HDU - 1002 A + B Problem II

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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

题解：很明显是大数相加，但是它全部都是正数那更容易了，输入之后全部反序变成数字，两个数组个个相加，把进位记录，差不多就是这样，不过要注意换行之类的

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int a[1010],b[1010],sum[1010];int len_a,len_b,len_sum;void changeInt(char str[],int num[],int *l){    *l = strlen(str) - 1;    int index = 0;    for(int i = *l;i >= 0 ;i--){        num[i] = str[index++] - '0';    }}void add(int a[],int b[],int s[]){    int c = 0;    int m = 0;    while(len_a>=m || len_b>=m){        if(len_a<m)a[m] = 0;        if(len_b<m)b[m] = 0;        s[m] = a[m]+b[m]+c;        c = s[m]/10;        if(s[m]>9){            s[m]=s[m]%10;        }        m++;    }    if(c) s[m] = c;    len_sum = m - 1 + c;}void write(int num[],int l){    for(int i = l;i >= 0;i--)printf("%d",num[i]);}int main(){    int T;cin>>T;int count = 0;    for(int i = 0; i < T ; i++){        char m[1010],n[1010];        if(count != T && count)printf("\n");        cin>>m>>n;        changeInt(m,a,&len_a);changeInt(n,b,&len_b);        add(a,b,sum);        printf("Case %d:\n",++count);        write(a,len_a);        printf(" + ");        write(b,len_b);        printf(" = ");        write(sum,len_sum);        printf("\n");    }    return 0;}

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