poj 3983 快算24

 這道題目一看就應該知道是枚舉了。一共4個數，只需要3個運算子，直接迴圈才64次。再加上括弧，每次又要5次。

 

(a b) (c d)

((a b) c) d

(a (b c)) d

a ((b c) d)

a (b (c d))

 

求一次才運算500多次，直接暴力解決了。

#include <stdio.h>#include <string.h>#include <stdlib.h>//四種符號char operators[4] = {'+', '-', '*', '/'};//符號運算double calc( double a, int operatorr, double b){switch (operators[operatorr]){case '+':return a + b;break;case '-':return a - b;break;case '*':return a * b;break;case '/':return a / b;break;}}int calculator(int i, int j, int k, double a, double b, double c, double d){ if (calc(calc(a, i, b),j,calc(c, k, d)) - 24.0 == 0) { printf("(%.0lf%c%.0lf)%c(%.0lf%c%.0lf)\n",a, operators[i], b, operators[j], c, operators[k], d); return 1; } if (calc(calc(calc(a, i, b), j, c), k, d) - 24.0 == 0) { printf("((%.0lf%c%.0lf)%c%.0lf)%c%.0lf)\n",a, operators[i], b, operators[j], c, operators[k], d); return 1; } if (calc(calc(a, i, calc(b, j, c)), k, d) - 24.0 == 0) { printf("(%.0lf%c(%.0lf%c%.0lf))%c%.0lf)\n",a, operators[i], b, operators[j], c, operators[k], d); return 1; } if (calc(a, i, calc(calc(b, j, c), k, d)) - 24.0 == 0) { printf("%.0lf%c((%.0lf%c%.0lf)%c%.0lf)\n",a, operators[i], b, operators[j], c, operators[k], d); return 1; } if (calc(a, i, calc(b, j, calc(c, k, d))) == 24.0) { printf("%.0lf%c(%.0lf%c(%.0lf%c%.0lf))\n",a, operators[i], b, operators[j], c, operators[k], d); return 1; } return 0;}int main(){double a,b,c,d;while (scanf("%lf %lf %lf %lf", &a, &b, &c, &d) != EOF){for (int i = 0; i < 4; ++ i){for(int j = 0; j < 4; ++ j){for(int k = 0; k < 4; ++ k){if(calculator(i,j,k,a,b,c,d))//題目中說有唯一解，找到後直接跳出goto success;}}}success:continue;}return 0;}