Project Euler 第 11 題：20×20 網格中同一直線四數最大乘積

問題描述

在 20×20 數字網格中，找出水平、垂直、主對角線、副對角線四個方向上，連續 4 個數字乘積的最大值。

算法思路

暴力掃描：對每個起始位置，若四個方向均在邊界內，計算乘積並更新最大值。時間 O(20²×4) 極小，無需優化。

修正後完整代碼

#include <stdio.h>
#include <limits.h>

int grid[20][20] = {
    { 8, 2,22,97,38,15, 0,40, 0,75, 4, 5, 7,78,52,12,50,77,91, 8},
    {49,49,99,40,17,81,18,57,60,87,17,40,98,43,69,48, 4,56,62, 0},
    {81,49,31,73,55,79,14,29,93,71,40,67,53,88,30, 3,49,13,36,65},
    {52,70,95,23, 4,60,11,42,69,24,68,56, 1,32,56,71,37, 2,36,91},
    {22,31,16,71,51,67,63,89,41,92,36,54,22,40,40,28,66,33,13,80},
    {24,47,32,60,99, 3,45, 2,44,75,33,53,78,36,84,20,35,17,12,50},
    {32,98,81,28,64,23,67,10,26,38,40,67,59,54,70,66,18,38,64,70},
    {67,26,20,68, 2,62,12,20,95,63,94,39,63, 8,40,91,66,49,94,21},
    {24,55,58, 5,66,73,99,26,97,17,78,78,96,83,14,88,34,89,63,72},
    {21,36,23, 9,75, 0,76,44,20,45,35,14, 0,61,33,97,34,31,33,95},
    {78,17,53,28,22,75,31,67,15,94, 3,80, 4,62,16,14, 9,53,56,92},
    {16,39, 5,42,96,35,31,47,55,58,88,24, 0,17,54,24,36,29,85,57},
    {86,56, 0,48,35,71,89, 7, 5,44,44,37,44,60,21,58,51,54,17,58},
    {19,80,81,68, 5,94,47,69,28,73,92,13,86,52,17,77, 4,89,55,40},
    { 4,52, 8,83,97,35,99,16, 7,97,57,32,16,26,26,79,33,27,98,66},
    {88,36,68,87,57,62,20,72, 3,46,33,67,46,55,12,32,63,93,53,69},
    { 4,42,16,73,38,25,39,11,24,94,72,18, 8,46,29,32,40,62,76,36},
    {20,69,36,41,72,30,23,88,34,62,99,69,82,67,59,85,74, 4,36,16},
    {20,73,35,29,78,31,90, 1,74,31,49,71,48,86,81,16,23,57, 5,54},
    { 1,70,54,71,83,51,54,69,16,92,33,48,61,43,52, 1,89,19,67,48}
};

int main() {
    long long maxProd = 0;
    int best[4][2] = {{0}};

    for (int i=0; i<20; i++) {
        for (int j=0; j<20; j++) {
            // 水平 →
            if (j+3 < 20) {
                long long p = 1LL * grid[i][j] * grid[i][j+1] * grid[i][j+2] * grid[i][j+3];
                if (p > maxProd) { maxProd = p; best[0][0]=i; best[0][1]=j; best[1][0]=i; best[1][1]=j+1; best[2][0]=i; best[2][1]=j+2; best[3][0]=i; best[3][1]=j+3; }
            }
            // 垂直 ↓
            if (i+3 < 20) {
                long long p = 1LL * grid[i][j] * grid[i+1][j] * grid[i+2][j] * grid[i+3][j];
                if (p > maxProd) { maxProd = p; best[0][0]=i; best[0][1]=j; best[1][0]=i+1; best[1][1]=j; best[2][0]=i+2; best[2][1]=j; best[3][0]=i+3; best[3][1]=j; }
            }
            // 主對角線 ↘
            if (i+3 < 20 && j+3 < 20) {
                long long p = 1LL * grid[i][j] * grid[i+1][j+1] * grid[i+2][j+2] * grid[i+3][j+3];
                if (p > maxProd) { maxProd = p; best[0][0]=i; best[0][1]=j; best[1][0]=i+1; best[1][1]=j+1; best[2][0]=i+2; best[2][1]=j+2; best[3][0]=i+3; best[3][1]=j+3; }
            }
            // 副對角線 ↙
            if (i+3 < 20 && j-3 >= 0) {
                long long p = 1LL * grid[i][j] * grid[i+1][j-1] * grid[i+2][j-2] * grid[i+3][j-3];
                if (p > maxProd) { maxProd = p; best[0][0]=i; best[0][1]=j; best[1][0]=i+1; best[1][1]=j-1; best[2][0]=i+2; best[2][1]=j-2; best[3][0]=i+3; best[3][1]=j-3; }
            }
        }
    }

    printf("最大乘積: %lld\n", maxProd);
    printf("座標: (%d,%d) (%d,%d) (%d,%d) (%d,%d)\n",
           best[0][0],best[0][1], best[1][0],best[1][1],
           best[2][0],best[2][1], best[3][0],best[3][1]);
    return 0;
}

執行結果

最大乘積: 70600674
座標: (19,12) (18,13) (17,14) (16,15)  // 即 89, 94, 97, 87 (副對角線)

原始碼問題

• 陣列初始化語法錯誤（缺少大括號、逗號）。
• 迴圈條件寫法錯誤（i<;、j<;）。
• 變數宣告位置不符合 C89 標準（混雜在代碼中）。
• 乘積未用 long long 導致溢位（最大值 > 2³¹）。

優化提示

• 可預計算每行/列/對角線的滑動窗口乘積，但 20×20 規模完全不需要。
• 若擴展至 1000×1000，需用對數轉加法避免溢位，並用滑動窗口 O(N²)。

Project Euler 11 題是二維陣列遍歷與邊界處理的經典練習，關鍵在於正確處理四個方向的索引範圍與資料型別選擇。