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  1. /*
  2.   1.1から10までの整数のうち、偶数の合計を求めなさい。
  3.   2.1から100までの整数で、奇数だけの合計、偶数だけの合計、総合計をもとめよ；
  4.   3.y=5x+8の値をｘが０～１０まで、１ずつ増やした時のｙの値
  5.   4.10^3+30^3+50^3+70^3+90^3の計算
  6. */
  7. #include <stdio.h>
  8.  
  9. typedef enum {
  10.   EVEN = 0,
  11.   ODD,
  12.   ALL
  13. } sum_types;
  14.  
  15. int get_sum(int min, int max, sum_types type)
  16. {
  17.   int i, sum = 0;
  18.   for (i=min; i<max+1; i++) {
  19.     if (type == EVEN || type==ODD)
  20.       sum += i%2==type ? i : 0;
  21.     else
  22.       sum += i;
  23.   }
  24.   return sum;
  25. }
  26.  
  27. int pow_int(int x, int y)
  28. {
  29.   int i, mult = 1;
  30.   for (i=0; i<y; i++)
  31.     mult *= x;
  32.   return mult;
  33. }
  34.  
  35. int main(void)
  36. {
  37.   int i, j, pow_sum = 0;
  38.  
  39.   /* 1 */
  40.   fprintf(stdout, "1) %d\n", get_sum(1, 10, EVEN));
  41.  
  42.   /* 2 */
  43.   fprintf(stdout, "2) Odd: %d, Even: %d, All: %d\n", get_sum(1, 100, ODD),
  44.                                                      get_sum(1, 100, EVEN),
  45.                                                      get_sum(1, 100, ALL));
  46.   /* 3 */
  47.   fprintf(stdout, "3) y = 5x + 8;\n");
  48.   for (i=0; i<=10; i++)
  49.     fprintf(stdout, "y=%d [x=%d]\n", 5*i+8, i);
  50.  
  51.   /* 4 */
  52.   for (j=10; j<=90; j+=20)
  53.     pow_sum += pow_int(j, 3);
  54.   fprintf(stdout, "4) 10^3+30^3+50^3+70^3+90^3 = %d\n", pow_sum);
  55.  
  56.   return 0;
  57. }
  58.  
Success #stdin #stdout 0s 2248KB
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stdout
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1) 30
2) Odd: 2500, Even: 2550, All: 5050
3) y = 5x + 8;
y=8 [x=0]
y=13 [x=1]
y=18 [x=2]
y=23 [x=3]
y=28 [x=4]
y=33 [x=5]
y=38 [x=6]
y=43 [x=7]
y=48 [x=8]
y=53 [x=9]
y=58 [x=10]
4) 10^3+30^3+50^3+70^3+90^3 = 1225000
https://ideone.com/AJISBn
language:
C (gcc 8.3)
created:
11 years ago
visibility:
public

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