Hdu 1002 A + B Problem II (高精度相加)

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 208611    Accepted Submission(s): 40103

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

大数相加
分别用字符串a，和字符串b，来存储两个不同的数，用字符串c进行保存
将字符串的每位通过 -'0'来得到相应的整数
从后往前，逐一对每位进行相加，如果结果>=10，则下一位的字符串+1，当前结果 -10，并保存
接下来对于剩下的位进行考虑，如果k>0则表明还有剩下，则把剩下的位，对c进行赋值

#include <stdio.h>#include <string.h>const int N = 1000;int main() {int n,p;//辅助变量char a[N],b[N],c[N]={0},d[N]={0};//c用于倒叙输出int len1,len2;int sum;scanf("%d",&n);for(int i=0;i<n;i++) {printf("Case %d:\n",i+1);scanf("%s%s",a,b);len1 = strlen(a);len2 = strlen(b);len1--;len2--;printf("%s + %s = ",a,b);p=0;while(len1 >=0 && len2>=0) {sum = c[p] + (a[len1]-'0')+(b[len2]-'0');if( sum >= 10) {c[p]=sum-10+'0';c[p+1]++;}else {c[p]=sum+'0';}p++;len1--;len2--;}//将剩下的位数补给cif(len1 >= 0) {while(len1 >=0 ) {sum = c[p]+a[len1]-'0';if( sum >= 10) {c[p]=sum-10+'0';c[p+1]++;}elsec[p]=sum+'0';p++;len1--;}}else if(len2 >= 0) {while(len2 >= 0) {sum = c[p]+b[len2]-'0';if( sum >= 10) {c[p]=sum-10+'0';c[p+1]=1;}elsec[p]=sum+'0';p++;len2--;}}for(int j=p-1,k=0;j>=0;j--,k++) {d[j]=c[k];}d[p]='\0';printf("%s\n",d);if( i+1 != n)//对于2组之间加空行printf("\n");memset(c,0,sizeof(c));//将数组c清零}return 0;}