ACER01

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ACER01
2天前

#include<iostream>
#include<cstdio>

using namespace std;

int main()
{
int n,num;  \\ n 为层数，num为最中间*的个数
cin >> n;
num = n;
int m; \\ m为下班部分图形的代换层数
if(n == 1) \\ 当n为1时不存在上下两个对称部分，所以单独列出
{
cout << "*" <<endl;
return 0;
}
for(int i = 1 ; i < n; ) \\ 上半层 * 的个数来循环直到中间层
{
m = i; \\ 让m在最后获取到中间层上一层的*数
for (int  j = 1; j <= (num - i) / 2; j++ ) \\ 计算每一行前半部分的空格数
{
cout << " ";
}
for (int  k = (num - i) / 2 + 1; k <= n - (num - i) /2; k++)
\\  从前半部分的空格数结束之后的后一个位置开始循环，直到*的最后一个位置结束
\\ （num - i）/2为后半部分空格的数量,由n减去后半部分空格数量则为*的最后位置
{
cout<< "*";
}
for(int z = n - (num - i) / 2 + 1; z <= n; z ++)
\\ 从*结束的后一个位置开始循环输出空格，直到空格和*的总数达到n
{
cout << " ";
}
cout << endl;
i = i + 2;
}
for(int j = 0; j < n; j++) \\ 循环输出最中间层的*
{
cout << "*";
}
cout << endl;
for(int i = m; i >= 0;)
\\ 因为下半部分图形是上半部分的对称，所以采取反方向递减来输出，让下半层第一层的点数从中间层上一层的点数开始递减
{
\\ 空格和*的计算输出方式和上半层一样，主要通过层数和*数来影响
for (int  j = 1; j <= (num - i) / 2; j++ )
{
cout << " ";
}
for (int  k = (num - i) / 2 + 1; k <= n - (num - i) /2; k++)
{
cout<< "*";
}
for(int z = n - (num - i) / 2 + 1; z <= n; z ++)
{
cout << " ";
}
cout << endl;
i = i - 2;  \\ 以每层减少2 * 来递减
}

return 0;
}



ACER01
3天前
#include<iostream>
#include<cstdio>

using namespace std;

int main()
{
int n,num;
cin >> n;
num = n;
int m;
if(n == 1)
{
cout << "*" <<endl;
return 0;
}
for(int i = 1 ; i < n; )
{
m = i;
for (int  j = 1; j <= (num - i) / 2; j++ )
{
cout << " ";
}
for (int  k = (num - i) / 2 + 1; k <= n - (num - i) /2; k++)
{
cout<< "*";
}
for(int z = n - (num - i) / 2 + 1; z <= n; z ++)
{
cout << " ";
}
cout << endl;
i = i + 2;
}
for(int j = 0; j < n; j++)
{
cout << "*";
}
cout << endl;
for(int i = m; i >= 0;)
{
for (int  j = 1; j <= (num - i) / 2; j++ )
{
cout << " ";
}
for (int  k = (num - i) / 2 + 1; k <= n - (num - i) /2; k++)
{
cout<< "*";
}
for(int z = n - (num - i) / 2 + 1; z <= n; z ++)
{
cout << " ";
}
cout << endl;
i = i - 2;
}

return 0;
}


ACER01
3天前
#include<iostream>
#include<cstdio>

using namespace std;

int main()
{
int n;
long long int x;
int flag;
long long int m;
cin >> n;
for (int i =0; i < n; i++)
{
m = 2;
flag = 0;
cin >> x;
while(m * m <= x)
{
if (x % m == 0)
{
cout << x << " is not prime" << endl;
flag = 1;
break;
}
m++;
}
if (flag == 0)
{
cout << x << " is prime" << endl;
}
}
return 0;
}


ACER01
3天前
#include<iostream>
#include<cstdio>

using namespace std;

int main()
{
int n;
long long int  x;
cin >> n;
while(n)
{
cin >> x;
if(x == 6 || x == 28 || x == 496 || x == 8128 || x == 33550336)
{
cout << x << " is perfect" << endl;
}
else
{
cout << x << " is not perfect" << endl;
}
n--;
}
return 0;
}


ACER01
3天前
#include<iostream>
#include<cstdio>

using namespace std;

int main()
{
int n, m, sum;
cin >> n >> m;
while(n > 0 && m > 0)
{
sum = 0;
if(m > n)
{
for(int i = n; i <= m; i++)
{
cout << i << " ";
sum += i;

}
cout << "Sum=" << sum << endl;
}
else if(m < n)
{
for(int i = m; i <= n; i++)
{
cout << i << " ";
sum += i;

}
cout << "Sum=" << sum << endl;
}
else
{
cout << n << " ";
cout << "Sum=" << m << endl;
}
cin >> n >> m;
}
return 0;
}


ACER01
3天前
#include<iostream>
#include<cstdio>

using namespace std;

int main()
{
int n, pre, now;
pre = 0;
now = 1;
cin >> n;
if(n == 1)
{
cout << pre << endl;
}
else if(n == 2)
{
cout <<pre << " " << now << endl;
}
else
{
cout << pre << " " << now;
for (int i = 0; i < n - 2; i++)
{
cout<< " " << pre + now;
int temp = pre;
pre = now;
now = temp + now;
}
cout << endl;
}

return 0;
}


ACER01
3天前
#include<iostream>
#include<cstdio>

using namespace std;

int main()
{
int n, x, y, sum;
sum = 0;
cin >> n;
while(n)
{
cin >> x >> y;
if (x > y)
{
for(int i = y + 1; i < x; i++)
{
if (i % 2 != 0)
{
sum += i;
}
}
cout << sum << endl;
sum = 0;
}
else if(x < y)
{
for(int i = x + 1; i < y; i++)
{
if (i % 2 != 0)
{
sum += i;
}
}
cout << sum << endl;
sum = 0;
}
else
{
cout << "0" << endl;
}
n--;
}

return 0;
}


ACER01
3天前
#include<iostream>
#include<cstdio>

using namespace std;

int main()
{
int n, x, y, num;
cin >> n;
x = 0;
y = 0;
while(n)
{
cin >> num;
if(num >= 10 && num <= 20)
{
x++;
}
else
{
y++;
}
n--;
}
cout << x << " in " << endl;
cout << y << " out " << endl;
return 0;
}


ACER01
3天前
#include<iostream>
#include<cstdio>

using namespace std;

int main()
{
int n, a;
char t;
int c = 0;
int r = 0;
int f = 0;
double percent;
cin >> n;
for (int i = 0; i < n; i++)
{
cin >> a >> t;
if( t == 'C')
{
c += a;
}
else if (t == 'R')
{
r += a;
}
else if (t == 'F')
{
f += a;
}
}
int sum = c + r + f;
cout << "Total: " << sum << " animals " << endl;
cout << "Total coneys: " << c << endl;
cout << "Total rats: " << r << endl;
cout << "Total frogs: " << f << endl;
percent = c *1.0 / sum;
printf("Percentage of coneys: %0.2lf %\n",percent * 100);
percent = r *1.0 / sum;
printf("Percentage of rats: %0.2lf %\n", percent * 100);
percent = f *1.0 / sum;
printf("Percentage of frogs: %0.2lf %\n", percent * 100);
return 0;
}


ACER01
3天前
#include<iostream>
#include<cstdio>

using namespace std;

int main()
{
int n;
cin >> n;
for (int i = 1; i <= 10; i++)
{
cout << i << " x " << n << " = " << i * n << endl;
}
return 0;
}