hdu1002(高精度加法運算)
A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 434867 Accepted Submission(s): 84639
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
這是一個高精度加法運算,已有的資料型別資料範圍不夠。只能另想它法。程式碼參考了老師另一個程式的程式碼,讓程式碼更加簡潔易懂。
程式思路:
1、用字元陣列儲存大數,此時使用scanf很方便。
2、注意讀入的大整數高位在陣列的低位。
3、另外一個結果陣列ans中首先將一個數組的值拷貝過來,ans陣列低位放數的低位,以便進位產生更高的位。
4、將另一個數組的數加到ans中,注意進位(carry)。
5、這兩個陣列的長度可能不一致。所以使用while(carry > 0){}(程式碼37行處),可以使高位產生的連續進位得到處理。
6、考慮到while(carry > 0){}中j++會產生多餘的位(為0)同時考慮到00…00 + 00…00 的情況,要去除前導0。(程式碼46行處)
#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
const int N = 1000;
const int BASE = 10;
char a[N+1],b[N+1],ans[N+1];
int main(){
int t,i,j,k,len,anslen,carry;
cin>>t;
for(k = 1;k <= t;k++){
scanf("%s%s",a,b);
memset(ans,0,sizeof(ans));
anslen = len = strlen(a);
for(i = len - 1,j =0;i >= 0;i--)
ans[j++] = a[i] - '0';
len = strlen(b);
if(len > anslen)
anslen = len;
carry = 0;
for(i = len-1,j=0;i >= 0;i--,j++){
ans[j] += b[i] - '0' + carry;
carry = ans[j] / BASE;
ans[j] %= BASE;
}
while(carry > 0){
ans[j] += carry;
carry = ans[j] / BASE;
ans[j++] %= BASE;
}
if(j > anslen)
anslen = j;
//去除前導0
for(i = anslen -1;i >= 0;i--)
if(ans[i])
break;
if(i < 0)
i = 0;
//輸出
cout<<"Case "<<k<<":"<<endl;
for(j = 0;j < strlen(a);j++)
cout<<a[j] - '0';
cout<<" + ";
for(j = 0;j < strlen(b);j++)
cout<<b[j] - '0';
cout<<" = ";
for(;i>=0;i--)
cout<<(int)ans[i];
cout<<endl;
if(k != t) cout<<endl;
}
return 0;
}
此時就可以AC了。