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

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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

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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。。。