【问题描述】
从文件中输入N（1≤N≤10000）个由小写英文字母组成的字符串，每个字符串的长度不超过200，统计它们出现的次数。
【输入】
第一行：N
从第二行到第N+1行，每行一个字符串。
【输出】
按输入顺序输出每个字符串及出现的次数。每行包括一个字符串和这个字符串出现的次数，它们之间用一个空格隔开

【输入样例】
10
abcde
abc
aabbc
abcde
abcde
abc
abc
abcde
aabbc
abc


【输出样例】
abcde 4
abc 4
aabbc 2

[ 本帖最后由 BlueGuy 于 2009-8-24 22:00 编辑 ]

就查 BlueGuy 是否在数组中

回复 3楼 zqy110007
是 strcmp

回复 7楼 wxjeacen
你写个输入就行了
像这样
for (i = 0; i < 1000; i++)
    gets(a[i]);

Perhaps the most obvious difference is that DirectX, as opposed to OpenGL, is more than just a graphics API. DirectX contains tools to deal with such components of a game as sound, music, input, networking, and multimedia. On the other hand, OpenGL is strictly a graphics API. So what aspect of OpenGL sets it apart from the DirectX graphics component?

Well, first things first: both APIs rely on the use of the traditional graphics pipeline. This is the same pipeline that has been used in computer games since the early days of computer graphics. Although it has been modified in slight ways to adapt with advancements in hardware, the basic idea remains intact.

Both OpenGL and DirectX describe vertices as a set of data consisting of coordinates in space that define the vertex location and any other vertex related data. Graphics primitives, such as points, lines, and triangles, are defined as an ordered set of vertices. There is a difference in how each API handles how vertices are combined to form primitives.

There are a bunch of differences in the DirectX and OpenGL APIs, so I will list a few of those for you. This chart is based off the book, OpenGL Game Programming and a few of these may now be incorrect as new DirectX versions are released. If you wish to correct me, please do via one of the ways listed at the end of this tutorial

我最喜欢的算法题目就是这道了   可是没人参与讨论

for (i = 0; a[i] == b[i]; i++)  
  if (a[i] == 0)     
   return 0;
return a[i] - b[i]; 

#include <stdio.h>
#include <string.h>
int main(void)
{
    printf("%d", strcmp("1", "1234"));
    return 0;
}

我把题目标准化了
欢迎参与讨论  共同学习

#include <stdio.h>
#include <string.h>

#define N 10010
#define M 10007
struct hashtabel
{
   
    char str[200] ;
    int num;
}a[N], b[N];

int hash(char *s)

{

    int h = 0, a = 127;
   
    for (; *s ; s++)
        
        h = (a*h + *s) % M;
   
    return h;
   
}

int getnum(char *s)
{
    int x;
   
    x = hash(s);
   
    while (strcmp(a[x].str, "") && strcmp(a[x].str, s)) // 该位置有值且值不匹配串
    {
        
        x = (x+1) % M;
        if (!(strcmp(a[x].str, "")))                    // 找一个没有值的位置
        {
            strcpy(a[x].str, s);
            return x;
        }
    }
    return x;
}

int main(void)
{
    char t[200];
    int i, j, n, total;
   
    for (i = 0; i < N; i++)
    {
        strcpy(a[i].str, "");
        a[i].num = 0;
    }
   
    scanf("%d\n", &n);
    for (i = 0, j = 0;  i < n; i++)
    {
        
        gets(t);
        total = getnum(t);
        
        if (a[total].num == 0)
        {
            strcpy(b[j].str, t);
            b[j].num = total;
            j++;
        }
        a[total].num++;
    }
   
    for (i = 0; i < j; i++)
        printf("%s %d\n", b[i].str, a[b[i].num].num);
   
    return 0;
}