时间限制：3000 ms | 内存限制：65535 KB
难度：3
描述
I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.
A,B must be positive.
输入
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.
输出
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.
样例输入
2 1 2 112233445566778899 998877665544332211样例输出
Case 1: 1 + 2 = 3 Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110来源
经典题目
#include<stdio.h> #include<string.h> int main() { char str1[1001],str2[1001]; int t,j=0; scanf("%d",&t); while(t--) { scanf("%s%s",str1,str2); int a[1001]= {0},b[1001]= {0},c[1001]= {0},i; int m,n,max1=0; m=strlen(str1); n=strlen(str2); max1=(m>n?m:n);//m,n大小，输出较大的数的长度 for(i=0; i<max1; i++) { a[m-i-1]=str1[i]-'0'; b[n-i-1]=str2[i]-'0'; } for(i=0; i<max1; i++) c[i]=a[i]+b[i]; for(i=0; i<max1; i++) { c[i+1]+=c[i]/10; c[i]=c[i]; } printf("Case %d:\n%s + %s = ",++j,str1,str2); //倒序输出和； if(c[max1]==0) max1--; for(i=max1; i>=0; i--) printf("%d",c[i]); printf("\n"); } return 0; }
