acm 总结之大数加法

以杭电1002为例

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 165579    Accepted Submission(s): 31635

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

 

Author

Ignatius.L

 

 

思路：计算机所能提供的加法很有限，不会超过100位，对于过于100位的甚至上千位上万位，我们只能寻求其他方法。这里我们采用数组存放各位数据，然后从个位开始模拟加法过程

代码如下：

#include "stdio.h"
#include "string.h"
#define N 1100
char s1[N],s2[N];         //存入要加的2个字符串
int a1[N],a2[N],a[N];   //a[N]存最后结果
int main()
{int i,j,k;
int m,n;
int q;
int r;
int l;
int x;
memset(s1,'0',sizeof(s1));  初始化s1
memset(s2,'0',sizeof(s2));  初始化s2
scanf("%d",&r);
q=0;
l=r;
while(r--)
{scanf("%s %s",s1,s2);
m=strlen(s1);
n=strlen(s2);

x=0;

memset(a1,0,sizeof(a1));
memset(a2,0,sizeof(a2));
memset(a,0,sizeof(a));

for(i=0,j=m-1;i<m;i++)     //采用倒叙存储2数组各个位
{a1[i]=s1[j]-'0';
j--;}
for(i=0,j=n-1;i<n;i++)
{a2[i]=s2[j]-'0';j--;
}

if(m>n)
k=m;
else
k=n;

for(i=0;i<=k;i++)        //注意：为防止溢出，我们在加时在最后多加一次，即i<=k 而不是i<k
{a[i]=a1[i]+a2[i]+x;
if(a[i]>=10)
{ a[i]-=10;
 x=1;

}
else
x=0;
}
q++;       //用来输出Case
printf("Case %d: ",q);

for(i=m-1;i>=0;i--)
printf("%d",a1[i]);

printf(" + ");

for(i=n-1;i>=0;i--)
printf("%d",a2[i]);

printf(" = ");

if(a[k]==0)k--;   //  特别注意：这里是用来判断 最后没有0输出的k值；即此步是为了在最后的结果输出时前边不输出0。 比如结果100，没有此步将输出0100

for(i=k;i>=0;i--)
printf("%d",a[i]);

if(q!=l)            //此处为判断输出的格式 ：相邻的两组数据空一行，最后一组数据的最后无空行
printf(" ");
else
printf(" ");

}

return 0;
}