周赛一 ACdream 1204 模拟

Suppose there are a polynomial which has n nonzero terms, please print the integration polynomial of the given polynomial.

The polynomial will be given in the following way, and you should print the result in the same way:

k[1] e[1] k[2] e[2] ... k[n] e[n]

where k[i] and e[i] respectively represent the coefficients and exponents of nonzero terms, and satisfies e[1] < e[2] < ... < e[n].

Note:

• Suppose that the constant term of the integration polynomial is 0.
• If one coefficient of the integration polynomial is an integer, print it directly.
• If one coefficient of the integration polynomial is not an integer, please print it by using fraction a/b which satisfies thata is coprime to b.

Input

There are multiple cases.

For each case, the first line contains one integer n, representing the number of nonzero terms.

The second line contains 2*n integers, representing k[1], e[1], k[2], e[2], ..., k[n], e[n]。

1 ≤ n ≤ 1000

-1000 ≤ k[i] ≤ 1000, k[i] != 0, 1 ≤ i ≤ n

0 ≤ e[i] ≤ 1000, 1 ≤ i ≤ n

Output

Print the integration polynomial in one line with the same format as the input.

Notice that no extra space is allowed at the end of each line.

Sample Input

31 0 3 2 2 4

Sample Output

1 1 1 3 2/5 5

Hint

f(x) = 1 + 3x² + 2x⁴

After integrating we get: ∫f(x)dx = x + x³ + (2/5)x⁵

---

#include<iostream>#include<algorithm>#include<cstring>using namespace std;struct node{    int data;    int num;} p[10000];//int cmp(node p1,node p2)//{//    return p1.num<p2.num;//}int gcd(int a,int b){    return b?gcd(b,a%b):a;}int main(){    int n;    while(cin>>n)    {        for(int i=1; i<=n; i++)        {            cin>>p[i].data>>p[i].num;        }//        sort(p,p+n,cmp);       //不用排序        int kk=1;        for(int i=1; i<=n; i++)        {            if(kk==1)                kk=0;            else                cout<<" ";            if(p[i].data>=0)            {                if(p[i].data%(p[i].num+1)==0)                    cout<<p[i].data/(p[i].num+1)<<" "<<p[i].num+1;                else                {                    int a=gcd(p[i].data,p[i].num+1);                   // cout<<a<<endl;                         cout<<p[i].data/a<<"/"<<(p[i].num+1)/a<<" "<<p[i].num+1;                }            }            else      {p[i].data*=-1;if(p[i].data%(p[i].num+1)==0)                    cout<<-1*p[i].data/(p[i].num+1)<<" "<<p[i].num+1;                else                {                    int a=gcd(p[i].data,p[i].num+1);                    cout<<-1*p[i].data/a<<"/"<<(p[i].num+1)/a<<" "<<p[i].num+1;                }}        }        cout<<endl;    }}//1.注意p[i].data为负数时//2. p[i].data与p[i].num可约分时