算法入门（代码模板）

电玩女神 2022-09-05 00:04 135阅读 0赞

## 高精度加法 ##

// C = A + B, A >= 0, B >= 0
    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;
    }

## 快速排序算法 ##

void quick_sort(int q[], int l, int r)
    { 
        if (l >= r) return;
    
        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]);
        }
        quick_sort(q, l, j), quick_sort(q, j + 1, r);
    }

## 归并排序算法 ##

void merge_sort(int q[], int l, int r)
    { 
        if (l >= r) return;
    
        int mid = l + r >> 1;
        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 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];
    }

## 整数二分算法 ##

bool check(int x) { /* ... */} // 检查x是否满足某种性质
    
    // 区间[l, r]被划分成[l, mid]和[mid + 1, r]时使用：
    int bsearch_1(int l, int r)
    { 
        while (l < r)
        { 
            int mid = l + r >> 1;
            if (check(mid)) r = mid;    // check()判断mid是否满足性质
            else l = mid + 1;
        }
        return l;
    }
    // 区间[l, r]被划分成[l, mid - 1]和[mid, r]时使用：
    int bsearch_2(int l, int r)
    { 
        while (l < r)
        { 
            int mid = l + r + 1 >> 1;
            if (check(mid)) l = mid;
            else r = mid - 1;
        }
        return l;
    }

## 浮点数二分算法 ##

bool check(double x) { /* ... */} // 检查x是否满足某种性质
    
    double bsearch_3(double l, double r)
    { 
        const double eps = 1e-6;   // eps 表示精度，取决于题目对精度的要求
        while (r - l > eps)
        { 
            double mid = (l + r) / 2;
            if (check(mid)) r = mid;
            else l = mid;
        }
        return l;
    }

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