Codeforces Round #207 (Div. 1) B. Xenia and Hamming

http://codeforces.ru/contest/356/problem/B

Solution

设lena为a串长度，lenb为b串长度，gcd，lcm，为lena和lenb的gcd,lcm… 你会发现一个同余关系，首先，lcm长度一次循环，在一次lcm长度内，a的每个字符，与b的每个同余位置的字符触碰且只触碰一次，然后就好办了。把答案求出来，再与lcm搞一搞。。

My code

//Hello. I'm Peter.
#include<cstdio>
#include<iostream>
#include<sstream>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<cctype>
#include<ctime>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
#define peter cout<<"i am peter"<<endl
#define input freopen("data.txt","r",stdin)
#define randin srand((unsigned int)time(NULL))
#define INT (0x3f3f3f3f)*2
#define LL (0x3f3f3f3f3f3f3f3f)*2
#define gsize(a) (int)a.size()
#define len(a) (int)strlen(a)
#define slen(s) (int)s.length()
#define pb(a) push_back(a)
#define clr(a) memset(a,0,sizeof(a))
#define clr_minus1(a) memset(a,-1,sizeof(a))
#define clr_INT(a) memset(a,INT,sizeof(a))
#define clr_true(a) memset(a,true,sizeof(a))
#define clr_false(a) memset(a,false,sizeof(a))
#define clr_queue(q) while(!q.empty()) q.pop()
#define clr_stack(s) while(!s.empty()) s.pop()
#define rep(i, a, b) for (int i = a; i < b; i++)
#define dep(i, a, b) for (int i = a; i > b; i--)
#define repin(i, a, b) for (int i = a; i <= b; i++)
#define depin(i, a, b) for (int i = a; i >= b; i--)
#define pi 3.1415926535898
#define eps 1e-6
#define MOD 1000000007
#define MAXN 1000010
#define N
#define M
ll n,m,lena,lenb,d,l,sum;
char a[MAXN],b[MAXN];
int c[MAXN][26];
int main(){
    cin>>n>>m;
    scanf("%s%s",a,b);
    lena=len(a),lenb=len(b);
    d=__gcd(lena,lenb);
    l=lena/d*lenb;
    int t=0;
    rep(i,0,lena){
        c[t++][a[i]-'a']+=1;
        if(t==d) t=0;
    }
    t=0;
    sum=l;
    rep(i,0,lenb){
        sum-=c[t++][b[i]-'a'];
        if(t==d) t=0;
    }
    sum=(n*lena/l)*sum;
    cout<<sum<<endl;
    return 0;
}