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  • hdu 5059 Help him

    Help him

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 480    Accepted Submission(s): 119


    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$leq a leq b leq 1000000000$
     

    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
     


    题解及代码:


           题意非常easy,给一个字符串。和一个区间,看这个字符串是否能构成一个合法的整数。而且其值是否在这个区间内(注意-0为非法数据)。

           首先直接排除-和-0的情况。然后推断前导0的情况,再推断0-9以外字符的情况,接着看其长度是否大于12,最后把它转化成整数,推断大小。

         

    #include <iostream>
    #include <cstdio>
    #include <cmath>
    #include <cstring>
    #include <algorithm>
    #include <map>
    using namespace std;
    typedef long long ll;
    
    
    bool judge(char s[],ll a,ll b)
    {
        ll num=0,d=1,i=0;
        int len=strlen(s),k=0;
        if(s[0]=='-') d=-1,k++;
        if(k==len||s[0]=='-'&&s[k]=='0') return false;
        for(i=k;i<len-1;i++)
        if(s[i]!='0') break;
    
        if(i!=k) return false;
        for(int i=k;i<len;i++)
        if(s[i]<'0'||s[i]>'9') return false;
    
        for(int i=k;i<len;i++)
        {
            num=num*10+s[i]-'0';
            if(i-k>11) return false;
        }
        //printf("%I64d
    ",num*d);
        if(num*d<a||num*d>b)  return false;
        return true;
    }
    
    int main()
    {
        ll a,b;
        char s[110];
        while(gets(s)!=NULL)
        {
           scanf("%I64d%I64d",&a,&b);
           if(a>b) swap(a,b);
    
           if(judge(s,a,b)) printf("YES
    ");
           else printf("NO
    ");
           getchar();
        }
        return 0;
    }






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  • 原文地址:https://www.cnblogs.com/mengfanrong/p/5076720.html
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