-3 + 1

Question details

There is a series where all numbers are non-negative integers. You can perform one operation as follows (note that the following two steps must be completed in one complete operation ):

(1) Select a number not less than 3 and reduce it by 3.

(2) increase the number by 1.

How many operations can be performed at most?

Input Format:

Multiple groups of data. The first row of each group of data is a positive integer N, indicating the number of numbers in the series. (1 <= n <= 20000)

The second row contains N non-negative integers separated by spaces. Each integer cannot exceed 1000000.

Output Format:

Output a row of data in each group, indicating the maximum number of operations that can be performed.

Input example:

1

10

2

10 11

Output example:

4

10

Solutions

For each number X: Use sum to record the numbers x/3 that can be added. Use an array to save the remainder (0, 1, 2) after each number % 3.

It is required that the operation be performed every 3-3 times in sum, and the maximum number of operations plus 1 in 0, 1, 2 is allowed.

The greedy idea is used to finish all 2 operations first, then finish all 1 operations, and finally consider the case of 0.

# Include <iostream> # include <cstdio> # include <cstring> # include <cstdlib> # include <cmath> using namespace STD; typedef long ll; ll sum, A [3]; int main () {// freopen ("C: \ Users \ Administrator \ Desktop \ kd.txt", "r", stdin ); ll N; while (~ Scanf ("% LLD", & N) {sum = 0; ll res = 0; A [1] = A [2] = 0; For (INT I = 0; I <n; I ++) {ll X; scanf ("% LLD", & X); sum + = x/3; X % = 3; A [x] ++; // statistics} res = sum; If (sum> 0) RES + = A [2]; while (1) // 1 {ll TMP = min (sum> 1, a [1]); Res + = TMP; sum-= TMP; A [1]-= TMP; if (sum <1 | A [1] <1) break;} while (1) // 0 {ll TMP = sum/3; sum % = 3; sum + = TMP; Res + = TMP; If (sum <= 2) break;} printf ("% LLD \ n", Res);} return 0 ;}