Arithmetic Expressions

Problem Statement :

5-year-old Shinchan had just started learning mathematics. Meanwhile, one of his studious classmates, Kazama, had already written a basic calculator which supports only three operations on integers: multiplication , addition , and subtraction . Since he had just learned about these operations, he didn't know about operator precedence, and so, in his calculator, all operators had the same precedence and were left-associative.

As always, Shinchan started to irritate him with his silly questions. He gave Kazama a list of  integers and asked him to insert one of the above operators between each pair of consecutive integers such that the result obtained after feeding the resulting expression in Kazama's calculator is divisible by . At his core, Shinchan is actually a good guy, so he only gave lists of integers for which an answer exists.

Can you help Kazama create the required expression? If multiple solutions exist, print any one of them.

Input Format

The first line contains a single integer  denoting the number of elements in the list. The second line contains  space-separated integers  denoting the elements of the list.

Output Format

Print a single line containing the required expressoin. You may insert spaces between operators and operands.


You are not allowed to permute the list.
All operators have the same precedence and are left-associative, e.g.,  is interpreted as 
Unary plus and minus are not supported, e.g., statements like , , or  are invalid.

Solution :


                            Solution in C :

In  C  :

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {
  int n;
  scanf("%d", &n);

  unsigned char * ops = malloc(n);  // '+' - 1, '-' - 2, '*' - 3
  memset(ops, 3, n);

  int * nums = malloc(n * sizeof(unsigned int));
  for (int i = 0; i < n; i ++) {
    scanf("%d", &nums[i]);

  unsigned char t[1000][101][2];

  int i;
  for (i = 0; i < n; i ++) {
    if (0 == i) {
      t[i][nums[i]][0] = 1;
      t[i][nums[i]][1] = 1;
    else {
      int j;
      for (j = 1; j < 101; j ++) {
        if (t[i - 1][j][0]) {
          int tmp = (j + nums[i]) % 101;
          t[i][tmp][0] = j;
          t[i][tmp][1] = 1;
          if (0 == tmp) {

          tmp = (101 + j - nums[i]) % 101;
          t[i][tmp][0] = j;
          t[i][tmp][1] = 2;
          if (0 == tmp) {

          tmp = (j * nums[i]) % 101;
          t[i][tmp][0] = j;
          t[i][tmp][1] = 3;
          if (0 == tmp) {

      if (j < 101) {

  int p = 0;
  for (int j = i; j > 0; j --) {
    ops[j] = t[j][p][1];
    p = t[j][p][0];

  for (int i = 0; i < n - 1; i ++) {
    printf("%d", nums[i]);
    switch (ops[i + 1]) {
      case 1:
      case 2:
  printf("%d\n", nums[n - 1]);


  return 0;

                        Solution in C++ :

In  C++  :

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std; 

typedef struct {
	string opers;
	int remainder; 
} expr; 

int main(void){
	int N; 
	cin >> N; // 2 - 10,000
	vector<int> vals(N); 
	for(int i=0; i<N; i++) cin >> vals[i]; // each 1 - 100

	expr init_expr; 
	init_expr.remainder = -1; 
	vector<expr> results(101, init_expr); 
	results[vals[0]].remainder = vals[0]; 
	for(int step=1; step<N; step++){
		vector<expr> res2(101, init_expr); 
		int stepval = vals[step]; 
		for(int j=0; j<=100; j++){
			int amount = results[j].remainder; 
			if (amount==-1) continue; 
			int ans1 = (amount+stepval)%101; 
			int ans2 = ((amount-stepval)+101)%101; 
			int ans3 = (amount*stepval)%101; 
			res2[ans1].remainder = ans1; 
			res2[ans2].remainder = ans2; 
			res2[ans3].remainder = ans3; 
			res2[ans1].opers = results[j].opers+"+"; 
			res2[ans2].opers = results[j].opers+"-"; 
			res2[ans3].opers = results[j].opers+"*"; 
	cout << vals[0]; 
	const string& opstring = results[0].opers; 
	for(int i=1; i<N; i++){
		cout << opstring[i-1] << vals[i];
	cout << endl; 
	return 0; 

                        Solution in Java :

In  Java :

import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {
    static void printResult(int[] arr, String result){
       for(int i=0; i<arr.length;i++){
    public static boolean calculation(int index, long cur, String result){
        if(cur%101 ==0){
            //System.out.println("optimize" + cur);
            for(int i=result.length(); i<arr.length-1; i++)
            return true;
        if(index == arr.length-1){
            if(((long)arr[index]*cur)%101 ==0){
                return true;
            else if(((long)arr[index]+cur)%101 ==0){
                return true;
            else if((cur-(long)arr[index])%101 == 0){
                return true;
                return false;
        if(!calculation(index+1, ((long)arr[index]*cur)%101,result+"*")){
            if(!calculation(index+1, ((long)arr[index]+cur)%101,result+"+")){
                if(!calculation(index+1, ((long)cur-arr[index])%101,result+"-")){
                    return false;

        return true;
    static int[] arr;
    public static void main(String[] args) {
        Scanner in = new Scanner(;
        int a;
        a = in.nextInt();
        arr = new int[a];
        for(int i=0; i<a; i++){
           arr[i] = in.nextInt(); 
        calculation(1, arr[0],"");//System.out.println(result);


                        Solution in Python : 
In  Python3 :

import sys


ops = [lambda x,y: x + y, lambda x,y: x * y, lambda x,y: x - y]
op2s = ['+', '*', '-', '']

def exp(i, value, l):
    if i == n:
        return l if value % 101 == 0 else None
    if len(cache[i]) > 0 and value in cache[i]:
        return cache[i][value]
    for k in range(3):
        if exp(i + 1, ops[k](value, a[i]), l) != None:
            l[i - 1] = k
            cache[i][value] = l
            return l
    cache[i][value] = None
    return None

n = int(input())
a = [int(x) for x in input().strip().split(' ')]

cache = [{}] * n

l = exp(1, a[0], [3] * n)

for i in range(n):

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