Codeforces 357D Xenia and Hamming【数学+思维】

B. Xenia and Hamming

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Xenia is an amateur programmer. Today on the IT lesson she learned about the Hamming distance.

The Hamming distance between two strings s = s₁s₂...s_n and t = t₁t₂...t_n of equal length n is value . Record[s_i ≠ t_i] is the Iverson notation and represents the following: ifs_i ≠ t_i, it is one, otherwise — zero.

Now Xenia wants to calculate the Hamming distance between two long strings a and b. The first string a is the concatenation of n copies of stringx, that is, . The second stringb is the concatenation of m copies of string y.

Help Xenia, calculate the required Hamming distance, given n, x, m, y.

Input

The first line contains two integers n andm (1 ≤ n, m ≤ 10¹²). The second line contains a non-empty stringx. The third line contains a non-empty stringy. Both strings consist of at most 10⁶ lowercase English letters.

It is guaranteed that strings a and b that you obtain from the input have the same length.

Output

Print a single integer — the required Hamming distance.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use thecin, cout streams or the%I64d specifier.

Examples

Input

100 10aaaaaaaaaaa

Output

0

Input

1 1abacabaabzczzz

Output

4

Input

2 3rzraz

Output

5

Note

In the first test case string a is the same as stringb and equals 100 letters a. As both strings are equal, the Hamming distance between them is zero.

In the second test case strings a and b differ in their 3-rd, 5-th, 6-th and 7-th characters. Thus, the Hamming distance equals 4.

In the third test case string a is rzrrzr and string b is azazaz. The strings differ in all characters apart for the second one, the Hamming distance between them equals 5.

题目大意：

给你两个字符串，长度不超过1e6，再给你两个数，表示第一个串重复的次数以及第二个串重复的次数。

定义一种距离其计算方式为：

两个字符串相同位子上的不同字符的个数。

思路：

1、首先我们计算出总长度，然后再算出两个字符串长度的lcm，那么对应总长度/lcm的值就是对应循环的次数，记做cont；

2、然后我们设定g=gcd（字符串A的长度，字符串B的长度）；

接着我们通过观察和思考发现，对应我们将字符串A和字符串B都分若干个子段，每个子段长度为g，接下来我们统计两个字符串在一个lcm长度之内相同的元素的个数，记做output，

那么不同的元素的个数为lcm-output；

已知其一共重复的次数为cont，那么结果就是cont*（lcm-output）

Ac代码：

#include<stdio.h>#include<string.h>#include<iostream>using namespace std;#define ll __int64char a[5000000];char b[5000000];ll vis[1005000][26];ll gcd(ll x,ll y){    return y==0?x:gcd(y,x%y);}int main(){    ll n,m;    while(~scanf("%I64d%I64d",&n,&m))    {        scanf("%s%s",a,b);        ll lena=strlen(a);        ll lenb=strlen(b);        ll zonglen=n*lena;        ll g=gcd(lena,lenb);        ll lcm=lena*lenb/g;        ll cont=zonglen/lcm;        ll output=0;        for(int i=0;i<lena;i++)        {            vis[i%g][a[i]-'a']++;        }        for(int i=0;i<lenb;i++)        {            output+=vis[i%g][b[i]-'a'];        }        output=lcm-output;        output*=cont;        printf("%I64d\n",output);    }}