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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
#543633#7789. Outro: True Love Waitsucup-team956TL 473ms7512kbC++202.3kb2024-09-01 17:37:582024-09-01 17:37:59

Judging History

你现在查看的是最新测评结果

  • [2024-09-01 17:37:59]
  • 评测
  • 测评结果:TL
  • 用时:473ms
  • 内存:7512kb
  • [2024-09-01 17:37:58]
  • 提交

answer

#include <bits/stdc++.h>

using namespace std;

#define SZ(x) ((int)((x).size()))
#define lb(x) ((x) & (-(x)))
#define bp(x) __builtin_popcount(x)
#define bpll(x) __builtin_popcountll(x)
#define mkp make_pair
#define pb push_back
#define fi first
#define se second
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, int> pli;
typedef pair<ll, ll> pll;
typedef pair<double, int> pdi;

const int N = 1e6;
const int Mod = 1e9 + 7;

ll ksm(ll x, ll y) {
    ll res = 1;
    while (y) {
        if (y & 1) {
            res = res * x % Mod;
        }
        x = x * x % Mod;
        y >>= 1;
    }
    return res;
}
ll inv(ll x) {
    return ksm(x, Mod - 2);
}

void solve() {
    string s, t; int k;
    cin >> s >> t >> k;

    reverse(s.begin(), s.end());
    reverse(t.begin(), t.end());

    if (SZ(s) < SZ(t)) {
        swap(s, t);
    }

    vector<int> p;
    p.resize(max(SZ(s), SZ(t)));
    for (int i = 0; i < SZ(t); i++) {
        p[i] = (s[i] - '0') ^ (t[i] - '0');
    }
    for (int i = SZ(t); i < SZ(s); i++) {
        p[i] = s[i] - '0';
    }

    int cnt0 = 0;
    while (cnt0 < SZ(p) && !p[cnt0]) {
        cnt0++;
    }
    if (cnt0 == SZ(p)) {
        int ans = 4 * (ksm(4, k - 1) - 1) % Mod * inv(3) % Mod;
        cout << ans << '\n';
        return ;
    }

    int c0 = cnt0 / 2 + 1;
    if (c0 < k) {
        cout << -1 << '\n';
        return ;
    }

    vector f(N + 1, 0);
    f[0] = 0; f[1] = 1;
    for (int i = 2; i <= N; i++) {
        if (i % 2 == 0) {
            f[i] = 2 * f[i - 1] % Mod;
        } else {
            f[i] = (2 * f[i - 1] + 1) % Mod;
        }
    }

    ll pos = 0;
    for (int i = SZ(p) - 1; i >= 0; i--) {
        if (p[i] == 0) {
            continue;
        }

        if (i % 2 == 0) {
            pos = (pos + f[i + 1]) % Mod;
        } else {
            pos = (pos + f[i + 1]) % Mod;
            p[i - 1] ^= 1;
        }
    }

    // cout << "pos = " << pos << '\n';
    
    ll ans = (pos + 4 * (ksm(4, k - 1) - 1) % Mod * inv(3) % Mod) % Mod;
    cout << ans << '\n';
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int T;
    cin >> T;
    while (T--) solve();
    return 0;
}

详细

Test #1:

score: 100
Accepted
time: 7ms
memory: 7512kb

input:

4
1 10 1
1 10 2
100 0 2
11 11 3

output:

2
-1
9
20

result:

ok 4 number(s): "2 -1 9 20"

Test #2:

score: 0
Accepted
time: 0ms
memory: 3576kb

input:

1
0 0 1

output:

0

result:

ok 1 number(s): "0"

Test #3:

score: 0
Accepted
time: 473ms
memory: 7496kb

input:

100
110111 11111 1
10110 101101 1
11010 111111 1
100110 1 1
10010 11010 1
1100 10111 1
100100 111110 1
101110 101100 1
1011 10110 1
110100 1110 1
11010 11000 1
11110 1000 1
111000 11101 1
110 1001 1
101010 11000 1
10 111110 1
110001 101000 1
1010 1000 1
10101 11 1
111011 11010 1
110001 100000 1
1100...

output:

78
59
69
70
15
38
39
3
32
60
3
29
69
12
45
52
37
3
29
64
22
39
54
69
65
27
33
76
34
18
57
13
81
15
23
70
69
36
18
23
29
42
69
54
6
0
63
3
29
15
10
16
80
24
37
59
71
13
23
31
21
34
23
48
21
47
7
44
42
3
37
75
59
29
55
39
29
28
29
70
55
16
54
47
24
18
79
60
8
26
64
58
32
6
8
37
2
68
42
44

result:

ok 100 numbers

Test #4:

score: 0
Accepted
time: 402ms
memory: 7384kb

input:

100
10011111 111 2
1011101100 1000000100 1
100011111 1001001111 1
1001100101 1100100001 1
10101000 10000100 1
1011110101 100011101 1
110100001 111011010 1
1101001100 1111101101 1
1001101 11011010 1
1101110110 1101011000 1
110011001 1100001111 2
1001111001 1011001111 1
1001110 1101110100 2
1110110100...

output:

295
248
788
431
73
930
144
319
283
76
-1
305
-1
-1
86
-1
312
293
1293
433
1179
0
884
963
1215
576
-1
1132
499
811
864
949
1322
406
526
862
-1
447
1203
1238
873
-1
-1
1131
1108
438
134
359
80
740
1057
752
31
950
1093
1261
650
235
996
876
504
925
1344
450
1010
273
-1
1144
1041
717
-1
164
-1
11
798
419...

result:

ok 100 numbers

Test #5:

score: -100
Time Limit Exceeded

input:

1000
1010011001 1100000000 1
1111001110 100100011 1
10000001 1110100110 1
1001000010 1111011110 1
11110001 101101110 1
10110001 110010 1
110111100 1111011111 1
1010101010 1111110000 1
11010110 11000110 1
1101101100 10001101 1
1101000110 111100110 3
1101100 10110 1
1001101001 10010001 1
1000110100 11...

output:


result: