Codeforces Round #366 (Div. 2) A. Hulk

Hulk的情感表达
博客介绍了一个有趣的问题,即根据层数n来确定Hulk复杂情感的表达方式,每增加一层情感就会在爱与恨之间切换,并提供了一个C++实现代码样例。

A. Hulk

time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

Dr. Bruce Banner hates his enemies (like others don’t). As we all know, he can barely talk when he turns into the incredible Hulk. That’s why he asked you to help him to express his feelings.

Hulk likes the Inception so much, and like that his feelings are complicated. They have n layers. The first layer is hate, second one is love, third one is hate and so on…

For example if n = 1, then his feeling is “I hate it” or if n = 2 it’s “I hate that I love it”, and if n = 3 it’s “I hate that I love that I hate it” and so on.

Please help Dr. Banner.
Input

The only line of the input contains a single integer n (1 ≤ n ≤ 100) — the number of layers of love and hate.
Output

Print Dr.Banner’s feeling in one line.
Examples
Input

1

Output

I hate it

Input

2

Output

I hate that I love it

Input

3

Output

I hate that I love that I hate it

ac代码:

/* ***********************************************
Author       : AnICoo1
Created Time : 2016-08-20-16.08 Saturday
File Name    : D:\MyCode\2016-8月\2016-8-20.cpp
LANGUAGE     : C++
Copyright  2016 clh All Rights Reserved
************************************************ */
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<map>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 1010000
#define LL long long
#define ll __int64
#define INF 0xfffffff
#define mem(x,y) memset(x,(y),sizeof(x))
#define PI acos(-1)
#define gn (sqrt(5.0)+1)/2
#define eps 1e-8
using namespace std;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
double dpow(double a,ll b){double ans=1.0;while(b){if(b%2)ans=ans*a;a=a*a;b/=2;}return ans;}
//head
int main()
{
    int n;cin>>n;
    for(int i=1;i<=n;i++)
    {
        if(i%2)
            printf("I hate ");
        else
            printf("I love ");
        if(i==n)
            printf("it");
        else
            printf("that ");
    }
    printf("\n");
    return 0;
}
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