C语言代码编程题汇总：显示表达式1*2+3*4+...+99*100的表示形式（采取交互的形式）

Posted 2020-09-21

显示表达式1*2+3*4+...+99*100的表示形式（采取交互的形式）

程序源代码如下：

 1 /*
 2     2017年6月8日08:03:38
 3     功能：采取交互的形式输出以下的表达式
 4 */
 5 #include "stdio.h"
 6 #include "string.h"
 7 
 8 int main()
 9 {
10     char seq[1000];
11     int i, j = 0, first, second, num;
12     printf("please input a odd number: ");
13     scanf("%d",&num);
14     if(num <= 10)
15     {
16         for (i = 1; i <= num; i++)
17         {
18              if (i % 2 != 0)
19                 seq[j++] = ‘0‘ + i,seq[j++] = ‘*‘;
20              else if(i % 2 == 0 && i != num)
21                 seq[j++] = ‘0‘ + i,seq[j++] = ‘+‘;
22              else if (i == num)     
23                 seq[j++] = ‘0‘ + 1,seq[j++] = ‘0‘;
24         }
25     }
26     else if(num > 10 && num < 100)
27     {
28         for (i = 1; i <= num; i++)    
29         {
30             if (i < 10)
31                  if (i % 2 != 0)
32                      seq[j++] = ‘0‘ + i,seq[j++] = ‘*‘;
33                  else
34                      seq[j++] = ‘0‘ + i,seq[j++] = ‘+‘;
35              else if (i >= 10 && i < num)
36              {
37                  first = i % 10;
38                  second = (i - first) / 10;
39                  if (i % 2 != 0)
40                      seq[j++] = ‘0‘ + second,seq[j++] = ‘0‘ + first,seq[j++] = ‘*‘;
41                  else if (i % 2 == 0 && i != num)
42                      seq[j++] = ‘0‘ + second,seq[j++] = ‘0‘ + first,seq[j++] = ‘+‘;
43              }
44              else if (i == num)
45                  seq[j++] = ‘0‘ + second ,seq[j++] = ‘0‘ + (first + 1) ;
46         }
47 
48     }
49     else if(num == 100)
50     {
51         for (i = 1; i <= 100; i++)    
52         {
53             if (i < 10)
54                  if (i % 2 != 0)
55                      seq[j++] = ‘0‘ + i,seq[j++] = ‘*‘;
56                  else
57                      seq[j++] = ‘0‘ + i,seq[j++] = ‘+‘;
58              else if (i >= 10 && i < 100)
59              {
60                  first = i % 10;
61                  second = (i - first) / 10;
62                  if (i % 2 != 0)
63                      seq[j++] = ‘0‘ + second,seq[j++] = ‘0‘ + first,seq[j++] = ‘*‘;
64                  else
65                      seq[j++] = ‘0‘ + second,seq[j++] = ‘0‘ + first,seq[j++] = ‘+‘;
66              }
67              else if (i == 100)
68                  seq[j++] = ‘0‘ + 1,seq[j++] = ‘0‘,seq[j++] = ‘0‘;
69         }
70     }
71     seq[j] = ‘\0‘;
72     puts(seq);
73 }
74 /*
75     总结：
76     在VC++6.0中显示的结果：
77     -----------------------------------------------------------------
78     please input a odd number: 98
79     1*2+3*4+5*6+7*8+9*10+11*12+13*14+15*16+17*18+19*20+21*22+23*24+25*26+27*28+29*30
80     +31*32+33*34+35*36+37*38+39*40+41*42+43*44+45*46+47*48+49*50+51*52+53*54+55*56+5
81     7*58+59*60+61*62+63*64+65*66+67*68+69*70+71*72+73*74+75*76+77*78+79*80+81*82+83*
82     84+85*86+87*88+89*90+91*92+93*94+95*96+97*98
83 
84     please input a odd number: 100
85     1*2+3*4+5*6+7*8+9*10+11*12+13*14+15*16+17*18+19*20+21*22+23*24+25*26+27*28+29*30
86     +31*32+33*34+35*36+37*38+39*40+41*42+43*44+45*46+47*48+49*50+51*52+53*54+55*56+5
87     7*58+59*60+61*62+63*64+65*66+67*68+69*70+71*72+73*74+75*76+77*78+79*80+81*82+83*
88     84+85*86+87*88+89*90+91*92+93*94+95*96+97*98+99*100
89     -----------------------------------------------------------------
90     注意：将整数型数据转化成字符型数据的格式
91             char m;
92             int n;
93             m = n + ‘0‘;（像10、100之类的数据除外）
94 */