当n=？时，含9的项之和开始大于不含9的项之和

gxqcn

在我的电脑上，gxq 56楼的程序和我这个程序性能相当，只需7秒就得到结果。gxq，能否在你的电脑上测试一下我的这个程序？ liangbch 发表于 2010-11-17 15:03

你的要快多了！ 我的56#结果：

n = 1 Wed Nov 17 16:46:05 2010 s0 = 2.717857142857143 s9 = 0.111111111111111 s0/s9 = 24.460714285714285 s0+s9 = 2.828968253968254 n = 2 Wed Nov 17 16:46:05 2010 s0 = 2.053991622224917 s9 = 0.294417641446448 s0/s9 = 6.976455663912790 s0+s9 = 2.348409263671365 n = 3 Wed Nov 17 16:46:05 2010 s0 = 1.817871425200977 s9 = 0.489221917709748 s0/s9 = 3.715842155460231 s0+s9 = 2.307093342910725 n = 4 Wed Nov 17 16:46:05 2010 s0 = 1.633364212583175 s9 = 0.669670962910865 s0/s9 = 2.439054853869453 s0+s9 = 2.303035175494040 n = 5 Wed Nov 17 16:46:05 2010 s0 = 1.469783389239973 s9 = 0.832846704579078 s0/s9 = 1.764770612837813 s0+s9 = 2.302630093819051 n = 6 Wed Nov 17 16:46:05 2010 s0 = 1.322783057766752 s9 = 0.979806535235522 s0/s9 = 1.350045146870536 s0+s9 = 2.302589593002275 n = 7 Wed Nov 17 16:46:06 2010 s0 = 1.190502772693381 s9 = 1.112082770300745 s0/s9 = 1.070516336091988 s0+s9 = 2.302585542994125 n = 8 Wed Nov 17 16:46:07 2010 s0 = 1.071452317287522 s9 = 1.231132820706344 s0/s9 = 0.870297907152530 s0+s9 = 2.302585137993867 n = 9 Wed Nov 17 16:46:27 2010 s0 = 0.964307069526290 s9 = 1.338278027967131 s0/s9 = 0.720558097326824 s0+s9 = 2.302585097493420

你的79#结果：

n = 1 Wed Nov 17 16:46:46 2010 s0 = 2.717857142857143 s9 = 0.111111111111111 s0/s9 = 24.460714285714285 n = 2 Wed Nov 17 16:46:46 2010 s0 = 2.053991622224917 s9 = 0.294417641446448 s0/s9 = 6.976455663912790 n = 3 Wed Nov 17 16:46:46 2010 s0 = 1.817871425200977 s9 = 0.489221917709748 s0/s9 = 3.715842155460231 n = 4 Wed Nov 17 16:46:46 2010 s0 = 1.633364212583175 s9 = 0.669670962910865 s0/s9 = 2.439054853869453 n = 5 Wed Nov 17 16:46:46 2010 s0 = 1.469783389239973 s9 = 0.832846704579078 s0/s9 = 1.764770612837813 n = 6 Wed Nov 17 16:46:46 2010 s0 = 1.322783057766752 s9 = 0.979806535235522 s0/s9 = 1.350045146870536 n = 7 Wed Nov 17 16:46:46 2010 s0 = 1.190502772693381 s9 = 1.112082770300745 s0/s9 = 1.070516336091988 n = 8 Wed Nov 17 16:46:47 2010 s0 = 1.071452317287522 s9 = 1.231132820706344 s0/s9 = 0.870297907152530 n = 9 Wed Nov 17 16:46:55 2010 s0 = 0.964307069526290 s9 = 1.338278027967131 s0/s9 = 0.720558097326824

gxqcn

看来是 liangbch 的预处理起了很大作用。

liangbch

60# 无心人 我的方法不新鲜，无心人在60#就提到这个想法，只不过我成功的实现了而已。

chyanog

根据我的测试，liangbch 和gxqcn两位大牛的代码效率高低不好比较啊。 我用的VC10（装VS2010还不到一个月），在默认Debug的情况下，前者11s，后者22s ，用Release+“使速度最大化”选项编译，多次比较竟然速度一致，均是11s！ 不过，用gcc编译，gxqcn的代码+优化选项却不起作用。。。 在Linux上暂未测试

gxqcn

我的代码已经相当底层了，所以编译器优化不大起作用。

chyanog

又实现了一种方法，以前想到过，但没写出来。代码比较丑，效率还算不错，在我的笔记本上运行14s。应该是我目前写的行数最多的程序了，感觉肯定能写精简些，但我还简化不了，哪位大牛有空了帮我看看,给点建议。

3. #include <stdio.h>
4. #include <time.h>
5. #include <math.h>
7. void disp(double s0,double s9)
8. {
9. printf( "s0 = %.15f\ns9 = %.15f\ns0/s9 = %.15f\n\n\n", s0, s9, s0/s9);
10. }
12. double calcS(int n)
13. {
14. double s=0.0;
15. int i,k;
16. k=(int)pow(10.0,n-1);
17. for(i=k;i<10*k;i++)
18. s+=1.0/i;
19. return s;
20. }
21. //------------------------------
22. void search_1()
23. {
24. double s9,s0;
25. s9=s0=0.0;
26. int a;
27. for (a=1;a<=8;a++)
28. s0+=1.0/a;
29. s9=calcS(1)-s0;
30. disp(s0,s9);
31. }
33. void search_2()
34. {
35. double s9,s0;
36. s9=s0=0.0;
37. int a,b;
38. for (a=1;a<=8;a++)
39. for (b=0;b<=8;b++)
40. s0+=1.0/(10*a+b);
41. s9=calcS(2)-s0;
42. disp(s0,s9);
43. }
45. void search_3()
46. {
47. double s9,s0,s;
48. s9=s0=0.0;
49. int a,b,c;
50. for (a=1;a<=8;a++)
51. for (b=0;b<=8;b++)
52. for (c=0;c<=8;c++)
53. s0+=1.0/(100*a+10*b+c);
54. s9=calcS(3)-s0;
55. disp(s0,s9);
56. }
57. void search_4()
58. {
59. double s9,s0,s;
60. s9=s0=0.0;
61. int a,b,c,d;
62. for (a=1;a<=8;a++)
63. for (b=0;b<=8;b++)
64. for (c=0;c<=8;c++)
65. for (d=0;d<=8;d++)
66. s0+=1.0/(1000*a+100*b+10*c+d);
68. s9=calcS(4)-s0;
69. disp(s0,s9);
70. }
72. void search_5()
73. {
74. double s9,s0,s;
75. s9=s0=0.0;
76. int a,b,c,d,e;
77. for (a=1;a<=8;a++)
78. for (b=0;b<=8;b++)
79. for (c=0;c<=8;c++)
80. for (d=0;d<=8;d++)
81. for (e=0;e<=8;e++)
82. s0+=1.0/(10000*a+1000*b+100*c+10*d+e);
83. s9=calcS(5)-s0;
84. disp(s0,s9);
85. }
87. void search_6()
88. {
89. double s9,s0,s;
90. s9=s0=0.0;
91. int a,b,c,d,e,f;
92. for (a=1;a<=8;a++)
93. for (b=0;b<=8;b++)
94. for (c=0;c<=8;c++)
95. for (d=0;d<=8;d++)
96. for (e=0;e<=8;e++)
97. for (f=0;f<=8;f++)
98. s0+=1.0/(100000*a+10000*b+1000*c+100*d+10*e+f);
100. s9=calcS(6)-s0;
101. disp(s0,s9);
102. }
104. void search_7()
105. {
106. double s9,s0,s;
107. s9=s0=0.0;
108. int a,b,c,d,e,f,g;
109. for (a=1;a<=8;a++)
110. for (b=0;b<=8;b++)
111. for (c=0;c<=8;c++)
112. for (d=0;d<=8;d++)
113. for (e=0;e<=8;e++)
114. for (f=0;f<=8;f++)
115. for(g=0;g<=8;g++)
116. s0+=1.0/(1000000*a+100000*b+10000*c+1000*d+100*e+10*f+g);
118. s9=calcS(7)-s0;
119. disp(s0,s9);
120. }
122. void search_8()
123. {
124. double s9,s0,s;
125. s9=s0=0.0;
126. int a,b,c,d,e,f,g,h;
127. for (a=1;a<=8;a++)
128. for (b=0;b<=8;b++)
129. for (c=0;c<=8;c++)
130. for (d=0;d<=8;d++)
131. for (e=0;e<=8;e++)
132. for (f=0;f<=8;f++)
133. for(g=0;g<=8;g++)
134. for(h=0;h<=8;h++)
135. s0+=1.0/(10000000*a+1000000*b+100000*c+10000*d+1000*e+100*f+10*g+h);
137. s9=calcS(8)-s0;
138. disp(s0,s9);
139. }
141. void search_9()
142. {
143. double s9,s0,s;
144. s9=s0=0.0;
145. int a,b,c,d,e,f,g,h,i;
146. for (a=1;a<=8;a++)
147. for (b=0;b<=8;b++)
148. for (c=0;c<=8;c++)
149. for (d=0;d<=8;d++)
150. for (e=0;e<=8;e++)
151. for (f=0;f<=8;f++)
152. for(g=0;g<=8;g++)
153. for(h=0;h<=8;h++)
154. for(i=0;i<=8;i++)
155. s0+=1.0/(100000000*a+10000000*b+1000000*c+100000*d+10000*e+1000*f+100*g+10*h+i);
157. s9=calcS(9)-s0;
158. disp(s0,s9);
159. }
161. int main()
162. {
163. double t0=clock();
164. search_1();
165. search_2();
166. search_3();
167. search_4();
168. search_5();
169. search_6();
170. search_7();
171. search_8();
172. search_9();
173. printf("Elapsed time: %f s\n",(clock()-t0)/CLOCKS_PER_SEC);
174. return 0;
175. }

