C++11(clang++ 3.9) 解法, 执行用时: 3ms, 内存消耗: 504K, 提交时间: 2020-08-04 16:33:15

#include <cstdio>
#define mod 1000000007

int n,m,k,a,b;
int ksm(int x,int y){int re=1;for(;(y&1?re=1ll*re*x%mod:0),y;x=1ll*x*x%mod,y>>=1);return re;}
#define inv(x) ksm(x,mod-2)

int main()
{
	scanf("%d %d %d %d %d",&n,&m,&k,&a,&b);
	int p=1ll*a*inv(b)%mod,T=1ll*(n+m)*inv(n+m+1)%mod;
	printf("%d",(1ll*ksm(T,k)*n%mod+1ll*p*T%mod*(ksm(T,k)-1+mod)%mod*inv(T-1)%mod)%mod);
}

pypy3(pypy3.6.1) 解法, 执行用时: 46ms, 内存消耗: 31084K, 提交时间: 2020-08-03 20:31:25

MOD = 1000000007
def fpow(a, b, p=1000000007):
  return pow(a, b, p)

n, m, k, a, b = map(int, input().split())
p = a * fpow(b, MOD - 2) % MOD
s = (n + m) * fpow(n + m + 1, MOD - 2) % MOD
ans = n * fpow(s, k) + p * (s - fpow(s, k + 1) + MOD * 100) * fpow(1 - s, MOD - 2)
ans %= MOD
print(ans)

Python3(3.5.2) 解法, 执行用时: 26ms, 内存消耗: 3552K, 提交时间: 2020-08-03 20:32:50

n, m, k, a, b = map(int, input().split())
mod = 10**9+7
m0 = pow(n+m, k, mod)
m1 = pow(n+m+1, k, mod)
left = b*m1
right = b*n*m0+a*(n+m)*(m1-m0)
ans = pow(left, mod-2, mod)*right
print(ans % mod)