两个大数相加问题！
代码如下所示。不知哪里出了什么逻辑上的问题，在处理进位上不正常。
求各位朋友帮忙指出。在此先行谢谢！
#include <iostream>
using namespace std;
#define ARRAY_SIZE 50
//Enter a big number, and store it as a string into an array ch,
//the size is the numbers of char.
void inputNumbers(char ch[], int& size);
//Reverse the elements of the array ch.
void reverseArray(char ch[], int size);
//Adding two big numbers, and the result will be stored in the array ch3,
//and return the size of the array ch3.
void computeAdding(char ch1[], int size1, char ch2[], int size2, char ch3[], int& size3);
//show the adding result.
void displayResult(char ch[], int size);
int main()
{
char ch1[ARRAY_SIZE], ch2[ARRAY_SIZE], result[ARRAY_SIZE];
int size1 = 0, size2 = 0, resultSize = 0;
cout << "Enter the first big number, ending with an enter:" << endl;
inputNumbers(ch1, size1);
cout << "Enter the second big number, ending with an enter:" << endl;
inputNumbers(ch2, size2);
reverseArray(ch1, size1);
reverseArray(ch2, size2);
computeAdding(ch1, size1, ch2, size2, result, resultSize);
displayResult(result, resultSize);
system("pause");
return 0;
}
//Function inputNumbers
void inputNumbers(char ch[], int& size)
{
char next;
cin.get(next);
while (next != '\n' && size < ARRAY_SIZE)
{
ch[size++] = next;
cin.get(next);
}
}//inputNumbers
//Function reverseArray
void reverseArray(char ch[], int size)
{
int i = 0, j = size-1;
char temp;
while (i <= j)
{
temp = ch[i];
ch[i] = ch[j];
ch[j] = temp;
i ++;
j --;
}
}//end reverseArray function
//function computeAdding
void computeAdding(char ch1[], int size1, char ch2[], int size2, char ch3[], int& size3)
{
int i = 0, j = 0;
char result;
while (i < size1 && j < size2)
{
result = ch1[i] + ch2[j] - 48;
if (result >= '10')
{
if (size1 == i && j < size2) //If size1 < size2
ch2[j] = ch2[j] + 1;
else
ch1[i] = ch1[i] + 1;
ch3[size3++] = result - '10' + '0';
}
else
ch3[size3++] = result;
i ++;
j++;
}
if (size1 < size2)//If the ch1 has more bits, execute this clause
{
for (i = size1; i < size2; i++)
{
ch3[size3++] = ch2[i] ;
}
}
else if (size1 > size2)
{
for (i = size2; i < size1; i++)
{
ch3[size3++] = ch2[i] ;
}
}
}//End reverseArray
//function displayResult
void displayResult(char ch[], int size)
{
reverseArray(ch, size);//make the number to be normal
cout << "The adding result is:" ;
for (int i = 0; i < size; i++)
cout << ch[i] ;
cout << endl;
}
[解决办法]
楼主，你大数相加那函数写的，有点乱，我重新给你写了个，你参考下：
void computeAdding(char ch1[], int size1, char ch2[], int size2, char ch3[], int& size3){ int i,cf,tmp; for( i=0,cf=0; i<size1&&i<size2; i++) { tmp = (ch1[i]-'0') + (ch2[i]-'0') + cf; ch3[i]= tmp%10+'0'; cf = tmp/10; } while( i<size1 ) { tmp = (ch1[i]-'0')+cf; ch3[i] = tmp%10+'0'; cf = tmp/10; i++; } while( i<size2 ){ tmp = (ch2[i]-'0') +cf; ch3[i] = tmp%10 +'0'; cf = tmp/10; i++; } if( cf ) { ch3[i]= cf+'0'; i++;} ch3[i]='\0'; size3 = i;}//End reverseArray
[解决办法]
//建议LZ不要用char类型进行相加运算，采用intvoid computeAdding(char ch1[], int size1, char ch2[], int size2, char ch3[], int& size3){ int i = 0, j = 0; char result; int result; int carry; while (i < size1 && j < size2) { result = ch1[i] + ch2[j] - 48; //这个地方出错了，如果是8+4得到的是12，这个时候已经不是数字字符 //这个地方需要判断进位 //最好使用int型进行操作,字符型不好判断进位 cout<<"reuslt="<<result<<endl; if (result >= '10') //这个地方也不对，没有10这个字符的 { if (size1 == i && j < size2) //If size1 < size2 ch2[j] = ch2[j] + 1; else ch1[i] = ch1[i] + 1; ch3[size3++] = result - '10' + '0'; } else ch3[size3++] = result; i++; j++; } if (size1 < size2)//If the ch1 has more bits, execute this clause { for (i = size1; i < size2; i++) { ch3[size3++] = ch2[i] ; } } else if (size1 > size2) { for (i = size2; i < size1; i++) { ch3[size3++] = ch2[i] ; } }}//End reverseArray
[解决办法]
//按照整数进行改写了一下void computeAdding(char ch1[], int size1, char ch2[], int size2, char ch3[], int& size3){ int i = 0, j = 0; int result = 0; int carry = 0;//初始进位为0 while(i < size1 && j < size2) { //计算两个数对应位的和，带进位的计算 result = (ch1[i]^0x30) + (ch2[j]^0x30)+carry; //取其最低位 ch3[size3++] = (result%10)^0x30; //进位 carry = result/10; i++; j++; } //第一个数位数大于第二个位数 while(i<size1) { int ret = (ch1[i]^0x30) +carry; ch3[size3++] = (ret%10)^0x30; carry = ret/10; i++; } //第二个数的位数大于第一个位数 while(j<size2) { int ret = (ch2[j]^0x30) +carry; ch3[size3++] = (ret%10)^0x30; carry = ret/10; j++; } //所以计算完成看是否有进位 if(i==size1 &&j==size2 && carry!=0) ch3[size3++] = carry^0x30; ch3[size3] = '\0'; }//End reverseArray
[解决办法]
#include <stdio.h>#include <string.h>#define MAXLEN 1000char a1[MAXLEN];char a2[MAXLEN];static int v1[MAXLEN];static int v2[MAXLEN];static int v3[MAXLEN];int i,j,n,L,z;void main(void) { scanf("%d",&n); for (j=0;j<n;j++) { scanf("%s%s",a1,a2); L=strlen(a1); for (i=0;i<L;i++) v1[i]=a1[L-1-i]-'0'; L=strlen(a2); for (i=0;i<L;i++) v2[i]=a2[L-1-i]-'0'; for (i=0;i<MAXLEN;i++) v3[i]=v1[i]+v2[i]; for (i=0;i<MAXLEN;i++) { if (v3[i]>=10) { v3[i+1]+=v3[i]/10; v3[i]=v3[i]%10; } } printf("Case %d:\n", j+1); printf("%s + %s = ", a1, a2); z=0; for (i=MAXLEN-1;i>=0;i--) { if (z==0) { if (v3[i]!=0) { printf("%d",v3[i]); z=1; } } else { printf("%d",v3[i]); } } if (z==0) printf("0"); printf("\n"); }}//Sample Input//3//0 0//1 2//112233445566778899 998877665544332211////Sample Output//Case 1://0 + 0 = 0//Case 2://1 + 2 = 3//Case 3://112233445566778899 + 998877665544332211 = 1111111111111111110