快算24 一副牌 c语言程序设计
答案:2 悬赏:50
解决时间 2021-02-27 14:26
- 提问者网友:呐年旧曙光
- 2021-02-27 08:17
快算24 一副牌 c语言程序设计
最佳答案
- 二级知识专家网友:梦中风几里
- 2021-02-27 09:10
//主程序
function funMain()
{
var m = new Array();
//四种运算符
m[0] = "+";
m[1] = "-";
m[2] = "*";
m[3] = "/";
//11种表达式
var exp1 = "a m1 b m2 c m3 d;";
var exp2 = "(a m1 b) m2 c m3 d;";
var exp3 = "(a m1 b m2 c) m3 d;";
var exp4 = "((a m1 b) m2 c) m3 d;";
var exp5 = "(a m1 (b m2 c)) m3 d;";
var exp6 = "a m1 (b m2 c) m3 d;";
var exp7 = "a m1 (b m2 c m3 d);";
var exp8 = "a m1 ((b m2 c) m3 d);";
var exp9 = "a m1 (b m2 (c m3 d));";
var exp10 = "a m1 b m2(c m3 d);";
var exp11 = "(a m1 b) m2 (c m3 d);";
var a,b,c,d;//四个数字
var m1,m2,m3;//三个运算符
for (var i=0;i<4;i++)
{
a = n[i];
for (var j=0;j<4;j++)
{
if ( i == j ) continue;//从未选的三个数字中选择一个数字
b = n[j];
for (var x=0;x<4;x++)
{
if ( x == j || x == i ) continue;//从未选的两个数字中选择一个数字
c = n[x];
for (var y=0;y<4;y++)
{
if ( y == x || y == j || y == i ) continue;//从未选的一个数字中选择一个数字
d = n[y];
for (var ta=0;ta<4;ta++)
{
m1 = m[ta];
for (var tb=0;tb<4;tb++)
{
m2 = m[tb];
for (var tc=0;tc<4;tc++)
{
m3 = m[tc];
for (var k=1;k<12;k++)
{
eval("test(exp"+k+",a,b,c,d,m1,m2,m3);");
}
function funMain()
{
var m = new Array();
//四种运算符
m[0] = "+";
m[1] = "-";
m[2] = "*";
m[3] = "/";
//11种表达式
var exp1 = "a m1 b m2 c m3 d;";
var exp2 = "(a m1 b) m2 c m3 d;";
var exp3 = "(a m1 b m2 c) m3 d;";
var exp4 = "((a m1 b) m2 c) m3 d;";
var exp5 = "(a m1 (b m2 c)) m3 d;";
var exp6 = "a m1 (b m2 c) m3 d;";
var exp7 = "a m1 (b m2 c m3 d);";
var exp8 = "a m1 ((b m2 c) m3 d);";
var exp9 = "a m1 (b m2 (c m3 d));";
var exp10 = "a m1 b m2(c m3 d);";
var exp11 = "(a m1 b) m2 (c m3 d);";
var a,b,c,d;//四个数字
var m1,m2,m3;//三个运算符
for (var i=0;i<4;i++)
{
a = n[i];
for (var j=0;j<4;j++)
{
if ( i == j ) continue;//从未选的三个数字中选择一个数字
b = n[j];
for (var x=0;x<4;x++)
{
if ( x == j || x == i ) continue;//从未选的两个数字中选择一个数字
c = n[x];
for (var y=0;y<4;y++)
{
if ( y == x || y == j || y == i ) continue;//从未选的一个数字中选择一个数字
d = n[y];
for (var ta=0;ta<4;ta++)
{
m1 = m[ta];
for (var tb=0;tb<4;tb++)
{
m2 = m[tb];
for (var tc=0;tc<4;tc++)
{
m3 = m[tc];
for (var k=1;k<12;k++)
{
eval("test(exp"+k+",a,b,c,d,m1,m2,m3);");
}
全部回答
- 1楼网友:一把行者刀
- 2021-02-27 10:11
我暂时保留我的看法!
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