【CF】207 Div.1 B.Xenia and Hamming

这题目一看很牛逼，其实非常easy。求求最小公倍数，最大公约数，均摊复杂度其实就是O(n)。

 /* 356B */

 #include <iostream>

 #include <string>

 #include <map>

 #include <queue>

 #include <set>

 #include <stack>

 #include <vector>

 #include <deque>

 #include <algorithm>

 #include <cstdio>

 #include <cmath>

 #include <ctime>

 #include <cstring>

 #include <climits>

 #include <cctype>

 #include <cassert>

 #include <functional>

 #include <iterator>

 #include <iomanip>

 using namespace std;

 //#pragma comment(linker,"/STACK:102400000,1024000")

 #define sti                set<int>

 #define stpii            set<pair<int, int> >

 #define mpii            map<int,int>

 #define vi                vector<int>

 #define pii                pair<int,int>

 #define vpii            vector<pair<int,int> >

 #define rep(i, a, n)     for (int i=a;i<n;++i)

 #define per(i, a, n)     for (int i=n-1;i>=a;--i)

 #define clr                clear

 #define pb                 push_back

 #define mp                 make_pair

 #define fir                first

 #define sec                second

 #define all(x)             (x).begin(),(x).end()

 #define SZ(x)             ((int)(x).size())

 #define lson            l, mid, rt<<1

 #define rson            mid+1, r, rt<<1|1

 const int maxn = 1e6+;

 char s[maxn], d[maxn];

 int cs[];

 int cd[];

 int main() {

     ios::sync_with_stdio(false);

     #ifndef ONLINE_JUDGE

         freopen("data.in", "r", stdin);

         freopen("data.out", "w", stdout);

     #endif

     __int64 n, m;

     scanf("%I64d %I64d", &n, &m);

     scanf("%s %s", s, d);

     int slen = strlen(s);

     int dlen = strlen(d);

     int g = __gcd(slen, dlen);

     __int64 lcm = 1LL * slen / g * dlen;

     __int64 n_ = lcm / slen;

     __int64 m_ = lcm / dlen;

     #ifndef ONLINE_JUDGE

         printf("n_ = %I64d, m_ = %I64d\n", n_, m_);

     #endif

     __int64 t = n / n_;

     __int64 tot;

     __int64 ans = ;

     int i, j, k;

     for (i=; i<g; ++i) {

         // count s

         memset(cs, , sizeof(cs));

         for (j=i,tot=; j<slen; j+=g,++tot) {

             ++cs[s[j]-'a'];

         }

         for (j=i; j<dlen; j+=g) {

             ans += (tot - cs[d[j]-'a']);

         }

     }

     ans *= t;

     printf("%I64d\n", ans);

     #ifndef ONLINE_JUDGE

         printf("time = %d.\n", (int)clock());

     #endif

     return ;

 }