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其实就是解2元一次方程啊……~~#include<iostream>
#include<cmath>
using namespace std;
int main()
{
int n ;
int sum;
int i ;
while(cin>>n)
{
sum = 0;
i = ceil(sqrt(1+8*n))/2.0;
sum = (i)*(i+1)/2;
if(sum ==n)
{
if(i%2 ==0) cout<<"TERM "<<n<<" IS "<<i<<"/"<<"1"<<endl;
else cout<<"TERM "<<n<<" IS "<<"1"<<"/"<<i<<endl;
}
else
{
sum -= i;
i--;
if(i%2==0) cout<<"TERM "<<n<<" IS "<<i+1-n+sum+1<<"/"<<n-sum<<endl;
else cout<<"TERM "<<n<<" IS "<<n-sum<<"/"<<i+1-n+sum+1<<endl;
}
}
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