文章目录
一、Sum of Round Numbers
总结
一、Sum of Round Numbers
本题链接：Sum of Round Numbers
题目：
A. Sum of Round Numbers
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
A positive (strictly greater than zero) integer is called round if it is of the form d00...0. In other words, a positive integer is round if all its digits except the leftmost (most significant) are equal to zero. In particular, all numbers from 1 to 9 (inclusive) are round.
For example, the following numbers are round: 4000, 1, 9, 800, 90.The following numbers are not round: 110, 707, 222, 1001.
You are given a positive integer n (1≤n≤1e4). Represent the number n as a sum of round numbers using the minimum number of summands (addends). In other words, you need to represent the given number n as a sum of the least number of terms, each of which is a round number.
Input
The first line contains an integer t (1≤t≤1e4) — the number of test cases in the input. Then t test cases follow.
Each test case is a line containing an integer n (1≤n≤1e4).
Output
Print t answers to the test cases. Each answer must begin with an integer k — the minimum number of summands. Next, k terms must follow, each of which is a round number, and their sum is n. The terms can be printed in any order. If there are several answers, print any of them.
Example
input
5
5009
7
9876
10000
10
output
2
5000 9
1
7
4
800 70 6 9000
1
10000
1
10
本博客给出本题截图：
题意：把一个数的每一位按照如下例子拆分出来，例如9872 = 9000 + 800 + 70 + 2，输出拆分后的数字的个数和这些数字，对于这个例子我们需要输出4（拆开后4个数），9000 800 70 2
AC代码
#include <iostream> #include <string> using namespace std; int main() { int t; cin >> t; while (t -- ) { string a; cin >> a; int cnt = 0; for (int i = 0; i < a.size(); i ++ ) if (a[i] != '0') cnt ++; cout << cnt << endl; for (int i = 0; i < a.size(); i ++ ) { if (a[i] == '0') continue; cout << a[i]; for (int j = i; j < a.size() - 1; j ++ ) cout << 0; cout << ' '; } puts(""); } return 0; }
总结
水题，不解释
