(3)IBM Minus One

最新推荐文章于 2024-03-31 10:09:12 发布

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Input

The input starts with the integer n on a line by itself - this is the number of strings to follow. The following n lines each contain one string of at most 50 upper-case letters.

Output

For each string in the input, first output the number of the string, as shown in the sample output. The print the string start is derived from the input string by replacing every time by the following letter in the alphabet, and replacing 'Z' by 'A'.

Print a blank line after each test case.

Sample Input

2
HAL
SWERC

Sample Output

String #1
IBM

String #2
TXFSD

#include <stdio.h>
#include <string.h>
int main(){
	char s[51];
	int z, zz, len, i;
	scanf("%d", &zz);
	for (z=1; z<=zz; z++){
		scanf("%s", s);
		printf("String #%d\n", z);
		len = strlen(s);
		for (i=0; i<len; i++){
			printf("%c", s[i]!='Z' ? s[i]+1 : 'A');
		}
		printf("\n\n");
	}
	return 0;
}

/**************************c语言************************/

初做题时，一直提示编译错误，可能我哪多个空行回车一直报错后来也没找到今天索性贴出c（别人） c++代码。警示自己。

#include <iostream>
#include <string>
using namespace std;
int main()
{

	long num,count;
	string str;
	cin>>num;
	for(int i=1;i<=num;i++)
	{
        cin>>str;
		//getline(cin,str);
		for(count=0;count<str.length();count++)
		{
			if(str[count]=='Z')
				str[count]='A';
			else
				str[count]=str[count]+1;
		}
		cout<<"String"<<'\t'<<"#"<<i<<'\n'<<str;
		//if(i<num)
		    cout<<'\n'<<'\n';
		//else	cout<<'\n';	
	   //str="0";
	}
	return 0;
}

c++**********************************************************************/