SSL2694 August 15, 2017 raise group T1 string (math, combined number modulo)

August 15, 2017 raise group T1 string

Description

There are two string s,t of length n and consist only of lowercase letters, satisfying s and T happen to have K-bit difference. Q in all strings that happen to have K-bit different from S, T is sorted by dictionary order. Because the answer can be large, the die 10^9+7 output.

Input

The first row of two integers n,k.
The second line is a string s.
The third line is a string T.

Output

An integer line represents the answer.

Analysis: We start with the first bit of statistics:
The first bit of the first string must not be greater than the first bit of T
So if the first bit of S is smaller than the first bit of T, then the first contribution to the answer is
C (n-1,m-1) 25^ (m-1) (t[1]-' a '-1) +c (n-1,m) *25^m and then M –
If the first bit of S is larger than the first bit of T, then the first contribution to the answer is C (n-1,m-1) 25^ (m-1) (t[1]-' a ') and then M –
If the first bit of S equals the first bit of T, then the contribution to the answer is C (n-1,m-1) 25^ (m-1) (t[1]-' a ')
And then the next one to deal with it just fine.
As for the combined number modulus: a/b%c=a*b^ (c-2)%c;

Code

#include <cstdio> #include <iostream> #include <cstring> #include <string> #define MO 1000000007

#define MAXN 200000 using namespace std;
Long Long F[maxn],g[maxn],l[maxn],ans;
int n,k;

Char S[MAXN],T[MAXN];
    Long long power (int a,int b) {long long r=1,base=a;
        while (b) {if (b&1) r*=base;
        R%=mo;
        Base*=base;
        Base%=mo;
    b>>=1;
} return R;
    } long long C (int x,int y) {long long sum=f[x]*l[x-y]%mo*l[y]%mo;
return f[x]*l[x-y]%mo*l[y]%mo;
    } int main () {scanf ("%d%d", &n,&k);
    cin>>s+1;
    cin>>t+1;
    F[0]=1;g[0]=1;l[0]=1;
        for (int i=1;i<=n;i++) {f[i]=f[i-1]*i%mo;
        G[i]=g[i-1]*25%mo;
    L[i]=power (f[i],mo-2);
    } Ans=1;
            for (int i=1;i<=n;i++) {if (S[i]==t[i]) {ans+= (t[i]-' a ') *g[k-1]%mo*c (n-i,k-1)%mo;
        Ans%=mo; } if (S[i]>t[i]) {ans+=(t[i]-' a ') *g[k-1]%mo*c (n-i,k-1)%mo;
            Ans%=mo;
        k--;
            } if (S[i]<t[i]) {ans+= (t[i]-' a '-1) *g[k-1]%mo*c (n-i,k-1)%mo;
            Ans%=mo;
            if (n-i>=k) ans+=c (n-i,k)%mo*g[k]%mo;
            Ans%=mo;
        k--;
}} printf ("%lld", ans); }