HDU 1002 A + B Problem II

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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

复制代码

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   2
1 2
112233445566778899 998877665544332211

Sample Output

复制代码

Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

我勒个去，差点被虐死，这么简单的题，开始我写的一个内联函数max和algorithm里面的函数写得不一样
  
  
  
  
   
   交3次才AC，WA的我给你们几个数据：
  
  
  
  
   
   0001 1000
  
  
  
  
   
   0 0
  
  
  
  
   
   000 0000
  
  
  
  
   
   9999 1
  
  
  
  
   
   1 9999
  
  
  
  
   
   99900 00999
  
  
  
  
   
   00999 99900
  
  
  
  
   
   这几个数据和样例全过的话，应该可以AC了

LANGUAGE:C++
  
  
  
  
   
   CODE:
  
  
  
  
   
   #include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stdlib.h>

#define maxn 1005

using namespace std;

char ans[maxn];
int anslen;

void plus(char s1[],char s2[])
{
	int len1=strlen(s1);
	int len2=strlen(s2);

for(int i=0;i<len1;i++)
		s1[i]-='0';
	for(int i=0;i<len2;i++)
		s2[i]-='0';

int mid1=len1>>1;
	int mid2=len2>>1;

//printf("%d %d %d %dn",len1,len2,mid1,mid2);

for(int i=0;i<mid1;i++)
		swap(s1[i],s1[len1-1-i]);
	for(int i=0;i<mid2;i++)
		swap(s2[i],s2[len2-1-i]);

anslen=max(len1,len2);
	
	//printf("n anslen= %dn",anslen);

int s=0,sum;
	for(int i=0;i<=anslen;i++)
	{
		sum=(s1[i]+s2[i]+s);
		ans[i]=sum%10;
		s=sum/10;
	}
}

void printans(char ans[],int anslen)
{
	//if(ans[anslen+1]==0)
	while(anslen>0&&(ans[anslen]==0))anslen--;
	//else printf("%d",ans[anslen+1]);
	if(anslen==0)
	{
		printf("%dn",ans[0]);
		return;
	}
	for(int i=anslen;i>-1;i--)
	{
		printf("%d",ans[i]);
	}
	printf("n");
}

int main()
{
	int cas;
	char s1[maxn],s2[maxn];
	scanf("%d",&cas);
	for(int i=1;i<=cas;i++)
	{
		memset(s1,0,sizeof(s1));
		memset(s2,0,sizeof(s2));
		scanf("%s%s",s1,s2);
		printf("Case %d:n",i);
		printf("%s + %s = ",s1,s2);
		plus(s1,s2);
		printans(ans,anslen);
		if(i!=cas)printf("n");
	}
	return 0;
}

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   Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110
  
  
  
  

  
  
  
  
   
   我勒个去，差点被虐死，这么简单的题，开始我写的一个内联函数max和algorithm里面的函数写得不一样
  
  
  
  
   
   交3次才AC，WA的我给你们几个数据：
  
  
  
  
   
   0001 1000
  
  
  
  
   
   0 0
  
  
  
  
   
   000 0000
  
  
  
  
   
   9999 1
  
  
  
  
   
   1 9999
  
  
  
  
   
   99900 00999
  
  
  
  
   
   00999 99900
  
  
  
  
   
   这几个数据和样例全过的话，应该可以AC了
  
  
  
  

  
  
  
  
   
   LANGUAGE:C++
  
  
  
  
   
   CODE:
  
  
  
  
   
   复制代码#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stdlib.h>

#define maxn 1005

using namespace std;

char ans[maxn];
int anslen;

void plus(char s1[],char s2[])
{
	int len1=strlen(s1);
	int len2=strlen(s2);

	for(int i=0;i<len1;i++)
		s1[i]-='0';
	for(int i=0;i<len2;i++)
		s2[i]-='0';

	int mid1=len1>>1;
	int mid2=len2>>1;

	//printf("%d %d %d %dn",len1,len2,mid1,mid2);

	for(int i=0;i<mid1;i++)
		swap(s1[i],s1[len1-1-i]);
	for(int i=0;i<mid2;i++)
		swap(s2[i],s2[len2-1-i]);

	anslen=max(len1,len2);
	
	//printf("n anslen= %dn",anslen);

	int s=0,sum;
	for(int i=0;i<=anslen;i++)
	{
		sum=(s1[i]+s2[i]+s);
		ans[i]=sum%10;
		s=sum/10;
	}
}

void printans(char ans[],int anslen)
{
	//if(ans[anslen+1]==0)
	while(anslen>0&&(ans[anslen]==0))anslen--;
	//else printf("%d",ans[anslen+1]);
	if(anslen==0)
	{
		printf("%dn",ans[0]);
		return;
	}
	for(int i=anslen;i>-1;i--)
	{
		printf("%d",ans[i]);
	}
	printf("n");
}

int main()
{
	int cas;
	char s1[maxn],s2[maxn];
	scanf("%d",&cas);
	for(int i=1;i<=cas;i++)
	{
		memset(s1,0,sizeof(s1));
		memset(s2,0,sizeof(s2));
		scanf("%s%s",s1,s2);
		printf("Case %d:n",i);
		printf("%s + %s = ",s1,s2);
		plus(s1,s2);
		printans(ans,anslen);
		if(i!=cas)printf("n");
	}
	return 0;
}#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stdlib.h>

#define maxn 1005

using namespace std;

char ans[maxn];
int anslen;

void plus(char s1[],char s2[])
{
	int len1=strlen(s1);
	int len2=strlen(s2);

	for(int i=0;i<len1;i++)
		s1[i]-='0';
	for(int i=0;i<len2;i++)
		s2[i]-='0';

	int mid1=len1>>1;
	int mid2=len2>>1;

	//printf("%d %d %d %dn",len1,len2,mid1,mid2);

	for(int i=0;i<mid1;i++)
		swap(s1[i],s1[len1-1-i]);
	for(int i=0;i<mid2;i++)
		swap(s2[i],s2[len2-1-i]);

	anslen=max(len1,len2);
	
	//printf("n anslen= %dn",anslen);

	int s=0,sum;
	for(int i=0;i<=anslen;i++)
	{
		sum=(s1[i]+s2[i]+s);
		ans[i]=sum%10;
		s=sum/10;
	}
}

void printans(char ans[],int anslen)
{
	//if(ans[anslen+1]==0)
	while(anslen>0&&(ans[anslen]==0))anslen--;
	//else printf("%d",ans[anslen+1]);
	if(anslen==0)
	{
		printf("%dn",ans[0]);
		return;
	}
	for(int i=anslen;i>-1;i--)
	{
		printf("%d",ans[i]);
	}
	printf("n");
}

int main()
{
	int cas;
	char s1[maxn],s2[maxn];
	scanf("%d",&cas);
	for(int i=1;i<=cas;i++)
	{
		memset(s1,0,sizeof(s1));
		memset(s2,0,sizeof(s2));
		scanf("%s%s",s1,s2);
		printf("Case %d:n",i);
		printf("%s + %s = ",s1,s2);
		plus(s1,s2);
		printans(ans,anslen);
		if(i!=cas)printf("n");
	}
	return 0;
}