复制代码

gxqcn

86# chyanog 重新整理了一下，但测试起来，效率似乎没有明显的提升。

1. #include <stdio.h>
2. #include <time.h>
4. void disp(double s0,double s9)
5. {
6. printf( "s0 = %.15f\ns9 = %.15f\ns0/s9 = %.15f\n\n\n", s0, s9, s0/s9);
7. }
9. double calcS(int n)
10. {
11. static int b, e=1;
12. double s=0.0;
13. int k;
14. b=e, e*=10;
15. for(k=b;k<e;k++)
16. s+=1.0/k;
17. return s;
18. }
19. //------------------------------
20. void search_1()
21. {
22. double s9,s0=0.0;
23. int k;
24. for (k=1;k<=8;k++)
25. s0+=1.0/k;
26. s9=calcS(1)-s0;
27. disp(s0,s9);
28. }
30. void search_2()
31. {
32. double s9,s0=0.0;
33. int a,b,k=10;
34. for (b=1;b<=8;b++,k+=1)
35. for (a=0;a<=8;a++,k++)
36. s0+=1.0/k;
37. s9=calcS(2)-s0;
38. disp(s0,s9);
39. }
41. void search_3()
42. {
43. double s9,s0=0.0;
44. int a,b,c,k=100;
45. for (c=1;c<=8;c++,k+=10)
46. for (b=0;b<=8;b++,k+=1)
47. for (a=0;a<=8;a++,k++)
48. s0+=1.0/k;
49. s9=calcS(3)-s0;
50. disp(s0,s9);
51. }
53. void search_4()
54. {
55. double s9,s0=0.0;
56. int a,b,c,d,k=1000;
57. for (d=1;d<=8;d++,k+=100)
58. for (c=0;c<=8;c++,k+=10)
59. for (b=0;b<=8;b++,k+=1)
60. for (a=0;a<=8;a++,k++)
61. s0+=1.0/k;
62. s9=calcS(4)-s0;
63. disp(s0,s9);
64. }
66. void search_5()
67. {
68. double s9,s0=0.0;
69. int a,b,c,d,e,k=10000;
70. for (e=1;e<=8;e++,k+=1000)
71. for (d=0;d<=8;d++,k+=100)
72. for (c=0;c<=8;c++,k+=10)
73. for (b=0;b<=8;b++,k+=1)
74. for (a=0;a<=8;a++,k++)
75. s0+=1.0/k;
76. s9=calcS(5)-s0;
77. disp(s0,s9);
78. }
80. void search_6()
81. {
82. double s9,s0=0.0;
83. int a,b,c,d,e,f,k=100000;
84. for (f=1;f<=8;f++,k+=10000)
85. for (e=0;e<=8;e++,k+=1000)
86. for (d=0;d<=8;d++,k+=100)
87. for (c=0;c<=8;c++,k+=10)
88. for (b=0;b<=8;b++,k+=1)
89. for (a=0;a<=8;a++,k++)
90. s0+=1.0/k;
91. s9=calcS(6)-s0;
92. disp(s0,s9);
93. }
95. void search_7()
96. {
97. double s9,s0=0.0;
98. int a,b,c,d,e,f,g,k=1000000;
99. for (g=1;g<=8;g++,k+=100000)
100. for (f=0;f<=8;f++,k+=10000)
101. for (e=0;e<=8;e++,k+=1000)
102. for (d=0;d<=8;d++,k+=100)
103. for (c=0;c<=8;c++,k+=10)
104. for (b=0;b<=8;b++,k+=1)
105. for (a=0;a<=8;a++,k++)
106. s0+=1.0/k;
107. s9=calcS(7)-s0;
108. disp(s0,s9);
109. }
111. void search_8()
112. {
113. double s9,s0=0.0;
114. int a,b,c,d,e,f,g,h,k=10000000;
115. for (h=1;h<=8;h++,k+=1000000)
116. for (g=0;g<=8;g++,k+=100000)
117. for (f=0;f<=8;f++,k+=10000)
118. for (e=0;e<=8;e++,k+=1000)
119. for (d=0;d<=8;d++,k+=100)
120. for (c=0;c<=8;c++,k+=10)
121. for (b=0;b<=8;b++,k+=1)
122. for (a=0;a<=8;a++,k++)
123. s0+=1.0/k;
124. s9=calcS(8)-s0;
125. disp(s0,s9);
126. }
128. void search_9()
129. {
130. double s9,s0=0.0;
131. int a,b,c,d,e,f,g,h,i,k=100000000;
132. for (i=1;i<=8;i++,k+=10000000)
133. for (h=0;h<=8;h++,k+=1000000)
134. for (g=0;g<=8;g++,k+=100000)
135. for (f=0;f<=8;f++,k+=10000)
136. for (e=0;e<=8;e++,k+=1000)
137. for (d=0;d<=8;d++,k+=100)
138. for (c=0;c<=8;c++,k+=10)
139. for (b=0;b<=8;b++,k+=1)
140. for (a=0;a<=8;a++,k++)
141. s0+=1.0/k;
142. s9=calcS(9)-s0;
143. disp(s0,s9);
144. }
146. int main()
147. {
148. double t0=clock();
149. search_1();
150. search_2();
151. search_3();
152. search_4();
153. search_5();
154. search_6();
155. search_7();
156. search_8();
157. search_9();
158. printf("Elapsed time: %f s\n",(clock()-t0)/CLOCKS_PER_SEC);
159. return 0;
160. }

