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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
#628397#8049. Equal SumsMisty7ML 1ms3760kbC++204.2kb2024-10-10 19:58:362024-10-10 19:58:48

Judging History

你现在查看的是最新测评结果

  • [2024-10-10 19:58:48]
  • 评测
  • 测评结果:ML
  • 用时:1ms
  • 内存:3760kb
  • [2024-10-10 19:58:36]
  • 提交

answer

#include <bits/stdc++.h>

using i64 = long long;

template<class T>
constexpr T power(T a, i64 b) {
    T res {1};
    for (; b; b /= 2, a *= a) {
        if (b % 2) {
            res *= a;
        }
    }
    return res;
}
 
constexpr i64 mul(i64 a, i64 b, i64 p) {
    i64 res = a * b - i64(1.L * a * b / p) * p;
    res %= p;
    if (res < 0) {
        res += p;
    }
    return res;
}
 
template<i64 P>
struct MInt {
    i64 x;
    constexpr MInt() : x {0} {}
    constexpr MInt(i64 x) : x {norm(x % getMod())} {}
    
    static i64 Mod;
    constexpr static i64 getMod() {
        if (P > 0) {
            return P;
        } else {
            return Mod;
        }
    }
    constexpr static void setMod(i64 Mod_) {
        Mod = Mod_;
    }
    constexpr i64 norm(i64 x) const {
        if (x < 0) {
            x += getMod();
        }
        if (x >= getMod()) {
            x -= getMod();
        }
        return x;
    }
    constexpr i64 val() const {
        return x;
    }
    constexpr MInt operator-() const {
        MInt res;
        res.x = norm(getMod() - x);
        return res;
    }
    constexpr MInt inv() const {
        return power(*this, getMod() - 2);
    }
    constexpr MInt &operator*=(MInt rhs) & {
        if (getMod() < (1ULL << 31)) {
            x = x * rhs.x % int(getMod());
        } else {
            x = mul(x, rhs.x, getMod());
        }
        return *this;
    }
    constexpr MInt &operator+=(MInt rhs) & {
        x = norm(x + rhs.x);
        return *this;
    }
    constexpr MInt &operator-=(MInt rhs) & {
        x = norm(x - rhs.x);
        return *this;
    }
    constexpr MInt &operator/=(MInt rhs) & {
        return *this *= rhs.inv();
    }
    friend constexpr MInt operator*(MInt lhs, MInt rhs) {
        MInt res = lhs;
        res *= rhs;
        return res;
    }
    friend constexpr MInt operator+(MInt lhs, MInt rhs) {
        MInt res = lhs;
        res += rhs;
        return res;
    }
    friend constexpr MInt operator-(MInt lhs, MInt rhs) {
        MInt res = lhs;
        res -= rhs;
        return res;
    }
    friend constexpr MInt operator/(MInt lhs, MInt rhs) {
        MInt res = lhs;
        res /= rhs;
        return res;
    }
    friend constexpr std::istream &operator>>(std::istream &is, MInt &a) {
        i64 v;
        is >> v;
        a = MInt(v);
        return is;
    }
    friend constexpr std::ostream &operator<<(std::ostream &os, const MInt &a) {
        return os << a.val();
    }
    friend constexpr bool operator==(MInt lhs, MInt rhs) {
        return lhs.val() == rhs.val();
    }
    friend constexpr bool operator!=(MInt lhs, MInt rhs) {
        return lhs.val() != rhs.val();
    }
    friend constexpr bool operator<(MInt lhs, MInt rhs) {
        return lhs.val() < rhs.val();
    }
};
 
template<>
i64 MInt<0>::Mod = 998244353;
 
constexpr int P = 998244353;
using Z = MInt<P>;

constexpr int B = 501;

int main () {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);

    int n, m;
    std::cin >> n >> m;

    std::vector<int> a(n), b(n), c(m), d(m);
    for (int i = 0; i < n; i++) {
        std::cin >> a[i] >> b[i];
    }
    for (int i = 0; i < m; i++) {
        std::cin >> c[i] >> d[i];
    }

    std::vector f(n + 1, std::vector (m + 1, std::vector<Z> (2 * B)));
    f[0][0][B] = 1;
    f[0][0][B + 1] = -1;

    for (int i = 0; i <= n; i++) {
        for (int j = 0; j <= m; j++) {
            for (int k = 0; k < 2 * B; k++) {
                if (k > 0) {
                    f[i][j][k] += f[i][j][k - 1];
                }
                if (k < B && i < n) {
                    f[i + 1][j][k + a[i]] += f[i][j][k];
                    f[i + 1][j][k + b[i] + 1] -= f[i][j][k];
                } else if (k >= B && j < m) {
                    f[i][j + 1][k - d[j]] += f[i][j][k];
                    f[i][j + 1][k - c[j] + 1] -= f[i][j][k];
                }
            }
        }
    }

    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            std::cout << f[i][j][B] << " \n"[j == m];
        }
    }

    return 0;
}

详细

Test #1:

score: 100
Accepted
time: 1ms
memory: 3760kb

input:

2 3
1 2
2 3
1 4
2 2
1 3

output:

2 0 0
3 4 4

result:

ok 6 numbers

Test #2:

score: -100
Memory Limit Exceeded

input:

500 500
19 458
1 480
7 485
50 461
12 476
15 461
48 466
40 453
46 467
9 458
27 478
26 472
46 459
29 490
6 500
17 487
48 484
28 472
28 459
25 480
4 491
29 481
36 460
2 491
44 499
22 473
20 458
4 483
27 471
2 496
11 461
43 450
2 478
37 466
15 459
42 482
7 451
19 455
2 453
47 475
48 450
1 474
46 471
9 4...

output:

411 79401 9145270 673005095 180581065 984223118 586589234 293043270 404363796 865361724 665487988 118838806 926189944 226338288 521479857 808644951 786041288 340769021 177100 21 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ...

result: