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.
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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.
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.
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.
2
1 2
112233445566778899 998877665544332211
Case 1:
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110
#include<stdio.h> #include<string.h> int main() { int i,j,k,m,n,t,a[1002],len; char s1[1001],s2[1001]; scanf("%d",&t); for(n=0;n<t;n++) { scanf("%s%s",s1,s2); len=(strlen(s1)>=strlen(s2)?strlen(s1):strlen(s2)); for(i=0;i<len+1;i++) a[i]=0; m=0; for(i=strlen(s1)-1,j=strlen(s2)-1;i>=0&&j>=0;i--,j--){ k=s1[i]+s2[j]-2*‘0‘; if(k>=10) { a[m++]+=k%10; a[m]++; } else { a[m++]+=k; if(a[m-1]>=10) { a[m]++; a[m-1]%=10; } } } if(i>=0) { for(j=i;j>=0;j--) { k=s1[j]-‘0‘; if(k>=10) { a[m++]+=k%10; a[m]++; } else{ a[m++]+=k; if(a[m-1]>=10) { a[m]++; a[m-1]%=10; } } } } else if(j>=0) { for(i=j;i>=0;i--){ k=s2[i]-‘0‘; if(k>=10) { a[m++]+=k%10; a[m]++; } else { a[m++]+=k; if(a[m-1]>=10) { a[m]++; a[m-1]%=10; } } } } printf("Case %d:\n",n+1); printf("%s + %s = ",s1,s2); if(a[m]>0) m++; for(i=m-1;i>=0;i--) printf("%d",a[i]); if(n!=t-1) printf("\n"); } return 0; }
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原文地址：http://www.cnblogs.com/tong69/p/5677206.html
