hdu1002 ,大數相加,一點都不難,嘿嘿嘿
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
上程式碼
#include<iostream> #include<string.h> using namespace std; char a[1001],b[1001]; int n1[1001],n2[1001]; int main() { int x,w=0; cin>>x; while(x--) { int l1,l2,i,j,max; for(int l=0;l<1001;l++) { n1[l]=0; n2[l]=0; } cin>>a>>b; l1=strlen(a); l2=strlen(b); if(l1>l2)max=l1; else max=l2; for(i=l1-1,j=0; i>=0 ;i--) { n1[j]=a[i]-48; j++; } for(i=l2-1,j=0; i>=0 ;i--) { n2[i]=b[j]-48; j++; } for(i=0;i<max;i++) { n1[i]=n1[i]+n2[i]; if(n1[i]>9) { n1[i+1]=n1[i+1]+1; n1[i]=n1[i]-10; } } w++; cout<<"Case "<<w<<":"<<endl; cout<<a<<" + "<<b<<" = "; if(n1[i]) for(j=i;j>-1;j--) cout<<n1[j]; else for(j=i-1;j>-1;j--) cout<<n1[j]; cout<<endl; if(x) cout<<endl; } return 0; }