 2 2

output:

no
yes

result:

ok 2 token(s): yes count is 1, no count is 1

Test #3:

score: 0
Accepted
time: 440ms
memory: 6132kb

input:

1000 1000
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ...

output:

yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
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yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
...

result:

ok 1000 token(s): yes count is 1000, no count is 0

Test #4:

score: -100
Time Limit Exceeded

input:

100000 100000
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ...

output:

yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
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yes
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yes
yes
yes
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yes
yes
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yes
yes
yes
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yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
...

result: