Codeforces 509C. Sums of Digits 贪心枚举


贪心枚举,代码里的注释很详细

C. Sums of Digits
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Vasya had a strictly increasing sequence of positive integers a1, ..., an. Vasya used it to build a new sequence b1, ..., bn, where bi is the sum of digits of ai's decimal representation. Then sequence ai got lost and all that remained is sequence bi.

Vasya wonders what the numbers ai could be like. Of all the possible options he likes the one sequence with the minimum possible last number an. Help Vasya restore the initial sequence.

It is guaranteed that such a sequence always exists.

Input

The first line contains a single integer number n (1 ≤ n ≤ 300).

Next n lines contain integer numbers b1, ..., bn  — the required sums of digits. All bi belong to the range 1 ≤ bi ≤ 300.

Output

Print n integer numbers, one per line — the correct option for numbers ai, in order of following in sequence. The sequence should be strictly increasing. The sum of digits of the i-th number should be equal to bi.

If there are multiple sequences with least possible number an, print any of them. Print the numbers without leading zeroes.

Sample test(s)
input
3
1
2
3
output
1
2
3
input
3
3
2
1
output
3
11
100

/* ***********************************************
Author        :CKboss
Created Time  :2015年02月06日 星期五 15时31分06秒
File Name     :CF509C.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

int n;
int b[333];
int pos[333];
int num[333][500];

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);

	scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%d",b+i);
   	pos[0]=1; 

	/// 第一个数 贪心
	int temp=b[1];
	if(temp==0) num[1][pos[1]++]=0;
	while(temp>0)
	{
		if(temp>=9) { temp-=9; num[1][pos[1]++]=9; }
		else { num[1][pos[1]++]=temp; temp=0; }
	}

	/// 剩下的数 贪心
	for(int i=2;i<=n;i++)
	{
		int lastsum=b[i-1];/// 上一位的已经确定的数字和
		bool flag=false;
		for(int j=0;flag==false;j++) /// 枚举位置
		{
			if(j-1>=0) lastsum-=num[i-1][j-1];

			///枚举所在的位加多少
			for(int k=num[i-1][j]+1;k<=9&&flag==false;k++)
			{
				int add = k-num[i-1][j];
				/// 剩下的数的范围可以在 0 ~ 9*j 之间
				if(j-1>=0)
				{
					if(lastsum+add<=b[i]&&lastsum+add+9*j>=b[i]) /// 成功
					{
						flag=true;
						pos[i]=max(j+1,pos[i-1]);
						/// 贪心部分的值 val
						int val = b[i]-lastsum-add;
						int tn=0;
						while(val>0)
						{
							if(val>=9) { val-=9; num[i][tn++]=9; }
							else { num[i][tn++]=val; val=0; }
						}
						/// 枚举的那一位
						num[i][j]=k;
						/// 如果存在的不变部分
						for(int kk=j+1;kk<pos[i-1];kk++)
						{
							num[i][kk]=num[i-1][kk];
						}
					}
				}
				else /// 在j==0时特殊考虑
				{
					if(lastsum+add==b[i]) /// 成功
					{
						flag=true;
						pos[i]=pos[i-1];
						for(int kk=0;kk<pos[i];kk++)
						{
							num[i][kk]=num[i-1][kk];
						}
						num[i][j]=k;
					}
				}
			}
		}
	}

	for(int i=1;i<=n;i++)
	{
		for(int j=pos[i]-1;j>=0;j--) 
			printf("%d",num[i][j]); 
		putchar(10);
	}
	
    return 0;
}



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