Codeforces Round #640 (Div. 4)---题目+详解+代码(全题目）

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≤104). 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≤104) — 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≤104).

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
inputCopy
5
5009
7
9876
10000
10
outputCopy
2
5000 9
1
7
4
800 70 6 9000
1
10000
1
10

题意：
题意就是按权分解，给你一个数，需要你把这个数各个位上的数按权拿出来，比如1234 分解出1000+200+30+4。

思路：
将上述9876各个位置的数取出，6,7,8,9,同时别忘了，还要还原出具体位置，6(6),7(70),8(800),9(9000).
题目比较简单，就不多说了，直接贴代码啦

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
using namespace std;
int main()
{
   
	int n, m, i, j[10], k, t, l, o;
	scanf("%d", &n);
	while (n--)
	{
   
		scanf("%d", &m);
		for (i = 0, t = 0, l = 1; m>0; i++)
		{
   
			j[i] = (m % 10)*l;
			if (j[i])
			{
   
				t++;
			}
			l *= 10;
			m = m / 10;
		}
		o = i;
		printf("%d\n", t);
		for (i = 0; i < o; i++)
		{
   
			if (j[i])
			{
   
				printf("%d ", j[i]);
			}
		}
		printf("\n");
	}
	return 0;
}

B. Same Parity Summands

time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are given two positive integers n (1≤n≤109) and k (1≤k≤100). Represent the number n as the sum of k positive integers of the same parity (have the same remainder when divided by 2).

In other words, find a1,a2,…,ak such that all ai>0, n=a1+a2+…+ak and either all ai are even or all ai are odd at the same time.

If such a representation does not exist, then report it.

Input
The first line contains an integer t (1≤t≤1000) — the number of test cases in the input. Next, t test cases are given, one per line.

Each test case is two positive integers n (1≤n≤109) and k (1≤k≤100).

Output
For each test case print:

YES and the required values ai, if the answer exists (if there are several answers, print any of them);
NO if the answer does not exist.
The letters in the words YES and NO can be printed in any case.

Example
input
8
10 3
100 4
8 7
97 2
8 8
3 10
5 3
1000000000 9
output
YES
4 2 4
YES
55 5 5 35
NO
NO
YES
1 1 1 1 1 1 1 1
NO
YES
3 1 1
YES
111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111120

题意：
问一个数n能否分成k个大于0的奇数或k个大于0的偶数

思路：
如果能分成k个大于0的奇数的话，一定能够分出（k-1）个1并且剩下的数一定大于0且是奇数。

代码

#include <iostream>
using namespace std;
int main() {
   
	int t;
	cin >> t;
	while (t--) {
   
		int n, k;
		cin >> n >> k;
		int n1 = n - (k - 1);
		//把k-1个1去掉看是不是奇数 
		if (n1 > 0 && n1 % 2 == 1) {
   
			cout << "YES" << endl;
			for (int i = 0; i < k - 1; i ++ ) cout << "1 ";
			cout << n1 << endl;
			continue;
		}
		int n2 = n - 2 * (k - 1);
		//把k-1个2去掉看是不是偶数 
		if (n2 > 0 && n2 % 2 == 0) {
   
			cout << "YES" << endl;
			for (int i = 0; i < k - 1; i ++