HDU - 1002 大数A+B

朴灿烈づ我的快乐病毒、 2022-05-16 22:46 150阅读 0赞

#                                    A + B Problem II #

### 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 ###

1 2

112233445566778899 998877665544332211

### Sample Output ###

Case 1:

1 + 2 = 3

Case 2:

112233445566778899 + 998877665544332211 = 1111111111111111110

#include<stdio.h>
    #include<string.h>
    char a[1010],b[1010];
    int sum[1010];
    int main()
    {
    	int i,j,k,s,lena,lenb,t,d;
    	while(scanf("%d",&t)!=EOF)
    	{
    		d=t;
    		s=0;
    		while(t--)
    		{
    			s++;
    			scanf("%s%s",a,b);
    			lena=strlen(a);
    			lenb=strlen(b);
    			memset(sum,0,sizeof(sum));
    			k=0;
    			for(i=lena-1,j=lenb-1;i>=0&&j>=0;i--,j--)
    			{
    				sum[k]=a[i]-'0'+b[j]-'0'+sum[k];
    				if(sum[k]>=10)
    				{
    				sum[k]=sum[k]-10;
    				sum[k+1]=sum[k+1]+1;
    				}
    				k++;
    			}
    			while(j>=0)
    			{
    				sum[k]=b[j--]-'0'+sum[k];
    				if(sum[k]>=10)
    				{
    					sum[k]=sum[k]-10;
    					sum[k+1]=sum[k+1]+1;
    				}
    				k++;
    			}
    				
    			while(i>=0)
    			{
    				sum[k]=a[i--]-'0'+sum[k];
    				if(sum[k]>=10)
    				{
    					sum[k]=sum[k]-10;
    					sum[k+1]=sum[k+1]+1;
    				}
    				k++;
    			}
    				
    			k=1009;
    			while(k--)
    			{
    				if(sum[k]!=0)
    					break;
    			}
    			printf("Case %d:\n",s);
    			printf("%s + %s = ",a,b);
    			if(k==-1)
    				printf("0");
    			for(i=k;i>=0;i--)
    				printf("%d",sum[i]);
    			printf("\n");
    			if(s!=d)
    				printf("\n");
    		}
    	}
    	return 0;
    }