ChinoWithClanguage

In C language, there are two functions for copying arrays, ${memcpy }$ and ${memmove}$ . These two C language functions have a same purpose is copying the contents of a certain section of memory to a section of memory in units of ${Bytes}$ .

The difference between these two functions is that ${memcpy}$ does not check whether the source address range and the destination address range in memory have overlapping parts. It follows the logic of copying each ${Byte}$ sequentially from left to right, so that can cause the data copied is not equal to original data.

When using the ${memmove}$ function, if the source address range and the destination address range have overlapping part, ${memmove}$ will do special processing to ensure that the memory of the destination address range is filled with original data.

Now suppose that both the ${memcpy}$ and ${memmove}$ function calls require three parameters, $p_1, p_2, l$ , which respectively represent the source address, destination address, and the number of bytes copied. For example, given a string ${"abcdefg"}$ , if you call ${memcopy(1,3,5)}$ , the result will be ${"abababa"}$ , if you call ${memmove(1,3,5)}$ , the result will be It is ${"ababcde"}$ .
Now give you a string ${S}$ , and three call parameters, $p_1, p_2, l$ . Please tell Chino what the calling results of ${memcopy}$ and ${memmove}$ respectively are.

The first line of input data contains only a positive integer $N(1 \leq N \leq 10^5)$ indicates the length of the string.
The second line of input data contain a string ${S}$ of length ${N}$ . ${S}$ includes only lowercase English letters.
The third line of input data contain three integers $p_1,p_2,l(1 \leq p_1,p_2,l \leq n)$
And guarantee $1 \leq p_1+l-1,p_2+l-1 \leq n$ represents the parameters of calling two functions .

n = int(input()) s = input() p1, p2, l = map(int, input().split()) ss = list(s) for i in range(l): ss[i + p2 - 1] = ss[i + p1 - 1] print(''.join(ss)) s = list(s) # print(s) # print(s[p2 - 1: p2 + l - 1]) # print(s[p1 - 1: p1 + l - 1]) s[p2 - 1: p2 + l - 1] = s[p1 - 1: p1 + l - 1] if len(s) > n: s = s[:n] print(''.join(s))

#include <iostream> using namespace std; int main () { int n,a,b,c,i; char s[100010],t[100010]; cin>>n>>s+1>>a>>b>>c; for(i=1;i<=n+1;i++) t[i]=s[i]; for(i=0;i<c;i++) t[b+i]=s[a+i]; for(i=0;i<c;i++) s[b+i]=s[a+i]; cout<<s+1<<"\n"<<t+1<<"\n"; return 0; }

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