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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[];

int main()

{

    while(gets(str)){

        long long a,b;

        scanf("%lld%lld",&a,&b);

        getchar();

        if(strcmp(str,"")==){

            if(a<=&&b>=) printf("YES\n");

            else printf("NO\n");

            continue;

        }

        int len = strlen(str);

        if(len>){

            printf("NO\n");

            continue;

        }

        int s = ;

        bool flag = false,is_nag = false;

        if(str[]=='-') {

            s++;

            is_nag = true;

        }

        long long sum = ;

        if(str[s]==''||!isdigit(str[s])) flag = true;

        for(int i=s;i<len&&!flag;i++){

            if(isdigit(str[i])){

                sum = sum* + str[i]-'';

            }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 ;

}

巴特西

hdu 5059(模拟)

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