九度OJ 1037：Powerful Calculator（强大的计算器）（大整数运算）

时间限制：1 秒

内存限制：32 兆

特殊判题：否

提交：1821

解决：528

题目描述：: Today, facing the rapid development of business, SJTU recognizes that more powerful calculator should be studied, developed and appeared in future market shortly. SJTU now invites you attending such amazing research and development work.

    In most business applications, the top three useful calculation operators are Addition (+), Subtraction (-) and Multiplication (×) between two given integers. Normally, you may think it is just a piece of cake. However, since some integers for calculation
in business application may be very big, such as the GDP of the whole world, the calculator becomes harder to develop.

    For example, if we have two integers 20 000 000 000 000 000 and 4 000 000 000 000 000, the exact results of addition, subtraction and multiplication are:

    20000000000000000 + 4000000000000000 = 24 000 000 000 000 000

    20000000000000000 - 4000000000000000 = 16 000 000 000 000 000

    20000000000000000 × 4000000000000000 = 80 000 000 000 000 000 000 000 000 000 000

    Note: SJTU prefers the exact format of the results rather than the float format or scientific remark format. For instance, we need "24000000000000000" rather than 2.4×10^16.

    As a programmer in SJTU, your current task is to develop a program to obtain the exact results of the addition (a + b), subtraction (a - b) and multiplication (a × b) between two given integers a and b.

输入：: Each case consists of two separate lines where the first line gives the integer a and the second gives b (|a| <10^400 and |b| < 10^400).

输出：: For each case, output three separate lines showing the exact results of addition (a + b), subtraction (a - b) and multiplication (a × b) of that case, one result per lines.

样例输入：

20000000000000000

4000000000000000

样例输出：

24000000000000000

16000000000000000

80000000000000000000000000000000

来源：: 2007年上海交通大学计算机研究生机试真题

思路：

考察大整数运算中的加减乘法，涉及符号，更繁琐一些。

我的思路用一个含202个int的数组来表示一个数，每个数只存储4位。

（1）只存4位的原因是：如果5位，乘法运算时能到到10位，超过了int的表示范围。

（2）用202个而不是101个的原因是：乘法运算最大可能位数加倍。

最高位表示符号位，其他如果没有数就全置零。

加减法运算选择要看两个数的符号是否相同，并不给予原来算式中的加减法。

代码：

#include <stdio.h>

#include <stdlib.h>

#include <string.h>

#define N 202

#define M 10000

int prase(char s[], int a[])

{

    memset(a, 0, N*sizeof(int));

    int n = strlen(s);

    int flag = 0;

    if (s[0] == '-' || s[0] == '+')

    {

        if (s[0] == '-')

            flag = 1;

        s++;

        n--;

    }   

    int i, j;

    char t[5];

    j = 0;

    for (i=n-1; i>=0; i-=4)

    {

        int jj = 0;

        for (int k=i-3; k<=i; k++)

        {

            if (k>=0)

                t[jj++] = s[k];

        }

        t[jj] = '\0';

        a[j++] = atoi(t);

    }

    a[N-1] = flag;

    return j;

}       

void printA(int a[], int n)

{

    int i;

    for (i=n-1; i>=0; i--)

    {

        if (a[i] != 0)

            break;

    }

    if (i < 0)

    {

        printf("0\n");

        return ;

    }

    int flag = a[N-1];

    if (flag == 1)

        printf("-");

    printf("%d", a[i]);

    for (int j=i-1; j>=0; j--)

        printf("%04d", a[j]);

    printf("\n");

}

int sub(int a[], int na, int b[], int nb, int c[])

{

    int i;

    for (i=0; i<na; i++)

    {

        if (a[i] < b[i])

        {

            a[i+1] --;

            a[i] += M;

        }

        c[i] += a[i]-b[i];

    }

    for (i=na-1; i>=0; i--)

    {

        if (c[i] != 0)

            break;

    }

    return i+1;

}

int plus(int a[], int na, int flag, int b[], int nb, int c[])

{

    memset(c, 0, N*sizeof(int));

    int nm = (na > nb) ? na : nb;

    int fa = a[N-1];

    int fb = (b[N-1]+flag)%2;

    int i;

    if (fa == fb)

    {

        for (i=0; i<=nm; i++)

        {

            c[i] += a[i] + b[i];

            if (c[i] >= M)

            {

                c[i+1] ++;

                c[i] -= M;

            }

        }

        c[N-1] = fa;

        if (c[nm] != 0)

            return nm + 1;

        else

            return nm;

    }

    int ns;

    for (i=nm; i>=0; i--)

    {

        if (a[i] != b[i])

            break;

    }

    if (i < 0)

        return 0;

    if (a[i] > b[i])

    {

        ns = sub(a, na, b, nb, c);

        c[N-1] = fa;

    }

    else

    {

        ns = sub(b, nb, a, na, c);

        c[N-1] = fb;

    }

    return ns;

}   

int mult(int a[], int na, int b[], int nb, int c[])

{

    memset(c, 0, N*sizeof(int));

    int i, j, k;

    for (i=0; i<na; i++)

    {

        for (j=0; j<nb; j++)

        {

            k = i+j;

            c[k] += a[i]*b[j];

            if (c[k] >= M)

            {

                c[k+1] += c[k]/M;

                c[k] %= M;

            }

        }

    }

    c[N-1] = (a[N-1]+b[N-1])%2;

    for (i=na+nb; i>=0; i--)

    {

        if (c[i] != 0)

            break;

    }

    return i+1;

}

int main(void)

{

    char sa[2*N], sb[2*N];

    int na, nb;

    int a[N], b[N];

    int np, ns, nm;

    int p[N], s[N], m[N];

    while (scanf("%s%s", sa, sb) != EOF)

    {

        na = prase(sa, a);

        nb = prase(sb, b);

        //printA(a, na);

        //printA(b, nb);

        np = plus(a, na, 0, b, nb, p);

        printA(p, np);

        ns = plus(a, na, 1, b, nb, s);

        printA(s, ns);

        nm = mult(a, na, b, nb, m);

        printA(m, nm);

    }       

    return 0;

}

/**************************************************************

    Problem: 1037

    User: liangrx06

    Language: C

    Result: Accepted

    Time:0 ms

    Memory:916 kb

****************************************************************/

巴特西

九度OJ 1037：Powerful Calculator（强大的计算器）（大整数运算）

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巴特西

九度OJ 1037：Powerful Calculator（强大的计算器） （大整数运算）

最新文章

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九度OJ 1037：Powerful Calculator（强大的计算器）（大整数运算）