2023 rcpc

2023/rcpc/d2/editorial/i-ksumt-586.cpp

@@ -0,0 +1,81 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+
+#define int long long
+
+const int modulo = 1e9 + 7;
+
+int k,s,t;
+int fact[5000005],invfact[5000005];
+
+int C(int x,int y)
+{
+    if (x < y)
+        return 0;
+    if (y < 0)
+        return 0;
+    return fact[x] * invfact[y] % modulo * invfact[x - y] % modulo;
+}
+
+int f(int x,int y)
+{
+    if (x < y)
+        return 0;
+    if (x == 0 and y == 0)
+        return 1;
+    if (y == 0)
+        return 0;
+    ///in cate moduri pot distribui x in y numere pozitive
+    ///in cate moduri pot pune y - 1 bare in x - 1 pozitii cu oricare doua bare diferite
+    ///C(x - 1,y - 1)
+    return C(x - 1,y - 1);
+}
+
+int lgpow(int x,int y)
+{
+    int z = 1;
+    while (y != 0)
+    {
+        if (y % 2 == 1)
+            z = z * x % modulo;
+        x = x * x % modulo;
+        y /= 2;
+    }
+    return z;
+}
+
+void prec()
+{
+    fact[0] = invfact[0] = 1;
+    for (int i = 1; i <= 5e6; i++)
+        fact[i] = i * fact[i - 1] % modulo;
+    invfact[(int)5e6] = lgpow(fact[(int)5e6],modulo - 2);
+    for (int i = 5e6 - 1; i >= 1; i--)
+        invfact[i] = invfact[i + 1] * (i + 1) % modulo;
+}
+
+signed main()
+{
+    prec();
+    cin >> k >> s >> t;
+    int d = k / t,r = k % t;
+    int ans = 0;
+    for (int x = 0; x <= s / (d + 1); x++)
+    {
+        if ((s - (d + 1) * x) % d == 0)
+            ans = (ans + f(x,r) * f((s - (d + 1) * x) / d,t - r)) % modulo;
+    }
+    cout << ans;
+    return 0;
+}
+
+/**
+int d = k / t,r = k % t
+(d + 1)(a[1] + a[2] + ... + a[r]) + d(a[r + 1] + ... + a[k])
+fie x = a[1] + ... + a[r]
+pentru fiecare x posibil, avem:
+0 daca (s - (d + 1)x) % d != 0
+f(x,r) * f((s - (d + 1)x) / d,k - r) altfel
+si avem suma din asta
+**/