A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 317528    Accepted Submission(s): 61692 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   Author

Ignatius.L

原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=1002

大数相加思路非常简单，只要用数组将大数的每一位都存起来，注意进位。。。。。

AC代码：

#include<stdio.h> #include<string.h> #define Max 101 void print(char sum[]); void bigNumAdd(char a[],char b[],char sum[]); int main() { char a[Max]; char b[Max]; char sum[Max]; gets(a); gets(b); bigNumAdd(a,b,sum); print(sum); return 0; } void bigNumAdd(char a[],char b[],char sum[]) { int i=0; int c=0;//表示进位 //初始化,对以后位运算有很大帮助！ char m[Max]={0}; char n[Max]={0}; memset(sum,0,Max*sizeof(char)); //这里不能写成memset(sum,0,sizeof(sum));原因见注意事项1 //字符串反转且字符串变数字 int lenA=strlen(a); int lenB=strlen(b); for (i=0;i<lenA;i++) { m[i]=a[lenA-i-1]-'0'; } for (i=0;i<lenB;i++) { n[i]=b[lenB-i-1]-'0'; } //位运算 for (i=0;i<lenA||i<lenB;i++) { sum[i]=(m[i]+n[i]+c)%10+'0';//得到末位 c=(m[i]+n[i]+c)/10;//得到进位 } } void print(char sum[]) { int i=0; int j=0; int len = strlen(sum); for (i=len-1;sum[i]==0;i--); //找到第一个不为零的位置，方便输出 for (j=i;j>=0;j--) { printf("%c",sum[j]); } }