hdu 5059 Help him（模拟）

Help him

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2510    Accepted Submission(s): 520

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

题意：给你一个字符串，先判断字符串是否是合法的数字，在判断是否是给定区间里的数

思路：直接模拟，注意字符串会有空格，注意用long long 十位数组可能会有50E超过int范围

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;int main(){    char s[510];    long long a,b;    while(gets(s))    {        scanf("%lld %lld",&a,&b);        int len=strlen(s);        int flag=0;        if(s[0]!='-'&&(s[0]<'0'||s[0]>'9'))            flag=1;        for(int i=1; i<len; i++)            if(s[i]<'0'||s[i]>'9')                flag=1;        if(s[0]=='-'&&(len==1||s[1]=='0'))            flag=1;        if(len>1&&s[0]=='0')            flag=1;        if(flag) printf("NO\n");        else        {            if(len>11) printf("NO\n");            else            {                long long ans=0;                if(s[0]=='-')                {                    for(int i=1;i<len;i++)                        ans=ans*10+s[i]-'0';                    ans=-ans;                }                else                     for(int i=0;i<len;i++)                        ans=ans*10+s[i]-'0';                if(ans>=a&&ans<=b) printf("YES\n");                else printf("NO\n");            }        }        getchar();    }    return 0;}