遗失的赋值(Lost Assignment) 题解

思路分析

按题目要求使用对应数据结构与算法模拟/递推求解，注意数据范围边界与时间复杂度。

参考代码

#include <algorithm>
#include <iostream>
#include <utility>
#include <vector>
using namespace std;

const long long MOD = 1000000007;

long long qpow(long long a, long long b) {
    long long res = 1;
    a %= MOD;
    while (b > 0) {
        if (b & 1) res = res * a % MOD;
        a = a * a % MOD;
        b >>= 1;
    }
    return res;
}

void solve() {
    long long n, m, v;
    cin >> n >> m >> v;
    vector<pair<long long, long long>> c(m);
    for (int i = 0; i < m; ++i) cin >> c[i].first >> c[i].second;
    sort(c.begin(), c.end());
    c.erase(unique(c.begin(), c.end()), c.end());
    for (size_t i = 0; i + 1 < c.size(); ++i) {
        if (c[i].first == c[i + 1].first && c[i].second != c[i + 1].second) {
            cout << 0 << "\n";
            return;
        }
    }
    long long ans = qpow(v, 2LL * (c.front().first - 1)) * qpow(v, 2LL * (n - c.back().first)) % MOD;
    for (size_t i = 0; i + 1 < c.size(); ++i) {
        long long L = c[i + 1].first - c[i].first;
        long long tot = qpow(v, 2LL * L);
        long long bad = qpow(v, L - 1) * ((v - 1) % MOD) % MOD;
        long long ways = (tot - bad + MOD) % MOD;
        ans = ans * ways % MOD;
    }
    cout << ans << "\n";
}

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