題干：

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.?

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

解題報告：

? ?別忘特盤一下n==0的情況，因為這種情況fit函數處理不了。

AC代碼：

#include<cstdio> #include<queue> #include<cstring> #include<cmath> #include<iostream> #include<algorithm> using namespace std; const int MAX = 100000 +5; int a[1000]; int n,x,p; void fit(int x) {while(x) {a[++p] = x%10;x/=10;} } int main() {int t,flag;cin>>t;while(t--) {p = 0;flag = 0;scanf("%d",&n);if(n<0) flag=1,n=-n;if(n == 0) {puts("0");continue;}fit(n);int i=0;for(i = 1; i<=p; i++) {if(a[i]!=0) break;}int cnt0 = i-1;if(flag == 1) putchar('-');for(int j = i; j<=p; j++) {printf("%d",a[j]);}for(int j = cnt0; j>0; j--) putchar('0');printf("\n");}return 0; }

?

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