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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
#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;
}

详细

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

result: