HDU A + B Problem II 大数据

import static java.lang.System.out;
import java.math.BigInteger;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        BigInteger a, b;
        Scanner sc = new Scanner(System.in);
        int T = sc.nextInt();
        for (int i = 1; i <= T; i++) {
            a = sc.nextBigInteger();
            b = sc.nextBigInteger();
            out.println("Case " + i + ":");
            out.println(a + " + " + b + " = " + a.add(b));
            if (i < T)
                out.println();
        }
    }
}

java中更加快捷

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#define sc(x) scanf("%d",&x)
#define pf(x) printf("%d
",x)
#define PF(x) printf("%d",x)
#define CL(x,y) memset(x,y,sizeof(x))
using namespace std;
const int MAX = 1005;
char a[MAX], b[MAX];
int aLen, bLen, n;
int T, k, i, j;
void Add(char* a, char* b);
int main()
{
    sc(T);
    for(i = 1; i <= T; i++)
    {
        CL(a, 0);
        CL(b, 0);
        cin >> a >> b;
        printf("Case %d:
", i);
        cout << a << " + " << b << " = ";
        Add(a, b);
        if(i < T) printf("
");
    }
    return 0;
}
void Add(char* a, char* b)
{
    aLen = strlen(a);
    bLen = strlen(b);
    n = MAX - (aLen > bLen ? aLen : bLen);//设置等位线
    for(j = aLen-1, k = MAX; j >= 0; --j, --k)
    {
        a[k] = a[j]-'0';
        a[j] = 0;
    }
    for(j = bLen-1, k = MAX; j >= 0; --j, --k)
    {
        b[k] = b[j]-'0';
        b[j] = 0;
    }
    for(k = MAX; k >= n; --k)
        a[k] += b[k];
    for(k = MAX; k >= n; --k)
    {
        int temp = a[k]/10;//进位
        a[k-1] += temp;
        a[k] %= 10;
    }
    for(k = n+1; k <= MAX; k++)
//            cout << a[k] << endl;
        PF(a[k]);
        cout << endl;
}