杭电ACM1002初级题 求助

[sunyuke]

A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22627 Accepted Submission(s): 4035


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

[waizqfor]

[acdbxzyw]

用数组

[执假以为真]

简单写了一个，你自己稍微改下

#include <cstdlib>
#include <iostream>

using namespace std;

void Reverse(char array[], int size)
{
	int i = 0, j = size-1;
	char tmp;
	while(i<j)
	{
		tmp = array[i];
		array[i] = array[j];
		array[j] = tmp;
		i++;
		j--;
	}
}

void Add(char * a, char * b, char * c)
{
	int i = 0;
	int tmp, r = 0;
	while(a[i]!='\0' && b[i]!='\0')
	{
		tmp = a[i] - '0' + b[i] - '0' + r;
		if(tmp > 9)
		{
			r=1;
			tmp -= 10;
		}
		else
			r=0;
		
		c[i] = tmp + '0';
		i++;
	}
	if(a[i] == '\0' && b[i] == '\0')
	{
		if(r==1)
			c[i] = '1';
	}
	else if(a[i] == '\0')
	{
		if(r==0)
		{
			strcpy(c+i, b+i);
		}
		else
		{
			Add("1", b+i, c+i);
		}
	}
	else //  b[i] == '\0'
	{
		if(r==0)
		{
			strcpy(c+i, a+i);
		}
		else
		{
			Add("1", a+i, c+i);
		}
	}
	
	Reverse(c, strlen(c));	
}

int main()
{
	char a[] = {'1','1','2','2','3','3','4','4','5','5','6','6','7','7','8','8','9','9','\0'};
	char b[] = {'9','9','8','8','7','7','6','6','5','5','4','4','3','3','2','2','1','1','\0'};
	
	char c[1000];
	memset(c, 1000, 0);	
	Add(a, b, c);	
	cout<<c<<endl;	
	cin.get();
}

[deerwin1986]

大数啊 用INT数组快些
或者用CHAR数组占空间小些...

[sunyuke]

[Quote=引用 3 楼 nirendao 的回复:]
简单写了一个，你自己稍微改下


C/C++ code
#include <cstdlib>
#include <iostream>

using namespace std;

void Reverse(char array[], int size)
{
int i = 0, j = size-1;
char tmp;
while(i<j)
{
tmp = array[i];
array[i] = array[j];
array[j] = tmp;
i++;
j--;
}
}

void Add(char * a, char * b, char * c)
{
int i = 0;…
[/Quote]

VS2008一调试 满眼的烫烫烫烫烫烫

[sunyuke]

我自己刚写的 大家评点一下

#include <iostream>
using namespace std;
void add(char *large,char *small,const int llenght,const int slenght)
{
	char *answer = large;
	char *pLarge = &large[llenght]; 
	char *pSmall = &small[slenght]; 
	int count = llenght;
	while(pSmall != small)
	{
		answer[count] = *pLarge + *pSmall;
		pLarge--;
		pSmall--;
		count--;
	}
	answer[count] =*pLarge + *pSmall;
	for(count = llenght;count > 0;count--)
	{
			if(answer[count] > 9)
		{
			answer[count] -= 10;
			answer[count-1] += 1;
		}
	}
	count = 0;
	while(count < llenght)
		cout<<(int)answer[count++];
}
int main()
{ 
	int times;
	cin>>times;	
	char numA[1002];	
	char numB[1002];
	for(int i = 1;i <= times;i++)
	{
		cin>>numA;
		cin>>numB;
		cout<<"Case "<<i<<':'<<endl;
		cout<<numA<<' '<<'+'<<' '<<numB<<' '<<'='<<' ';
		int alenght = strlen(numA);
		int blenght = strlen(numB);
		int count = 0;
		for(;count < alenght;count++)
			numA[count] -= 48;
		count = 0;
		for(;count < blenght;count++)
			numB[count] -= 48;
		if(alenght > blenght)
			add(numA,numB,alenght,blenght);
		else
			add(numB,numA,blenght,alenght);
		cout<<endl;
		if(i != times)
			cout<<endl;
	}
	return 0;
}

[执假以为真]

我用的是DEV-CPP，用VS2008的话不清楚，但应该很好改吧

[执假以为真]

把
char c[1000];
memset(c, 1000, 0);

改成
char c[1000] = {0};

试试看， 我在g++下也通过了。

[sk_orc]

垃圾，递交上去绝对错误，题目都没看清楚！