A+B Problem II——大數定理
A+B Problem II
時間限制:3000 ms | 記憶體限制:65535 KB 難度: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.
- 輸入
- 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.
- 輸出
- 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.
- 樣例輸入
-
2 1 2 112233445566778899 998877665544332211
- 樣例輸出
-
Case 1: 1 + 2 = 3 Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110
-
這道題簡單一看只是大數定理,但其實比較坑爹的是 輸入00008+2應該輸出10,需要過濾前導0,這道題如果不是前導0我只用了半小時就做對了,還去拉了個大便,結果看討論的說有前導零,又費了一個小時,唉!!
#include<iostream> #include<string> #include<cstdio> using namespace std; int i,j,y,n,k,h,p,lena,lenb; int a[1000],b[1000],sum[1000],f[1000]; int main() { string a1,b1; cin>>n; for(y=1;y<=n;y++) { cin>>a1>>b1; lena=a1.length(); lenb=b1.length(); for(i=0;i<1000;i++) { a[i]=0;b[i]=0;f[i]=0; } for(i=lena-1;i>=0;i--) a[lena-1-i]=a1[i]-'0'; for(i=lenb-1;i>=0;i--) b[lenb-1-i]=b1[i]-'0'; k=0; for(i=0;i<lenb||i<lena;i++) { h=a[i]+b[i]+k; f[i]=h%10; k=h/10; } if(k!=0) f[i++]=k; cout<<"Case "<<y<<":"<<endl<< a1 <<" + "<< b1 <<" = "; p=0; for(j=i-1;j>=0;j--) { if(p==0&&f[j]==0) { continue; } else { p=1; cout<<f[j]; } } cout<<endl; } return 0; }
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