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  1. #include <stdio.h>
  2. #include <string.h>
  3.  
  4. #define NUM 6561 // 3^8
  5. const char kigou[] = {'+','-','*'};
  6.  
  7. void print(int en[8])
  8. {
  9.     int i;
  10.     for (i = 0; i < 8; i++)
  11.         printf("%1d%1c", i+1,kigou[en[i]]);
  12.     printf("9=100\n");
  13. }
  14.  
  15. void eval(int en[8])
  16. {
  17.     int num[9] = {1,2,3,4,5,6,7,8,9};
  18.     int i, pm, sum;
  19.     for (i = 0; i < 8; i++) {
  20.         if (en[i] == 2) {
  21.             num[i+1] = num[i] * num[i+1];
  22.             num[i] = 0;
  23.         }
  24.     }
  25.     for (i = 0; i < 8; i++) {
  26.         if (num[i] != 0) {
  27.             sum = num[i];
  28.             break;
  29.         }
  30.     }
  31.     for (; i < 8; i++) {
  32.         if (en[i] == 0) {
  33.             pm = 1;
  34.         } else if (en[i] == 1) {
  35.             pm = -1;
  36.         }
  37.         sum += num[i+1] * pm;
  38.     }
  39.     if (sum == 100) {
  40.         print(en);
  41.     }
  42. }
  43. void form(int en[8], int num)
  44. {
  45.     int i;
  46.     for (i = 0; i < 8; i++) {
  47.         int y = num / 3;
  48.         en[i] = num % 3;
  49.         num = y;
  50.     }
  51. }
  52.  
  53. int main()
  54. {
  55.     int en[8];
  56.     int i;
  57.     for (i=0; i<NUM; i++) {
  58.         memset(en, 0, sizeof(en));
  59.         form(en,i);
  60.         eval(en);
  61.     }
  62.     return 0;
  63. }
  64.  
Success #stdin #stdout 0.02s 2724KB
comments (?)
stdin 
Standard input is empty
stdout 
1*2*3*4+5+6+7*8+9=100
1-2+3*4*5+6*7+8-9=100
1-2+3*4*5-6+7*8-9=100
1+2+3+4+5+6+7+8*9=100
1*2*3+4+5+6+7+8*9=100
1-2*3+4*5+6+7+8*9=100
1+2*3+4*5-6+7+8*9=100
1-2*3-4+5*6+7+8*9=100
1+2-3*4+5*6+7+8*9=100
1*2*3*4+5+6-7+8*9=100
1-2*3-4-5+6*7+8*9=100
1+2-3*4-5+6*7+8*9=100
1+2+3-4*5+6*7+8*9=100
1*2*3-4*5+6*7+8*9=100
https://ideone.com/55maI
language:
C++ (gcc 8.3)
created:
12 years ago
visibility:
public

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