C - 曼联

Description

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.

题意：输入n，n组数，每组两个数，前者代表x的系数，后者代表x的幂数，求该函数

原函数x的系数和幂数。

Sample Input

3

1 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

<span style="font-family:SimSun;font-size:18px;">#include <cstdio>#include <algorithm>#include <cmath>using namespace std;int n,key;struct node{    int k,e;} st[1002];int main(){    while(~scanf("%d",&n))    {        for(int i=1; i<=n; i++)        {            int cc;            scanf("%d %d",&st[i].k,&cc);            st[i].e=cc+1;        }        for(int i=1; i<=n; i++)        {            if(st[i].k%st[i].e==0)                printf("%d %d%c",st[i].k/st[i].e,st[i].e,i==n?'\n':' ');            else            {                int pp;                if(st[i].k<0)                       //求两个数能同时整除的数时用两个数的正值求                pp=-st[i].k;                else                pp=st[i].k;                key=__gcd(pp,st[i].e);                printf("%d/%d %d%c",st[i].k/key,st[i].e/key,st[i].e,i==n?'\n':' ');            }        }    }    return 0;}</span>