【CF908D】New Year and Arbitrary Arrangement

Problem

Description

Given three numbers \ (K, PA, Pb \) , each with a \ (\ frac {pa} { pa + pb} \) is added to the back of a probability awith \ (\ frac {pb} { pa + pb} \) is added to the back of a probability that b, when there is a \ (K \) th shaped as aba sub-sequence (not continuous) stop.

Seeking the last sub-sequence abnumber expected.

The answer to \ (9 + 10 ^ 7 \) modulo.

Sample

Input 1

1 1 1

Output 1

2

Input 2

3 1 4

Output 2

370000006

Range

\(k\le1000,p_a,p_b\le10^6\)

Algorithm

\ (DP \) , the probability and expectation

Mentality

Set \ (f_ {i, j} \) represents the currently \ (I \) a \ (A \) , \ (J \) subsequences \ (ab & \) , the sub-sequence at the end of the entire sequence \ (ab & \) of the desired number. The first dimension may be found infinite, considered backwards.

\ (f_ {i, j} \) transfer, there are two cases, one is added at the end \ (A \) , proceeds to \ (. 1 + F_ {I, J} \) , but joined \ (B \ ) was transferred to a \ (F_ {I, I + J} \) . Then push down the very obvious equation:
\ [F_ {I, J} = \ {FRAC P_A} + {P_A P_B F_ {I}. 1 +, + J} \ {FRAC P_A P_B} {+} F_ {P_B i, j + 1} \]
but the first infinite dimension of the problem is still not resolved border problems must be considered.

We found that when \ (i + j \ ge k \) when, if we add \ (b \) , the entire string will be terminated immediately.

Then for a state \ (f_ {i, j} , (i + j \ ge k) \) , a state is provided on this continuous addition \ (X \) a \ (A \) was added a \ (B \) , then:
\ [F_ {I, J} = \ FRAC {P_B} {P_A + P_B} \ sum_ {X = 0} ^ \ infty (I + J + X) (\ FRAC {P_A} {P_A + p_b}) ^ x \]
this is a geometric sequence, then we direct geometric series summation enough.

Calculated to obtain:
\ [F_ {I, J} = I + J + \ {FRAC P_A P_B} {} \]
credited directly to search on a star.

Code

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <vector>
using namespace std;
long long read() {
  long long x = 0, w = 1;
  char ch = getchar();
  while (!isdigit(ch)) w = ch == '-' ? -1 : 1, ch = getchar();
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * w;
}
const int Max_n = 1e3 + 5, mod = 1e9 + 7;
int n, a, b;
int pa, pb, pp;
int f[Max_n][Max_n];
int ksm(int a, int b) {
  int res = 1;
  for (; b; b >>= 1, a = 1ll * a * a % mod)
    if (b & 1) res = 1ll * res * a % mod;
  return res;
}
int DP(int a, int ab) {
  int &res = f[a][ab];
  if (res != -1) return res;
  if (a + ab >= n) {
    res = (a + ab + pp) % mod;
    return res;
  }
  return res =
             (1ll * pa * DP(a + 1, ab) % mod + 1ll * pb * DP(a, ab + a) % mod) %
             mod;
}
int main() {
#ifndef ONLINE_JUDGE
  freopen("D.in", "r", stdin);
  freopen("D.out", "w", stdout);
#endif
  n = read(), a = read(), b = read();
  pa = 1ll * a * ksm(a + b, mod - 2) % mod;
  pb = 1ll * b * ksm(a + b, mod - 2) % mod;
  pp = 1ll * a * ksm(b, mod - 2) % mod;
  memset(f, -1, sizeof(f));
  printf("%d\n", DP(1, 0));
}