排序

快速排序

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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

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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];
}

查找

二分查找

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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;
}

LeetCode.35

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class Solution {
    public int searchInsert(int[] nums, int target) {
        int l = 0;
        int r = nums.length - 1;
        while (l <= r) {
            int mid = l + r >> 1;
            if (nums[mid] == target)
                return mid;
            if (nums[mid] >= target)
                r = mid - 1;
            else
                l = mid + 1;
        }
        return l;
    }
}

浮点数二分法解题模板

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double l=a,r=b;
while(r-l>c){
    double mid=(l+r)/2;
    if(check(mid)) l=mid;
    else r=mid;
}

不考虑mid-1和mid+1的边界问题模板

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int L=-1,R=n;
while(L+1!=R)
{
	int mid=L+R>>1;
	if(check()) L=mid;
	else R=mid;
	//最后根据你所分左右两边区间的结果
	//选取L或者R作为结果
}

高精度

高精度加法

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// 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;
}

高精度减法

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// C = A - B, 满足A >= B, A >= 0, B >= 0
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;
}

高精度乘低精度

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// C = A * b, A >= 0, b >= 0
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;
    }

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

    return C;
}

高精度除低精度

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// A / b = C ... r, A >= 0, b > 0
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;
}

java版本

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package basic.highPrecision;

import java.io.*;
import java.math.BigInteger;


public class highPrecision {

    public static void main(String[] args) {

    }

    /**
     * 791.高精度加
     */
    public static void add() throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        BigInteger a = new BigInteger(br.readLine());
        BigInteger b = new BigInteger(br.readLine());
        System.out.println(a.add(b));
        br.close();
    }

    /**
     * 792.高精度减
     */
    public static void subtract() throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        BigInteger a = new BigInteger(br.readLine());
        BigInteger b = new BigInteger(br.readLine());
        System.out.println(a.subtract(b));
        br.close();
    }

    /**
     * 793.高精度乘
     */
    public static void multiply() throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        BigInteger a = new BigInteger(br.readLine());
        BigInteger b = new BigInteger(br.readLine());
        System.out.println(a.multiply(b));
        br.close();
    }

    /**
     * 795.高精度除
     */
    public static void divideAndReminder() throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        BigInteger a = new BigInteger(br.readLine());
        BigInteger b = new BigInteger(br.readLine());
        BigInteger[] c = a.divideAndRemainder(b);
        System.out.println(c[0]);   //a/b
        System.out.println(c[1]);   //a%b
        br.close();
    }

    /**
     * 高精度取mod
     */
    public static void mod() throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        BigInteger a = new BigInteger(br.readLine());
        BigInteger b = new BigInteger(br.readLine());
        BigInteger c = a.mod(b);
        System.out.println(c);
        br.close();
    }
}