hdu 1102A + B Problem II

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
这类型的题以前做过，就没做了，今天学弟让我找bug,找了老半天结果是输出问题，无语。我发现我们找bug总是先找算法错误，这也算是个提醒吧，先看数据结构，再看输出，最后看算法！

思路：把两个字符型的从对应的最后位置开始相加，开个数组保留中间结果，思路应该还是好想到的！

#include<stdio.h>#include<string.h> int main(){    int lena, lenb, m, i, j, l, jzs, c, wsh, k[10006], sum, ll;    char a[10006], b[10006];    scanf("%d",&sum);    for(ll = 0; ll < sum; ll++)    {        scanf("%s",a);        scanf("%s",b);        memset(k,0,sizeof(k));        lena = strlen(a);        lenb = strlen(b);        if(lena < lenb)//对长点的字串控制下        {                 char t[10006];             strcpy(t, b);            strcpy(b, a);            strcpy(a,t);            m = lenb;            lenb = lena;            lena = m;        }        jzs = 0;//进位        l = 0;        for(i = lena-1, j = lenb-1; j != -1; i--, j--)        {            wsh = (a[i]-'0')+(b[j]-'0')+jzs;//本位            if(wsh > 9)            {                jzs = 1;                wsh = wsh%10;            }            else jzs = 0;            k[l] = wsh;            l++;        }        for(; i != -1; i--)        {            wsh = (a[i] - '0')+jzs;            if(wsh > 9)            {                jzs = 1;                wsh = wsh%10;            }            else jzs = 0;            k[l] = wsh;            l++;        }        if(jzs == 1)        {            k[l] = 1;            l++;        }        printf("Case %d:\n",ll+1);        printf("%s + %s = ",a,b);        for(c = l-1; c != -1; c--)        {            printf("%d",k[c]);         }        printf("\n");        if(ll!=sum-1)        printf("\n");    }    return 0 ;}