[1002]: A + B Problem II

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 281561 Accepted Submission(s): 54201

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

#include<stdio.h>#include<string.h>#define N 1005int common(int a, int b){    if(a<b){        return a;    }    else{        return b;    }}int rest(int a, int b){    int temp;    if(b>a){        temp = b;         b = a;        a = temp;    }    return (a-b-1);}int main(){    char num1[N], num2[N], sum1[N], sum2[N]; // num存储输入的数,sum存储结果     int n;    while(scanf("%d", &n)!=EOF){ // n为要输入的行数         int k, i = 0, add=0;        for(; i<n; i++){            if(i){ // Output a blank line between two test cases.                printf("\n");            }            scanf("%s%s", num1, num2);            int len1 = strlen(num1); // 计算两个输入数的位数             int len2 = strlen(num2);            int long1 = len1-1, long2 = len2-1; // 对应的下标             if( len1 >= len2 ){                k = long2;                for(; k>=0; k--){                    sum2[k] = (num1[long1]-'0') + (num2[long2]-'0')+add;                    if(sum2[k]>=10){                        sum2[k] -= 10;                        add = ((num1[long1]-'0' )+ (num2[long2]-'0')+add)/10;                    }                    else{                        add = 0;                    }                    long1--;                    long2--;                }                k = len1-len2-1;                for(; k>=0; k--){                    sum1[k] = (num1[k]-'0')+add;                    if(sum1[k]>=10 && k>0){                        sum1[k] -= 10;                        add = ((num1[k]-'0')+add)/10;                    }                     else{                        add = 0;                    }                }            }            else{                k = long1;                for(; k>=0; k--){                    sum2[k] = (num1[long1]-'0') + (num2[long2]-'0')+add;                    if(sum2[k]>=10){                        sum2[k] -= 10;                        add = ((num1[long1]-'0' )+ (num2[long2]-'0')+add)/10;                    }                    else{                        add = 0;                    }                     long1--;                    long2--;                }                k = len2-len1-1;                for(; k>=0; k--){                    sum1[k] = (num2[k]-'0')+add;                    if(sum1[k]>=10 && k>0){                        sum1[k] -= 10;                        add = ((num2[k]-'0')+add)/10;                    }                     else{                        add = 0;                    }                }            }            int lenSum1 = rest(len1, len2);            int lenSum2 = common(len1, len2);            printf("Case %d:\n", i+1);            printf("%s + %s = ", num1, num2);            k = 0;            for(; k<=lenSum1; k++){                printf("%d", sum1[k]);            }            k = 0;            for(; k<lenSum2; k++){                printf("%d", sum2[k]);            }            printf("\n");        }    }    return 0;}

我的博客里面有两篇是关于这道题目的，主要是第一种思路是借鉴他人的（由于地址不明，所以没有注明），今天这种思路（个人解题结果：）是逆着第一种的思路