HDU 5059 Help him (模拟)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5059

Problem Description

As you know, when you want to hack someone's program, you must submit your test data. However sometimes you will submit invalid data, so we need a data checker to check your data. Now small W has prepared a problem for BC, but he is too busy to write the data checker. Please help him to write a data check which judges whether the input is an integer ranged from a to b (inclusive).
Note: a string represents a valid integer when it follows below rules.
1. When it represents a non-negative integer, it contains only digits without leading zeros.
2. When it represents a negative integer, it contains exact one negative sign ('-') followed by digits without leading zeros and there are no characters before '-'.
3. Otherwise it is not a valid integer.

 

Input

Multi test cases (about 100), every case occupies two lines, the first line contain a string which represents the input string, then second line contains a and b separated by space. Process to the end of file.

Length of string is no more than 100.
The string may contain any characters other than '\n','\r'.
-1000000000≤a≤b≤1000000000

 

Output

For each case output "YES" (without quote) when the string is an integer ranged from a to b, otherwise output "NO" (without quote).

 

Sample Input

10-100 1001a0-100 100

 

Sample Output

YESNO

 

Source

BestCoder Round #12

题意：

　　给定一个字符串，如果这个字符串是一个整数，并这个整数在[a,b]的范围之内，输出YES，其它的输出NO。

这个字符串是整数的条件：

1、如果它是正整数，它只包含前导不是0的数（这个数前面没有零）。

2、如果它是负整数，只包含一个'-'符号，任然没有前导0。

3、除此之外都不是非法的

PS:

注意前导零，和非法字符！

代码如下：

#include <cstdio>#include <cstring>const int maxn = 517;char str[maxn], ans[maxn];typedef __int64 LL;int flag = 0, mark = 0;LL ATOL(char s[]){    int len = strlen(s);    LL tt = 0;    int i = 0;    if(s[0] == '-')//负号    {        mark = 1;        i++;    }    for(; i < len; i++)    {        if(s[i]<'0' || s[i]>'9')//非法字符        {            flag = 1;            break;        }        if(tt == 0 && s[i]=='0' && len > 1)        {            //前导零            flag = 1;            break;        }        tt = tt*10+s[i]-'0';    }    return tt;}int main(){    LL a, b;    LL tt;    while (gets(str))    {        flag = 0;        mark = 0;        scanf("%I64d%I64d", &a, &b);        getchar();        int len = strlen(str);        if (len >= 13 || len == 0)        {            printf("NO\n");            continue;        }        //tt = atol(str);        tt = ATOL(str);        //printf("tt:: %I64d\n",tt);        if(mark && tt == 0)//-0        {            printf("NO\n");            continue;        }        if(flag)        {            printf("NO\n");            continue;        }        if(mark)//负数            tt = -tt;        if (a <= tt && tt <= b)            printf("YES\n");        else            printf("NO\n");    }    return 0;}