题目1474: DotA
题目描述
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<n≤1000000000).
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 [+]
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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<stdio.h>
#include<string.h>
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<stdio.h>
#include<string.h>
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;
}
补充：软件开发 , C++ ,
