程序设计综合实践 1.8

无符号大数加、减运算。程序设计中经常遇到无符号大数加、减运算问题,请在样例程序Ex1.4基础上实现无符号大数减运算。题目要求输入两个无符号大数,保证一个大数不小于第二个大数,输出它们的和、差。

输入格式:
两个无符号大数,前一个大于等于第二个。

输出格式:
第1行为两个无符号大数相加结果,后一行为两个无符号大数相减结果。

输入样例:1234567890987654321333888999666
147655765659657669789687967867
输出样例:1234567890987654321333888999666+147655765659657669789687967867=1382223656647311991123576967533
1234567890987654321333888999666-147655765659657669789687967867=1086912125327996651544201031799
作者李卫明
单位杭州电子科技大学
代码长度限制16 KB
时间限制400 ms
内存限制64 MB

Java版本:

package TestCode.class2;

import java.util.ArrayList;
import java.util.Scanner;

public class BigNumber {
    public static int[] split_Number(String str){
        str = new StringBuffer(str).reverse().toString();

        char[] str_arr = str.toCharArray();
        int count=0;
        int segment = str.length()/8;//java会自动向下取整
        int[] number_Array =new int[segment+1];
        String input_str="";
        for (int i = 0; i < segment; i++) {
            for (int j = 0; j < 8; j++) {
                int index = 8*i+j;
                input_str = input_str.concat(str_arr[index]+"");
            }
            input_str = new StringBuffer(input_str).reverse().toString();
            number_Array[i]=Integer.parseInt(input_str);
            input_str="";

            if(i==(segment-1)){
                input_str="";
                for (int j = 0; j < str.length()-8*(i+1); j++) {
                    int index = 8*(i+1)+j;
                    input_str = input_str.concat(str_arr[index]+"");
                }
                input_str = new StringBuffer(input_str).reverse().toString();
                number_Array[i+1]=Integer.parseInt(input_str);
            }
        }
        return number_Array;
    }
    public static ArrayList input_number(){
        //8位一段
        Scanner sc=new Scanner(System.in);
        String str="";
        ArrayList arr = new ArrayList();
        while (sc.hasNext()){
            str = sc.nextLine();
            arr.add(str);
        }
        return arr;
    }
    public static void bigNumber_add(int[] int_arr1,int[] int_arr2){
        int minSegment = (int_arr1.length>int_arr2.length) ?int_arr2.length:int_arr1.length;
        int maxSegment = (int_arr1.length>int_arr2.length) ?int_arr1.length:int_arr2.length;
        int segmentLength=8;
        int[] resSet = new int[maxSegment];

        for (int i = 0; i < maxSegment; i++) {
            int count=0;
            if(i<=minSegment){
                if(count==0){
                int number_int = (int_arr1[i]+int_arr2[i]); //* ( (int)Math.pow(10,i*8) );
                    //有进位
                count = (new String(number_int+"")).length() > int_arr1.length?1:0;
                resSet[i]=number_int;
                }else{
                   int number_int =  int_arr1[i]+int_arr2[i]+1;
                   count = (new String(number_int+"")).length() > int_arr1.length?1:0;
                   resSet[i]=number_int;
                }
            }else{
                int number_int = (int_arr1.length>int_arr2.length) ?int_arr1[i] : int_arr2[i];
                resSet[i]=number_int;
            }
        }
        for (int i = resSet.length-1; i >=0; i--) {
            System.out.print(resSet[i]);
        }
        System.out.println();

    }
    public static String sub_Normalization(int number){
        String str="";
        if(number<100000000) str=""+number;
        if(number<10000000) str="0"+number;
        if(number<1000000) str="00"+number;
        if(number<100000) str="000"+number;
        if(number<10000) str="0000"+number;
        if(number<1000) str="00000"+number;
        if(number<100) str="000000"+number;
        if(number<10) str="0000000"+number;
        if(number==0) str="00000000";
        return str;
    }
    public static void bigNumber_sub(int[] int_arr1,int[] int_arr2){
        int minSegment = (int_arr1.length>int_arr2.length) ?int_arr2.length:int_arr1.length;
        int maxSegment = (int_arr1.length>int_arr2.length) ?int_arr1.length:int_arr2.length;
        int segmentLength=8;
        int[] resSet = new int[maxSegment];
        int count=0;
        for (int i = 0; i < maxSegment; i++) {
            if(i<=minSegment){
                if(count==0){
                    if( (int_arr1[i]-int_arr2[i]) <0 ){
                        resSet[i] = int_arr1[i]-int_arr2[i]+100000000;
                        count=1;
                    }
                    else {
                        int number_int = (int_arr1[i]-int_arr2[i]); //* ( (int)Math.pow(10,i*8) );
                        resSet[i] = number_int;
                    }
                }else{
                    if( (int_arr1[i]-int_arr2[i]-1) <0 ){
                        resSet[i] = int_arr1[i]-int_arr2[i]-1+100000000;
                        count=1;
                    } else {
                        int number_int = (int_arr1[i]-int_arr2[i]-1); //* ( (int)Math.pow(10,i*8) );
                        resSet[i] = number_int;
                    }
                }

