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Reverse Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9922 Accepted Submission(s): 4406
Problem Description
Welcome to 2006'4 computer college programming contest!
Specially, I give my best regards to all freshmen! You are the future of HDU ACM! And now, I must tell you that ACM problems are always not so easy, but, except this one... Ha-Ha!
Give you an integer; your task is to output its reverse number. Here, reverse number is defined as follows:
1. The reverse number of a positive integer ending without 0 is general reverse, for example, reverse (12) = 21;
2. The reverse number of a negative integer is negative, for example, reverse (-12) = -21;
3. The reverse number of an integer ending with 0 is described as example, reverse (1200) = 2100.
Specially, I give my best regards to all freshmen! You are the future of HDU ACM! And now, I must tell you that ACM problems are always not so easy, but, except this one... Ha-Ha!
Give you an integer; your task is to output its reverse number. Here, reverse number is defined as follows:
1. The reverse number of a positive integer ending without 0 is general reverse, for example, reverse (12) = 21;
2. The reverse number of a negative integer is negative, for example, reverse (-12) = -21;
3. The reverse number of an integer ending with 0 is described as example, reverse (1200) = 2100.
Input
Input file contains multiple test cases. There is a positive integer n (n<100) in the first line, which means the number of test cases, and then n 32-bit integers follow.
Output
For each test case, you should output its reverse number, one case per line.
Sample Input
3 12 -12 1200
Sample Output
21 -21 2100
Author
lcy
Source
HDU 2006-4 Programming Contest
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lxj | We have carefully selected several similar problems for you: 1279 1256 1302 1276 1283
这题就是需要注意以下几种输入情况
输入 0 000 00021 -00021 12000 -12000 -000 输出 0 0 12 -12 21000 -21000 -000
其他就没啥了
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<stack>
using namespace std;
int main()
{int n;while(~scanf("%d",&n)){getchar();while(n--){char num[1000];char num2[1000];scanf("%s",num);int len=strlen(num);int flag=0;for(int i=0; i<len; i++){if(num[i]<='9'&&num[i]>'0')///先排除只含零的情况{flag=1;break;}}if(flag==0){printf("%s\n",num);continue;}int ed=len-1;for(int i=len-1; i>=0; i--){if(num[i]>'0'&&num[i]<='9'){ed=i;break;}}for(int i=0; i<len; i++){if(num[i]=='-'){printf("%c",num[i]);}else if(num[i]>'0'&&num[i]<='9'){for(int j=ed;j>=i;j--){printf("%c",num[j]);}printf("%s\n",num+ed+1);break;}}}}return 0;
}