关于一个运算优化的问题

wayne

才一个主题贴，就奉上这么长的代码，着实让我吃惊。。虽然我能看得懂代码，但毕竟内功远远不够，

wayne

我开始醒悟，我既非理科，又非计算机专业的，进了这地方，是一个很大的冒失啊。 向你们致敬，想你们学习了！

无心人

用我给出的表，直接查表，似乎还能加速 你的代码已经比我给出的表数据多了

liangbch

我从0.342优化到0.171,如果优化后速度再提高一倍，就是第一名了。按照楼主的思路，将XOR和log10这2步操作合在一起，用汇编语言写成一个函数，速度应该可以更快。等有时间我将尝试一下。

无心人

呵呵 我第一次是0.468， 普通算法 不如shshsh_0510的普通算法 按照楼主的思路写的代码 现在是0.234 看来还需要优化啊

无心人

目前的笨代码 和各位无法比啊

2. #include <stdlib.h>
3. #include <stdio.h>
4. #include <string.h>
6. typedef struct
7. {
8. unsigned int d[4];
9. } UINT128;
11. unsigned int CompTable[128][2] = {0,0, 0,0, 0,0, 0,1, 1,0, 1,0, 1,1, 2,0, 2,0, 2,1, 3,0, 3,0, 3,0, 3,1,
12. 4,0,4,0,4,1,5,0,5,0,5,1,6,0,6,0,6,0,6,1,7,0,7,0,7,1,8,0,8,0,
13. 8,1,9,0,9,0,9,0,9,1,10,0,10,0,10,1,11,0,11,0,11,1,12,0,12,
14. 0,12,0,12,1,13,0,13,0,13,1,14,0,14,0,14,1,15,0,15,0,15,0,15,
15. 1,16,0,16,0,16,1,17,0,17,0,17,1,18,0,18,0,18,0,18,1,19,0,19,
16. 0,19,1,20,0,20,0,20,1,21,0,21,0,21,0,21,1,22,0,22,0,22,1,23,
17. 0,23,0,23,1,24,0,24,0,24,0,24,1,25,0,25,0,25,1,26,0,26,0,26,
18. 1,27,0,27,0,27,0,27,1,28,0,28,0,28,1,29,0,29,0,29,1,30,0,30,
19. 0,30,1,31,0,31,0,31,0,31,1,32,0,32,0,32,1,33,0,33,0,33,1,34,
20. 0,34,0,34,0,34,1,35,0,35,0,35,1,36,0,36,0,36,1,37,0,37,0,37,0,37,1,38,0};
22. UINT128 pow10[40] = {1,0,0,0,
23. 10,0,0,0,
24. 100,0,0,0,
25. 1000,0,0,0,
26. 10000,0,0,0,
27. 100000,0,0,0,
28. 1000000,0,0,0,
29. 10000000,0,0,0,
30. 100000000,0,0,0,
31. 1000000000,0,0,0,
32. 1410065408,2,0,0,
33. 1215752192,23,0,0,
34. 3567587328,232,0,0,
35. 1316134912,2328,0,0,
36. 276447232,23283,0,0,
37. 2764472320,232830,0,0,
38. 1874919424,2328306,0,0,
39. 1569325056,23283064,0,0,
40. 2808348672,232830643,0,0,
41. 2313682944,2328306436,0,0,
42. 1661992960,1808227885,5,0,
43. 3735027712,902409669,54,0,
44. 2990538752,434162106,542,0,
45. 4135583744,46653770,5421,0,
46. 2701131776,466537709,54210,0,
47. 1241513984,370409800,542101,0,
48. 3825205248,3704098002,5421010,0,
49. 3892314112,2681241660,54210108,0,
50. 268435456,1042612833,542101086,0,
51. 2684354560,1836193738,1126043566,1,
52. 1073741824,1182068202,2670501072,12,
53. 2147483648,3230747430,935206946,126,
54. 0,2242703233,762134875,1262,
55. 0,952195850,3326381459,12621,
56. 0,932023908,3199043520,126217,
57. 0,730304488,1925664130,1262177,
58. 0,3008077584,2076772117,12621774,
59. 0,16004768,3587851993,126217744,
60. 0,160047680,1518781562,1262177448,
61. 0, 0, 0, 0};
64. inline int _log2(unsigned int r)
65. {
66. __asm
67. {
68. bsr eax,r
69. }
70. }
72. inline int log2( UINT128 n)
73. {
