1037C_ Equalize(字符串)

modify 改变

C. Equalize

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given two binary strings $\mathrm{aa and bb of\; the\; same\; length.\; You\; can\; perform\; the\; following\; two\; operations\; on\; the\; string aa:}$

• Swap any two bits at indices $\mathrm{ii and jj respectively\; (1\le i,j\le n1\le i,j\le n),\; the\; cost\; of\; this\; operation\; is |i-j||i-j|,\; that\; is,\; the\; absolute\; difference\; between ii and jj.}$
• Select any arbitrary index $\mathrm{ii (1\le i\le n1\le i\le n)\; and\; flip\; (change 00 to 11 or 11 to 00)\; the\; bit\; at\; this\; index.\; The\; cost\; of\; this\; operation\; is 11.}$

Find the minimum cost to make the string $\mathrm{aa equal\; to bb.\; It\; is\; not\; allowed\; to\; modify\; string bb.}$

Input

The first line contains a single integer $\mathrm{nn (1\le n\le 1061\le n\le 106) —\; the\; length\; of\; the\; strings aa and bb.}$

The second and third lines contain strings $\mathrm{aa and bb respectively.}$

Both strings $\mathrm{aa and bb have\; length nn and\; contain\; only\; \text{'}0\text{'}\; and\; \text{'}1\text{'}.}$

Output

Output the minimum cost to make the string $\mathrm{aa equal\; to bb.}$

Examples

input

Copy

3
100
001

output

Copy

2

input

Copy

4
0101
0011

output

Copy

1

Note

In the first example, one of the optimal solutions is to flip index $\mathrm{11 and\; index 33,\; the\; string aa changes\; in\; the\; following\; way:\; "100" \to \to "000" \to \to "001".\; The\; cost\; is 1+1=21+1=2.}$

The other optimal solution is to swap bits and indices $\mathrm{11 and 33,\; the\; string aa changes\; then\; "100" \to \to "001",\; the\; cost\; is\; also |1-3|=2|1-3|=2.}$

In the second example, the optimal solution is to swap bits at indices $\mathrm{22 and 33,\; the\; string aa changes\; as\; "0101" \to \to "0011".\; The\; cost\; is |2-3|=1|2-3|=1.}$

 1 #include
 2 #include
 3 #include
 4 using namespace std;
 5 int main()
 6 {
 7     int n;
 8     char a[1000006],b[1000005];
 9     while(~scanf("%d",&n))
10     {
11         scanf("%s",a+1);
12         scanf("%s",b+1);
13         int ans=0;
14         for(int i=1;i<=n;i++)
15         {
16             if(a[i]!=b[i])
17             {
18                 if(a[i+1]!=b[i+1]&&a[i+1]!=a[i])
19                 a[i+1]=b[i+1];
20                 ans++;
21             }
22         }
23         printf("%d\n",ans);
24     }
25     return 0;
26 }

 

 

转载于:https://www.cnblogs.com/LLLAIH/p/9723883.html