Design program according to the order from big to small order output function f (a, b)=2 * a * a * b * b the minimum 100 function value and the corresponding values of two parameters, in which a and b are natural Numbers,
Requirements: (1) as a function of storage structure should be as far as possible to save space,
(2) the designed algorithm and time complexity of the process should be as small as possible,

CodePudding user response:

Fyi:

 # include & lt; Stdio. H> 
#include  

Struct data 
{
int a; 
int b; 
}; 

Int my_cmp (const void * I, const void * j) 
{
Struct data * d1=(struct data *) I; 
Struct data * d2=(struct data *) j; 
Int x=2 * d1 - & gt; A * d1 - & gt; A + d1 - & gt; B * d1 - & gt; b; 
Int y=2 * d2 - & gt; A * d2 - & gt; A + d2 - & gt; B * d2 - & gt; b; 
Return the x - y; 
} 

Int main (void) 
{
Int I, j, value, c=0; 
Struct data d [200]; 

for(i=1; i<=10; I++) 
{
for(j=1; j<=20; J++) 
{
D [c]. A=I; 
D [c]. B=j; 
C + +; 
} 
} 

Tree (d., 200, sizeof (struct data), my_cmp); 

For (c=0; C<100; C + +) 
{
Value=https://bbs.csdn.net/topics/2 * d [c] [c]. A * d a + d [c]. [c]. B * d b; 
Printf (" % d: % d \ \ t t t % d % d \ \ n ", c, d [c]. A, d [c]. B, value); 
} 

return 0; 
} 

Results:

 E: \ Workspace> TCC - run the demo. C 
0:1 1 3 
1:1 2 6 
2:2 1 9 
3:1, 3, 11 
4:2 2 12 
5:2, 3, 17 
6:1 April 18 
7:3 1 19 
8:2 3 22 
9:2 April 24 
10:1 5 27 
11:3 3 and 27 
12:33 4 1 
13:2 5 33 
14:3, 4, 34 
15:4 2 36 
16:1 6 38 
17:4 3 41 
18:3 5 43 
19:6 44 2 
20:4 4 48 
21: five 51 
 122:1 7 51 
23:3 6 54 
24:5 2 54 
2 July 25:57 
26:4, 5 57 
27:5 3 59 
28:1 8 66 
29:5 April 66 
30:3 July 67 
31:4 6 68 
32:2 8 72 
33: June 1 73 
34:5 5 75 
35: June 2, 76 
36:6 3 81 
37:4 July 81 
38:3 8 82 
39:1 9 83 
40:5 6 86 
41:6 4 88 
42:2 September 89 
43:4 8 96 
44: June 5, 97 
45:3 September 99 
46:5, 7, 99 
47: July 1, 99 
48: July 2, 102 
49:1 10 102 
50: July 3, 107 
51: June 6, 108 
52:2 to 10 108 
53:4 September 113 
54: July 4 114 
55:5 August 114 
56:3 118 
57:6 7 121 
58: July 5, 123 
59:1 11 123 
60:2 11 129 
61:8 1 129 
62:5 September 131 
63:8 2 132 
64:4, 10, 132 
65:7 June 134 
66:6 8 136 
67:8 March 137 
68:3 November 139 
69:8 April 144 
12, 146, 70:1 
71:7 July 147 
72:10 150 
73:2 12 152 
74:8 5 153 
75: June 9 153 
76:4 November 153 
77:7 8 162 
78:3 12 162 
79:9 1 163 
80:8 June 164 
81:9 2 166 
82:5 November 171 
83:9 3 171 
84:1 13 171 
85:6 to 10 172 
86:4, 12, 176 
87:2 13 177 
88:8 July 177 
89:9 April 178 
90:7 September 179 
91:3 13 187 
92:9 5 187 
93:8 August 192 
94:6 November 193 
95: May 12 194 
96:7 to 10 198 
14, 198, 97:1 
98:9 6 198 
99:10 1 201 

CodePudding user response:

Can use the method of insertion sort side traverse a, b the result of the increasing f, save minimum 100 values

 int main () 
{
Int a, b, f [3], [100], t, I, j; 
for (i=0; i<100; I++) f [2] [I]=3 * 100 * 100; 
For (a=0; a<10; {+) 
For (b=0; b<15. B++) {
T=a + 2 * a * b * b; 
for (i=0; T> F [2] [I] & amp; & i<100; i++);//insert find insertion position 
If (i<100) {
For (j=99; J> i; J -) {
F [0] [j]=f [0] [1]; 
F [1] [j]=f [1] [1]; 
F [2] [j]=f [2] [1]; 
} 
F [0] [I]=a; 
F [1] [I]=b; 
F [2] [I]=t; 
} 
} 
} 
for (i=0; nullnullnullnullnullnull