在获取测试用例输入后，它不会等待字符串输入。我试过fgets（）;但是，fgets（）需要用户的换行符，我认为不会被在线评委接受。那么，这里的问题是什么？这是我的代码。
#include<stdio.h>
int main ()
{
int t, j;
int cnt = 0;
scanf("%d", &t);
while(t--)
{
cnt++;
int i, count = 0;
char s[101];
scanf("%[^\n]s", s);
for(i=0;s[i] != '\0'; i++)
{
if(s[i] == 'a' || s[i] == 'd' || s[i] == 'g' || s[i] == 'j' || s[i] == 'm' || s[i] == 'p' || s[i] == 't' || s[i] == 'w' || s[i] == ' ')
{
count += 1;
continue;
}
else if(s[i] == 'b' || s[i] == 'e' || s[i] == 'h' || s[i] == 'k' || s[i] == 'n' || s[i] == 'q' || s[i] == 'u' || s[i] == 'x')
{
count += 2;
continue;
}
else if(s[i] == 'c' || s[i] == 'f' || s[i] == 'i' || s[i] == 'l' || s[i] == 'o' || s[i] == 'r' || s[i] == 'v' || s[i] == 'y')
{
count += 3;
continue;
}
else if(s[i] == 's' || s[i] == 'z')
{
count += 4;
continue;
}
}
printf("Case #%d: %d\n", cnt, count);
}
return 0;
}
答案 0 :(得分：4)
更改
scanf("%[^\n]s", s);
到
scanf(" %100[^\n]", s);
100确保它不会尝试写过字符串的结尾。s，因为这会尝试匹配输入中的文字s。 [^\n]不是%s格式运算符的修饰符，它是一个完全不同的格式运算符。答案 1 :(得分：1)
如果您包含指向特定问题的链接，那将会非常有用。然后我们可以建议您是否正确回答了问题。
例如，大写字母，标点符号等等？
例如，问题是否保证每个输入字符串将<= 100个字符，或者是否存在不超过前100个字符的条件？
如果我们不知道问题是什么，我们无法确定发布的代码是否正在回答问题。
大多数I / O的时间/ CPU周期非常昂贵，尤其是scanf()和printf()
大多数在线问题都有执行时间，超过允许的时间限制将导致建议的答案（您提供的）失败。
适当的快速＆＃39;通过查看先前问题的答案，您可以轻松找到I / O功能。以下是一些功能齐全的示例，可帮助您入门：
#include <stdio.h>
void fastRead( size_t *a );
void fastWrite( size_t a );
inline void fastRead(size_t *a)
{
int c=0;
// note: 32 is space character
while (c<33) c=getchar_unlocked();
// initialize result value
*a=0;
// punctuation parens, etc are show stoppers
while (c>47 && c<58)
{
*a = (*a)*10 + (size_t)(c-48);
c=getchar_unlocked();
}
//printf( "%s, value: %lu\n", __func__, *a );
} // end function: fastRead
inline void fastWrite(size_t a)
{
char snum[20];
//printf( "%s, %lu\n", __func__, a );
int i=0;
do
{
// 48 is numeric character 0
snum[i++] = (char)((a%10)+(size_t)48);
a=a/10;
}while(a>0);
i=i-1; // correction for overincrement from prior 'while' loop
while(i>=0)
{
putchar_unlocked(snum[i--]);
}
putchar_unlocked('\n');
} // end function: fastWrite
大多数情况下（您可能需要进行一些实验）在线代码判断在每个测试用例（包括最后一个测试用例）之后需要将换行符输出到stdout。这通常通过以下方式完成：
putchar_unlocked('\n');
当值永远不会小于0时，（通常）最好使用size_t而不是int，
使用指示content或usage（或更好，两者）的变量名称是一种很好的编程习惯。
代码更具可读性和可理解性
将上述所有内容放在一起会导致：
#include <stdio.h> // putchar_unlocked(), getchar_unlocked()
void fastRead( size_t *a );
void fastWrite( size_t a );
inline void fastRead(size_t *a)
{
int c=0;
// note: 32 is space character
while (c<33) c=getchar_unlocked();
// initialize result value
*a=0;
// punctuation parens, etc are show stoppers
while (c>47 && c<58)
{
*a = (*a)*10 + (size_t)(c-48);
c=getchar_unlocked();
}
//printf( "%s, value: %lu\n", __func__, *a );
} // end function: fastRead
inline void fastWrite(size_t a)
{
char snum[20];
//printf( "%s, %lu\n", __func__, a );
int i=0;
do
{
// 48 is numeric character 0
snum[i++] = (char)((a%10)+(size_t)48);
a=a/10;
}while(a>0);
i=i-1; // correction for overincrement from prior 'while' loop
while(i>=0)
{
putchar_unlocked(snum[i--]);
}
putchar_unlocked('\n');
} // end function: fastWrite
int main ( void )
{
size_t numTestCases;
fastRead( &numTestCases );
for( size_t testCase =1; testCase <= numTestCases; testCase++)
{
size_t count = 0;
int ch;
while( (ch = getchar_unlocked()) != '\n' )
{
if( ch == 'a'
|| ch == 'd'
|| ch == 'g'
|| ch == 'j'
|| ch == 'm'
|| ch == 'p'
|| ch == 't'
|| ch == 'w'
|| ch == ' ')
{
count += 1;
}
else if( ch == 'b'
|| ch == 'e'
|| ch == 'h'
|| ch == 'k'
|| ch == 'n'
|| ch == 'q'
|| ch == 'u'
|| ch == 'x')
{
count += 2;
}
else if( ch == 'c'
|| ch == 'f'
|| ch == 'i'
|| ch == 'l'
|| ch == 'o'
|| ch == 'r'
|| ch == 'v'
|| ch == 'y')
{
count += 3;
}
else if( ch == 's'
|| ch == 'z')
{
count += 4;
}
}
printf("Case #%lu: %lu\n", testCase, count);
putchar_unlocked( '\n' );
}
return 0;
}
使用对printf()和fastWrite()
putchar_unlocked()的通话速度
此外，除非问题特别要求在printf()的调用中使用多余的字符，否则代码可能会说：
fastWrite( testCase );
putchar_unlocked( ' ' );
fastWrite( count );
putchar_unlocked( '\n' );
还非常希望在代码顶部列出问题的参数。这既可以帮助您和其他任何人阅读您的代码。
以下是此类文档的一个简单示例：
/*
* criteria 1
* Using two characters: . (dot) and * (asterisk)
* print a frame-like pattern
*
* criteria 2
* You are given t - the number of test cases
*
* criteria 3
* for each of the test cases input two positive integers:
* l - the number of lines and
* c - the number of columns of a frame.
