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
Comments