复制代码

G-Spider

整理了一下，瓶颈在n=9时，用时最多，效率没有提升，或许将calcS()通过筛选方式只算与9有关的可能会好一点，(calcS(9)再做减法，代价非常大，占整体时间的一半以上)。还有依赖性(比如:static不是好的开始)，这个好消除。

1. #include <stdio.h>
2. #include <time.h>
4. void disp(double s0,double s9,int n)
5. {
6. printf( "s0[%d] = %.15f\ns9[%d] = %.15f\ns0/s9 = %.15f\n\n\n", n,s0,n,s9,s0/s9);
7. }
9. double calcS(int n)//n=1,2,3,...,9
10. {
11. int e[10]={1,10,100,1000,10000,100000,1000000,10000000,100000000,1000000000};
12. int k;
13. double s=0.0;
14. for(k=e[n-1];k<e[n];k++)
15. s+=1.0/k;
16. return s;
17. }
19. //------------------------------
20. double search_1()
21. {
22. double s9,s0=0.0;
23. int k;
24. for (k=1;k<=8;k++)
25. s0+=1.0/k;
26. return s0;
28. }
30. double search_2()
31. {
32. double s9,s0=0.0;
33. int a,b,k=10;
34. for (b=1;b<=8;b++,k+=1)
35. for (a=0;a<=8;a++,k++)
36. s0+=1.0/k;
37. return s0;
39. }
41. double search_3()
42. {
43. double s9,s0=0.0;
44. int a,b,c,k=100;
45. for (c=1;c<=8;c++,k+=10)
46. for (b=0;b<=8;b++,k+=1)
47. for (a=0;a<=8;a++,k++)
48. s0+=1.0/k;
49. return s0;
51. }
53. double search_4()
54. {
55. double s9,s0=0.0;
56. int a,b,c,d,k=1000;
57. for (d=1;d<=8;d++,k+=100)
58. for (c=0;c<=8;c++,k+=10)
59. for (b=0;b<=8;b++,k+=1)
60. for (a=0;a<=8;a++,k++)
61. s0+=1.0/k;
62. return s0;
64. }
66. double search_5()
67. {
68. double s9,s0=0.0;
69. int a,b,c,d,e,k=10000;
70. for (e=1;e<=8;e++,k+=1000)
71. for (d=0;d<=8;d++,k+=100)
72. for (c=0;c<=8;c++,k+=10)
73. for (b=0;b<=8;b++,k+=1)
74. for (a=0;a<=8;a++,k++)
75. s0+=1.0/k;
76. return s0;
78. }
80. double search_6()
81. {
82. double s9,s0=0.0;
83. int a,b,c,d,e,f,k=100000;
84. for (f=1;f<=8;f++,k+=10000)
85. for (e=0;e<=8;e++,k+=1000)
86. for (d=0;d<=8;d++,k+=100)
87. for (c=0;c<=8;c++,k+=10)
88. for (b=0;b<=8;b++,k+=1)
89. for (a=0;a<=8;a++,k++)
90. s0+=1.0/k;
91. return s0;
93. }
95. double search_7()
96. {
97. double s9,s0=0.0;
98. int a,b,c,d,e,f,g,k=1000000;
99. for (g=1;g<=8;g++,k+=100000)
100. for (f=0;f<=8;f++,k+=10000)
101. for (e=0;e<=8;e++,k+=1000)
102. for (d=0;d<=8;d++,k+=100)
103. for (c=0;c<=8;c++,k+=10)
104. for (b=0;b<=8;b++,k+=1)
105. for (a=0;a<=8;a++,k++)
106. s0+=1.0/k;
107. return s0;
109. }
111. double search_8()
112. {
113. double s9,s0=0.0;
114. int a,b,c,d,e,f,g,h,k=10000000;
115. for (h=1;h<=8;h++,k+=1000000)
116. for (g=0;g<=8;g++,k+=100000)
117. for (f=0;f<=8;f++,k+=10000)
118. for (e=0;e<=8;e++,k+=1000)
119. for (d=0;d<=8;d++,k+=100)
120. for (c=0;c<=8;c++,k+=10)
121. for (b=0;b<=8;b++,k+=1)
122. for (a=0;a<=8;a++,k++)
123. s0+=1.0/k;
124. return s0;
126. }
128. double search_9()
129. {
130. double s9,s0=0.0;
131. int a,b,c,d,e,f,g,h,i,k=100000000;
132. for (i=1;i<=8;i++,k+=10000000)
133. for (h=0;h<=8;h++,k+=1000000)
134. for (g=0;g<=8;g++,k+=100000)
135. for (f=0;f<=8;f++,k+=10000)
136. for (e=0;e<=8;e++,k+=1000)
137. for (d=0;d<=8;d++,k+=100)
138. for (c=0;c<=8;c++,k+=10)
139. for (b=0;b<=8;b++,k+=1)
140. for (a=0;a<=8;a++,k++)
141. s0+=1.0/k;
142. return s0;
144. }
145. int main()
146. {
147. double s0[10]={0};
148. double s9[10]={0};
149. double t0=clock();
151. s0[1]=search_1();
152. s0[2]=search_2();
153. s0[3]=search_3();
154. s0[4]=search_4();
155. s0[5]=search_5();
156. s0[6]=search_6();
157. s0[7]=search_7();
158. s0[8]=search_8();
159. s0[9]=search_9();
162. s9[1]=calcS(1)-s0[1];
163. disp(s0[1],s9[1],1);
164. s9[2]=calcS(2)-s0[2];
165. disp(s0[2],s9[2],2);
166. s9[3]=calcS(3)-s0[3];
167. disp(s0[3],s9[3],3);
168. s9[4]=calcS(4)-s0[4];
169. disp(s0[4],s9[4],4);
170. s9[5]=calcS(5)-s0[5];
171. disp(s0[5],s9[5],5);
172. s9[6]=calcS(6)-s0[6];
173. disp(s0[6],s9[6],6);
174. s9[7]=calcS(7)-s0[7];
175. disp(s0[7],s9[7],7);
176. s9[8]=calcS(8)-s0[8];
177. disp(s0[8],s9[8],8);
178. s9[9]=calcS(9)-s0[9];
179. disp(s0[9],s9[9],9);
181. printf("Elapsed time: %f s\n",(clock()-t0)/CLOCKS_PER_SEC);
182. system("Pause");
183. return 0;
184. }

