(公约数问题) Xenia and Hamming (356B)
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题意:给出两个字符串(等长),求出其相同位置字符不同的个数。
其中:字符串是一个重复的字符串,先给出一个比较短的,然后再重复N次。
解法:自己体会吧
#include<iostream>#include<algorithm>#include<cstring>using namespace std;#define N 1000010int sum[N][26]={0};char a[N],b[N];int gcd(int a,int b){if (a>b)swap(a,b);return b%a==0?a:gcd(b%a,a);}int main( ){ int i,j,k;long long n,m;int lena,lenb;cin>>n>>m>>a>>b;lena=strlen(a);lenb=strlen(b);int G=gcd(lena,lenb);long long L = 1LL*lena*lenb/G;long long ans=0;for (i=0;i<lena;i++)sum[i%G][a[i]-'a']++;for (i=0;i<lenb;i++)ans+=sum[i%G][b[i]-'a'];cout<<(1LL*n*lena)/L*(L-ans)<<endl; return 0;}
Xenia is an amateur programmer. Today on the IT lesson she learned about the Hamming distance.
The Hamming distance between two strings s = s1s2... sn and t = t1t2... tn of equal length n is value . Record [si ≠ ti] is the Iverson notation and represents the following: if si ≠ ti, 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 string x, that is, . The second string b is the concatenation of m copies of string y.
Help Xenia, calculate the required Hamming distance, given n, x, m, y.
The first line contains two integers n and m (1 ≤ n, m ≤ 1012). The second line contains a non-empty string x. The third line contains a non-empty string y. 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.
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 the cin, cout streams or the %I64dspecifier.
100 10aaaaaaaaaaa
0
1 1abacabaabzczzz
4
2 3rzraz
5
In the first test case string a is the same as string b 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.
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