【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 ;
 }