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Coinisi.

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活动打卡代码 AcWing 3729. 改变数组元素（每日一题）

Coinisi.
2个月前

                //这里填你的代码^^
//注意代码要放在两组三个点之间，才可以正确显示代码高亮哦~
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <cmath>
#include <cstring>
#include <unordered_map>
#include <unordered_set>

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
#define IOS std::ios::sync_with_stdio(false)
#define pb push_back
#define inf 0x3f3f3f3f
#define YES cout << "YES" << endl;
#define yes cout << "yes" << endl;
#define no cout << "no" << endl;
#define NO cout << "NO" << endl;
#define int long long
#define x first
#define y second
#define cmp [&](PII a, PII b){ return a.y < b.y; }

const int N = 5e5+10, mod = 1e9+7, M = 1e6+5, K = 1e5+10, Z = 2e5+7, Q = 1e4+7;

using namespace std;
typedef long long LL;
typedef priority_queue<int> PQI; 
typedef priority_queue <int, vector<int>, greater<>> PQGI;
typedef pair<int, int> PII;

int a[Z];

void solve() 
{
    int n; cin >> n;
    memset(a, 0, sizeof a);
    for(int i = 1; i <= n; i ++)
    {
        int x; cin >> x;
        x = min(x, i);
        int l = i - x + 1, r = i;
        a[l] ++, a[r + 1] --;
    }
    for(int i = 1; i <= n; i ++) a[i] += a[i - 1];
    for(int i = 1; i <= n; i ++) 
        if(a[i] >= 1) cout << 1 << ' ';
        else cout << '0' << ' ';
    cout << endl;
}

signed main()
{
    int T = 1;
    cin >> T;
    while(T --) solve();
    return 0;
}

            

新鲜事原文

Coinisi.
2个月前

2023年三月28日，Acwimg APP内测，第一次登录！
    

活动打卡代码 AcWing 4874. 约数

Coinisi.
2个月前

                //这里填你的代码^^
#include <iostream>
#include <cmath>
#include <algorithm>
#define int long long
using namespace std;

const int N = 1e6 + 7;

int primes[N], cnt;
bool st[N];

void get_primes(int n)
{
    for(int i = 2; i <= n; i ++)
    {
        if(st[i]) continue;
        primes[cnt ++] = i;
        for(int j = i + i; j <= n; j += i)
            st[j] = true;
    }
}

void solve(int u)
{
    // cout << (int)sqrt(u) << "---" << st[(int)sqrt(u)] << "---" << endl;
    if((int)sqrt(u) * (int)sqrt(u) == u && !st[(int)sqrt(u)]) cout << "YES" << endl;
    else cout << "NO" << endl;
    return;

}

signed main()
{
    get_primes(N-7);
    int n; cin >> n;
    for(int i = 1; i <= n; i ++)
    {
        int x; cin >> x;
        if(x == 1) {cout << "NO" << endl; continue;}
        solve(x);
    }
    return 0;
}
//注意代码要放在两组三个点之间，才可以正确显示代码高亮哦~

            

活动打卡代码 AcWing 4873. 简单计算

Coinisi.
2个月前

                //这里填你的代码^^
#include <iostream>

using namespace std;

int main()
{
    int x1, y1, x2, y2; cin >> x1 >> y1 >> x2 >> y2;
    cout << max(abs(x1 - x2), abs(y1 - y2)) << endl;
    return 0;
}
//注意代码要放在两组三个点之间，才可以正确显示代码高亮哦~

            

题解 AcWing 4867. 整除数

Coinisi.
2个月前

                #include <iostream>

using namespace std;

int main()
{
    int n, k; cin >> n >> k;

    int t = n / k;

    for(int i = t; ; i ++)
        if(k * i > n) {cout << k * i << endl; return 0;}

    return 0;
}

活动打卡代码 AcWing 797. 差分（算法基础课）

Coinisi.
2个月前

                //这里填你的代码^^
#include <iostream>

using namespace std;

const int N = 100000 + 8;

int a[N], b[N];

void insert(int l, int r, int w)
{
    b[l] = b[l] + w, b[r + 1] = b[r + 1] - w;
}

int main()
{
    int n, m; cin >> n >> m;

    for(int i = 1; i <= n; i ++) 
        cin >> a[i];

    for(int i=1;i<=n;i++) insert(i,i,a[i]);

    for(int i = 1; i <= m; i ++)
    {
        int l, r, c; cin >> l >> r >> c;
        insert(l, r, c);
    }
    for(int i = 1; i <= n; i ++)
        b[i] = b[i - 1] + b[i];

    for(int i = 1; i <= n; i ++)
        cout << b[i] << ' ';

    cout << endl;
    return 0;
}
//注意代码要放在两组三个点之间，才可以正确显示代码高亮哦~

            

活动打卡代码 AcWing 99. 激光炸弹（算法提高课）

Coinisi.
3个月前

                #include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <cmath>
#include <cstring>
#include <unordered_map>
#include <unordered_set>

