40 lines
2.3 KiB
JavaScript
40 lines
2.3 KiB
JavaScript
// ★ 第八刀的代价读数：【放宽到老板的判据之后，有多少人会被多问一句「你走哪条」】
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// 老板上一刀要的那个数逐字：「⚠ 而必须先说清代价：实测汉有 169,984 个 profile（32.4%）
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// 至少有一条输在 _idx 的路 —— 即三个汉朝用户里有一个会被多问一句。」⇒ 现在这个数变了。
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import { DYNASTIES, computeRank, computeAxes } from '../src/dynasties.js';
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import { SUB_REGISTRY } from '../src/talents.js';
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const SUBS = SUB_REGISTRY.map((s) => s.id);
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const WANGS = [0, 29, 30, 54, 55, 74, 75, 100];
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const TOT = (1 << 16) * WANGS.length;
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const P = {};
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for (const d of DYNASTIES) P[d.id] = { name: d.name, ask: 0, n2: 0, n3: 0, idxOnly: 0, hasLex: 0, noRoad: 0 };
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for (let m = 0; m < (1 << 16); m++) {
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const subs = []; for (let i = 0; i < 16; i++) if (m & (1 << i)) subs.push(SUBS[i]);
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const axes = computeAxes(subs);
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for (const wang of WANGS) for (const d of DYNASTIES) {
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const r = computeRank({ axes, wang }, d.id);
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const s = P[d.id];
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if (!r.channel) { s.noRoad++; continue; }
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const tie = r.zelu.tie || [];
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if (tie.length <= 1) continue;
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s.ask++;
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if (tie.length === 2) s.n2++; else s.n3++;
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const kinds = tie.filter((t) => !t.isDefault).map((t) => t.lostAt);
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if (kinds.every((k) => k === 'idx')) s.idxOnly++;
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if (kinds.includes('lex')) s.hasLex++;
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}
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}
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const N = (n) => n.toLocaleString();
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const pad = (s, n) => String(s).padEnd(n), num = (s, n) => String(s).padStart(n);
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console.log(`\n══ 【放宽后：有多少人会被多问一句「你走哪条」】每朝 ${N(TOT)} 个 profile ══\n`);
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console.log(pad('朝', 8) + num('要问', 10) + num('占比', 8) + num('2 选 1', 9) + num('≥3 选', 8) +
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num('★ 只含 idx', 12) + num('★ 含 lex', 10) + num('一条路都不开', 13));
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console.log('─'.repeat(80));
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let A = 0;
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for (const d of DYNASTIES) {
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const s = P[d.id]; A += s.ask;
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console.log(pad(s.name, 8) + num(N(s.ask), 10) + num((s.ask / TOT * 100).toFixed(1) + '%', 8) +
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num(N(s.n2), 9) + num(N(s.n3), 8) + num(N(s.idxOnly), 12) + num(N(s.hasLex), 10) + num(N(s.noRoad), 13));
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}
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console.log(`\n【总计】${N(TOT * DYNASTIES.length)} 格中 ${N(A)} 格（${(A / (TOT * DYNASTIES.length) * 100).toFixed(1)}%）会被多问一句。`);
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