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#2
ytunx2008-07-09 04:29
这是我写的，不知道对你有没有帮助
#include <iostream.h>
#include <string.h> #define MAX 200 class HugeInt { //重载运算符<< friend ostream &operator << (ostream &output, HugeInt &h); //重载运算符>> friend istream &operator >> (istream &input, HugeInt &h); public: HugeInt(char *str = ""); //重载运算符" = " HugeInt & operator = (HugeInt &h); //重载运算符" + " HugeInt & operator + (HugeInt &h); public: unsigned int num[MAX]; }; HugeInt::HugeInt(char *str) { for (int i=0; i<MAX; i++) num[i] = 0; //i为数组起始位置 i = MAX - strlen(str) ; for (; i<MAX; i++) num[i] = *str++ - '0'; } /////////////////////////////////////////////////////////// //定义重载函数 // << ostream & operator << (ostream &output, HugeInt &h) { for (int i=0; i<MAX; i++) if (h.num[i] != 0) break; //从第一个不为零的位置开始输出 for (; i<MAX; i++) output<<h.num[i]; return output; } // >> istream & operator >> (istream &input, HugeInt &h) { char buff[1000]; int i, j; cin>>buff; //计算输入的长度和数组开始位置 i = MAX - strlen(buff); for (j=0; i<MAX; i++, j++) h.num[i] = buff[j] - '0'; return input; } // = HugeInt & HugeInt::operator = (HugeInt &h) { //检查是否自我赋值 if (&h == this) return *this; for (int i=0; i<MAX; i++) num[i] = h.num[i]; return *this; } // + HugeInt & HugeInt::operator + (HugeInt &h) { static HugeInt sum; int n = 0; //进位 for (int i=MAX-1; i>0; i--) { //从最后一位开始相加,满10进1(n=1) sum.num[i] = (num[i] + h.num[i] + n)%10; n = (num[i] + h.num[i] + n)/10; } //如果最高位相加满10，数据溢出 if ((num[i] + h.num[i] + n)/10 == 1) { cerr<<"数据溢出!"<<endl; } else sum.num[i] = (num[i] + h.num[i] + n)%10; return sum; } ////////////////////////////////////////////////// //主函数 int main() { HugeInt a[20], b[20]; HugeInt sum[20]; int count; cin>>count; //case for (int i=0; i<count; i++) { cin>>a[i]>>b[i]; sum[i] = a[i] + b[i]; } cout<<endl; for (i=0; i<count; i++) { cout<<"case "<<i+1<<":"<<endl; cout<<a[i]<<" + "<<b[i]<<" = "<<sum[i]<<endl<<endl; } return 0; } |
ACM种有这么一题：
Problem Description
I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.
Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000.
Output
For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line is the an equation "A + B = Sum", Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases.
Sample Input
2
1 2
112233445566778899 998877665544332211
Sample Output
Case 1:
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110
我写了这么一个程序：
#include <iostream>
#include <string.h>
using namespace std;
char a[1001],b[1001];
char* add(char *a,char *b){
int i,j,k = 0,tmp[105],l1 = strlen(a),l2 = strlen(b);
for (i = l1-1,j = l2-1;i >= 0 && j >= 0;--i,--j)
tmp[k++] = a[i]+b[j]-'0'-'0';
for (;i >= 0;--i)
tmp[k++] = a[i]-'0';
for (;j >= 0;--j)
tmp[k++] = b[j]-'0';
tmp[k] = 0;
for (i = 0;i < k;++i){
tmp[i+1] += tmp[i]/10;
tmp[i] %= 10;
}
if (!tmp[k])
--k;
for (i = 0;i <= k;++i)
a[i] = tmp[k-i] + '0';
a[k+1] = '\0';
return a;
}
int main(){
char *r;
int t;
int i;
cin>>t;
if(t>=1&&t<=20)
{
for(i=1;i<=t;i++)
{
scanf("%s%s",a,b);
r = add(a,b);
cout<<"Case "<<i<<":"<<endl;
cout<<a<<" + "<<b<<" = "<<r<<endl;
if(i!=t)
cout<<endl;
}
}
return 0;
}
但提交时出现了以下错误：
Runtime Error
(ACCESS_VIOLATION
哪位大虾帮我下，不知道还有什么好的方法解决这题，快疯了。。。