[JavaScript] 纯文本查看 复制代码
function type__1475(N) {
var g = 'https://miniapp.qmai.cn/web/cmk-center/receive/takePartInReceive'
g += N;
console.log(g)
N = (((((sig(g) + '|') + 0) + '|') + new Date()["getTime"]()) + '|1'),g = ua(N, true)
return encodeURIComponent(g);
}
function sig(L) {
for (var N = 0, g = encodeURIComponent(L), B = 0; B < g['length']; B++)
N = (((N << 7) - N) + 398) + g['charCodeAt'](B), N |= 0;
return N;
};
function ua(E, H) {
var W = ['3', '4', '2', '1', '0'],
P = 0;
while (!![]) {
switch (W[P++]) {
case '0':
switch (M['length'] % 4) {
default:
case 0:
return M;
case 1:
return M + '===';
case 2:
return M + '==';
case 3:
return M + '=';
}
continue;
case '1':
if (H)
return M;
continue;
case '2':
var M = uu(E, 6, function (L) {
return V["uGGDj"]["charAt"](L);
});
continue;
case '3':
var K = {};
K["uGGDj"] = 'DGi0YA7BemWnQjCl4+bR3f8SKIF9tUz/xhr2oEOgPpac=61ZqwTudLkM5vHyNXsVJ';
var V = K;
continue;
case '4':
if (null == E)
return '';
continue;
}
break;
}
};
function uu(E, H, W) {
if (null == E)
return '';
for (var P, M, K, V, L = {}, N = {}, B = '', Q = 2, R = 3, I = 2, T = [], w = 0, k = 0, v = 0; v < E['length']; v += 1)
if (K = E['charAt'](v),
Object['prototype']['hasOwnProperty']['call'](L, K) || (L[K] = R++,
N[K] = true),
V = B + K,
Object['prototype']['hasOwnProperty']['call'](L, V))
B = V;
else {
if (Object['prototype']['hasOwnProperty']['call'](N, B)) {
if (B['charCodeAt'](0) < 256) {
for (P = 0; P < I; P++)
w <<= 1, k == H - 1 ? (k = 0,
T['push'](W(w)),
w = 0) : k++;
for (M = B['charCodeAt'](0), P = 0; P < 8; P++)
w = w << 1 | 1 & M,
k == H - 1 ? (k = 0, T['push'](W(w)), w = 0) : k++, M >>= 1;
} else {
for (M = 1, P = 0; P < I; P++)
w = w << 1 | M, k == H - 1 ? (k = 0, T['push'](W(w)), w = 0) : k++, M = 0;
for (M = B['charCodeAt'](0), P = 0; P < 16; P++)
w = w << 1 | 1 & M, k == H - 1 ? (k = 0, T['push'](W(w)), w = 0) : k++, M >>= 1;
}
0 == --Q && (Q = Math['pow'](2, I), I++), delete N[B];
} else {
for (M = L[B], P = 0; P < I; P++)
w = w << 1 | 1 & M,
k == H - 1 ? (k = 0, T['push'](W(w)), w = 0) : k++, M >>= 1;
}
0 == --Q && (Q = Math['pow'](2, I), I++),
L[V] = R++,
B = String(K);
}
if ('' !== B) {
if (Object['prototype']['hasOwnProperty']['call'](N, B)) {
if (B['charCodeAt'](0) < 256) {
for (P = 0; P < I; P++)
w <<= 1, k == H - 4097 ? (k = 0, T['push'](W(w)),
w = 0) : k++;
for (M = B['charCodeAt'](0), P = 0; P < 8; P++)
w = w << 1 | 1 & M,
k == H - 1 ? (k = 0,
T['push'](W(w)), w = 0) : k++, M >>= 1;
} else {
for (M = 1, P = 0; P < I; P++)
w = w << 1 | M, k == H - 1 ? (k = 0, T['push'](W(w)), w = 0) : k++, M = 0;
for (M = B['charCodeAt'](0), P = 0; P < 16; P++)
w = w << 1 | 1 & M,
k == H - 1 ? (k = 0, T['push'](W(w)), w = 0) : k++, M >>= 1;
}
0 == --Q && (Q = Math['pow'](2, I), I++), delete N[B];
} else {
for (M = L[B], P = 0; P < I; P++)
w = w << 1 | 1 & M,
k == H - 1 ? (k = 0, T['push'](W(w)), w = 0) : k++, M >>= 1;
}
0 == --Q && (Q = Math['pow'](2, I), I++);
}
for (M = 2, P = 0; P < I; P++)
w = w = w << 1 | 1 & M,
k == H - 1 ? (k = 0, T['push'](W(w)), w = 0) : k++, M >>= 1;
for (;;) {
if (w <<= 1, k == H - 1) {
T['push'](W(w));
break;
}
k++;
}
return T['join']('');
};
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