9.4 小红书笔试-题解

第一题循环节

#include <algorithm>
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

int n;
int m;
long long k;
int a[1234];

int main() {
    cin >> n >> m >> k;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
    }
    int pos = k % (2 * n);
    if (pos == 0) {
        pos = n;
    }

    if (pos <= n) {
        cout << a[pos] << endl;
    }
    else {
        cout << a[n - (pos - n) + 1] << endl;
    }

    return 0;
}

第二题乘积为7的最小操作数

把所有数都转换为正数，枚举每个数变为7时，需要的总操作数，取最小作为答案

#include <algorithm>
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

int n;
int a[200005];

int main() {
    cin >> n;
    bool flag = false;
    int downCount = 0;
    long long sum = 0;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        if (a[i] == 0) {
            flag = true;
        }
        if (a[i] < 0) {
            downCount++;
        }
        if (a[i] < 0) {
            a[i] = abs(a[i]);
        }
        sum += abs(a[i] - 1);
    }

    if (!flag && downCount % 2 == 0) {
        flag = true;
    }

    long long ans = 1e17;
    for (int i = 1; i <= n; i++) {
        long long tmp = sum - abs(a[i] - 1) + abs(a[i] - 7);
        ans = min(ans, tmp);
    }

    if (!flag) {
        ans += 2;
    }

    cout << ans << endl;

    return 0;
}

第三题最短路

二分+dijkstra

#include <algorithm>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <vector>
using namespace std;

const long long Maxn = 2000065;
typedef long long ll;

priority_queue<pair<ll, ll>> q;
long long ans, n, m, tot, ver[Maxn], head[Maxn], Next[Maxn], edge[Maxn], d[Maxn], v[Maxn];
int U[Maxn], V[Maxn];
void add(long long x, long long y, long long z) {
    ver[++tot] = y;
    Next[tot] = head[x];
    head[x] = tot;
    edge[tot] = z;
}
long long dij(long long s, long long e, long long T) {
    memset(d, 0x7f, sizeof(d));
    memset(v, 0, sizeof(v));
    d[s] = 0;
    q.push(make_pair(0, s));
    while (q.size()) {
        long long x = q.top().second;
        q.pop();
        if (v[x])
            continue;
        v[x] = 1;
        for (int i = head[x]; i; i = Next[i]) {
            ll y = ver[i], z = edge[i];
            ll dis = 1;
            if (z > T)
                dis = 10000000000000;
            if (d[y] > d[x] + dis) {
                d[y] = d[x] + dis;
                q.push(make_pair(-d[y], y));
            }
        }
    }
    return d[e];
}
int main() {
    int n, m, k;
    cin >> n >> m >> k;
    for (int i = 1; i <= m; i++)
        cin >> U[i];
    for (int i = 1; i <= m; i++)
        cin >> V[i];
    for (int i = 1; i <= m; i++) {
        long long d;
        cin >> d;
        add(U[i], V[i], d);
        add(V[i], U[i], d);
    }
    int l = 1, r = 100000, ans = 0;
    while (l <= r) {
        int mid = (l + r) >> 1;
        int dis = dij(1, n, mid);
        if (dis <= k) {
            r = mid - 1;
            ans = mid;
        }
        else {
            l = mid + 1;
        }
    }
    cout << ans << endl;
    return 0;
}

#笔试##小红书#