Help him

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 248    Accepted Submission(s): 58

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$\leq a \leq b \leq 1000000000$
 
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
 
Recommend
heyang   |   We have carefully selected several similar problems for you:  5061 5060 5057 5056 5053 
 

SB了。。

以后 注意点用atoi...  可能过long long 用 strtoll

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
char S[200];
char C[200];
int len;
long long a,b;
int getans(char *A)
{
int i;
int lena=strlen(A);
if(lena==0) return 0;
if(lena!=1&&A[0]=='0') return 0;
for(i=0;i<lena;i++)
if(!('0'<=A[i]&&A[i]<='9')) return 0;
return 1;
}
int main()
{
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
int ans;
long long k;
while(gets(S)!=NULL)
{
scanf("%I64d%I64d",&a,&b);
gets(C);
len=strlen(S);
if(S[0]=='-'&&S[1]=='0') { printf("NO\n");continue;}
if(len>11||len==0) { printf("NO\n");continue;}
if(len==11&&S[0]!='-') { printf("NO\n");continue;}
if(S[0]=='-')
ans=getans(S+1);
else ans=getans(S);
if(!ans) { printf("NO\n");continue;}
else
{
k=strtoll(S,NULL,10);
if(a<=k&&k<=b) printf("YES\n");
else printf("NO\n");
}
memset(S,0,sizeof(S));
}
return 0;
}

学长的十分美好的代码

const int N = 105;
char s[N], t[N];
int a, b, c;
bool solve(){
if(sscanf(s, "%d", &c) != 1) return 0;
sprintf(t, "%d", c);
if(strcmp(s, t) != 0) return 0;
return a<=c && c<=b;
}
int main(){
//freopen("in.txt", "r", stdin);
while(gets(s)){
scanf("%d%d", &a, &b);
getchar();
bool ans = solve();
puts(ans ? "YES" : "NO");
}
return 0;
}

Bestcoder HDU5059 Help him 字符串处理的更多相关文章

  1. 字符串处理 BestCoder Round #43 1001 pog loves szh I

    题目传送门 /* 字符串处理:是一道水题,但是WA了3次,要注意是没有加'\0'的字符串不要用%s输出,否则在多组测试时输出多余的字符 */ #include <cstdio> #incl ...

  2. hdu 4908 BestCoder Sequence 发现M中值是字符串数, 需要预处理

    BestCoder Sequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Other ...

  3. BestCoder Round #89 02单调队列优化dp

    1.BestCoder Round #89 2.总结:4个题,只能做A.B,全都靠hack上分.. 01  HDU 5944   水 1.题意:一个字符串,求有多少组字符y,r,x的下标能组成等比数列 ...

  4. BestCoder 1st Anniversary B.Hidden String DFS

    B. Hidden String Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://bestcoder.hdu.edu.cn/contests/co ...

  5. BestCoder Round #88

    传送门:BestCoder Round #88 分析: A题统计字符串中连续字串全为q的个数,预处理以下或加个cnt就好了: 代码: #include <cstdio> #include ...

  6. hdu 5311 Hidden String (BestCoder 1st Anniversary ($))(深搜)

    http://acm.hdu.edu.cn/showproblem.php?pid=5311 Hidden String Time Limit: 2000/1000 MS (Java/Others)  ...

  7. BestCoder Round #14

    Harry And Physical Teacher Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Ja ...

  8. hdu5635 BestCoder Round #74 (div.2)

    LCP Array  Accepts: 131  Submissions: 1352  Time Limit: 4000/2000 MS (Java/Others)  Memory Limit: 13 ...

  9. 【BestCoder】【Round#41】

    枚举+组合数?+DP+数学问题 http://bestcoder.hdu.edu.cn/contests/contest_show.php?cid=582 QAQ许久没打过比赛,来一发BC,结果还是只 ...

随机推荐

  1. [Regular Expressions] Find the Start and End of Whole Words

    Regular Expression Word Boundaries allow to perform "whole word only" searches within our ...

  2. 男同胞爱小秘籍--作为爱他的女朋友了几天C规划

    各位男同胞,不知道你的女朋友没有在过去的一问天,你这个问题~~ 场景重现: 女友:"今天天气不错." 你们:"对" 女友:"今天是我们知道它的最初几天 ...

  3. scrapy使用crontab定时任务不能自动执行的调试

    在用crontab进行定时任务时,发现任务并没有执行.而手动bash yourshell.sh时可以正常的执行程序.以下是个人的解决流程. 一.将错误打印打out.log */10 * * * * b ...

  4. E514:write error(file system full?)

    vi编辑某文件,保存时报错,提示:E514: write error (file system full?)---写入错误,磁盘满了? 查看磁盘空间:df -h根目录磁盘空间已满,used%100. ...

  5. Android学习笔记--JNI的使用方法

    1.JNI是什么 JNI是Java Native Interface的缩写,它提供若干的API实现Java与其他语言之间的通信.而Android Framework由基于Java语言的的Java层与基 ...

  6. ios 获取屏幕的属性

    屏幕尺寸     CGRect screen = [UIscreen mainScreen].bounds 状态栏尺寸  CGRect rect = [[UIApplication sharedApp ...

  7. .Net Memory -- Windbg基本命令

    命令 解释 .cls 清空命令窗口屏幕 .load dllfullpath 加载debugger扩展dll如SOS sosex psscor. .loadby dll moduleName 加载deb ...

  8. 轻量级jquery框架之--面板(panel)

    面板需求: (1)支持可拖拽,面板将作为后期的布局组件.window组件.alert组件的基础. (2)支持自定义工具栏,工具栏位置定义在面板底部,工具栏依赖toolbar组件. (3)支持加载JSO ...

  9. 让操作javascript对象数组像.net lamda表达式一样

    让操作javascript对象数组像.net lamda表达式一样 随着web应用程序的富客户端化.ajax的广泛使用及复杂的前端业务逻辑.对js对象数组.json数组的各种操作越来越多.越来越复杂. ...

  10. 利用raspberry pi搭建typecho笔记(三) typecho nginx sqlite FAQ

    前言 这是一个汇总文,用来总结我在整个配置过程中遇到的各种问题.因为我在解决这些问题的过程中发现,typecho被部署在这种需要完全自己配置的平台上的情况是比较少的,相关的资料也比较少,所以我的解决过 ...