牛客小白月赛81

牛客小白月赛81

A.小辰打比赛

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n , x , y , sum;
    cin >> n >> x; sum = 0;
    for(int i = 1 ; i <= n ; ++i)
    {
        cin >> y;
        if(y < x)
            sum += y;
    }
    cout << sum << '\n'; return 0;
}

B.小辰的圣剑

#include<iostream>
using namespace std;
const int N = 5010;
int n;
long long m , u;
int A[N] , B[N];
int main()
{
    int Max = 0;
    cin >> n >> m >> u;
    for(int i = 1 ; i <= n ; ++i)
        cin >> A[i];
    for(int i = 1 ; i <= n ; ++i)
        cin >> B[i];
    for(int i = 1 ; i <= n ; ++i)
    {
        int j = i;
        long long res_u = 0ll , res_m = m;
        while(j <= n)
        {
            if(res_m >= A[j] && res_u + B[j] <= u)
            {
                res_m -= A[j]; res_u += B[j];
            }
            else
                break;
            j++;
        }
        Max = max(Max , j - i);
    }
    cout << Max << '\n';
}

C.陶陶学算术

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1e5+10;
int tot;
int prime[MAXN] , vis[MAXN] , Output1[MAXN] , Output2[MAXN];
void Init()
{
    int n = 1e5;
    for(int i = 2 ; i <= n ; ++i)
    {
        if(!vis[i]) prime[++tot] = i;
        for(int j = 1 ; j <= tot && i * prime[j] <= n ; ++j)
        {
            vis[i * prime[j]] = 1;
            if(i * prime[j] == 0)
                break;
        }
    }
}
 
void Solve(int f[] , int &flag)
{
    int m , op , x;
    flag = 0;
    cin >> m;
    while(m--)
    {
        cin >> op >> x;
        if(x < 0)
            x = -x , flag ^= 1;
        for(int i = 1 ; prime[i] * prime[i] <= x ; ++i)
        {
            if(x % prime[i] == 0)
            {
                int cnt = 0;
                while(x % prime[i] == 0) x /= prime[i] , cnt++;
                if(op == 1)
                    f[i] += cnt;
                else
                    f[i] -= cnt;
            }
        }
        if(x > 1)
        {
            int y = lower_bound(prime + 1 , prime + 1 + tot , x) - prime;
            if(op == 1)
                f[y]++;
            else
                f[y]--;
        }
    }
}
 
int main()
{
    int flag1 , flag2;
    Init();
    Solve(Output1 , flag1);
    Solve(Output2 , flag2);
    if(flag1 != flag2) { cout << "NO" << '\n'; return 0;  }
    for(int i = 1 ; i <= tot ; ++i)
        if(Output1[i] != Output2[i]) { cout << "NO" << '\n'; return 0; }
    cout << "YES" << '\n';
    return 0;
}

D.小辰的借钱计划

#include<bits/stdc++.h>
using namespace std;
 
int Solve()
{
    int m , a , tot_case;
    long long Sum;
    cin >> m >> a;
    tot_case = 0; Sum = 0ll;
    for(int i = a + a ; i + a <= m ; i += a)
        Sum += i , tot_case++;
    for(int i = 1 ; i * i <= a ; ++i)
    {
        if(a % i == 0)
        {
            if(a + i <= m)
                tot_case++ , Sum += i;
            if(i * i != m && a + a / i <= m)
                tot_case++ , Sum += a / i;
        }
    }
    if(Sum > 1ll * a * tot_case)
        cout << "YES" << '\n';
    else
        cout << "NO" << '\n';
    return 0;
}
 
int main()
{
    int T; cin >> T; while(T--) Solve();
    return 0;
}

E.小辰的智慧树

Error
人菜但题水，第54行的判断应该是不会成立的，但是交上去之后发现只过了20%多一点。
但是注释掉之后居然能过。
先留个坑，若有大佬路过，感谢指正。

#include<bits/stdc++.h>
using namespace std;
const int N = 1e6+10;
int n;
long long m;
int p[N];
struct Node{
    int h , c;
}A[N];
 
bool cmp(int x , int y) { return A[x].h > A[y].h; }
 
long long Check(int mid)
{
    long long res = 0ll;
    for(int i = 1 ; i <= n ; ++i)
        res += max(A[i].h - max(A[i].c , mid) , 0);
    return res;
}
 
int main()
{
    ios::sync_with_stdio(false); cin.tie(0);
    int l , r , mid , tot;
    long long res , Answer;
    cin >> n >> m;
    for(int i = 1 ; i <= n ; ++i)
        cin >> A[i].h >> A[i].c;
    l = 0; r = 1e6;
    while(l < r)
    {
        mid = (l + r) >> 1;
        if(Check(mid) <= m)
            r = mid;
        else
            l = mid + 1;
    }
    Answer = res = 0ll;
    for(int i = 1 ; i <= n ; ++i)
    {
        if(A[i].h > max(l , A[i].c))
        {
            int x = A[i].h - max(l , A[i].c);
            Answer += 1ll * x * (2 * A[i].h - x);
            res += x;
        }
    }
    res = m - res;
    tot = 0;
    for(int i = 1 ; i <= n ; ++i)
        if(l > A[i].c)
            p[++tot] = i;
    sort(p + 1 , p + 1 + tot , cmp);
    if(Check(l - 1) <= m)
        cout << "NO" << '\n';
    for(long long i = 1 ; i <= min(res , 1ll * tot) ; ++i)
    {
        int x = A[p[i]].h - l;
        Answer -= 1ll * x * (2 * A[p[i]].h - x);
        Answer += 1ll * (x + 1) * (2 * A[p[i]].h - (x + 1));
    }
    cout << Answer << '\n';
    return 0;
}

F.小辰刚学 gcd

这题目中数据范围$1 <= a_i <= 2^{31}$，good。刚好比int最大值大1.

#pragma GCC optimize(2)
#include<bits/stdc++.h>
using namespace std;
const int N = 6e5+10;
#define int long long
int n , m;
long long Array[N];
vector< pair<int,int> > Gcd[N];
 
signed main()
{
    ios::sync_with_stdio(false); cin.tie(0);
    unordered_map<long long,int> Visit;
    int l , r , tmp;
    cin >> n >> m;
    for(int i = 1 ; i <= n ; ++i)
        cin >> Array[i];
    for(int i = 1 ; i <= n ; ++i)
    {
        Gcd[i].push_back(make_pair(Array[i],i));
        Visit[Array[i]] = i;
        for(auto x:Gcd[i-1])
        {
            tmp = __gcd(x.first , Array[i]);
            if(!Visit.count(tmp) || Visit[tmp] != i)
            {
                Visit[tmp] = i;
                Gcd[i].push_back(make_pair(tmp , x.second));
            }
        }
    }
    while(m--)
    {
        int cnt = 0;
        cin >> l >> r;
        for(auto x:Gcd[r])
            if(x.second >= l)
                cnt++;
            else
                break;
        cout << cnt << '\n';
    }
    return 0;
}
/*
6 1
3 3 3 6 5 6
1 6
*/