CodeForces - 1517A Sum of 2050

A. Sum of 2050
time limit per test1 second
memory limit per test256 megabytes

A number is called 2050-number if it is 2050, 20500, …, (2050⋅10k for integer k≥0).

Given a number n, you are asked to represent n as the sum of some (not necessarily distinct) 2050-numbers. Compute the minimum number of 2050-numbers required for that.

Input
The first line contains a single integer T (1≤T≤1000) denoting the number of test cases.

The only line of each test case contains a single integer n (1≤n≤1018) denoting the number to be represented.

Output
For each test case, output the minimum number of 2050-numbers in one line.

If n cannot be represented as the sum of 2050-numbers, output −1 instead.

Example
input
6
205
2050
4100
20500
22550
25308639900
output
-1
1
2
1
2
36

Note
In the third case, 4100=2050+2050.

In the fifth case, 22550=20500+2050.

问题链接：CodeForces - 1517A Sum of 2050
问题简述：（略）
问题分析：（略）
AC的C++语言程序如下：

/* CodeForces - 1517A Sum of 2050 */#include <bits/stdc++.h>using namespace std;const int M = 2050;int main()
{int t;scanf("%d", &t);while (t--) {long long n, ans = -1;scanf("%lld", &n);if (n % M == 0) {ans = 0;n /= M;while (n) {ans += n % 10;n /= 10;}}printf("%lld\n", ans);}return 0;
}```

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