A--A Repeating Characters

A--A Repeating Characters

时间： 1ms        内存：128M

描述：

For this problem,you will write a program that takes a string of characters,S,and creates a new string of characters,T,with each character repeated R times.That is,R copies of the first character of S,followed by R copies of the second character of S,and so on.Valid characters for S are the QR
Code “alphanumeric” characters:
     0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ$%*+-. /:

输入：

The first line of input contains a single integer P,(1<=P<=1000),which is the number of data sets that follow. Each data set is single line of input consisting of the data set number N,followed by a space,followed by the repeat count R,(1<=R<=8),followed by a space ,followed by the string S.The length of string S will always be at least one and  no more than 20 characters.All the characters will be from the set of characters shown above.

输出：

For each data set there is one line of output. It contains the data set number,
N, followed by a single apace which is then followed by the new string T,which is made of each character in S repeated R times.

示例输入：

2
1 3 ABC
2 5 /HTP

示例输出：

1 AAABBBCCC
2 /////HHHHHTTTTTPPPPP

提示：

参考答案（内存最优[1032]）：

/*
 * A - Repeating Characters
 * Greater NY Regional, Oct 30, 2011
 */
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define	MAX_REPEAT	8
#define	MAX_CHARS	20

int main(int argc, char **argv)
{
	char szBuf[64], szStr[32], szNew[MAX_REPEAT * MAX_CHARS + 1];
	int n, np, ds, r, i, io, idx, nLen;

	if(::fgets(&(szBuf[0]), sizeof(szBuf)-1, stdin) != NULL){
		np = ::atoi(&(szBuf[0]));
		for(n = 1; n <= np; n++){
			if(::fgets(&(szBuf[0]), sizeof(szBuf)-1, stdin) == NULL){
				break;
			}
			if(::sscanf(&(szBuf[0]), "%d %d %s", &(ds), &(r), &(szStr[0])) != 3){
				break;
			}
			if(ds != n){
				::fprintf(stderr, "Problem number mismatch: %d != %d\n", n, ds);
				break;
			}
			nLen = ::strlen(&(szStr[0]));
			for(io = 0, idx = 0; idx < nLen; idx++){
				for(i = 0; i < r; i++){
					szNew[io++] = szStr[idx];
				}
			}
			szNew[io] = '\0';
			::fprintf(stdout, "%d %s\n", n, &(szNew[0]));
		}
	}
	return(0);
}

参考答案（时间最优[0]）：

/*
 * A - Repeating Characters
 * Greater NY Regional, Oct 30, 2011
 */
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define	MAX_REPEAT	8
#define	MAX_CHARS	20

int main(int argc, char **argv)
{
	char szBuf[64], szStr[32], szNew[MAX_REPEAT * MAX_CHARS + 1];
	int n, np, ds, r, i, io, idx, nLen;

	if(::fgets(&(szBuf[0]), sizeof(szBuf)-1, stdin) != NULL){
		np = ::atoi(&(szBuf[0]));
		for(n = 1; n <= np; n++){
			if(::fgets(&(szBuf[0]), sizeof(szBuf)-1, stdin) == NULL){
				break;
			}
			if(::sscanf(&(szBuf[0]), "%d %d %s", &(ds), &(r), &(szStr[0])) != 3){
				break;
			}
			if(ds != n){
				::fprintf(stderr, "Problem number mismatch: %d != %d\n", n, ds);
				break;
			}
			nLen = ::strlen(&(szStr[0]));
			for(io = 0, idx = 0; idx < nLen; idx++){
				for(i = 0; i < r; i++){
					szNew[io++] = szStr[idx];
				}
			}
			szNew[io] = '\0';
			::fprintf(stdout, "%d %s\n", n, &(szNew[0]));
		}
	}
	return(0);
}

题目和答案均来自于互联网，仅供参考，如有问题请联系管理员修改或删除。