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G-Spider

考虑到一个现象，发现做减法有个好处，我们知道欧拉公式，lim(1+1/2+1/3+……+1/n) =In(n)+ C 从n=8开始，可以满足本题的精度要求，所以有如下估计式 1/10000000+1/10000001+...+1/99999999=ln(99999999/9999999) 这个常数用系统自带的计算器就可以算出来 同理n=9以可以算出。 这样一来，时间缩短了1/3。以前上面的程序在我机子上要运行9.16s ，现在只需要2.59s (小数点前6位均准确，所以下面的结果只输出了前6位(这无效率影响，上面的9.16s也是只输出前6位的时间))

1. #include <stdio.h>
2. #include <time.h>
4. void disp(double s0,double s9,int n)
5. {
6. printf( "s0[%d] = %.6f\ns9[%d] = %.6f\ns0/s9 = %.6f\n\n\n", n,s0,n,s9,s0/s9);
7. }
9. double calcS(int n)//n=1,2,3,...,7
10. {
11. int e[10]={1,10,100,1000,10000,100000,1000000,10000000,100000000,1000000000};
12. int k;
13. double s=0.0;
14. for(k=e[n-1];k<e[n];k++)
15. s+=1.0/k;
16. return s;
17. }
19. //------------------------------
20. double search_1()
21. {
22. double s9,s0=0.0;
23. int k;
24. for (k=1;k<=8;k++)
25. s0+=1.0/k;
26. return s0;
28. }
30. double search_2()
31. {
32. double s9,s0=0.0;
33. int a,b,k=10;
34. for (b=1;b<=8;b++,k+=1)
35. for (a=0;a<=8;a++,k++)
36. s0+=1.0/k;
37. return s0;
39. }
41. double search_3()
42. {
43. double s9,s0=0.0;
44. int a,b,c,k=100;
45. for (c=1;c<=8;c++,k+=10)
46. for (b=0;b<=8;b++,k+=1)
47. for (a=0;a<=8;a++,k++)
48. s0+=1.0/k;
49. return s0;
51. }
53. double search_4()
54. {
55. double s9,s0=0.0;
56. int a,b,c,d,k=1000;
57. for (d=1;d<=8;d++,k+=100)
58. for (c=0;c<=8;c++,k+=10)
59. for (b=0;b<=8;b++,k+=1)
60. for (a=0;a<=8;a++,k++)
61. s0+=1.0/k;
62. return s0;
64. }
66. double search_5()
67. {
68. double s9,s0=0.0;
69. int a,b,c,d,e,k=10000;
70. for (e=1;e<=8;e++,k+=1000)
71. for (d=0;d<=8;d++,k+=100)
72. for (c=0;c<=8;c++,k+=10)
73. for (b=0;b<=8;b++,k+=1)
74. for (a=0;a<=8;a++,k++)
75. s0+=1.0/k;
76. return s0;
78. }
80. double search_6()
81. {
82. double s9,s0=0.0;
83. int a,b,c,d,e,f,k=100000;
84. for (f=1;f<=8;f++,k+=10000)
85. for (e=0;e<=8;e++,k+=1000)
86. for (d=0;d<=8;d++,k+=100)
87. for (c=0;c<=8;c++,k+=10)
88. for (b=0;b<=8;b++,k+=1)
89. for (a=0;a<=8;a++,k++)
90. s0+=1.0/k;
91. return s0;
93. }
95. double search_7()
96. {
97. double s9,s0=0.0;
98. int a,b,c,d,e,f,g,k=1000000;
99. for (g=1;g<=8;g++,k+=100000)
100. for (f=0;f<=8;f++,k+=10000)
