HDOJ-1002大数相加

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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

#include <iostream>#include<cstdio>#include<string>#include<string.h>void add(char a[],char b[],char c[]);int main(int argc, char *argv[]) {int x;scanf("%d",&x);for(int i=1;i<=x;i++){char a[1005],b[1005],c[1005];a[0]=b[0]='0';//数组首位置0，方便最高位进位 scanf("%s",&a[1]);scanf("%s",&b[1]);printf("Case %d:\n",i);add(a,b,c);printf("%s + %s = ",&a[1],&b[1]);if(c[0]!='0')printf("%c",c[0]);printf(i==x?"%s\n":"%s\n\n",&c[1]);}return 0;}void add(char a[],char b[],char c[]){int lenA,lenB,lenC;lenA=strlen(a);lenB=strlen(b);lenC=(lenA>lenB)?lenA:lenB;//数组c长度为a，b较长长度。 int carry=0,t=0;c[lenC]='\0';//字符串以'\0'做结束标志。 for(int i=lenC-1;i>=0;i--){lenA--;lenB--;if(lenA>=0&&lenB>=0){t=a[lenA]-'0'+b[lenB]-'0'+carry;carry=0;//a, b数组的末位开始相加，取其数值做判断加完进位后carry置0 if(t>9){c[i]=t-10+'0';//数值大于10 ，取其个位，进位carry置1. carry=1;}else c[i]=t+'0';}if(lenA>=0&&lenB<0){// a数组位数较长，将a数组多余字符赋给c。 t=a[lenA]-'0'+carry;carry=0;if(t>9){c[i]=t-10+'0';carry=1;}else c[i]=t+'0';}if(lenB>=0&&lenA<0){//b数组较长 t=b[lenB]-'0'+carry;carry=0;if(t>9){c[i]=t-10+'0';carry=1;}else c[i]=t+'0';}}}

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