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HDU 5059 Help him (模拟)

2024-07-26 18:00:54   220人阅读

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
-100 100
1a0
-100 100
 

Sample Output
YES
NO
 

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;
}


HDU 5059 Help him (模拟)


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