joj1013 Polynomial Multiplication

 1013: Polynomial Multiplication

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Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	4660	807	Standard

A polynomial is a sum of terms, where each term is a coefficient multiplied by the variable x raised to some nonnegative integer power. The largest such power is referred to as the degree of the polynomial. For example: x^3+x^2+x+1, x^3+x, x^2, and 1 are all polynomials.

Input

A positive integer denoting the number of cases.
(The degree of polynomial 1) followed by the value of each coefficient for polynomial 1.
(The degree of polynomial 2) followed by the value of each coefficient for polynomial 2.
No polynomial will have degree greater than five. All coefficients will be greater than or equal to zero and less than 65536. Each polynomial will be on a separate line. A blank line will separate each case.

Output

The product of polynomial 1 and polynomial 2. The caret symbol (^) should be used to denote exponentiation. That is, x^2 means x*x. But, instead of x^1, just use x, and instead of x^0, use nothing. Terms where the coefficient is 0 must be omitted.

Example

In this example we multiply the following two sets of polynomials:
x^4+2x^3+x^2+5x and x^3+2x^2+1
x^3+x+1 and 2x^3+x+2

Input

2
4 1 2 1 5 0
3 1 2 0 1

3 1 0 1 1
3 2 0 1 2

Output

x^7 + 4x^6 + 5x^5 + 8x^4 + 12x^3 + x^2 + 5x
2x^6 + 3x^4 + 4x^3 + x^2 + 3x + 2

 

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This problem is used for contest: 187  198 

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多项式乘法，就是开两个数组遍历相乘结果存在第三个数组里。

注意输出格式很变态...刚开始写本地运行能过，一提交就超时...

由于最初设想处理“+”时把从第二个起全部写成“ + %dx^%d"，于是在for里开了个while找第一项。果断超时。

后来把“+”处理和数字处理分开，AC。

#include <stdio.h>
#include <string.h>

int main()
{
    freopen("in.txt", "r", stdin);

    int sample, high1, high2, i, j, k, t;
    int a[30], b[30], c[60];

    scanf("%d", &sample);

    while (sample--)
    {
        memset(c, 0, sizeof(a));

        scanf("%d", &high1);
        for (i=high1; i>=0; --i)
        {
            scanf("%d", &a[i]);
        }
        scanf("%d", &high2);
        for (i=high2; i>=0; --i)
        {
            scanf("%d", &b[i]);
        }

        for (i=0; i<high1+1; ++i)
        {
            for (j=0; j<high2+1; ++j)
            {
                c[i+j] += a[i]*b[j];
            }
        }


        //处理是0
        int flagprint = 1;
        for (k=high1+high2; k>=0; --k)
        {
            if (c[k])
            {
                flagprint = 0;
            }
        }
        if (flagprint)
        {
            printf("0");
        }

        int flag = 0;
        for (k=high1+high2; k>=0; --k)
        {
            if (0 != c[k])
            {
                flag = 1;
                if (1 != c[k] && 0 != k)
                {
                    printf("%d", c[k]);
                }
                else if (0 == k)
                {
                    printf("%d", c[k]);
                }

                if (k != 0)
                {
                    printf("x");
                }
                if (k > 1)
                {
                    printf("^%d", k);
                }
            }
            if (0 != k && flag)
            {
                if (0 != c[k-1])
                {
                    printf(" + ");
                }

            }
        }

        printf("\n");
    }

    return 0;
}