101. for (e=0;e<=8;e++,k+=1000)
102. for (d=0;d<=8;d++,k+=100)
103. for (c=0;c<=8;c++,k+=10)
104. for (b=0;b<=8;b++,k+=1)
105. for (a=0;a<=8;a++,k++)
106. s0+=1.0/k;
107. return s0;
109. }
111. double search_8()
112. {
113. double s9,s0=0.0;
114. int a,b,c,d,e,f,g,h,k=10000000;
115. for (h=1;h<=8;h++,k+=1000000)
116. for (g=0;g<=8;g++,k+=100000)
117. for (f=0;f<=8;f++,k+=10000)
118. for (e=0;e<=8;e++,k+=1000)
119. for (d=0;d<=8;d++,k+=100)
120. for (c=0;c<=8;c++,k+=10)
121. for (b=0;b<=8;b++,k+=1)
122. for (a=0;a<=8;a++,k++)
123. s0+=1.0/k;
124. return s0;
126. }
128. double search_9()
129. {
130. double s9,s0=0.0;
131. int a,b,c,d,e,f,g,h,i,k=100000000;
132. for (i=1;i<=8;i++,k+=10000000)
133. for (h=0;h<=8;h++,k+=1000000)
134. for (g=0;g<=8;g++,k+=100000)
135. for (f=0;f<=8;f++,k+=10000)
136. for (e=0;e<=8;e++,k+=1000)
137. for (d=0;d<=8;d++,k+=100)
138. for (c=0;c<=8;c++,k+=10)
139. for (b=0;b<=8;b++,k+=1)
140. for (a=0;a<=8;a++,k++)
141. s0+=1.0/k;
142. return s0;
144. }
145. int main()
146. {
147. double s0[10]={0};
148. double s9[10]={0};
149. //由欧拉公式lim(1+1/2+...+1/n)=lnn+C (n->无穷,c为常数)
150. //从第n=8开始，n值以达到其极限的精度要求
151. //完全可以用估计式ln( n/m )=1/n +1/(n-1)+...+1/(m+1)
152. //这样明显减少计算时间
153. double s_8=2.3025851829940506340183244547094;
154. double s_9=2.3025851019940457335179917876844;
156. double t0=clock();
158. s0[1]=search_1();
159. s0[2]=search_2();
160. s0[3]=search_3();
161. s0[4]=search_4();
162. s0[5]=search_5();
163. s0[6]=search_6();
164. s0[7]=search_7();
165. s0[8]=search_8();
166. s0[9]=search_9();
169. s9[1]=calcS(1)-s0[1];
170. disp(s0[1],s9[1],1);
171. s9[2]=calcS(2)-s0[2];
172. disp(s0[2],s9[2],2);
173. s9[3]=calcS(3)-s0[3];
174. disp(s0[3],s9[3],3);
175. s9[4]=calcS(4)-s0[4];
176. disp(s0[4],s9[4],4);
177. s9[5]=calcS(5)-s0[5];
178. disp(s0[5],s9[5],5);
179. s9[6]=calcS(6)-s0[6];
180. disp(s0[6],s9[6],6);
181. s9[7]=calcS(7)-s0[7];
182. disp(s0[7],s9[7],7);
183. //==================
184. s9[8]=s_8-s0[8];
185. disp(s0[8],s9[8],8);
186. s9[9]=s_9-s0[9];
187. disp(s0[9],s9[9],9);
189. printf("Elapsed time: %f s\n",(clock()-t0)/CLOCKS_PER_SEC);
190. system("Pause");
191. return 0;
192. }

