A + B Problem II

InputThe 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.

2
1 2
112233445566778899 998877665544332211

Sample Output

Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

題解：大整數加法，字符串輸入，保存，轉為int型，存入對應數組，滿十進一

AC代碼：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using
 
 namespace std;
char a[1004],b[1004];
int aa[1004],bb[1004];
int main()
{
    int t;
    scanf("%d",&t);
    int flag=0;
    while(t--)
    {
        flag++;
        printf("Case %d:\n",flag);
        cin>>a>>b;
        memset(aa,0,sizeof(aa));
        memset(bb,0,sizeof(bb));
        for(int i=strlen(a)-1,k=0;i>=0;i--,k++)
        {
            aa[k]=a[i]-‘0‘;
        }
        for(int i=strlen(b)-1,k=0;i>=0;i--,k++)
        {
            bb[k]=b[i]-‘0‘;
        }
        int max=strlen(a)>strlen(b)?strlen(a):strlen(b);
        for(int i=0;i<max;i++)
        {
            aa[i+1]=aa[i+1]+(aa[i]+bb[i])/10;
            aa[i]=(aa[i]+bb[i])%10;
        }
        printf("%s + %s = ",a,b);
        int str=aa[max]>0?max:max-1;
        for(int i=str;i>=0;i--)
        {
            printf("%d",aa[i]);
        }
        printf("\n");
        if(t!=0)
            printf("\n");
    }
    return 0;
}

今天也是元氣滿滿的一天！good luck！

HDU 1002 B - A + B Problem II