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Eric2020


访客:1197

离线:15小时前


活动打卡代码 AcWing 796. 子矩阵的和

#include <iostream>
using namespace std;
int n,m,q;
int a[1010][1010],s[1010][1010];
int main()
{
    cin>>n>>m>>q;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
            scanf("%d",&a[i][j]);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
            s[i][j]=s[i-1][j]+s[i][j-1]-s[i-1][j-1]+a[i][j];
    for(int i=0;i<q;i++)
    {
        int x1,y1,x2,y2;
        scanf("%d %d %d %d",&x1,&y1,&x2,&y2);
        printf("%d\n",s[x2][y2]-s[x1-1][y2]-s[x2][y1-1]+s[x1-1][y1-1]);
    }
}


活动打卡代码 AcWing 795. 前缀和

#include <iostream>

using namespace std;

const int N=100010;
int n,m;
int a[N],s[N];

int main()
{
    cin>>n>>m;
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    s[1]=a[1];
    for(int i=2;i<=n;i++)
        s[i]=s[i-1]+a[i];
    while(m--)
    {
        int l,r;
        scanf("%d%d",&l,&r);
        printf("%d\n",s[r]-s[l-1]);
    }
    return 0;
}


活动打卡代码 AcWing 794. 高精度除法

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

vector<int> div(vector<int> &A,int b,int &r)
{
    vector<int> C;
    r=0;
    for(int i=A.size()-1;i>=0;i--)
    {
        r=r*10+A[i];
        C.push_back(r/b);
        r%=b;
    }
    reverse(C.begin(),C.end());
    while(C.size()>1&&C.back()==0)
        C.pop_back();
    return C;
}

int main()
{
    string a;
    int b;
    cin>>a>>b;
    vector<int> A;
    for(int i=a.size()-1;i>=0;i--)
        A.push_back(a[i]-'0');
    int r;
    auto C=div(A,b,r);
    for(int i=C.size()-1;i>=0;i--)
        printf("%d",C[i]);
    cout<<endl<<r;
    return 0;
}


活动打卡代码 AcWing 793. 高精度乘法

#include <iostream>
#include <vector>

using namespace std;

vector<int> mul(vector<int> &A,int b)
{
    vector<int> C;
    int t=0;
    for(int i=0;i<A.size()||t;i++)
    {
        if(i<A.size()) t+=A[i]*b;
        C.push_back(t%10);
        t/=10;
    }
    for(int i=C.size()-1;i>=0;i--)
        if(C[i]==0&&i) C.pop_back();
        else break;
    return C;
}

int main()
{
    string a;
    int b;
    cin>>a>>b;
    vector<int> A;
    for(int i=a.size()-1;i>=0;i--)
        A.push_back(a[i]-'0');
    auto C=mul(A,b);
    for(int i=C.size()-1;i>=0;i--)
        printf("%d",C[i]);
    return 0;
}


活动打卡代码 AcWing 792. 高精度减法

#include <iostream>
#include <vector>

using namespace std;

bool cmp(vector<int> &A, vector<int> &B)
{
    if (A.size() != B.size()) return A.size() > B.size();

    for (int i = A.size() - 1; i >= 0; i -- )
        if (A[i] != B[i])
            return A[i] > B[i];

    return true;
}

vector<int> sub(vector<int> &A, vector<int> &B)
{
    vector<int> C;
    for (int i = 0, t = 0; i < A.size(); i ++ )
    {
        t = A[i] - t;
        if (i < B.size()) t -= B[i];
        C.push_back((t + 10) % 10);
        if (t < 0) t = 1;
        else t = 0;
    }

    while (C.size() > 1 && C.back() == 0) C.pop_back();
    return C;
}

int main()
{
    string a, b;
    vector<int> A, B;
    cin >> a >> b;
    for (int i = a.size() - 1; i >= 0; i -- ) A.push_back(a[i] - '0');
    for (int i = b.size() - 1; i >= 0; i -- ) B.push_back(b[i] - '0');

    vector<int> C;

    if (cmp(A, B)) C = sub(A, B);
    else C = sub(B, A), cout << '-';

    for (int i = C.size() - 1; i >= 0; i -- ) cout << C[i];
    cout << endl;

    return 0;
}


活动打卡代码 AcWing 791. 高精度加法

#include <iostream>
#include <vector>

using namespace std;

vector<int> add(vector<int> &A, vector<int> &B)
{
    if (A.size() < B.size()) return add(B, A);

    vector<int> C;
    int t = 0;
    for (int i = 0; i < A.size(); i ++ )
    {
        t += A[i];
        if (i < B.size()) t += B[i];
        C.push_back(t % 10);
        t /= 10;
    }

    if (t) C.push_back(t);
    return C;
}

int main()
{
    string a, b;
    vector<int> A, B;
    cin >> a >> b;
    for (int i = a.size() - 1; i >= 0; i -- ) A.push_back(a[i] - '0');
    for (int i = b.size() - 1; i >= 0; i -- ) B.push_back(b[i] - '0');

    auto C = add(A, B);

    for (int i = C.size() - 1; i >= 0; i -- ) cout << C[i];
    cout << endl;

    return 0;
}


活动打卡代码 AcWing 790. 数的三次方根

#include <iostream>

using namespace std;

int main()
{
    double x;
    cin >> x;

    double l = -100, r = 100;
    while (r - l > 1e-8)
    {
        double mid = (l + r) / 2;
        if (mid * mid * mid >= x) r = mid;
        else l = mid;
    }

    printf("%.6lf\n", l);
    return 0;
}


活动打卡代码 AcWing 789. 数的范围

#include <iostream>

using namespace std;

const int N = 100010;

int n, m;
int q[N];

int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 0; i < n; i ++ ) scanf("%d", &q[i]);

    while (m -- )
    {
        int x;
        scanf("%d", &x);

        int l = 0, r = n - 1;
        while (l < r)
        {
            int mid = l + r >> 1;
            if (q[mid] >= x) r = mid;
            else l = mid + 1;
        }

        if (q[l] != x) cout << "-1 -1" << endl;
        else
        {
            cout << l << ' ';

            int l = 0, r = n - 1;
            while (l < r)
            {
                int mid = l + r + 1 >> 1;
                if (q[mid] <= x) l = mid;
                else r = mid - 1;
            }

            cout << l << endl;
        }
    }

    return 0;
}


活动打卡代码 AcWing 788. 逆序对的数量

#include <iostream>
using namespace std;
typedef long long LL;
const int N=1e5+10;
int a[N],tmp[N];
LL merge_sort(int q[],int l,int r)
{
    if (l>=r) return 0;
    int mid=l+r>>1;
    LL res=merge_sort(q,l,mid)+merge_sort(q,mid+1,r);
    int k= 0,i=l,j=mid+1;
    while(i<=mid&&j<=r)
        if(q[i]<=q[j])tmp[k++]=q[i++];
        else
        {
            res+=mid-i+1;
            tmp[k++]=q[j++];
        }
    while(i<=mid) tmp[k++]=q[i++];
    while(j<=r) tmp[k++]=q[j++];
    for (i=l,j=0;i<=r;i++,j++) q[i]=tmp[j];
    return res;
}
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;i++) scanf("%d",&a[i]);
    cout<<merge_sort(a,0,n-1)<<endl;
    return 0;
}


活动打卡代码 AcWing 786. 第k个数

#include <iostream>

using namespace std;

const int N = 100010;

int q[N];

int quick_sort(int q[], int l, int r, int k)
{
    if (l >= r) return q[l];

    int i = l - 1, j = r + 1, x = q[l + r >> 1];
    while (i < j)
    {
        do i ++ ; while (q[i] < x);
        do j -- ; while (q[j] > x);
        if (i < j) swap(q[i], q[j]);
    }

    if (j - l + 1 >= k) return quick_sort(q, l, j, k);
    else return quick_sort(q, j + 1, r, k - (j - l + 1));
}

int main()
{
    int n, k;
    scanf("%d%d", &n, &k);

    for (int i = 0; i < n; i ++ ) scanf("%d", &q[i]);

    cout << quick_sort(q, 0, n - 1, k) << endl;

    return 0;
}