周赛-Integration of Polynomial

Integration of Polynomial
Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others)
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Problem 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 that a 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

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
数学的不定积分加GCD

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <list>
#include <algorithm>
#define LL long long
#define RR freopen("output.txt","r",stdoin)
#define WW freopen("input.txt","w",stdout)

using namespace std;

const int MAX = 100100;

const int MOD = 1000000007;

int k[1100],e[1100];

int num[1100];
int GCD(int a,int b)
{
    return b==0?a:GCD(b,a%b);
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        memset(num,0,sizeof(num));
        for(int i=0;i<n;i++)
        {
            scanf("%d %d",&k[i],&e[i]);
            e[i]++;
            if(k[i]%e[i]==0)
            {
                k[i]/=e[i];
            }
            else
            {
                num[i]=e[i];
                int ans=GCD(abs(k[i]),num[i]);
                k[i]/=ans;
                num[i]/=ans;
            }
        }
        for(int i=0;i<n;i++)
        {
            if(i)
            {
                printf(" ");
            }
            if(num[i])
            {
                printf("%d/%d %d",k[i],num[i],e[i]);
            }
            else
            {
                printf("%d %d",k[i],e[i]);
            }
        }
        printf("\n");
    }

    return 0;
}