复制代码

结果： s0[1] = 2.717857 s9[1] = 0.111111 s0/s9 = 24.460714 s0[2] = 2.053992 s9[2] = 0.294418 s0/s9 = 6.976456 s0[3] = 1.817871 s9[3] = 0.489222 s0/s9 = 3.715842 s0[4] = 1.633364 s9[4] = 0.669671 s0/s9 = 2.439055 s0[5] = 1.469783 s9[5] = 0.832847 s0/s9 = 1.764771 s0[6] = 1.322783 s9[6] = 0.979807 s0/s9 = 1.350045 s0[7] = 1.190503 s9[7] = 1.112083 s0/s9 = 1.070516 s0[8] = 1.071452 s9[8] = 1.231133 s0/s9 = 0.870298 s0[9] = 0.964307 s9[9] = 1.338278 s0/s9 = 0.720558 Elapsed time: 2.599000 s 请按任意键继续. . .

gxqcn

根据欧拉定理，所有的n位正整数的倒数和$=ln(10^n)-ln(10^(n-1))=ln10=2.3025850929940456840179914546844$ 上述式子当n比较大时成立，也即 无心人 的 $s0+s9$ gxqcn 发表于 2010-5-19 14:59

楼上的与我在22#想的一致，所以的我程序特意输出了“s0+s9”