NYOJ-A+B Problem II

A+B Problem II
时间限制：3000 ms | 内存限制：65535 KB
难度：3
描述
I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.

A,B must be positive.

输入
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.
输出
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.
样例输入

2
1 2
112233445566778899 998877665544332211

样例输出

Case 1:
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

AC

#include <stdio.h>#include <string.h>int main(){    int n,i,j,k,q,m;    int len1,len2;    scanf("%d",&n);    for(i=1;i<=n;i++)    {        char a[10000]={0},b[10000]={0};        int c[10000]={0},x[10000]={0},y[10000]={0}; //！！！！！！清0         scanf("%s",a);        getchar();        scanf("%s",b);        len1=strlen(a);len2=strlen(b);        k=0;q=0;        for(j=len1-1;j>=0;j--)        {            x[k]=a[j]-'0';             k++;        }        for(j=len2-1;j>=0;j--)        {            y[q]=b[j]-'0';            q++;        }        m=k>q?k:q;        for(j=0;j<m;j++)        {            c[j]=x[j]+y[j];            if(c[m-1]>=10)//最高位如果大于10，要多加一位             {                m=m+1;            }            if(c[j]>=10)//进位             {                x[j+1]+=1;//进到下一位的某个数上                 c[j]-=10;            }        }        printf("Case %d:\n%s + %s = ",i,a,b);        for(j=m-1;j>=0;j--)        {printf("%d",c[j]);}        printf("\n");    }    return 0;}