如下，想要求(a+b)和(a-b)的精确值，并显示计算结果？

Question

设输入数据的最大长度不超过70位，且其中不得包含数字以外的其它符号。如果输入的数据不正确则显示“Input error.”。例如，如果输入是：
a=1234567890
b=112233445566778899
则显示结果为：
a+b=112233446801346789
a-b=-112233444332211009

Answer 1

#include <iostream>#include <algorithm>#include<cstdlib>#include<string>#include<cstring>using namespace std;inline int max(int a,int b){return a>b?a:b;}class LONG{public:static const int size=1000,base=10;int len,num[size];//len存储位长，num[i]存储第i位的值bool flag;//true=+,false=-friend int cmp(LONG &a,LONG &b);//必须加friend，否则编译器以为cmp为内部函数LONG()bool iszero(){return len==1&&num[0]==0;}LONG(char *s){//以char类表示的数值转换为 LONG类；注意基都是base,//并且s必须以'\0'结尾，因为要用到strlen（）函数memset(num,0,sizeof(num));int i;len=strlen(s);&nbsp;&nbsp;for(i=0; i<len; i++)&nbsp;num[i]=s[len-i-1]-'0';while(len>0&&num[len-1]==0) len--;if(len==0)len++;if(len>1&&(s[0]=='-'||s[0]=='+')){len--;&nbsp;&nbsp;num[len]=0;}if(s[0]=='-')flag=false;else flag=true;}void print(int k=0){//输出int i;if(k!=0&&flag)printf("+");if(!flag)printf("-");printf("%d",num[len-1]);for(i=len-2; i>=0; i--) printf("%d",num[i]);}LONG operator+(LONG &b){LONG operator-(LONG &b);if((flag^b.flag)==true){if(b.flag==false){LONG c(b);c.flag=true;c=*this-c;return c;}else {LONG c(*this);c.flag=true;c=b-c;return c;}}LONG c;int i,t;if(len>b.len) c.len=len; else c.len=b.len;for(i=t=0; i<c.len; i++){t+=num[i]+b.num[i];c.num[i]=t%base;t/=base;}if(t>0)c.flag=flag;return c;}LONG operator-(LONG &b){if(b.iszero())return *this;if(this->iszero()){LONG c=b;c.flag=1^b.flag;return c;}&nbsp;&nbsp;if((flag^b.flag)==true){if(b.flag==false){LONG c(b);c.flag=true;return c+*this;}else {LONG c(b);c.flag=false;return c+*this;}}LONG c;int i,t;int key=cmp(*this,b);if(key==0)return LONG("0");c.len=max(len,b.len);if(key<0){c.flag=false;if(b.flag==false){for(i=t=0; i<c.len; i++){t=num[i]-b.num[i]-t;if(t>=0) c.num[i]=t,t=0;else c.num[i]=t+base,t=1;}}else{for(i=t=0; i<c.len; i++){t=b.num[i]-num[i]-t;if(t>=0) c.num[i]=t,t=0;else c.num[i]=t+base,t=1;}}}else {c.flag=true;if(flag==false){for(i=t=0; i<c.len; i++){t=b.num[i]-num[i]-t;if(t>=0) c.num[i]=t,t=0;else c.num[i]=t+base,t=1;}}else{for(i=t=0; i<c.len; i++){t=num[i]-b.num[i]-t;if(t>=0) c.num[i]=t,t=0;else c.num[i]=t+base,t=1;}}}&nbsp;&nbsp;while(c.len>1&&c.num[c.len-1]==0)c.len--;return c;}void chx(){//规格化while(len>1&&num[len-1]==0)len--;}};inline int cmp(LONG &a,LONG &b){if((a.flag^b.flag)==true){if(a.flag==true)return 1;return -1;}if(a.flag==true){if(b.len>a.len)return -1;if(b.len<a.len)return 1;for(int i=a.len-1;i>=0;i--){if(b.num[i]>a.num[i])return -1;if(b.num[i]<a.num[i])return 1;}}if(a.flag==false){if(b.len>a.len)return 1;if(b.len<a.len)return -1;for(int i=a.len-1;i>=0;i--){if(b.num[i]>a.num[i])return 1;if(b.num[i]<a.num[i])return -1;}}return 0;}&nbsp;int main(){char *s1=new char[71];char *s2=new char[71];//题目要求不超过70位LONG a,b,c;for(printf("请输入a、b：\n");scanf("%s%s",s1,s2);printf("\n请输入a、b：\n")){&nbsp;&nbsp;a=LONG(s1);b=LONG(s2);&nbsp;&nbsp;printf("a+b=");c=a+b;c.print();printf("\na-b=");c=a-b;c.print();}return 0;}

