2025年河南工业大学2025新生周赛（5）

A:A-B

简单的的高精度减法，数组模拟算术减法即可，注意范围。

#include<bits/stdc++.h>
using namespace std;
const int N = 1e6;
int a[N], b[N], c[N];
int main() 
{
    string s1, s2;
    cin >> s1 >> s2;
    if (s1.size() < s2.size() || (s1.size() == s2.size() && s1 < s2))
    {
        string s = s1;
        s1 = s2;
        s2 = s;
        cout << '-';
    }
    reverse(s1.begin(), s1.end());
    reverse(s2.begin(), s2.end());
    int lena = s1.size(), lenb = s2.size();
    int lenc = max(lena, lenb) + 1;
    for (int i = 0; i < lena; i++) a[i] = s1[i] - '0';
    for (int i = 0; i < lenb; i++) b[i] = s2[i] - '0';
    for (int i = 0; i < lenc; i++)
    {
        if (a[i] < b[i]) a[i + 1]--, a[i] += 10;
        c[i] = a[i] - b[i];
    }
    while (lenc > 1 && !c[lenc - 1]) lenc--;
    for (int i = lenc - 1; i >= 0; i--) cout << c[i];
    return 0;
}

B:圣诞节前前前夕

签到题，注意int范围不够，开long long 即可，两种解法。

#include<bits/stdc++.h>
using namespace std;
const int N = 2e5+10;
typedef long long ll;
 
int main()
{
 
    ll ans = 0;
    for(ll i  =1;i<=20251125;i++)  ans+=i;
    cout<<ans;
 
 
    return 0;
}

#include<bits/stdc++.h>
using namespace std;
const int N = 2e5+10;
typedef long long ll;
 
int main()
{
 
    ll n = 20251125;
    ll res = (n + 1) * n / 2;
    cout << res << endl;
 
 
    return 0;
}

 C:NASA的食物补给

01背包题

#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
#define ull unsigned long long
#define ll long long
#define PII pair<int,int>
const int N = 2e3+10;
const int mod = 1e9+7;
const int P = 1;
int n, m;
int h, t;
int f[N][N], c[N];
int v[N], w[N];
 
void solve()
{
    cin >> h >> t;
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        int hh, tt, kk;
        cin >> hh >> tt >> kk;
        for (int j = h; j >= hh; j--)
            for (int k = t; k >= tt; k--)
                f[j][k] = max(f[j][k], f[j - hh][k - tt] + kk);
    }
    cout << f[h][t];
}
 
 
 
 
int main() 
{
    ios::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    int T = 1;
    //cin >> T;
    while (T--) solve();
    return 0;
}

D:接水

模拟题，s数组充当水龙头，模拟接水过程即可。

#include<iostream>
#include<algorithm>
using namespace std;
int a[10005];
int n, m;
int s[105];
int t;
int main(void)
{
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> a[i];
    for (int i = 1; i <= n; i++)
    {
        t = 1;
        for (int j = 2; j <= m; j++)
            if (s[t] > s[j]) t = j;
        s[t] += a[i];
    }
    int mx = 0;
    for (int i = 1; i <= m; i++)
    {
        mx = max(s[i], mx);
    }
    cout << mx;
}

E:阳台接雨

线性dp,两个数组分别存储前后最大值，两边最大值的最小值与当前值相减即可。

#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
#define ull unsigned long long
#define ll long long
#define PII pair<int,int>
const ll N = 1e4+10;
const int mod = 80112002;
const int P = 1;

void solve()
{
    int n;
    cin>>n;
    vector<int> h(n);
    for(int i=0;i<n;i++) cin>>h[i];
    vector<int> ma1(n+1,0),ma2(n+1,0);
    int sum1=0,sum2=0;
    for(int i=0;i<n;i++)
    {
        if(!i) ma1[i]=h[i];
        else ma1[i]=max(ma1[i-1],h[i]);
    }
    for(int i=n-1;i>=0;i--)
    {
        ma2[i]=max(ma2[i+1],h[i]);
    }
    for(int i=0;i<n;i++)
    {
        sum1+=min(ma1[i],ma2[i]);
        sum2+=h[i];
    }
    cout<< sum1-sum2<<endl;
}




int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    int T = 1;
    //cin >> T;
    while (T--) solve();
    return 0;
}

F:领导领导

lca板子题，有兴趣可以去了解，这里提供一个倍增法。

#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
#define ull unsigned long long
#define ll long long
#define PII pair<int,int>
const ll N = 1e4+10;
const int mod = 80112002;
const int P = 1;
int n, m, s, a, b;
vector<int> e[N];
int dep[N], fa[N][22];
void dfs(int u, int father)
{
    dep[u] = dep[father] + 1;
    fa[u][0] = father;
    for (int i = 1; i <= 20; i++)
        fa[u][i] = fa[fa[u][i - 1]][i - 1];
    for (int v : e[u])
        if (v != father) dfs(v, u);
}
int lca(int u, int v)
{
    if (dep[u] < dep[v]) swap(u, v);
    for (int i = 20; i >= 0; i--)
        if (dep[fa[u][i]] >= dep[v])
            u = fa[u][i];
    if (u == v) return v;
    for (int i = 20; i >= 0; i--)
    {
        if (fa[u][i] != fa[v][i])
            u = fa[u][i], v = fa[v][i];
    }
    return fa[u][0];
}
void solve()
{
    cin >> n >> m >> s;
    for (int i = 1; i < n; i++)
    {
        int x, y;
        cin >> x >> y;
        e[x].push_back(y);
        e[y].push_back(x);
    }
    dfs(s, 0);
    while (m--)
    {
        int u, v;
        cin >> u >> v;
        cout << lca(u, v) << endl;
    }
}




int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    int T = 1;
    //cin >> T;
    while (T--) solve();
    return 0;
}

G:校园文化节宣传标语

双指针，剔除无关元素，前后对比即可。

#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
#define ull unsigned long long
#define ll long long
#define PII pair<int,int>
const ll N = 5e5 + 10;
const ll Mod = 1e9 + 7;
const int P = 1;

bool check(const string& s)
{
    int l = 0, r = s.size() - 1;
    while (l < r) 
    {
        while (l < r && !isalpha(s[l])) l++;
        while (l < r && !isalpha(s[r])) r--;
        if (tolower(s[l++]) != tolower(s[r--]))
            return false;
    }
    return true;
}
 
void solve()
{
    string s;
    getline(cin,s);
    cout<<(check(s)?"true":"false");
}

int main()
{
ios::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
int T = 1;
//cin >> T;
while (T--) solve();
return 0;
}

H:合并果子

贪心加堆，可用两个数组模拟顺序队列，每次最小的两个相加，加完重新插入排序，重复过程到只剩一个元素，总值即为答案。

#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
#define ull unsigned long long
#define ll long long
#define PII pair<int,int>
const ll N = 1e5 + 10;
const ll Mod = 1e9 + 7;
const int P = 1;

void solve()
{
    priority_queue<int, vector<int>, greater<int>> pq;
    int n;
    cin >> n;
    for (int i = 0; i < n; i++)
    {
        int x;
        cin >> x;
        pq.push(x);
    }
    int sum = 0;
    while (pq.size() > 1)
    {
        int a = pq.top(); pq.pop();
        int b = pq.top(); pq.pop();
        int temp = a + b;
        sum += temp;
        pq.push(temp);
    }
    cout << sum << endl;
}




int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    int T = 1;
    //cin >> T;
    while (T--) solve();
    return 0;
}