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By Jack_sparrow_06, history, 3 months ago, In English

For the problem 1827A - Counting Orders, when in the question they mentioned that two reorderings are different, if the resultant array is different. When I check the submitted code with the below testcase, I got answer 4. But when I dry run for the testcase in the note, I found the resultant array are same. Hence the unique ways of reordering must be 2. Anyone plz explain and don't downvote for the question. Thanks in advance.

CODE

                int n;
		cin >> n;
		vector<int> a(n), b(n);
		for(int i = 0; i < n; i++) cin >> a[i];
		for(int i = 0; i < n; i++) cin >> b[i];
		sort(a.begin(), a.end());
		sort(b.begin(), b.end());
		ll ans = 1ll;
		bool flag = false;
		if(a[n - 1] <= b[n - 1]) flag = true;
		else {
			int i = 0, j = 0, cnt = 0;
			while(i < n) {
				int val = a[i];
				while(j < n && val > b[j]) {
					j++;
				}
				ll res = (j - i);
				ans *= (res % mod);
				ans %= mod;
				i++;
			}
		}

		if(flag) cout << 0 << endl;
		else cout << ans << endl;
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3 months ago, # |
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Auto comment: topic has been updated by Jack_sparrow_06 (previous revision, new revision, compare).

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3 months ago, # |
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Bro ,you have to reorder array A not array B ... then you can reorder B as follows : 2 4 5 2 5 4 4 2 5 5 2 4