杭电OJ100题——2050-2054（C++版）折线分割平面BitsetPictureSwitch GameA == B ?

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概述

折线分割平面

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 41525 Accepted Submission(s): 27496

Problem Description

我们看到过很多直线分割平面的题目，今天的这个题目稍微有些变化，我们要求的是n条折线分割平面的最大数目。比如，一条折线可以将平面分成两部分，两条折线最多可以将平面分成7部分，具体如下所示。

Input

输入数据的第一行是一个整数C,表示测试实例的个数，然后是C 行数据，每行包含一个整数n(0<n<=10000),表示折线的数量。

Output

对于每个测试实例，请输出平面的最大分割数，每个实例的输出占一行。

Sample Input

2 1 2

Sample Output

2 7

#include<iostream>
using namespace std;
int main(){
int c;
cin>>c;
while(c--){
int n;
cin>>n;
long long *arr = new long long[10000];
arr[0] = 2;
arr[1] = 7;
long k = 5;
for(int i = 2;i<10000;i++){
arr[i] = arr[i-1] + (k+4);
k += 4;
}
cout<<arr[n-1]<<endl;
}
return 0;
}

Bitset

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 30691 Accepted Submission(s): 22535

Problem Description

Give you a number on base ten,you should output it on base two.(0 < n < 1000)

Input

For each case there is a postive number n on base ten, end of file.

Output

For each case output a number on base two.

Sample Input

1 2 3

Sample Output

1 10 11

#include<iostream>
using namespace std;
int main(){
int n;
while(cin>>n){
char arr[15];
int a_size = 0;
do{
arr[a_size++] = n%2+'0';
n /= 2;
}while(n!=0);
for(int i = a_size-1;i>0;i--)
cout<<arr[i];
cout<<arr[0]<<endl;
}
return 0;
}

Picture

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 40391 Accepted Submission(s): 19651

Problem Description

Give you the width and height of the rectangle,darw it.

Input

Input contains a number of test cases.For each case ,there are two numbers n and m (0 < n,m < 75)indicate the width and height of the rectangle.Iuput ends of EOF.

Output

For each case,you should draw a rectangle with the width and height giving in the input.
after each case, you should a blank line.

Sample Input

3 2

Sample Output

+---+

| |

+---+

#include<stdio.h>
#include<iostream>
using namespace std;
int main(){
int width,height;
while(scanf("%d%d",&width,&height)!=EOF){
//输出第一行
cout<<"+";
for(int i=0;i<width;i++)
cout<<"-";
cout<<"+"<<endl;
//输出高
for(int i=0;i<height;i++){
cout<<"|";
for(int i=0;i<width;i++)
cout<<" ";
cout<<"|"<<endl;
}
//输出最后一行
cout<<"+";
for(int i=0;i<width;i++)
cout<<"-";
cout<<"+"<<endl;
cout<<endl;
}
}

Switch Game

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 23954 Accepted Submission(s): 14579

Problem Description

There are many lamps in a line. All of them are off at first. A series of operations are carried out on these lamps. On the i-th operation, the lamps whose numbers are the multiple of i change the condition ( on to off and off to on ).

Input

Each test case contains only a number n ( 0< n<= 10^5) in a line.

Output

Output the condition of the n-th lamp after infinity operations ( 0 - off, 1 - on ).

Sample Input

1 5

Sample Output

1 0

#include<iostream>
using namespace std;
int main(){
int n;
while(cin>>n){
int result = 0;
for(int i=1;i<=n;i++)
if(n%i==0){
if(result==0)
result = 1;
else
result = 0;
}
cout<<result<<endl;
}
return 0;
}

A == B ?

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 135808 Accepted Submission(s): 21674

Problem Description

Give you two numbers A and B, if A is equal to B, you should print "YES", or print "NO".

Input

each test case contains two numbers A and B.

Output

for each case, if A is equal to B, you should print "YES", or print "NO".

Sample Input

1 2

2 2

3 3

4 3

Sample Output

YES

#include<iostream>
using namespace std;
#include<string>
void trim0(string& b)
{
int len = b.length();
//如果是小数就去除末尾的0，如2.0000和2
if(b.find('.')!=string::npos)
{
for(int i=len-1;b[i]=='0';i--)
len--;
b=b.substr(0,len);
}
if(b[len-1]=='.')
b=b.substr(0,len-1);
}
int main()
{
string a,b;
while(cin>>a>>b)
{
trim0(a);
trim0(b);
if(a==b)
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
}
}