*
* criteria 4
* Use one line break in between successive patterns
*/
答案 2 :(得分：0)
鉴于您提供的最新信息，以下是我建议的答案。
如果仍未通过判决，至少你有一个非常坚实的基础，可以通过实验来确定正确的答案。
/*
* Cell phones have become an essential part of modern life.
* In addition to making voice calls,
* cell phones can be used to send text messages,
* which are known as SMS for short.
* Unlike computer keyboards, most cell phones have limited number of keys.
* To accommodate all alphabets, letters are compacted into single key.
* Therefore, to type certain characters,
* a key must be repeatedly pressed until that character is shown on the display
panel.
*
* In this problem we are interested in finding out the number of times
* keys on a cell phone must be pressed to type a particular message.
* In this problem we will assume that the key pad of our cell phone is arranged as follows.
* ---------------------
* |1 | abc | def |
* ---------------------
* | ghi | jkl | mno |
* ---------------------
* | pqrs | tuv | wxyz |
* ---------------------
* | | <SP>| |
* ---------------------
* In the above grid each cell represents one key.
* Here <SP> means a space.
* In order to type the letter `a', we must press that key once,
* however to type `b' the same key must be pressed twice
* and for `c' three times.
* In the same manner, one key press for `d',
* two for `e'
* and three for `f'.
* This is also applicable for the remaining keys and letters.
* Note that it takes a single press to type a space.
*
* Input
* The first line of input will be a positive integer T
* where T denotes the number of test cases.
* T lines will then follow each containing only spaces and lower case letters.
* Each line will contain at least 1 and at most 100 characters.
*
* Output
* For every case of input there will be one line of output.
* It will first contain the case number
* followed by the number of key presses required to type the message of that case.
* Look at the sample output for exact formatting.
*
* Sample Input
* 2
* welcome to ulab
* good luck and have fun
*
* Sample Output
*
* Case #1: 29
* Case #2: 41
*/
#include <stdio.h> // putchar_unlocked(), getchar_unlocked(), puts(), sprintf()
#include <string.h> // strcpy(), strcat()
#define MAX_OUTPUT_LEN 50
void fastRead( size_t *a );
void fastWrite( size_t a );
inline void fastRead(size_t *a)
{
int c=0;
// note: 32 is space character
while (c<33) c=getchar_unlocked();
// initialize result value
*a=0;
// punctuation parens, etc are show stoppers
while (c>47 && c<58)
{
*a = (*a)*10 + (size_t)(c-48);
c=getchar_unlocked();
}
//printf( "%s, value: %lu\n", __func__, *a );
} // end function: fastRead
inline void fastWrite(size_t a)
{
char snum[20];
//printf( "%s, %lu\n", __func__, a );
int i=0;
do
{
// 48 is numeric character 0
snum[i++] = (char)((a%10)+(size_t)48);
a=a/10;
}while(a>0);
i=i-1; // correction for overincrement from prior 'while' loop
while(i>=0)
{
putchar_unlocked(snum[i--]);
}
putchar_unlocked('\n');
} // end function: fastWrite
const size_t keyPressCounts[] =
{
1, 2, 3, // a b c
1, 2, 3, // d e f
1, 2, 3, // g h i
1, 2, 3, // j k l
1, 2, 3, // m n o
1, 2, 3, 4,// p q r s
1, 2, 3, // t u v
1, 2, 3, 4 // w x y z
};
int main ( void )
{
size_t numTestCases;
fastRead( &numTestCases );
for( size_t testCase =1; testCase <= numTestCases; testCase++)
{
size_t keypresses = 0;
int ch;
while( (ch = getchar_unlocked()) != '\n' )
{
if( ' ' == ch )
keypresses += 1;
else
keypresses += keyPressCounts[ ch - 'a' ];
} // end while()
char outputLine[ MAX_OUTPUT_LEN ];
strcpy( outputLine, "Case #" );
sprintf( &outputLine[ strlen(outputLine) ], "%lu", testCase );
strcat( outputLine, ": " );
sprintf( &outputLine[ strlen(outputLine) ], "%lu", keypresses );
puts( outputLine );
} // end for()
return 0;
} // end function: main