DotA(Defence of the Ancients) is very popular in Zhejiang University.Now a new hero appears in DotA.This hero has a miraculous skill.If the target's HP n is an even number,then it will be cut in half after skill-using.Otherwise the targe's HP will plus one after skill-using.Given a target's HP,we want to decrease the target's HP from n to 1 only with this new skill.
输入
The input consists of multiple test cases.Each case contain one line with an integer n(1
输出
For each test case,output the target's HP's changing process on a single line.The HP values connect with '-' character.
样例输入
5
21
样例输出
5-6-3-4-2-1
21-22-11-12-6-3-4-2-1
提示 [+]
*** 提示已隐藏，点击上方 [+] 可显示 ***
来源
2013年浙江大学复试机试模拟题
[cpp]
/*********************************
* 日期：2013-3-25
* 作者：SJF0115
* 题号: 题目1474: DotA
* 来源：http://acmclub.com/problem.php id=1474
* 结果：AC
* 来源：2013年浙江大学复试机试模拟题
* 总结：
**********************************/
#include
#include
int main()
{
long long int N;
int first,i;
//freopen("C:\\Users\\SJF\\Desktop\\acm.txt","r",stdin);
while(scanf("%lld",&N)!=EOF){
first = 1;
printf("%lld",N);
while(N > 1){
if(N % 2 == 0){
N = N / 2;
}
else{
N = N + 1;
}
printf("-");
printf("%lld",N);
}
printf("\n");
}
return 0;
}
/*********************************
* 日期：2013-3-25
* 作者：SJF0115
* 题号: 题目1474: DotA
* 来源：http://acmclub.com/problem.php id=1474
* 结果：AC
* 来源：2013年浙江大学复试机试模拟题
* 总结：
**********************************/
#include
#include
int main()
{
long long int N;
int first,i;
//freopen("C:\\Users\\SJF\\Desktop\\acm.txt","r",stdin);
while(scanf("%lld",&N)!=EOF){
first = 1;
printf("%lld",N);
while(N > 1){
if(N % 2 == 0){
N = N / 2;
}
else{
N = N + 1;
}
printf("-");
printf("%lld",N);
}
printf("\n");
}
return 0;
}