Winter-1-C A + B II 解题报告及测试数据

Time Limit:1000MS

Memory Limit:32768KB

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

以下是代码：


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#include <iostream>

#include <cstring>

#include <cstdio>

using namespace std;

char a[1000],b[1000];

int c[1100],n;

void cal(char a[],char b[],int r)

{

    int i = strlen(a),j = strlen(b);

    int k=0,t;

    i--;j--;

    memset(c,0,sizeof(c));

    while(i>=0 || j >=0){

        if(i>=0 && j>=0)t=a[i]+b[j]+c[k]-‘0‘-‘0‘;

        else if (i<0)t=b[j]+c[k]-‘0‘;

        else t=a[i]+c[k]-‘0‘;

        if(t>=10){

            c[k++]=t%10;c[k]+=1;

        }else c[k++]=t;

        i--;j--;

    }

    while(c[k]==0)k--;

    printf("Case %d:\n%s + %s = ",r,a,b);

    for(i=k;i>=0;i--)printf("%d",c[i]);

    printf("\n");

    if(r!=n)printf("\n");

}

int main(){

    int len1,len2;

    cin >> n;

    for(int i=0;i<n;i++){

        scanf("%s%s",a,b);

        cal(a,b,i+1);

    }

}
时间： 2024-10-13 16:09:18

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