A + B Problem II（利用字符串进行四则运算）

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 337902 Accepted Submission(s): 65560

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

由于超过整型能计算的范围，所以需要用字符串来特殊处理

AC代码

#include <stdio.h>
#include <string.h>

char s1[1000+10];
char s2[1000+10];
char s[1000+10];
int main()
{
    int n;
    scanf("%d",&n);
    int k=0;//输出样例序号
    int flag=0;//输出格式控制标记
    while(n--)
    {
        memset(s2,0,sizeof s1);
        memset(s2,0,sizeof s2);
        memset(s,0,sizeof s);
        scanf("%s",s1);
        getchar();
        scanf("%s",s2);
        getchar();
        int len1=strlen(s1);
        int len2=strlen(s2);
        int cnt=0;
        int ok=0;//用来记录是否进位
        if(len1>len2)
        {
            while(len1)
            {
                int cp;
                if(len2!=0)
                {
                    cp=ok+s1[--len1]+s2[--len2]-'0'-'0';
                }
                else
                    cp=ok+s1[--len1]-'0';
                if(cp<10)
                    {s[cnt++]=cp+'0';ok=0;}
                else
                    {s[cnt++]=cp%10+'0';ok=1;}
              //  printf("%c++len1=%d len2=%d++\n",s[cnt],len1,len2);
            }
            if(ok==1)
            s[cnt++]='1';
            s[cnt]='\0';
        }
        else if(len1<=len2)
        {
            while(len2)
            {
                int cp;
                if(len1!=0)
                {
                    cp=ok+s1[--len1]+s2[--len2]-'0'-'0';
                }
                else
                    cp=ok+s2[--len2]-'0';
                if(cp<10)
                    {s[cnt++]=cp+'0';ok=0;}
                else
                    {s[cnt++]=cp%10+'0';ok=1;}
                //printf("%c++len1=%d len2=%d++\n",s[cnt],len1,len2);
            }
            if(ok==1)
            s[cnt++]='1';
            s[cnt]='\0';
        }
        if(flag++!=0)
            printf("\n");//输出格式控制
        printf("Case %d:\n",++k);
        printf("%s + %s = ",s1,s2);
        int i;
        for(i=cnt-1; i>=0; i--)//正向储存，反向输出
            printf("%c",s[i]);
        printf("\n");
    }
    return 0;
}

我基本是按思路来写，看了一下别人的代码，简洁不少，算法一样

#include<stdio.h>
#include<string.h>
int main()
{
    char a[1000],b[1000],c[1001];
    int i,j=1,p=0,n,n1,n2;
    scanf("%d",&n);
    while(n)
    {
        scanf("%s %s",a,b);
        printf("Case %d:\n",j);
        printf("%s + %s = ",a,b);
        n1=strlen(a)-1;
        n2=strlen(b)-1;
        for(i=0;n1>=0||n2>=0;i++,n1--,n2--)//把我的二合一了，很厉害
        {
            if(n1>=0&&n2>=0){c[i]=a[n1]+b[n2]-'0'+p;}
            if(n1>=0&&n2<0){c[i]=a[n1]+p;}
            if(n1<0&&n2>=0){c[i]=b[n2]+p;}
            p=0;
            if(c[i]>'9'){c[i]=c[i]-10;p=1;}
        }
        if(p==1){printf("%d",p);}
        while(i--)
        {printf("%c",c[i]);}
        j++;
        if(n!=1){printf("\n\n");}
        else {printf("\n");}
        n--;
    }
}

这题还有个思路，就是把两个串变成等长，短串前面补0，最后直接加
例如123456 + 546 ，可以变成123456 + 000456 ，这样直接加起来，不用考虑长度问题
但是呢又要多一步把短串补0，也挺麻烦
目前还米有更好的思路。。。