C_穷举法：求24点

输入4个十以内的整数，要求用加减乘除运算符和括号组成表达式，要求表达式的值等于24，打印出所有可能。

分析：

1，四个数字，中间需要填三个运算符，一共4 * 4 * 4 = 64种组合
2，四个数字运算时，是两两结合。考虑到括号的存在，所以结合方式总共有5种，如下：
((a # b) # c) # d
(a # (b # c)) # d
a # ((b # c) # d)
a # (b # (c # d))
(a # b) # (c # d)
3,所以组合情况变为了320种，我们只需要在这320种种找到值为24的表达式即可。

完整代码如下：

#define _CRT_SECURE_NO_WARNINGS# include<stdio.h># include<stdlib.h># define N 24/*运算器*/float calc(float a, float b, char op){    float result = 0;    switch (op)    {    case '+':        result = a + b;        break;    case '-':        result = a - b;        break;    case '*':        result = a * b;        break;    case '/':        result = a / b;        break;    }    return result;}/*第一种计算模式：((x#x)#x)#x*/float calc_modle_1(float a, float b, float c, float d, char op1, char op2, char op3)  {    float num1, num2, num3;    num1 = calc(a, b, op1);    num2 = calc(num1, c, op2);    num3 = calc(num2, d, op3);    return num3;}/*第二种计算模式：(x#(x#x))#x*/float calc_modle_2(float a, float b, float c, float d, char op1, char op2, char op3)  {    float num1, num2, num3;    num1 = calc(b,c,op2);    num2 = calc(a, num1, op1);    num3 = calc(num2, d, op3);    return num3;}/*第三种计算模式：x#((x#x)#x)*/float calc_modle_3(float a, float b, float c, float d, char op1, char op2, char op3)  {    float num1, num2, num3;    num1 = calc(b, c, op2);    num2 = calc(num1, d, op3);    num3 = calc(a, num2, op1);    return num3;}/*第四种计算模式：x#(x#(x#x))*/float calc_modle_4(float a, float b, float c, float d, char op1, char op2, char op3){    float num1, num2, num3;    num1 = calc(c, d, op3);    num2 = calc(b, num1, op2);    num3 = calc(a, num2, op1);    return num3;}/*第五种计算模式：(x#x)(x#x)*/float calc_modle_5(float a, float b, float c, float d, char op1, char op2, char op3){    float num1, num2, num3;    num1 = calc(a, b, op1);    num2 = calc(c, d, op3);    num3 = calc(num1, num2, op2);    return num3;}int get_result(float a,float b,float c,float d,char op[]){    int i, j, k, count = 0;    for (i = 0; i < 4; i++)    {        for (j = 0; j < 4; j++)        {            for (k = 0; k < 4; k++)            {                if (calc_modle_1(a, b, c, d, op[i], op[j], op[k]) == N)                {                    printf("((%0.1f %c %0.1f) %c %0.1f) %c %0.1f = %d\n", a, op[i], b, op[j], c, op[k], d,N);                    count++;                }                if (calc_modle_2(a, b, c, d, op[i], op[j], op[k]) == N)                {                    printf("(%0.1f %c(%0.1f %c %0.1f)) %c %0.1f = %d\n", a, op[i], b, op[j], c, op[k], d,N);                    count++;                }                if (calc_modle_3(a, b, c, d, op[i], op[j], op[k]) == N)                {                    printf("%0.1f %c((%0.1f %c %0.1f) %c %0.1f) = %d\n", a, op[i], b, op[j], c, op[k], d,N);                    count++;                }                if (calc_modle_4(a, b, c, d, op[i], op[j], op[k]) == N)                {                    printf("%0.1f %c (%0.1f %c (%0.1f %c %0.1f)) = %d\n", a, op[i], b, op[j], c, op[k], d,N);                    count++;                }                if (calc_modle_5(a, b, c, d, op[i], op[j], op[k]) == N)                {                    printf("(%0.1f %c %0.1f) %c (%0.1f %c %0.1f) = %d\n", a, op[i], b, op[j], c, op[k], d,N);                    count++;                }                //如果要打印所有的表达式及其值，注释掉上面的5个if语句，然后把下面代码解放                /*printf("((%0.1f %c %0.1f) %c %0.1f) %c %0.1f = %0.2f\n", a, op[i], b, op[j], c, op[k], d, calc_modle_1(a, b, c, d, op[i], op[j], op[k]));                printf("(%0.1f %c(%0.1f %c %0.1f)) %c %0.1f = %0.2f\n", a, op[i], b, op[j], c, op[k], d, calc_modle_2(a, b, c, d, op[i], op[j], op[k]));                printf("%0.1f %c((%0.1f %c %0.1f) %c %0.1f) = %0.2f\n", a, op[i], b, op[j], c, op[k], d, calc_modle_3(a, b, c, d, op[i], op[j], op[k]));                printf("%0.1f %c (%0.1f %c (%0.1f %c %0.1f)) = %0.2f\n", a, op[i], b, op[j], c, op[k], d, calc_modle_4(a, b, c, d, op[i], op[j], op[k]));                printf("(%0.1f %c %0.1f) %c (%0.1f %c %0.1f) = %0.2f\n", a, op[i], b, op[j], c, op[k], d, calc_modle_5(a, b, c, d, op[i], op[j], op[k]));*/            }        }    }    return count;}void main(){    int a, b, c, d, count = 0,judge = 1;    char op[4] = { '+', '-', '*', '/' };    printf("请输入十以内的整数，以空格间隔：   ");    while (judge)    {        scanf("%d %d %d %d", &a, &b, &c, &d);        getchar();        if (a<0 || a>10 || b<0 || b>10 || c<0 || c>10 || d<0 || d>10)        {            printf("您输入的数字不符合要求，请重新输入： \n");            continue;        }        judge = 0;    }    count = get_result(a, b, c, d, op);    if (count)    {        printf("\n\n符合计算结果的表达式有：%d 种", count);    }    else    {        printf("\n\n符合计算结果的表达式不存在！！！");    }    getchar();}

运行结果如下：