ACMjava求解最大连续和的三种方法 暴力枚举，S前缀，回溯法

2023年11月7日 172次阅读 来源: 回溯法 
package com.supermars.practice;

import java.io.BufferedInputStream;
import java.util.Scanner;

public class 最大连续和1 {
    /*
     * 暴力枚举
     */
    static Scanner cin = new Scanner(new BufferedInputStream(System.in));
    static int A[] = new int[1 << 7];

    public static void main(String[] args) {
        while (cin.hasNext()) {
            int n = cin.nextInt();
            for (int i = 0; i < n; i++) {
                A[i]=cin.nextInt();
            }
            int best=A[0];
            for (int i = 0; i < n; i++) {
                for (int j = i; j <n; j++) {
                    int sum=0;
                    for (int k = i; k <=j; k++) {
                        sum+=A[k];
                    }
                    best=best<sum?sum:best;
                }
            }
            System.out.println(best);
            
            
        }

    }
}

————————————–

package com.supermars.practice;

import java.io.BufferedInputStream;
import java.util.Scanner;

public class 最大连续和2 {
    /**
     * S[i]=A1+A2..+Ai Ai+..Aj=S[j]-S[i-1]
     */
    static Scanner cin = new Scanner(new BufferedInputStream(System.in));
    static int A[] = new int[1 << 7];
    static int S[] = new int[1 << 7];

    public static void main(String[] args) {
        while (cin.hasNext()) {
            int n = cin.nextInt();
            for (int i = 1; i <= n; i++) {
                A[i] = cin.nextInt();
            }

            // 求前缀S
            for (int i = 1; i <= n; i++) {
                S[i] += S[i - 1] + A[i];
            }
            // S[j]-S[i-1] 最大

            int best = 0;
            for (int i = 1; i <= n; i++) {
                int sum = 0;
                for (int j = i; j <= n; j++) {
                    best = best < (sum = (S[j] - S[i - 1])) ? sum : best;
                }
            }
            System.out.println(best);
        }

    }
}

—————————–

package com.supermars.practice;

import java.io.BufferedInputStream;
import java.util.Scanner;

public class 最大连续和3 {

    static Scanner cin = new Scanner(new BufferedInputStream(System.in));
    static int A[] = new int[1 << 7];

    public static void main(String[] args) {
        while (cin.hasNext()) {
            int n = cin.nextInt();
            for (int i = 0; i < n; i++) {
                A[i] = cin.nextInt();
            }

            System.out.println(maxsum(A, 0, n));
        }
    }

    private static int maxsum(int[] A, int x, int y) {
        int m = 0, max = 0, l, r;
        if (y - x == 1)
            return A[x]; // 只有一个元素时返回
        m = x + (y - x) / 2;// 划分[x,m)[m,y)
        l = maxsum(A, x, m);
        r = maxsum(A, m, y);
        max = l > r ? l : r;

        // 3合并(1)
        int v = 0, L = A[m - 1];
        for (int i = m - 1; i >= x; i--) {
            L = L < (v += A[i]) ? v : L;
        }
        int R = A[m];
        v = 0;
        // 3合并(2)
        for (int i = m; i < y; i++) {
            R = R < (v += A[i]) ? v : R;
        }

        return max > (L + R) ? max : (L + R);
    }
}
    原文作者：回溯法
    原文地址: https://blog.csdn.net/supera_li/article/details/23957551
    本文转自网络文章，转载此文章仅为分享知识，如有侵权，请联系博主进行删除。
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