Masala #1188
Massivlar soni
Uzunligi \(m\) ga teng butun sonlardan iborat \(a\) massividagi inversiyalar soni deb quyidagi shartlarni qanoatlantiruvchi \((i, j)\) juftliklar soniga aytiladi:
- \(1≤i<j≤m\)
- \(a_i >a_j\)
Siz \(1\) dan \(n\) gacha bo‘lgan butun sonlardan tashkil topgan \(n\) ta elementli (\(1\) dan \(n\) gacha bo‘lgan barcha butun sonlar bir martadan ishtirok etgan) inversiyalar soni \(k\) ga teng bo‘lgan massivlar sonini aniqlang. Bu son juda katta bo‘lishi mumkinligi sababli, uni \(10^9+7\) ga bo‘lgandagi qoldig‘ini toping.
Birinchi qatorda ikkita butun son - \(n(1 ≤ n ≤ 1000)\) va \(k(0 ≤ k ≤ 10000)\) kiritiladi.
Masala javobini \(10^9+7\) ga bo'lgandagi qoldig'ini chiqaring.
| # | input.txt | output.txt |
|---|---|---|
| 1 |
10 1 |
9 |
| 2 |
4 3 |
6 |
| 3 |
9 13 |
17957 |
1-testda mumkin bo'lgan holatlar
1 - (2, 1, 3, 4, 5, 6, 7, 8, 9, 10)
2 - (1, 3, 2, 4, 5, 6, 7, 8, 9, 10)
3 - (1, 2, 4, 3, 5, 6, 7, 8, 9, 10)
4 - (1, 2, 3, 5, 4, 6, 7, 8, 9, 10)
5 - (1, 2, 3, 4, 6, 5, 7, 8, 9, 10)
6 - (1, 2, 3, 4, 5, 7, 6, 8, 9, 10)
7 - (1, 2, 3, 4, 5, 6, 8, 7, 9, 10)
8 - (1, 2, 3, 4, 5, 6, 7, 9, 8, 10)
9 - (1, 2, 3, 4, 5, 6, 7, 8, 10, 9)
Yuqoridagi barcha holatlarda inversiyalar soni 1 ga teng. Osongina ko'rish mumkin-ki boshqa holat mavjud emas.
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