原题链接:数学的图表
看了第一个大佬的,感觉是我想复杂了
,不过还是可以值得借鉴一下的
#include "iostream"
#include "cstdio"
#include "cstring"
#include "string"
#include "sstream"
#include "cmath"
using namespace std;
string a[1001][1001];
int main()
{
int n;
while (cin >> n)
{
if (n == 1)
printf("1/1\n");
else
{
int m = sqrt(n);
m = m + 2;
for (int i = 0; i < m; i++)
{
for (int j = 0; j < m; j++)
{
stringstream ii, jj;
string s1, s2;
ii << i + 1;
ii >> s1;
jj << j + 1;
jj >> s2;
a[i][j] = s1;
a[i][j] += '/';
a[i][j] += s2;
}
}
int temp = 1;
int x = 0, y = 0;
loop:if (x == 0)
{
y = y + 1;
while (y >= 0)
{
if (temp == n - 1)
cout << a[x][y] << endl;
temp++;
if (temp == n)
{
goto loop1;
}
x++;
y--;
}
y++;
x--;
}
if (y == 0)
{
x = x + 1;
while (x >= 0)
{
if(temp==n-1)
cout << a[x][y] << endl;
temp++;
if (temp == n)
{
goto loop1;
}
x--;
y++;
}
x++;
y--;
}
loop1:if (temp < n)
goto loop;
}
}
return 0;
}附上第二次写的正确的答案
#include <iostream>
#include <cmath>
#include <string>
#include <cstdio>
using namespace std;
int main()
{
int n;
while (cin >> n)
{
int i = 0;
int s = 0;
while (s < n)
{
i++;
s = s + i;
}
if (i % 2 == 1)
{
s = s - i;
int a = i;
int b = 1;
for (int i = 1; i < n - s; i++)
{
a--;
b++;
}
cout << a << "/" << b << endl;
}
else
{
s = s - i;
int a = 1;
int b = i;
for (int j = 1; j < n - s; j++)
{
a++;
b--;
}
cout << a << "/" << b << endl;
}
}
return 0;
}附上第三种方法=.= ,可以优化下,懒得优化了
#include <iostream>
#include <cmath>
#include <string>
#include <cstdio>
using namespace std;
int main()
{
int n;
while(cin >> n)
{
int a, b;
if (n == 1)
cout << "1/1" << endl;
if (n == 2)
cout << "1/2" << endl;
else
{
a = 2; b = 1;
int flag = 1;
int flag1 = 1, flag2 = 1;
for (int i = 3; i < n; i++)
{
if (b == 1 && flag == 1)
{
a = a + 1;
flag = 0;
flag1 = 0;
flag2 = 1;
continue;
}
if (a == 1 && flag == 1)
{
b = b + 1;
flag = 0;
flag2 = 0;
flag1 = 1;
continue;
}
if (!flag1)
{
a = a - 1;
b = b + 1;
flag = 1;
continue;
}
if (!flag2)
{
a = a + 1;
b = b - 1;
flag = 1;
continue;
}
}
cout << a << "/" << b << endl;
}
}
return 0;
}0.0分
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