ACM: uva 1069 -&n…

Always an integer

Combinatorics is a branch of mathematics chiefly concerned with counting discrete objects. For instance, how many ways can you pick two people out of a crowd of n  n  1×1  n×n 

Many questions like these have answers that can be reduced to simple polynomials in n  n(n - 1)/2  (n2 - n)/2  (n4 - 6n3 + 23n2 - 18n + 24)/24  n(n + 1)(2n + 1)/6  (2n3 + 3n2 + n)/6 

These polynomials are answers to questions that can have integer answers only. But since they have fractional coefficients, they look as if they could produce non-integer results! Of course, evaluating these particular polynomials on a positive integer always results in an integer. For other polynomials of similar form, this is not necessarily true. It can be hard to tell the two cases apart. So that, naturally, is your task.

Input 

The input consists of multiple test cases, each on a separate line. Each test case is an expression in the form (P)/D  P  D  P  CnE  C  E 

1. E  0E100  E  CnE  C  E  CnE  Cn  C  CnE  n  - n 

2. C  C  E  CnE  nE  - nE 

3. Only non-negative C 

4. Exponents in consecutive terms are strictly decreasing.

5. C  D 

See the sample input for details.

Input is terminated by a line containing a single period.

Output 

For each test case, print the case number (starting with 1). Then print `Always an integer' if the test case polynomial evaluates to an integer for every positive integer n Not always an integer' otherwise. Print the output for separate test cases on separate lines. Your output should follow the same format as the sample output.

Sample Input 

Sample Output 

题意: 判断多项式是否恒为整数.

解题思路:

         1. 代入1～k+1判断是否能整除即可. k是多项式最高项的次数.

         2. 训练指南上有详细的数学归纳法证明.

代码:

#include
#include
#include
using namespace std;
#define MAX 3100
typedef long long ll;

struct node
{
    int c, e;
    bool flag;
}a[110];

char str[MAX];
char ch[15];
int D;

inline int max(int a, int b)
{
    return a > b ? a : b;
}

ll getValue(int num, int value, int D)
{
    ll ans = 1;
    for(int i = 0; i < a[num].e; ++i)
        ans = ans*value%D;
    ans = ans*a[num].c%D;
    return a[num].flag ? ans : -ans;
}

bool judge(int value, int D, int num)
{
    ll ans = 0;
    for(int i = 0; i < num; ++i)
        ans = (ans+getValue(i, value, D))%D;
    return ans == 0;
}

int main()
{
//    freopen("input.txt", "r", stdin);
    int num = 1;
    while(scanf("%s", str) != EOF)
    {
        if(str[0] == '.') break;
       
        int len = strlen(str);
        int index = 0;
        while(str[index] != '(') index++;
        index++;
       
        int cur = 0, maxn = 0;
        while( str[index] != ')' )
        {
            if(cur == 0) //flag
            {
                a[cur].flag = true;
                if(str[index] == '-')
                {
                    a[cur].flag = false;
                    index++;
                }
            }
            else
            {
                a[cur].flag = (str[index] == '+');
                index++;
            }
           
            // number;