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Project Euler 18–Maximum path sum I

Wu Yudong    August 12, 2018     欧拉计划   9   

By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.

3
7 4
2 4 6
8 5 9 3

That is, 3 + 7 + 4 + 9 = 23.

Find the maximum total from top to bottom of the triangle below:

75
95 64
17 47 82
18 35 87 10
20 04 82 47 65
19 01 23 75 03 34
88 02 77 73 07 63 67
99 65 04 28 06 16 70 92
41 41 26 56 83 40 80 70 33
41 48 72 33 47 32 37 16 94 29
53 71 44 65 25 43 91 52 97 51 14
70 11 33 28 77 73 17 78 39 68 17 57
91 71 52 38 17 14 91 43 58 50 27 29 48
63 66 04 68 89 53 67 30 73 16 69 87 40 31
04 62 98 27 23 09 70 98 73 93 38 53 60 04 23

NOTE: As there are only 16384 routes, it is possible to solve this problem by trying every route. However, Problem 67, is the same challenge with a triangle containing one-hundred rows; it cannot be solved by brute force, and requires a clever method! ;o)

题目大意:
从下面展示的三角形的顶端出发,不断移动到在下一行与其相邻的元素,能够得到的最大路径和是23。

3
7 4
2 4 6
8 5 9 3

如上图,最大路径和为 3 + 7 + 4 + 9 = 23。

求从下面展示的三角形顶端出发到达底部,所能够得到的最大路径和:

75
95 64
17 47 82
18 35 87 10
20 04 82 47 65
19 01 23 75 03 34
88 02 77 73 07 63 67
99 65 04 28 06 16 70 92
41 41 26 56 83 40 80 70 33
41 48 72 33 47 32 37 16 94 29
53 71 44 65 25 43 91 52 97 51 14
70 11 33 28 77 73 17 78 39 68 17 57
91 71 52 38 17 14 91 43 58 50 27 29 48
63 66 04 68 89 53 67 30 73 16 69 87 40 31
04 62 98 27 23 09 70 98 73 93 38 53 60 04 23

 注意: 在这个问题中,由于只有16384条路径,通过尝试所有的路径来解决问题是可行的。但是,对于第67题,虽然是一道相同类型的题目,但是三角形将拥有一百行,此时暴力破解将不能解决,而需要一个更加聪明的办法!;o)

#include<stdio.h>

#define N 15
int main()
{
    char t[5];
    int s[N][N]={0};
    FILE *f;
    int i,j;
    f = fopen("18.txt","r");
    for (i = 0; i < N; i++) {
        for (j = 0; j <= i; j++) {
            fgets(t,4,f);
            s[i][j] =atoi(t);
        }
    }
    fclose(f);
    for ( i = N-2; i >=0; i--) {
          for ( j = 0; j <= i; j++) {
               if (s[i+1][j] > s[i+1][j+1]) {
                s[i][j]+=s[i+1][j];
            } else {
                s[i][j]+=s[i+1][j+1];
            }
        }
    }
    printf("answer: %d\n",s[0][0]);
    return 0;
}

Answer:1074

Completed on Thu, 1 May 2014, 16:31

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