HDOJ--1266--Reverse Number

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:

  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.

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;
}
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