題目描述:
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:
- The reverse number of a positive integer ending without 0 is general reverse, for example, reverse (12) = 21;
- The reverse number of a negative integer is negative, for example, reverse (-12) = -21;
- The reverse number of an integer ending with 0 is described as example, reverse (1200) = 2100.
輸入描述:
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.
輸出描述:
For each test case, you should output its reverse number, one case per line.
輸入:
3
12
-12
1200
輸出:
21
-21
2100
題意:
翻轉整數
題解:
直接暴力
代碼:
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 1000 + 5;
char s[maxn];
int main(){
int t;
scanf("%d",&t);
while(t--){
int flag = 0,last,first;
scanf("%s",s);
int len = strlen(s);
char test = s[len-1];
last = len - 1;
while(test == '0'){
last --;
test = s[last];
}
if(s[0] == '-') flag = 1;
first = 0;
for(int i = 0;i < len - 1; i ++){
if(s[i] != '0' && s[i] != '-'){
first = i;
break;
}
}
if(flag) printf("-");
for(int i = last;i >= first; i --)
printf("%c",s[i]);
for(int i = last + 1;i <= len - 1; i ++)
printf("0");
printf("\n");
}
return 0;
}