HDU 1106 排序(atoi函数和strtok函数)

排序

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 48472 Accepted Submission(s): 13987

Problem Description
输入一行数字，如果我们把这行数字中的‘5’都看成空格，那么就得到一行用空格分割的若干非负整数（可能有些整数以‘0’开头，这些头部的‘0’应该被忽略掉，除非这个整数就是由若干个‘0’组成的，这时这个整数就是0）。

你的任务是：对这些分割得到的整数，依从小到大的顺序排序输出。

Input
输入包含多组测试用例，每组输入数据只有一行数字（数字之间没有空格），这行数字的长度不大于1000。

输入数据保证：分割得到的非负整数不会大于100000000；输入数据不可能全由‘5’组成。

Output
对于每个测试用例，输出分割得到的整数排序的结果，相邻的两个整数之间用一个空格分开，每组输出占一行。

Sample Input
0051231232050775

Sample Output
0 77 12312320

Source
POJ

题解：使用atoi()函数和strtok()函数搞定。。。。
strtok函数解析：
Syntax:
#include < string.h>
char *strtok( char *str1, const char *str2 );

The strtok() function returns a pointer to the next “token” in str1, where str2 contains the delimiters that determine the token. strtok() returns NULL if no token is found. In order to convert a string to tokens, the first call to strtok() should have str1 point to the string to be tokenized. All calls after this should have str1 be NULL.

For example:

char str[] = “now # is the time for all # good men to come to the # aid of their country”;
char delims[] = “#”;
char *result = NULL;
result = strtok( str, delims );
while( result != NULL ) {
printf( “result is \”%s\”\n”, result );
result = strtok( NULL, delims );
}

The above code will display the following output:

result is “now ”
result is ” is the time for all ”
result is ” good men to come to the ”
result is ” aid of their country”

AC代码：

#include<cstdio>#include<cstdlib>#include<string>#include<cstring> char s[1005];int a[1005];char * temp;int cmp(const void *a,const void *b){    return *(int *)a-*(int *)b;}int main(){    int c;    while(gets(s))    {        a[0]=atoi(strtok(s,"5"));        c=1;        while(temp=strtok(NULL,"5"))        {            a[c++]=atoi(temp);        }        qsort(a,c,sizeof a[0],cmp);        printf("%d",a[0]);        for(int i=1;i<c;i++)        {            printf(" %d",a[i]);        }        printf("\n");    }    return 0;}

orAC2代码：

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<cstdlib>using namespace std;const int maxx = 1001;char a[maxx][9]; //字符串 long b[maxx];  //数字 int cmp(const void *a,const void *b){    return *(int*)a-*(int*)b;}int main(){    string s;    while(cin>>s){        int i=0,j=0,k=0;        for(i=0,j=0,k=0;s[i]!='\0';i++,k++){            a[j][k]=s[i];            if(s[i]=='5'){                a[j][k]='\0';                if(strlen(a[j])==0){                    k=-1;continue;                }                b[j]=atoi(a[j]); //将字符串转换成数字                     j++;k=-1;            }        }            a[j][k]='\0';            b[j]=atoi(a[j]);            if(s[i-1]=='5')            j--;        qsort(b,j+1,sizeof(long),cmp);           for(i=0;i<=j;i++)          {            cout<<b[i];            if(i<j)                cout<<" ";          }        cout<<endl;    }      return 0;}