4个数字的混合运算得固定结果的可能性

作者佚名技术来源服务器技术浏览发布时间 2012-07-11

点击浏览该文件大家给帮忙测试下 : ) //主要的运算代码如下:(这种东西做起来太麻烦了..) //复制到第一桢,也能通过输出窗口进行测试 function all_r(num1, num2, num3, num4) {
s = ["+", "-", "*", "/"];
n = [];
for (i=0; i<4; i++) {
temp1 = num1+s[i]+num2;
for (j=0; j<4; j++) {
temp2 = temp1+s[j]+num3;
for (k=0; k<4; k++) {
temp3 = temp2+s[k]+num4;
n.push(temp3);
}
}
}
}
function Result(num1, num2, num3, num4, r) {
kong = false;
all_r(num1, num2, num3, num4);
var result_arr = [];
var result;
var num = 0;
for (z=0; zfor (i=0; inum1_len = new String(num1).length;
num2_len = new String(num2).length;
num3_len = new String(num3).length;
m1 = n[z].substr(num1_len, 1);
m2 = n[z].substr(num1_len+num2_len+1, 1);
m3 = n[z].substr(num1_len+num2_len+num3_len+2, 1);
if (m1 == "+" and m2 == "+" and m3 == "+") {
result = Number(num1)+Number(num2)+Number(num3)+Number(num4);
}
if (m1 == "+" and m2 == "+" and m3 == "-") {
result = Number(num1)+Number(num2)+Number(num3)-num4;
}
if (m1 == "+" and m2 == "+" and m3 == "*") {
result = Number(num1)+Number(num2)+num3*num4;
}
if (m1 == "+" and m2 == "+" and m3 == "/") {
result = Number(num1)+Number(num2)+num3/num4;
}
//------------------------------
if (m1 == "+" and m2 == "-" and m3 == "+") {
result = Number(num1)+Number(num2)-num3+Number(num4);
}
if (m1 == "+" and m2 == "-" and m3 == "-") {
result = Number(num1)+Number(num2)-num3-num4;
}
if (m1 == "+" and m2 == "-" and m3 == "*") {
result = Number(num1)+Number(num2)-num3*num4;
}
if (m1 == "+" and m2 == "-" and m3 == "/") {
result = Number(num1)+Number(num2)-num3/num4;
}
//-----------------------------
if (m1 == "+" and m2 == "*" and m3 == "+") {
result = Number(num1)+num2*num3+Number(num4);
}
if (m1 == "+" and m2 == "*" and m3 == "-") {
result = Number(num1)+num2*num3-num4;
}
if (m1 == "+" and m2 == "*" and m3 == "*") {
result = Number(num1)+num2*num3*num4;
}
if (m1 == "+" and m2 == "*" and m3 == "/") {
result = Number(num1)+num2*num3/num4;
}
//------------------------
if (m1 == "+" and m2 == "/" and m3 == "+") {
result = Number(num1)+num2/num3+Number(num4);
}
if (m1 == "+" and m2 == "/" and m3 == "-") {
result = Number(num1)+num2/num3-num4;
}
if (m1 == "+" and m2 == "/" and m3 == "*") {
result = Number(num1)+num2/num3*num4;
}
if (m1 == "+" and m2 == "/" and m3 == "/") {
result = Number(num1)+num2/num3/num4;
}
//------------------------
if (m1 == "-" and m2 == "+" and m3 == "+") {
result = num1-num2+Number(num3)+Number(num4);
}
if (m1 == "-" and m2 == "+" and m3 == "-") {
result = num1-num2+Number(num3)-num4;
}
if (m1 == "-" and m2 == "+" and m3 == "*") {
result = num1-num2+num3*num4;
}
if (m1 == "-" and m2 == "+" and m3 == "/") {
result = num1-num2+num3/num4;
}
//----------------------------------
if (m1 == "-" and m2 == "-" and m3 == "+") {
result = num1-num2-num3+Number(num4);
}
if (m1 == "-" and m2 == "-" and m3 == "-") {
result = num1-num2-num3-num4;
}
if (m1 == "-" and m2 == "-" and m3 == "*") {
result = num1-num2-num3*num4;
}
if (m1 == "-" and m2 == "-" and m3 == "/") {
result = num1-num2-num3/num4;
}
//----------------------

凌众科技专业提供服务器租用、服务器托管、企业邮局、虚拟主机等服务，公司网站：http://www.lingzhong.cn 为了给广大客户了解更多的技术信息，本技术文章收集来源于网络,凌众科技尊重文章作者的版权，如果有涉及你的版权有必要删除你的文章，请和我们联系。以上信息与文章正文是不可分割的一部分,如果您要转载本文章,请保留以上信息，谢谢!

1 2 3 下一页

分享到：更多

你可能对下面的文章感兴趣

上一篇: 3D球棍模型下一篇: 一个遮罩效果

关于4个数字的混合运算得固定结果的可能性的所有评论

随机推荐