zzulioj - 2624: Small strange addition of H

Topic links: http://acm.zzuli.edu.cn/problem.php?id=2624

Title Description

H is very small like algorithm, in particular the various additions. Yes addition contains a variety of such binary full adder, semi-additive and the like.
Full adders: two input data bits are added, a result output and a carry bit, adder has a carry input.
Half adder: the sum of the two input data bits, a result output and a carry bit, there is no carry input of the adder. XOR operation is half adder. C, '^' is the exclusive-OR operator
, but commonly are decimal addition, people used in decimal, and the default is every decimal digit. A few days ago, I heard a small H ADD star has a new addition, then went ADD star. He found in ADD star's world, each number can be different every hex, this strange and cumbersome number is called "Add a few." For example, "0687 ......" represents the least significant bit is on the 7-ary number, the second bit is the inverse of octal, 3 bit binary number is 6, No. 4 is 10 decimal number, and the like. Each bit is binary 0 or d (decimal), or an integer in the range [2,9] interval. In theory this should be included into the tabulation infinite number of digits, into the tabulation is not given, it is considered to be decimal.

Entry

Firstly, a row of N bits into a first tab (0 <N <= 20), ending with a carriage return. Subsequently two rows, each row is given not more than one N-bit non-negative number of Add.

Export

Add the number of output and two in a row.

Sample input  Copy

7170687
03236
445

Sample output  Copy

4104

prompt

For example, corresponds to a tab into "0687", how to calculate "3236 + 445" mean? First we have calculated the lowest level: 6 + 5 = 11; 7 because the least significant bit is a binary, 4 and so we get a carry. Penultimate 2 is: 4 + 3 + 1 (carry) = 8; because this is the octal digit, so we get 0 and a carry. Bit 3: 4 + 2 + 1 (carry) = 7; 6 because this bit is a decimal, so we get 1 and a carry. The first four are: 3 + 1 (carry) = 4; because this bit is a decimal, so we get 7. Finally, we get: 3236 + 445 = 4104.

 

In fact, the deformation is large integer addition, I am in another blog post and there was a template, you can reference https://www.cnblogs.com/shuitiangong/p/12063528.html

#include<set>
#include<map>
#include<stack>
#include<queue>
#include<cmath>
#include<cstdio>
#include<cctype>
#include<string>
#include<vector>
#include<climits>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define max(a, b) (a > b ? a : b)
#define min(a, b) (a < b ? a : b)
#define mst(a) memset(a, 0, sizeof(a))
#define _test printf("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\n")
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
const double eps = 1e-7;
const int INF = 0x3f3f3f3f;
const ll ll_INF = 0x3f3f3f3f3f3f3f;
const int maxn = 1e3+10;
char ls[20];
char num1[maxn], num2[maxn], res[maxn];
void add() { //高精度整数加法(改)
    mst(res);
    int len1 = strlen(num1);
    int len2 = strlen(num2);
    int len3 = strlen(ls);
    reverse(num1, num1+len1);
    reverse(num2, num2+len2);
    int len, carry;
    len = carry = 0;
    for (int i = 0; i<len1 || i<len2; ++i) {
        carry += i<len1 ? num1[i] - '0' : 0;
        carry = I + <LEN2 num2 [I] -? ' 0 ' : 0 ;
         IF (LS [LEN3-I- . 1 !] = ' 0 ' ) { // non ~ 9 0 2 is the binary 
            res [len] = % with Carry (LS [LEN3-I- . 1 ] - ' 0 ' ) + ' 0 ' ; 
            with Carry / = (LS [LEN3-I- . 1 ] - ' 0 ' ); 
        } 
        the else { // 0 10 to press into the prepared 
            RES [len] with Carry =% 10 + ' 0 '; 
            With Carry / = 10 ; 
        }
         ++ len; 
    } 
    IF (with Carry) // determined last is not a carry- 
        RES [len ++] with Carry + = ' 0 ' ; 
    RES [len] = 0 ;
     for ( int I = len - . 1 ; I> 0 ; Inc. (www.i-levelmedia.com)) { // exclude the influence of the pilot's attention is not directly 0 i> = 0, the result is exactly equal to 0 if the error occurs 
        IF (RES [I] == ' 0 ' ) 
            RES [ I] = 0 ;
         the else  
            BREAK  ;
    }
    reverse(res, res+strlen(res));
}   
int main(void) {
    scanf("%s %s %s", ls, num1, num2);
    add();
    printf("%s\n", res);
    return 0;
}