How does C++ realize the program of calculating 24:00

Most people do not understand the knowledge points of this article "how to achieve the 24:00 calculation program on C++", so the editor summarizes the following contents for you. The contents are detailed, the steps are clear, and they can be used for reference. I hope you can get something after reading this article. Let's take a look at this "C++ how to achieve the calculation of 24:00 program" article.

Simple program:

1. Run the program

two。 Enter 4 integers, for example: 3 3 7 8

3. Show all possible combinations

Code:

# include "assert.h" # include double operate (double num1, double num2, int op) {assert (op > = 0 & & op)

< 4); if(op == 0){ return num1 + num2; } else if(op == 1){ return num1 - num2; } else if(op == 2){ return num1 * num2; } else{ return num1/num2; }} int calculate(int num1, int num2, int num3, int num4){ char operators[] = "+-*/"; for(int i = 0; i < 4; i ++) { for(int j = 0; j < 4; j ++) { for (int k = 0; k < 4; k ++) { double ret = operate(num1, num2, i); ret = operate(ret, num3, j); ret = operate(ret, num4, k); if(abs(ret - 24) < 0.001){ printf("((%d %c %d) %c %d) %c %d = %f\n", num1, operators[i], num2, operators[j], num3, operators[k], num4, ret); } ret = operate(num1, num2, i); double ret2 = operate(num3, num4, k); ret = operate(ret, ret2, j); if(abs(ret - 24) < 0.001){ printf("(%d %c %d) %c (%d %c %d) = %f\n", num1, operators[i], num2, operators[j], num3, operators[k], num4, ret); } ret = operate(num2, num3, j); ret = operate(num1, ret, i); ret = operate(ret, num4, k); if(abs(ret - 24) < 0.001){ printf("(%d %c (%d %c %d)) %c %d = %f\n", num1, operators[i], num2, operators[j], num3, operators[k], num4, ret); } ret = operate(num2, num3, j); ret = operate(ret, num4, k); ret = operate(num1, ret, i); if(abs(ret - 24) < 0.001){ printf("%d %c ((%d %c %d) %c %d) = %f\n", num1, operators[i], num2, operators[j], num3, operators[k], num4, ret); } ret = operate(num3, num4, k); ret = operate(num2, ret, j); ret = operate(num1, ret, i); if(abs(ret - 24) < 0.001){ printf("%d %c (%d %c (%d %c %d)) = %f\n", num1, operators[i], num2, operators[j], num3, operators[k], num4, ret); } } } } return 0;} int main(int argc, char* argv[]){ int nums[4] = {0, 0, 0, 0}; std::cin >

> nums [0] > > nums [1] > > nums [2] > > nums [3]; for (int I = 0; I < sizeof (nums) / sizeof (nums [0]); I + +) {int num1 = nums [I]; int ret = num1; for (int j = 0; j < sizeof (nums) / sizeof (nums [0]); j + +) {if (j = I) continue; int num2 = nums [j]; for (int k = 0) K < sizeof (nums) / sizeof (nums [0]); knight +) {if (k = = I | | k = = j) continue; int num3 = nums [k]; for (int l = 0; l < sizeof (nums) / sizeof (nums [0]); l + +) {if (l = I | | l = j | l = k) continue; int num4 = nums [l]; calculate (num1, num2, num3, num4);} return 0 } the above is about the content of this article on "how to achieve the program of calculating 24:00 on C++". I believe we all have a certain understanding. I hope the content shared by the editor will be helpful to you. If you want to know more about it, please pay attention to the industry information channel.