# include <stdio.h>
# include <string.h>
# include <stdlib.h>
# include <iostream>
# include <fstream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <math.h>
# include <algorithm>
using namespace std;
# define pi acos(-1.0)
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define For(i,n,a) for(int i=n; i>=a; --i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define Fo(i,n,a) for(int i=n; i>a ;--i)
typedef long long LL;
typedef unsigned long long ULL;
char a[1000+5],b[1000+5],c[1000+5];
int main()
{
int t,sum=0;
cin>>t;
while(t--)
{
if(sum)cout<<endl;
char str1[1000+5],str2[1000+5];
mem(a,‘\0‘);
mem(b,‘\0‘);
mem(c,‘\0‘);//因为多组数据,所以清零
cin>>str1;
for(int i=sizeof(a),j=strlen(str1)-1; j>=0; j--,i--)
a[i]=str1[j]-48;
cin>>str2;
for(int i=sizeof(b),j=strlen(str2)-1; j>=0; j--,i--)//将字符串放在数组末尾,并且将字符转换成ASCLL码所对应的数字
b[i]=str2[j]-48;
int num=0,i;
for(i=sizeof(a); i>0; i--)
{
c[i]=a[i]+b[i]+num;
num=c[i]/10;
c[i]%=10;
}
printf("Case %d:\n%s + %s = ",++sum,str1,str2);
for(i=0;i<=10005;i++)if(c[i])break;
for(int j=i; j<=sizeof(c); j++)
printf("%d",c[j]);//将大数表示出来
cout<<endl;
}
return 0;
}
HDU 1002 A - A + B Problem II (大数问题)
原文:http://www.cnblogs.com/teble/p/7192089.html