A + B Problem II HDU

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A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 388064    Accepted Submission(s): 75108

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

 

Author

Ignatius.L

大数a+b，两次比赛都没有做出来，老是出错，昨晚一直PE，无奈，竟然找不出来哪错了，，，最后一次，以后再也不能错了！！！

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;char a[1005],b[1005];int c[1005],d[1005],e[1005];int main(){    int t,kk=1;    scanf("%d",&t);    while(t--)    {        memset(c,0,sizeof(c));        memset(d,0,sizeof(d));        memset(e,0,sizeof(e));        scanf("%s%s",a,b);        int la=strlen(a),lb=strlen(b),k=0;        for(int i=la-1;i>=0;i--)            c[k++]=a[i]-'0';        k=0;        for(int i=lb-1;i>=0;i--)            d[k++]=b[i]-'0';        int le=max(la,lb);        for(int i=0;i<le;i++)        {            if(c[i]+d[i]+e[i]>9)                e[i+1]++;            e[i]=(c[i]+d[i]+e[i])%10;        }        printf("Case %d:\n",kk++);        printf("%s",a);        printf(" + ");        printf("%s",b);        printf(" = ");        if(e[le])            printf("%d",e[le]);        for(int i=le-1;i>=0;i--)            printf("%d",e[i]);            printf("\n");        if(t)            printf("\n");    }}