hdu1002(高精度)

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 181565    Accepted Submission(s): 34671

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

21 2112233445566778899 998877665544332211

 

Sample Output

Case 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110

 

Author

Ignatius.L

 

Recommend

We have carefully selected several similar problems for you:  1063 1316 1007 1133 2064 

      本题较简单。涉及到大数相加，可用字符串朱未处理，也可用Java 大叔类。

 

1.Java大数类

import java.math.BigInteger;import java.util.Scanner;public class hdu1002 {static Scanner input=new Scanner(System.in);public static void main(String[] str){while(input.hasNext()){int tag=1;int i=0;int m=input.nextInt();BigInteger a,b,c;while(i<m){a=input.nextBigInteger();b=input.nextBigInteger();c=a.add(b);System.out.println("Case "+(tag++)+":");System.out.println(a+" + "+b+" = "+c);}}}}

2.C++用字符串写

#include <stdio.h>#include <string.h>#define MAXLEN 1001int my_max(int a, int b){    return a > b? a : b;}void revert_string(const char *in, char *out){    int n = strlen(in);    while (n-- > 0) {        *out++ = in[n];    }    *out=0;}void *do_int_add(char *sa, char *sb, char *out){    int shift;    int salen = strlen(sa);    int sblen = strlen(sb);    int i, n;        n = my_max(salen, sblen);        if (salen < sblen) {        while (salen < sblen) {            sa[salen++] = '0';        }        sa[salen] = 0;    } else {        while (sblen < salen) {            sb[sblen++] = '0';        }        sb[sblen] = 0;    }    shift = 0;    for (i=0; i<n; i++) {        int sum = sa[i] - '0' + sb[i] - '0' + shift;        out[i] = (sum % 10) + '0';        shift = sum >= 10 ? 1:0;    }        if (shift) {        out[i] = '1';        out[i+1] = 0;    } else {        out[i] = 0;    }}void int_add(char *sa, char *sb, char *out){    char rsa[MAXLEN], rsb[MAXLEN], rout[MAXLEN];    revert_string(sa, rsa);    revert_string(sb, rsb);    do_int_add(rsa, rsb, rout);    revert_string(rout, out);}int main(){    char a[MAXLEN], b[MAXLEN], out[MAXLEN];    int i, n;    scanf("%d", &n);        if (n<1 || n>20)        return 0;    for (i=0; i<n; i++) {        scanf("%s", a);        scanf("%s", b);        int_add(a, b, out);        printf("Case %d:\n", i+1);        printf("%s + %s = %s\n", a, b, out);        if (i+1<n)            printf("\n");    }    return 0;}