本文主要是介绍CF——Average Numbers，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
Description
You are given a sequence of positive integers a1, a2, ..., an. Find all such indices i, that the i-th element equals the arithmetic mean of all other elements (that is all elements except for this one).
Input
The first line contains the integer n (2 ≤ n ≤ 2·105). The second line contains elements of the sequence a1, a2, ..., an (1 ≤ ai ≤ 1000). All the elements are positive integers.
Output
Print on the first line the number of the sought indices. Print on the second line the sought indices in the increasing order. All indices are integers from 1 to n.
If the sought elements do not exist, then the first output line should contain number 0. In this case you may either not print the second line or print an empty line.
Sample Input
5 1 2 3 4 5
1 3
4 50 50 50 50
4 1 2 3 4
题解：给出n个正整数，从第一个数开始遍历，若第i个数等于其他数之和的平均数，计数器加1，最后，输出符合条件的个数以及按照遍历顺序输出该数的序号，要注意的是，整数的序号从1开始算起。
#include <iostream> #include <stdio.h> #include <string.h> #include <algorithm>using namespace std;int shu[200000],sh[200000];int main() {int n,i,j,k;int sum = 0;scanf("%d",&n);for(i = 0; i<n; i++){scanf("%d",&shu[i]);sum = sum + shu[i];}int num = 0,mm,nn;j = 0;for(i = 0; i<n; i++){mm = sum - shu[i];nn = (n-1) * shu[i];if(mm == nn){num++;sh[j++] = i+1;}}printf("%d\n",num);if(num == 0)printf("\n");else{for(i = 0;i<j;i++){if(i == 0)printf("%d",sh[i]);elseprintf(" %d",sh[i]);}printf("\n");}return 0; }
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