Answer 2

建议采用字符串方式的加减法代替整型数据的加减运算，这需要编程实现，并且减法需要考虑借位（先判别2个字符的大小，然后操作），加法需要考虑进位，并判别2字符串a[],b[]的长度：如if（a[k]<b[k]) { result[k]='9'-b[k]+a[k]+1; carry=TRUE; } //得到结果的第k位ASCII码。else { result[k]=a[k]-b[k]; carry=FALSE; }

Answer 3

#include <iostream>#include <algorithm>#include<cstdlib>#include<string>#include<cstring>using namespace std;inline int max(int a,int b){return a>b?a:b;}class LONG{public:static const int size=1000,base=10;int len,num[size];//len存储位长，num[i]存储第i位的值bool flag;//true=+,false=-friend int cmp(LONG &a,LONG &b);//必须加friend，否则编译器以为cmp为内部函数LONG()bool iszero(){return len==1&&num[0]==0;}LONG(char *s){//以char类表示的数值转换为 LONG类；注意基都是base,//并且s必须以'\0'结尾，因为要用到strlen（）函数memset(num,0,sizeof(num));int i;len=strlen(s);&nbsp;&nbsp;for(i=0; i<len; i++)&nbsp;num[i]=s[len-i-1]-'0';while(len>0&&num[len-1]==0) len--;if(len==0)len++;if(len>1&&(s[0]=='-'||s[0]=='+')){len--;&nbsp;&nbsp;num[len]=0;}if(s[0]=='-')flag=false;else flag=true;}void print(int k=0){//输出int i;if(k!=0&&flag)printf("+");if(!flag)printf("-");printf("%d",num[len-1]);for(i=len-2; i>=0; i--) printf("%d",num[i]);}LONG operator+(LONG &b){LONG operator-(LONG &b);if((flag^b.flag)==true){if(b.flag==false){LONG c(b);c.flag=true;c=*this-c;return c;}else {LONG c(*this);c.flag=true;c=b-c;return c;}}LONG c;int i,t;if(len>b.len) c.len=len; else c.len=b.len;for(i=t=0; i<c.len; i++){t+=num[i]+b.num[i];c.num[i]=t%base;t/=base;}if(t>0)c.flag=flag;return c;}LONG operator-(LONG &b){if(b.iszero())return *this;if(this->iszero()){LONG c=b;c.flag=1^b.flag;return c;}&nbsp;&nbsp;if((flag^b.flag)==true){if(b.flag==false){LONG c(b);c.flag=true;return c+*this;}else {LONG c(b);c.flag=false;return c+*this;}}LONG c;int i,t;int key=cmp(*this,b);if(key==0)return LONG("0");c.len=max(len,b.len);if(key<0){c.flag=false;if(b.flag==false){for(i=t=0; i<c.len; i++){t=num[i]-b.num[i]-t;if(t>=0) c.num[i]=t,t=0;else c.num[i]=t+base,t=1;}}else{for(i=t=0; i<c.len; i++){t=b.num[i]-num[i]-t;if(t>=0) c.num[i]=t,t=0;else c.num[i]=t+base,t=1;}}}else {c.flag=true;if(flag==false){for(i=t=0; i<c.len; i++){t=b.num[i]-num[i]-t;if(t>=0) c.num[i]=t,t=0;else c.num[i]=t+base,t=1;}}else{for(i=t=0; i<c.len; i++){t=num[i]-b.num[i]-t;if(t>=0) c.num[i]=t,t=0;else c.num[i]=t+base,t=1;}}}&nbsp;&nbsp;while(c.len>1&&c.num[c.len-1]==0)c.len--;return c;}void chx(){//规格化while(len>1&&num[len-1]==0)len--;}};inline int cmp(LONG &a,LONG &b){if((a.flag^b.flag)==true){if(a.flag==true)return 1;return -1;}if(a.flag==true){if(b.len>a.len)return -1;if(b.len<a.len)return 1;for(int i=a.len-1;i>=0;i--){if(b.num[i]>a.num[i])return -1;if(b.num[i]<a.num[i])return 1;}}if(a.flag==false){if(b.len>a.len)return 1;if(b.len<a.len)return -1;for(int i=a.len-1;i>=0;i--){if(b.num[i]>a.num[i])return 1;if(b.num[i]<a.num[i])return -1;}}return 0;}&nbsp;int main(){char *s1=new char[71];char *s2=new char[71];//题目要求不超过70位LONG a,b,c;for(printf("请输入a、b：\n");scanf("%s%s",s1,s2);printf("\n请输入a、b：\n")){&nbsp;&nbsp;a=LONG(s1);b=LONG(s2);&nbsp;&nbsp;printf("a+b=");c=a+b;c.print();printf("\na-b=");c=a-b;c.print();}return 0;}&nbsp;