回答編集履歴
2
バグの修正
answer
CHANGED
|
@@ -51,7 +51,7 @@
|
|
|
51
51
|
for (i = 1; i <= np2; i++) {
|
|
52
52
|
for (j = 1; j <= np1; j++)
|
|
53
53
|
if (epoint1[j][0] == epoint2[i][0] &&
|
|
54
|
-
epoint1[j][
|
|
54
|
+
epoint1[j][1] == epoint2[i][1] &&
|
|
55
55
|
epoint1[j][2] == epoint2[i][2]) break;
|
|
56
56
|
if (j > np1) { // 重複しなかったので
|
|
57
57
|
conv[i] = ++k; // 新規節点番号を登録
|
1
コードの改善
answer
CHANGED
|
@@ -39,7 +39,6 @@
|
|
|
39
39
|
int point2[1024][4];
|
|
40
40
|
double epoint1[1024][3];
|
|
41
41
|
double epoint2[1024][3];
|
|
42
|
-
double ecomb2[1024][3];
|
|
43
42
|
int conv[1024];
|
|
44
43
|
|
|
45
44
|
read_data("in.dat1", point1, epoint1);
|
|
@@ -49,8 +48,6 @@
|
|
|
49
48
|
int np2 = point2[0][0], ne2 = point2[0][1];
|
|
50
49
|
|
|
51
50
|
int i, j, k = np1;
|
|
52
|
-
for (i = 1; i <= np1; i++)
|
|
53
|
-
for (j = 0; j < 3; j++) ecomb2[i][j] = epoint1[i][j];
|
|
54
51
|
for (i = 1; i <= np2; i++) {
|
|
55
52
|
for (j = 1; j <= np1; j++)
|
|
56
53
|
if (epoint1[j][0] == epoint2[i][0] &&
|
|
@@ -58,7 +55,7 @@
|
|
|
58
55
|
epoint1[j][2] == epoint2[i][2]) break;
|
|
59
56
|
if (j > np1) { // 重複しなかったので
|
|
60
57
|
conv[i] = ++k; // 新規節点番号を登録
|
|
61
|
-
for (j = 0; j < 3; j++)
|
|
58
|
+
for (j = 0; j < 3; j++) epoint1[k][j] = epoint2[i][j];
|
|
62
59
|
}
|
|
63
60
|
else conv[i] = j; // 重複したので、登録済み節点番号 j を使う
|
|
64
61
|
}
|
|
@@ -67,7 +64,8 @@
|
|
|
67
64
|
|
|
68
65
|
point1[0][0] = k;
|
|
69
66
|
point1[0][1] = ne1 + ne2;
|
|
70
|
-
write_data("in.dat", point1,
|
|
67
|
+
write_data("in.dat", point1, epoint1);
|
|
71
68
|
}
|
|
69
|
+
|
|
72
70
|
```
|
|
73
71
|
理解できたかどうかのコメントをお願いします。
|