74. if (n.d[3] != 0) return (96 + _log2(n.d[3]));
75. if (n.d[2] != 0) return (64 + _log2(n.d[2]));
76. if (n.d[1] != 0) return (32 + _log2(n.d[1]));
77. return (_log2(n.d[0]));
78. }
80. inline int compare( const void * a, const void * b )
81. {
82. unsigned int * l, * r;
83. l = (unsigned int *)a; r = (unsigned int *)b;
84. if (*(l+3) != *(r + 3)) return (*(l + 3) > *(r+3) ? 1 : -1);
85. if (*(l+2) != *(r + 2)) return (*(l + 2) > *(r+2) ? 1 : -1);
86. if (*(l+1) != *(r + 1)) return (*(l + 1) > *(r+1) ? 1 : -1);
87. return (* l > * r ? 1 : (* l == * r ? 0 : -1));
89. }
91. inline int log_10(UINT128 n)
92. {
93. int i = log2(n), j = CompTable[i][0];
94. if (CompTable[i][1] == 0) return (j);
95. if (compare(&n, &pow10[j + 1]) >= 0) return (j + 1); else return j;
96. }
98. int main(int argc, char* argv[])
99. {
100. int count,result,i,j, k;
101. UINT128 data[5000];
102. unsigned int l2[5000];
103. unsigned int l10[5000];
104. UINT128 r;
106. scanf("%d",&count);
107. result=0;
108. i = 0;
109. while (i < count) {
110. scanf("%u", &data[i].d[3]);
111. scanf("%u", &data[i].d[2]);
112. scanf("%u", &data[i].d[1]);
113. scanf("%u", &data[i].d[0]);
114. i++;
115. }
117. qsort( (void *)data, (size_t)count, 16, compare );
119. for (i = 0; i < count; i ++)
120. {
121. l10[i] = log_10(data[i]);
122. l2[i] = log2(data[i]);
123. }
125. for (i= count - 1; i > 0; i --)
126. {
127. k = l2[i];
128. if (CompTable[k][1] == 0)
129. {
130. j = i - 1;
131. while ((l2[j] == k) && (j >= 0))
132. {
133. r.d[3] = data[i].d[3] ^ data[j].d[3];
134. r.d[2] = data[i].d[2] ^ data[j].d[2];
135. r.d[1] = data[i].d[1] ^ data[j].d[1];
136. r.d[0] = data[i].d[0] ^ data[j].d[0];
137. result += log_10(r);
138. j --;
139. }
140. if (j >= 0)
141. result += l10[i] * (j + 1);
142. }
143. else
144. for (j = i - 1; j >= 0; j --)
145. {
147. r.d[3] = data[i].d[3] ^ data[j].d[3];
148. r.d[2] = data[i].d[2] ^ data[j].d[2];
149. r.d[1] = data[i].d[1] ^ data[j].d[1];
150. r.d[0] = data[i].d[0] ^ data[j].d[0];
151. result += log_10(r);
152. }
153. }
154. printf("%u",result*2);
155. return 0;
156. }

复制代码

0.203

无心人

__declspec(naked) int log2(UINT128 * n) { __asm { mov edx, [esp + 4] bsr eax, [edx] bsr ecx, [edx + 4] add ecx, 32 cmp [edx + 4], 0 cmovne eax, ecx bsr ecx, [edx + 8] add ecx, 64 cmp [edx + 8], 0 cmovne eax, ecx bsr ecx, [edx + 12] add ecx, 96 cmp [edx + 12], 0 cmovne eax, ecx ret } } 128位求2的对数非跳转版本！！

wayne

C中内嵌汇编，

liangbch

我将我的101楼代码的阅读权限调低了。现在只有积分达到200分，就可以看我101楼的代码

无心人

147# OK了 下面汇编化 log10和log10xor 呵呵