ACM刷题之HDU————A + B Problem II

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A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3109 Accepted Submission(s): 1155

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

21 2112233445566778899 998877665544332211

Sample Output

Case 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110

一道高精度算法的题目，要自己手动模拟加法运算。

写完记得测试下998+2 之类的进位数据，有可能错在这里。

下面是ac代码（有点长，其实可以再缩的）

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<math.h>#include<algorithm>using namespace std;char a[5000];char b[5000];int  sum[20000];int q[5000];int w[5000]; int main(){int i,j,num,zu,zu1,len1,len2,c,x;scanf("%d",&zu);for(zu1=1;zu1<=zu;zu1++){scanf("%s",a);scanf("%s",b);len1=strlen(a);for(i=0;i<len1;i++)q[i]=a[i]-'0';len2=strlen(b);for(i=0;i<len2;i++)w[i]=b[i]-'0';i=len1;j=len2;x=0;memset(sum,0,sizeof(sum));if(len1<=len2){j--;x=0;c=0;for(i=i-1;i>=0;i--,j--){sum[x]=(q[i]+w[j]+c)%10;c=(q[i]+w[j]+c)/10;x++;}for(;j>=0;j--){sum[x]=(w[j]+c)%10;c=(w[j]+c)/10;x++;}if(c>0){sum[x]=c;printf("Case %d:\n",zu1);printf("%s + %s = ",a,b);for(x=x;x>0;x--)printf("%d",sum[x]);printf("%d\n",sum[0]);if(zu1<zu)printf("\n");continue;}printf("Case %d:\n",zu1);printf("%s + %s = ",a,b);for(x=x-1;x>0;x--)printf("%d",sum[x]);printf("%d\n",sum[0]);if(zu1<zu)printf("\n");}else{i--;x=0;c=0;for(j=j-1;j>=0;i--,j--){sum[x]=(q[i]+w[j]+c)%10;c=(q[i]+w[j]+c)/10;x++;}for(;i>=0;i--){sum[x]=(q[i]+c)%10;c=(q[i]+c)/10;x++;}if(c>0){sum[x]=c;printf("Case %d:\n",zu1);printf("%s + %s = ",a,b);for(x=x;x>0;x--)printf("%d",sum[x]);printf("%d\n",sum[0]);if(zu1<zu)printf("\n");continue;}printf("Case %d:\n",zu1);printf("%s + %s = ",a,b);for(x=x-1;x>0;x--)printf("%d",sum[x]);printf("%d\n",sum[0]);if(zu1<zu)printf("\n");}}}

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