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.
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≤a≤b≤1000000000
Length of string is no more than 100.
The string may contain any characters other than ' ',' '.
-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
题意：这道题的题意是真的很简单，就是第一行输入一个字符串，第二行输入两个整数-1000000000 ≤ a ≤ b ≤ 1000000000，然后他问这个字符串在不在a~b的范围内（也就是看这个字符串能否转化为数字，能的话在不在a~b的范围内，而且不能有前导0）。
那么问题就来了，这个字符串说是啥字符都有啊~~~~，还真是啥都有。。。（A完这道题，真的不知道说啥了，什么叫坑，这就是坑）
#include<stdio.h> #include<string.h> #include<queue> #include<math.h> #include<stdlib.h> #include<algorithm> using namespace std; const int N=1e2+10; const int INF=0x3f3f3f3f; const int MOD=1e9+7; typedef long long LL; int main () { char s[N]; LL a, b, c; int len, flag, i, mark, legal; while (gets(s) != 0) { scanf("%lld %lld%*c", &a, &b); len = strlen(s); mark = 0; ///标记负号 legal = 1; ///是否合法 c = 0; ///保存字符串的值 flag = 0; ///0表示在范围内 1表示不在 if (s[0] == '-') mark = 1; for (i = 0; i < len; i++) { if ((s[i] == '-' && i == 0) || (s[i] >= '0' && s[i] <= '9')) continue; ///除了‘-’ 剩下的都必须是数字且不含有前导0的才是合法的串 legal = 0; break; } i = 0; if (mark) i++; if (s[i] == '0') legal = 0; if (len > 15) legal = 0; ///串太长超出范围 if (legal) ///只处理合法的串 { i = 0; if (mark) i++; ///排除‘-’的影响 for ( ; i < len; i++) c = c*10+(s[i]-'0'); if (mark) c = -c; ///有‘-’ 取相反数 if (c >= a && c <= b) flag = 1; } if (len == 1 && s[0] == '0' && 0 >= a && 0 <= b) flag = 1; ///字符串只有一个‘0’如果在区间内也合法 if (s[0] == '-' && len == 1) flag = 0; ///只有‘-’ 也是不合法的（怎么想到。。。） if (len == 0) flag = 0; ///空串是不合法的！！！！！(空串啊。。。坑) if (flag == 0) printf("NO "); else printf("YES "); } return 0; }