[codeforces 509]C. Sums of Digits

[codeforces 509]C. Sums of Digits

试题描述

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

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

It is guaranteed that such a sequence always exists.

输入

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

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

输出

Print n integer numbers, one per line — the correct option for numbers a_i, 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 b_i.

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

输入示例

3
3
2
1

输出示例

3
11
100

数据规模及约定

见“输入”

题解

显然是贪心，尽量使当前数接近上一个数，且大于等于上一个数 + 1. 搞一个类似高精度运算的模拟。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <map>
#include <set>
using namespace std;

const int BufferSize = 1 << 16;
char buffer[BufferSize], *Head, *Tail;
inline char Getchar() {
    if(Head == Tail) {
        int l = fread(buffer, 1, BufferSize, stdin);
        Tail = (Head = buffer) + l;
    }
    return *Head++;
}
int read() {
    int x = 0, f = 1; char c = getchar();
    while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }
    while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }
    return x * f;
}

#define maxn 310
int n, B[maxn], num[maxn][maxn], len[maxn];

int main() {
	n = read();
	for(int i = 1; i <= n; i++) B[i] = read();
	
	len[0] = 1;
	for(int i = 1; i <= n; i++) {
		num[i-1][1]++;
		for(int j = 1; j <= len[i-1]; j++) {
			num[i-1][j+1] += num[i-1][j] / 10;
			num[i-1][j] %= 10;
		}
		for(int j = len[i-1] + 1; num[i-1][j]; j++) {
			num[i-1][j+1] += num[i-1][j] / 10;
			num[i-1][j] %= 10;
			len[i-1] = j;
		}
		int tot = B[i], s = 0;
		for(int j = 1; j <= len[i-1]; j++) s += num[i-1][j];
		if(s == tot) memcpy(num[i], num[i-1], sizeof(num[i-1])), len[i] = len[i-1];
		else {
			len[i] = 0;
			for(int j = len[i-1]; j; j--)
				if(tot > num[i-1][j]) {
					len[i] = max(len[i], j);
					num[i][j] = num[i-1][j];
					tot -= num[i][j];
				}
				else {
					len[i] = max(len[i], j + 1);
					num[i][j+1]++; tot--;
					for(int k = j; k; k--) num[i][k] = 0;
					break;
				}
			for(int j = 1; j <= len[i]; j++) {
				num[i][j+1] += num[i][j] / 10;
				num[i][j] %= 10;
			}
			for(int j = len[i] + 1; num[i][j]; j++) {
				num[i][j+1] += num[i][j] / 10;
				num[i][j] %= 10;
				len[i] = j;
			}
			tot = B[i];
			for(int j = 1; j <= len[i]; j++) tot -= num[i][j];
//			for(int j = len[i]; j; j--) printf("%d", num[i][j]); putchar('\n');
			for(int j = 1; tot; j++) {
//				printf("%d(%d) ", tot, num[i][j]);
				if(tot >= 9 - num[i][j]) tot -= (9 - num[i][j]), num[i][j] = 9;
				else num[i][j] += tot, tot = 0;
				len[i] = max(len[i], j);
			}
//			putchar('\n');
		}
		for(int j = len[i]; j; j--) printf("%d", num[i][j]); putchar('\n');
	}
	
	return 0;
}
/*
5
30
29
28
42
11
*/