解密算法： 使用 JavaScript 解密 Base64 编码的字符串

function(t, e) { for (var n = [], r = 0, i = "", o = "", a = 0; a < 256; a++) n[a] = a; for (var s = 0; s < 256; s++) r = (r + n[s] + e.charCodeAt(s % e.length)) % 256, i = n[s], n[s] = n[r], n[r] = i; var c = 0; r = 0; for (var d = 0; d < t.length; d++) c = (c + 1) % 256, r = (r + n[c]) % 256, i = n[c], n[c] = n[r], n[r] = i, o += String.fromCharCode(t.charCodeAt(d) ^ n[(n[c] + n[r]) % 256]); return o }

t=17630733570,e=7FED2719FC7E4D5602FB1D9D11AFA01B

运算一下

function(o) { var e, n, r, i, o, a, s = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/'; r = t.length, n = 0, e = ""; while (n < r) { if (i = 255 & t.charCodeAt(n++), n === r) { e += s.charAt(i >> 2), e += s.charAt((3 & i) << 4), e += "=="; break } if (o = t.charCodeAt(n++), n === r) { e += s.charAt(i >> 2), e += s.charAt((3 & i) << 4 | (240 & o) >> 4), e += s.charAt((15 & o) << 2), e += "="; break } a = t.charCodeAt(n++), e += s.charAt(i >> 2), e += s.charAt((3 & i) << 4 | (240 & o) >> 4), e += s.charAt((15 & o) << 2 | (192 & a) >> 6), e += s.charAt(63 & a) } return e

o为上面运算的结果，算出e

内容：e = 'RlRfJiB5JGBYRlEgQ3VleSB5b3UgdXNlIFhNTCBhcyB0aGF0IGFyZSBhdmFpbGFibGUuLi4='