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题目描述

可被三整除的最大和

思路

本题
通用2
通用解法 $1$ 其实就是对针对本题代码修改之后的一个通用写法！

针对本题代码

class Solution {
public:int maxSumDivThree(vector<int>& nums) {int n = nums.size();int dp[n + 1][3];memset(dp, 0, sizeof dp);dp[0][0] = 0;dp[0][1] = INT_MIN;dp[0][2] = INT_MIN;for(int i = 0; i < n; i ++ ) {int x = nums[i];if(x % 3 == 1) {dp[i + 1][0] = max(dp[i][2] + x, dp[i][0]);dp[i + 1][1] = max(dp[i][0] + x, dp[i][1]);dp[i + 1][2] = max(dp[i][1] + x, dp[i][2]);}else if(x % 3 == 2) {dp[i + 1][0] = max(dp[i][1] + x, dp[i][0]);dp[i + 1][1] = max(dp[i][2] + x, dp[i][1]);dp[i + 1][2] = max(dp[i][0] + x, dp[i][2]);}else { // x % 3 == 0dp[i + 1][0] = max(dp[i][0] + x, dp[i][0]);dp[i + 1][1] = max(dp[i][1] + x, dp[i][1]);dp[i + 1][2] = max(dp[i][2] + x, dp[i][2]);}}for(int i = 1; i <= n; i ++ ) {for(int j = 0; j < 3; j ++ ) {cout << dp[i][j] << ' ';}cout << endl;}return dp[n][0];}
};

通用解法代码

#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#include <limits.h>
#include <time.h>using namespace std;int maxSumDivN1(const vector<int> &a, int k);
int maxSumDivN2(const vector<int> &a, int k);const int N = 500, M = 500;
vector<int> arr(N);int main()
{srand((unsigned int)time(NULL));for(int i = 0; i < N; i ++ )    arr[i] = rand() % M;for(int i = 2; i <= 100; i ++ ) {// cout << maxSumDivN1(a, i) << ' ' << maxSumDivN2(a, i) << endl;int a = maxSumDivN1(arr, i);int b = maxSumDivN2(arr, i);if(a != b) {cout << "Wrong: " << a << ' ' << b << endl;}}return 0;
}int maxSumDivN1(const vector<int> &nums, int k) {vector<int> dp(k, INT_MIN);dp[0] = 0;for (int a : nums) {vector<int> dp2(k + 1, 0);for (int i = 0; i < k; ++i)dp2[(i + a) % k] = max(dp[(i + a) % k], dp[i] + a);dp = dp2;}return dp[0];
}int maxSumDivN2(const vector<int> &nums, int k) {int n = nums.size();int dp[n + 1][k];memset(dp, -0x3f, sizeof dp);for(int i = 0; i < k; i ++ )    dp[0][i] = 0;bool flag[k];memset(flag, false, sizeof flag);   for(int i = 0; i < n; i ++ ) {for(int j = 0; j < k; j ++ ) {int sum = dp[i][j] + nums[i];int d = sum % k;flag[d] = true;dp[i + 1][d] = max(sum, max(dp[i + 1][d], dp[i][d]));}for(int j = 0; j < k; j ++ ) {if(flag[j]) flag[j] = false;else    dp[i + 1][j] = dp[i][j];}}return dp[n][0];
}