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
#define IOS std::ios::sync_with_stdio(false)
#define pb push_back
#define inf 0x3f3f3f3f
#define YES cout << "YES" << endl;
#define yes cout << "yes" << endl;
#define no cout << "no" << endl;
#define NO cout << "NO" << endl;
// #define int long long
//#define x first
//#define y second
#define cmp [&](PII a, PII b){ return a.y < b.y; }

const int N = 5e5+10, mod = 1e9+7, M = 1e6+5, K = 1e5+10, Z = 2e5+7, Q = 1e4+7;

using namespace std;
typedef long long LL;
typedef priority_queue<int> PQI; 
typedef priority_queue <int, vector<int>, greater<>> PQGI;
typedef pair<int, int> PII;

int a[5002][5002];


void solve()
{
    int n, r; cin >> n >> r;
    r = min(r, 5001);
    for(int i = 1; i <= n; i ++)
    {
        int x, y, w; 
        cin >> x >> y >> w;
        x ++, y ++;
       // a[x + 1][y + 1] += w;
        a[x][y] += w;
    }

    for(int i = 1; i <= 5001; i ++)
        for(int j = 1; j <= 5001; j ++)
            a[i][j] = a[i][j] + a[i - 1][j] + a[i][j - 1] - a[i - 1][j - 1];

    int max_prize = 0;
    for(int i = r; i <= 5001; i ++)
        for(int j = r; j <= 5001; j ++)
            max_prize = max(max_prize, a[i][j] - a[i - r][j] - a[i][j - r] + a[i - r][j - r]);

    cout << max_prize << endl;
} 

signed main()
{
    IOS; cin.tie(nullptr), cout.tie(nullptr); 
    int T = 1;
    // cin >> T;
    while( T -- ) solve();
    return 0;
}
//这里填你的代码^^
//注意代码要放在两组三个点之间，才可以正确显示代码高亮哦~

            

活动打卡代码 AcWing 1230. K倍区间（蓝桥杯辅导课）

Coinisi.
3个月前

                //这里填你的代码^^
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <cmath>
#include <cstring>
#include <unordered_map>
#include <unordered_set>

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
#define IOS std::ios::sync_with_stdio(false)
#define pb push_back
#define inf 0x3f3f3f3f
#define YES cout << "YES" << endl;
#define yes cout << "yes" << endl;
#define no cout << "no" << endl;
#define NO cout << "NO" << endl;
#define int long long
#define x first
#define y second
#define cmp [&](PII a, PII b){ return a.y < b.y; }

const int N = 5e5+10, mod = 1e9+7, M = 1e6+5, K = 1e5+10, Z = 2e5+7;

using namespace std;
typedef long long LL;
typedef priority_queue<int> PQI; 
typedef priority_queue <int, vector<int>, greater<>> PQGI;
typedef pair<int, int> PII;

int a[K], s[K];
int modd[K];

void solve()
{
    int n, k; cin >> n >> k; 
    modd[0] = 1;
    for(int i = 1; i <= n; i ++)
    {
        cin >> a[i];
        s[i] = (s[i - 1] % k + a[i] % k) % k;
        modd[s[i]] ++;
    }

    int cnt = 0;
    for(int i = 0; i <= k - 1; i ++)
        cnt = cnt + modd[i] * (modd[i] - 1) / 2;

    cout << cnt << endl;
    return;
} 

signed main()
{
    IOS; cin.tie(nullptr), cout.tie(nullptr); 
    int T = 1;
    // cin >> T;
    while( T -- ) solve();
    return 0;
}
//注意代码要放在两组三个点之间，才可以正确显示代码高亮哦~

            

题解 AcWing 1230. K倍区间

Coinisi.
3个月前

                #include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <cmath>
#include <cstring>
#include <unordered_map>
#include <unordered_set>

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
#define IOS std::ios::sync_with_stdio(false)
#define pb push_back
#define inf 0x3f3f3f3f
#define YES cout << "YES" << endl;
#define yes cout << "yes" << endl;
#define no cout << "no" << endl;
#define NO cout << "NO" << endl;
#define int long long
#define x first
#define y second
#define cmp [&](PII a, PII b){ return a.y < b.y; }

const int N = 5e5+10, mod = 1e9+7, M = 1e6+5, K = 1e5+10, Z = 2e5+7;

using namespace std;
typedef long long LL;
typedef priority_queue<int> PQI; 
typedef priority_queue <int, vector<int>, greater<>> PQGI;
typedef pair<int, int> PII;

int a[K], s[K];
int modd[K];

void solve()
{
    int n, k; cin >> n >> k; 
    modd[0] = 1;
    for(int i = 1; i <= n; i ++)
    {
        cin >> a[i];
        s[i] = (s[i - 1] % k + a[i] % k) % k;
        modd[s[i]] ++;
    }

    int cnt = 0;
    for(int i = 0; i <= k - 1; i ++)
        cnt = cnt + modd[i] * (modd[i] - 1) / 2;

    cout << cnt << endl;
    return;
} 

signed main()
{
    IOS; cin.tie(nullptr), cout.tie(nullptr); 
    int T = 1;
    // cin >> T;
    while( T -- ) solve();
    return 0;
}

            

活动打卡代码 AcWing 796. 子矩阵的和（算法基础课）

Coinisi.
3个月前

                //这里填你的代码^^
//注意代码要放在两组三个点之间，才可以正确显示代码高亮哦~
#include <iostream>
#include <algorithm>
#include <map>
#include <vector>
#include <cmath>
#include <cstring>

#define IOS std::ios::sync_with_stdio(false)
#define NO cout << "NO" << endl
#define YES cout << "YES" << endl

using namespace std;
typedef long long LL;
typedef pair<LL, LL> PII;

const int N = 1010;

int n, m, q;
int a[N][N];

signed main()
{
    IOS; cin >> n >> m >> q;
    for(int i = 1; i <= n; i ++)
        for(int j = 1; j <= m; j ++)
            cin >> a[i][j];

    for(int i = 1; i <= n; i ++)
        for(int j = 1; j <= m; j ++)
            a[i][j] = a[i][j] + a[i - 1][j] + a[i][j - 1] - a[i - 1][j - 1];

    while(q -- )
    {
        int x1, x2, y1, y2; cin >> x1 >> y1 >> x2 >> y2;
        cout << a[x2][y2] - a[x1 - 1][y2] - a[x2][y1 - 1] + a[x1 - 1][y1 - 1] << endl;
    }

    return 0;
}

            

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