HDU1002 Problem II(大数相加)(C++题解)
A题 HDOJ1002(大数加法)
A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 483720 Accepted Submission(s): 93351
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
2 1 2 112233445566778899 998877665544332211
Sample Output
Case 1: 1 + 2 = 3 Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110
代码如下:
#include <iostream> #include <cstdio> #include <cstring> using namespace std; int main(){ char a[8000],b[8000]; int an[8000],bn[8000],sum[8000],pre,flag=1; int t; scanf("%d",&t); while(t--){ memset(an,0,sizeof(an)); memset(bn,0,sizeof(bn)); memset(sum,0,sizeof(sum)); scanf("%s%s",a,b); int len1=strlen(a); int len2=strlen(b); int len=(len1>len2)?len1:len2; pre=0; for(int i=0;i<len1;i++){ an[len1-i-1]=a[i]-'0'; } for(int j=0;j<len2;j++){ bn[len2-j-1]=b[j]-'0'; } for(int k=0;k<len;k++){ sum[k]=an[k]+bn[k]+pre/10; pre=sum[k]; } if(pre>9){ sum[len]=pre/10%10; len++; pre/=10; } printf("Case %d:\n",flag++); printf("%s + %s = ",a,b); for(int s=0;s<len;s++){ printf("%d",sum[len-s-1]%10); } printf("\n"); if(t){ printf("\n"); } } return 0; }