/*
A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 210207 Accepted Submission(s): 40482
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 1000 int main(){ int n; while(~scanf("%d",&n)){ //吞掉回车符 getchar(); int i,j,k,a,b,t; //数组定义 char schar1[N+10]; char schar2[N+10]; int sint1[N+10]; int sint2[N+10]; for(k=1;k<=n;k++){ memset(sint1,sizeof(sint1));//只能初始化int型 memset(sint2,sizeof(sint2)); scanf("%s %s",schar1,schar2); a=strlen(schar1); b=strlen(schar2); //反转赋值 for(i=0; i<a; i++) sint1[a-1-i]=schar1[i]-'0';//一层循环 for(i=0; i<b; i++) sint2[b-1-i]=schar2[i]-'0'; //将a定义为a,b中最大的 if(a<b){ t=a; a=b; b=t; } //判断是否进位 for(i=0;i<=a;i++){ sint1[i]+=sint2[i]; if(sint1[i]>9){ sint1[i]-=10; sint1[i+1]++; //此处不是i++ } } printf("Case %d:\n%s + %s = ",schar2); //判断是否有进位,不同情况不同输出 if(sint1[a]==0){ for(i=a-1;i>=0;i--) //逆序输出 printf("%d",sint1[i]); printf("\n"); //这个printf不在for循环内部 } if(sint1[a]!=0){ for(i=a;i>=0;i--) printf("%d",sint1[i]); printf("\n"); //这个printf不在for循环内部 } if(k!=n) printf("\n"); //最后一个不换行 } } return 0; }
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