HDU A + B Problem II 大数据
1 import static java.lang.System.out; 2 import java.math.BigInteger; 3 import java.util.Scanner; 4 5 public class Main { 6 7 public static void main(String[] args) { 8 BigInteger a, b; 9 Scanner sc = new Scanner(System.in); 10 int T = sc.nextInt(); 11 for (int i = 1; i <= T; i++) { 12 a = sc.nextBigInteger(); 13 b = sc.nextBigInteger(); 14 out.println("Case " + i + ":"); 15 out.println(a + " + " + b + " = " + a.add(b)); 16 if (i < T) 17 out.println(); 18 } 19 } 20 }
java中更加快捷
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #define sc(x) scanf("%d",&x) 6 #define pf(x) printf("%d\n",x) 7 #define PF(x) printf("%d",x) 8 #define CL(x,y) memset(x,y,sizeof(x)) 9 using namespace std; 10 const int MAX = 1005; 11 char a[MAX], b[MAX]; 12 int aLen, bLen, n; 13 int T, k, i, j; 14 void Add(char* a, char* b); 15 int main() 16 { 17 sc(T); 18 for(i = 1; i <= T; i++) 19 { 20 CL(a, 0); 21 CL(b, 0); 22 cin >> a >> b; 23 printf("Case %d:\n", i); 24 cout << a << " + " << b << " = "; 25 Add(a, b); 26 if(i < T) printf("\n"); 27 } 28 return 0; 29 } 30 void Add(char* a, char* b) 31 { 32 aLen = strlen(a); 33 bLen = strlen(b); 34 n = MAX - (aLen > bLen ? aLen : bLen);//设置等位线 35 for(j = aLen-1, k = MAX; j >= 0; --j, --k) 36 { 37 a[k] = a[j]-'0'; 38 a[j] = 0; 39 } 40 for(j = bLen-1, k = MAX; j >= 0; --j, --k) 41 { 42 b[k] = b[j]-'0'; 43 b[j] = 0; 44 } 45 for(k = MAX; k >= n; --k) 46 a[k] += b[k]; 47 for(k = MAX; k >= n; --k) 48 { 49 int temp = a[k]/10;//进位 50 a[k-1] += temp; 51 a[k] %= 10; 52 } 53 for(k = n+1; k <= MAX; k++) 54 // cout << a[k] << endl; 55 PF(a[k]); 56 cout << endl; 57 }
