HDOJ 1002 A + B Problem II(大数A+B)
发布时间:2020-12-14 02:47:08 所属栏目:大数据 来源:网络整理
导读:A + B Problem II Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 237322????Accepted Submission(s): 45702 Problem Description I have a very simple problem for you. Given two integers
A + B Problem II
Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 237322????Accepted Submission(s): 45702
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.
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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.
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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.
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Sample Input
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Sample Output
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解题方法: ? ? ? ? ? ? ?先将字符串反转存入整形数组,整形数组按位相加,若相加后大于十则前一位加一。相加结束后, ? ? ?倒叙逐个输出整形数组的每一位数。 注意:输出格式和南阳不一样,南洋的代码会PE. AC代码如下:
#include<stdio.h> #include<stdlib.h> #include<string.h> int max(int x,int y) { int z; z=x>y?x:y;//比较两个值大小 return z; } int main() { int lena,lenb,i,n,t=0,z,numa[1100],numb[1100]; char a[1100],b[1100]; scanf("%d",&n); while(n--) { memset(numa,sizeof(numa)); memset(numb,sizeof(numb));//int数组清零 scanf("%s%s",a,b); lena=strlen(a); lenb=strlen(b); for(i=0;i<lena;i++) numa[lena-1-i]=a[i]-'0'; for(i=0;i<lenb;i++) numb[lenb-1-i]=b[i]-'0'; z=max(lena,lenb); for(i=0;i<z;i++)//此处i值的范围就是两个字符串长度的最大者 { numa[i]=numa[i]+numb[i]; if(numa[i]>9)//处理相加大于十的情况 { numa[i]=numa[i]-10; numa[i+1]++; } } t++; printf("Case %d:n",t); printf("%s + %s = ",b); if(numa[z]==0)// 判断数组第z位是否为零 { for(i=z-1;i>=0;i--) printf("%d",numa[i]); printf("n"); } else { for(i=z;i>=0;i--) printf("%d",numa[i]); printf("n"); } if(n!=0)//每一次的输入遇上一次的输出之间空一行 ,n=0是输出的是最后一组数据,即不用空出一行 printf("n"); } return 0; } (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |