HDU 5059 Help him（BestCoder Round #12）

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 ' ',' '.
-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

-100 100

1a0

-100 100

 

Sample Output：

YES

NO

 

题意：这道题的题意是真的很简单，就是第一行输入一个字符串，第二行输入两个整数-1000000000 ≤ a ≤ b ≤ 1000000000，然后他问这个字符串在不在a~b的范围内（也就是看这个字符串能否转化为数字，能的话在不在a~b的范围内，而且不能有前导0）。

 

那么问题就来了，这个字符串说是啥字符都有啊~~~~，还真是啥都有。。。（A完这道题，真的不知道说啥了，什么叫坑，这就是坑）

#include<stdio.h>
#include<string.h>
#include<queue>
#include<math.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;

const int N=1e2+10;
const int INF=0x3f3f3f3f;
const int MOD=1e9+7;

typedef long long LL;

int main ()
{
    char s[N];
    LL a, b, c;
    int len, flag, i, mark, legal;

    while (gets(s) != 0)
    {
        scanf("%lld %lld%*c", &a, &b);
        len = strlen(s);
        mark = 0; ///标记负号
        legal = 1; ///是否合法
        c = 0; ///保存字符串的值
        flag = 0; ///0表示在范围内 1表示不在
        if (s[0] == '-') mark = 1;
        for (i = 0; i < len; i++)
        {
            if ((s[i] == '-' && i == 0) || (s[i] >= '0' && s[i] <= '9')) continue; ///除了‘-’ 剩下的都必须是数字且不含有前导0的才是合法的串
            legal = 0;
            break;
        }

        i = 0;
        if (mark) i++;
        if (s[i] == '0') legal = 0;

        if (len > 15) legal = 0; ///串太长超出范围

        if (legal) ///只处理合法的串
        {
            i = 0;
            if (mark) i++; ///排除‘-’的影响

            for ( ; i < len; i++)
                c = c*10+(s[i]-'0');

            if (mark) c = -c; ///有‘-’ 取相反数

            if (c >= a && c <= b) flag = 1;
        }

        if (len == 1 && s[0] == '0' && 0 >= a && 0 <= b) flag = 1; ///字符串只有一个‘0’如果在区间内也合法
        if (s[0] == '-' && len == 1) flag = 0; ///只有‘-’ 也是不合法的（怎么想到。。。）
        if (len == 0) flag = 0; ///空串是不合法的！！！！！(空串啊。。。坑)

        if (flag == 0) printf("NO
");
        else printf("YES
");
    }

    return 0;
}