hdoj1002A + B Problem II

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 314386    Accepted Submission(s): 60948

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

 

一个大数相加问题用字符数组储存然后转移int型计算。

#include<cstdio>#include<cstring>#include<cstdlib> int main(){    char str1[1010],str2[1010];    int a[1010] = {0},b[1010] = {0};    int length1,length2;    int z;    int t;    int l=1;    scanf("%d",&t);    while(t --){        scanf("%s%s",str1,str2);        memset(a,0,sizeof(a));//清零数组        memset(b,0,sizeof(b));        length1 = strlen(str1);        length2 = strlen(str2);        for(int i = 0;i < length1;i ++)//反向存储容易进位            a[length1-i-1] = str1[i] - '0';        for(int i = 0;i < length2;i ++)            b[length2-i-1] = str2[i]- '0';                if(length1 > length2)            z = length1;        else             z = length2;        int p=0,q=0;        for(int i = 0;i <= z;i ++){            q = a[i]+b[i]+p;//进位设置            a[i] = q % 10;            p = q / 10;        }        printf("Case %d:\n",l);        printf("%s + %s = ",str1,str2);        if( a[z] ){            printf ( "%d",a[z] );        }        for( int i = z - 1;i >= 0;i -- )            printf( "%d", a[i] );        printf( "\n" );        if(t)            printf("\n");        l++;    }    return 0;}