            }else{
                int number_int = (int_arr1.length>int_arr2.length) ?int_arr1[i] : int_arr2[i];
                resSet[i]=number_int;
            }
        }
        for (int i = resSet.length-1; i >=0; i--) {
            if(i==(resSet.length-1)) System.out.print(resSet[i]);
            else {
                System.out.print(sub_Normalization(resSet[i]));
            }
            }
        System.out.println();
    }

    public static void main(String[] args) {
        ArrayList arr = input_number();
        int[] int_arr1 = split_Number(arr.get(0).toString());
        int[] int_arr2 = split_Number(arr.get(1).toString());
        bigNumber_add(int_arr1,int_arr2);
        bigNumber_sub(int_arr1,int_arr2);
    }
}
大数是指两个超出计算机数据类型范围的大整数进行运算。一般情况下,我们需要使用字符串来存储这两个大整数,然后模拟手算的过程来得到最终结果。以下是一个简单的C语言实现大数的代码示例: ``` #include <stdio.h> #include <string.h> #define MAX_LEN 1000 void reverse(char *s) { int len = strlen(s); for (int i = 0; i < len / 2; i++) { char tmp = s[i]; s[i] = s[len - i - 1]; s[len - i - 1] = tmp; } } void add(char *a, char *b, char *result) { int carry = 0; int len_a = strlen(a); int len_b = strlen(b); int len = len_a > len_b ? len_a : len_b; for (int i = 0; i < len; i++) { int da = i < len_a ? a[i] - '0' : 0; int db = i < len_b ? b[i] - '0' : 0; int sum = da + db + carry; result[i] = sum % 10 + '0'; carry = sum / 10; } if (carry > 0) { result[len] = carry + '0'; result[len + 1] = '\0'; } else { result[len] = '\0'; } reverse(result); } int compare(char *a, char *b) { int len_a = strlen(a); int len_b = strlen(b); if (len_a > len_b) { return 1; } else if (len_a < len_b) { return -1; } else { return strcmp(a, b); } } void sub(char *a, char *b, char *result) { int borrow = 0; int len_a = strlen(a); int len_b = strlen(b); if (compare(a, b) < 0) { char *tmp = a; a = b; b = tmp; len_a = strlen(a); len_b = strlen(b); result[0] = '-'; } int i, j; for (i = len_a - 1, j = len_b - 1; i >= 0 && j >= 0; i--, j--) { int da = a[i] - '0'; int db = b[j] - '0'; int diff = da - db - borrow; if (diff < 0) { diff += 10; borrow = 1; } else { borrow = 0; } result[i] = diff + '0'; } for (; i >= 0; i--) { int da = a[i] - '0'; int diff = da - borrow; if (diff < 0) { diff += 10; borrow = 1; } else { borrow = 0; } result[i] = diff + '0'; } int len = strlen(result); int pos = 0; while (pos < len && result[pos] == '0') { pos++; } if (pos == len) { result[0] = '0'; result[1] = '\0'; } else { if (result[0] == '-') { pos--; len--; } for (int i = pos; i < len; i++) { result[i - pos] = result[i]; } result[len - pos] = '\0'; } } int main() { char a[MAX_LEN], b[MAX_LEN], result[MAX_LEN]; printf("输入两个大整数:\n"); scanf("%s%s", a, b); add(a, b, result); printf("法结果:%s\n", result); sub(a, b, result); printf("法结果:%s\n", result); return 0; } ``` 该程序中使用 `reverse` 函数将字符串翻转,方便从低位到高位进行计算;使用 `add` 函数实现大数法,使用 `sub` 函数实现大数法;使用 `compare` 函数比较两个大整数的大小,以确定法的顺序。在输入时,需要保证输入的字符串没有前导零。
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