CodeForces - 937D Sleepy Game （乘法逆元）

Fafa and Ancient Alphabettime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputAncient Egyptians are known to have used a large set of symbols 

 to write on the walls of the temples. Fafa and Fifa went to one of the temples and found two non-empty words S1 and S2 of equal lengths on the wall of temple written one below the other. Since this temple is very ancient, some symbols from the words were erased. The symbols in the set 

 have equal probability for being in the position of any erased symbol.Fifa challenged Fafa to calculate the probability that S1 is lexicographically greater than S2. Can you help Fafa with this task?You know that 

, i. e. there were m distinct characters in Egyptians' alphabet, in this problem these characters are denoted by integers from 1 to m in alphabet order. A word xis lexicographically greater than a word y of the same length, if the words are same up to some position, and then the word x has a larger character, than the word y.We can prove that the probability equals to some fraction 

, where P and Q are coprime integers, and 

. Print as the answer the value 

, i. e. such a non-negative integer less than 109 + 7, such that 

, where 

 means that a and b give the same remainders when divided by m.InputThe first line contains two integers n and m (1 ≤ n,  m ≤ 105) — the length of each of the two words and the size of the alphabet 

, respectively.The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ m) — the symbols of S1. If ai = 0, then the symbol at position i was erased.The third line contains n integers representing S2 with the same format as S1.OutputPrint the value 

, where P and Q are coprime and 

 is the answer to the problem.ExamplesinputCopy

1 2
0
1

output

500000004

inputCopy

1 2
1
0

output

0

inputCopy

7 26
0 15 12 9 13 0 14
11 1 0 13 15 12 0

output

230769233

NoteIn the first sample, the first word can be converted into (1) or (2). The second option is the only one that will make it lexicographically larger than the second word. So, the answer to the problem will be 

, that is 500000004, because 

.In the second example, there is no replacement for the zero in the second word that will make the first one lexicographically larger. So, the answer to the problem is 

, that is 0.
题意：求S1>S2的概率，字符串某些位置可能为空，空的话，里面的字符可以是任意的。

解题思路：先把公式推出来，公式很好推。分三种情况讨论，a[i]==b[i]，a[i]==0&&b[i]!=0，a[i]!=0&&b[i]==0 三种情况分别计算概率，然后相加即可！中间所有的除法，用乘法逆元代替！

#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
const int MOD=1e9+7;

ll qpow(ll a,ll b,ll c){
ll ans=1;
ll temp=a;
while(b!=0){
if(b%2)ans=ans*temp%c;
temp=temp*temp%c;
b/=2;
}
return ans;
}

ll a[100005];
ll b[100005];

int main()
{

ll N,M;
cin>>N>>M;
for(int i=0;i<N;i++)
cin>>a[i];
for(int i=0;i<N;i++)
cin>>b[i];

ll one=qpow(M,MOD-2,MOD);
ll two=qpow(M*M%MOD,MOD-2,MOD);
ll ans=0;
ll cur=1;

for(int i=0;i<N;i++){
if(a[i]==0&&b[i]==0){
ans=(ans+cur*((M*M-M)/2)%MOD*two%MOD)%MOD;
cur=cur*one%MOD;
}
else{
if(a[i]==0&&b[i]!=0){
ans=(ans+cur*(M-b[i])%MOD*one%MOD)%MOD;
cur=cur*one%MOD;
}
else{
if(a[i]!=0&&b[i]==0){
ans=(ans+cur*(a[i]-1)%MOD*one%MOD)%MOD;
cur=cur*one%MOD;
}
else{
if(a[i]>b[i]){
ans=(ans+cur)%MOD;
break;
}
else{
if(a[i]==b[i])
continue;
else
break;
}
}
}
}
}
cout<<ans<<endl;

return 0;
}