Help him

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2500    Accepted Submission(s): 518

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:判断一个字符串是否符合要求:

假设为正数,不能有前导0

假设为负数，最前面有 - 号，整数部分不能有前导0

这个串必须在 [a,b]之间

这个题坑的地方：判断 0 ,一定开longlong,我就被long long 坑死了。然后还有一点就是gets()读入。

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;

char str[105];
int main()
{
    while(gets(str)){
        long long a,b;
        scanf("%lld%lld",&a,&b);
        getchar();
        if(strcmp(str,"0")==0){
            if(a<=0&&b>=0) printf("YES\n");
            else printf("NO\n");
            continue;
        }
        int len = strlen(str);
        if(len>11){
            printf("NO\n");
            continue;
        }
        int s = 0;
        bool flag = false,is_nag = false;
        if(str[0]==‘-‘) {
            s++;
            is_nag = true;
        }
        long long sum = 0;
        if(str[s]==‘0‘||!isdigit(str[s])) flag = true;
        for(int i=s;i<len&&!flag;i++){
            if(isdigit(str[i])){
                sum = sum*10 + str[i]-‘0‘;
            }else{
                flag = true;
            }
        }
        if(flag){
            printf("NO\n");
        }else{
            if(is_nag) sum = -sum;
            if(sum>=a&&sum<=b){
                printf("YES\n");
            }else{
                printf("NO\n");
            }
        }
    }
    return 0;
}