NOI批量评测 - 记录详情

#	状态分数	耗时	内存占用

#1	Accepted 5	1ms	324 KiB
#2	Accepted 5	1ms	404 KiB
#3	Accepted 5	1ms	404 KiB
#4	Accepted 5	1ms	404 KiB
#5	Accepted 5	1ms	404 KiB
#6	Accepted 5	1ms	404 KiB
#7	Accepted 5	1ms	404 KiB
#8	Accepted 5	1ms	404 KiB
#9	Accepted 5	1ms	316 KiB
#10	Accepted 5	1ms	404 KiB
#11	Accepted 5	3ms	916 KiB
#12	Accepted 5	25ms	2.1 MiB
#13	Accepted 5	1ms	404 KiB
#14	Accepted 5	3ms	532 KiB
#15	Accepted 5	32ms	2 MiB
#16	Accepted 5	13ms	1.4 MiB
#17	Accepted 5	1ms	404 KiB
#18	Accepted 5	3ms	660 KiB
#19	Accepted 5	32ms	2.2 MiB
#20	Accepted 5	17ms	1.8 MiB

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <queue>
#include <set>
#include <map>
#include <cassert>
#include <random>
#include <chrono>
#include <ctime>
#include <numeric>

typedef long long ll;
typedef double lf;
typedef unsigned long long ull;

namespace FastIO
{
    const int MAXSIZE = (1 << 20);
    char buf[MAXSIZE], *p1, *p2;
    #define gc (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, MAXSIZE, stdin), p1 == p2) ? EOF : *p1++)
    template <typename T>
    inline void Read(T &x)
    {
        x = 0; char ch = gc; bool f = 0;
        while (ch < '0' || ch > '9') { if (ch == '-') f = 1; ch = gc; }
        while (ch >= '0' && ch <= '9') x = x * 10 + (ch ^ 48), ch = gc;
        if (f) x = -x;
    }
}

using FastIO::Read;
using namespace std;

const int MAXN = 1e5 + 10, mod = 1e9 + 7;

inline ll Qpow(ll x, ll y) { ll res = 1; while (y) (y & 1) && (res = res * x % mod), x = x * x % mod, y >>= 1; return res; }

int n, m; ll v;

pair <int, int> a[MAXN];

int main()
{
    // freopen("day0/assign/assign3.in", "r", stdin);
    freopen("assign.in", "r", stdin);
    freopen("assign.out", "w", stdout);
    ios::sync_with_stdio(0), cin.tie(0);
    int T;
    Read(T);
    while (T--)
    {
        Read(n), Read(m), Read(v);
        for (int i = 1; i <= m; i++) Read(a[i].first), Read(a[i].second);
        sort(a + 1, a + m + 1);
        bool flag = 0;
        for (int i = 1; i < m; i++)
            if (a[i].first == a[i + 1].first && a[i].second != a[i + 1].second) { flag = 1; break; }
        if (flag) { cout << "0\n"; continue; }
        ll ans = 1;
        for (int i = 1; i < m; i++)
            if (a[i].first != a[i + 1].first)
            {
                int k = a[i + 1].first - a[i].first - 1;
                ans = ans * (Qpow(v, 2 * (k + 1)) - Qpow(v, k) * (v - 1) % mod) % mod;
            }
        if (a[1].first != 1) ans = ans * Qpow(v, 2 * (a[1].first - 1)) % mod;
        if (a[m].first != n) ans = ans * Qpow(v, 2 * (n - a[m].first)) % mod;
        cout << (ans % mod + mod) % mod << "\n";
    }
    return 0;
}

递交者: BJ-0027
题目: NOIP2024BGF NOIP2024BGF
比赛: 【官方数据 BJ】 NOIP2024 批量测试
语言: C++14(O2)
代码长度: 2 KiB
递交时间: 2024-12-6 17:27:58
评测时间: 2024-12-6 17:28:03

分数: 100
总耗时: 142ms
峰值时间: 32ms
峰值内存: 2.2 MiB

100 Accepted

代码

信息

100 Accepted

代码

信息

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