Perfect Number 二分

版权声明：原创文章转载时请注明出处 https://blog.csdn.net/weixin_42856843/article/details/88877972

Perfect Number 二分

Time limit：2000 ms Memory limit：262144 kB Source: Codeforces Round #460 (Div. 2) Tags： binary search brute force dp implementation number theory *1100 Editorial： Announcement Tutorial

描述

We consider a positive integer perfect, if and only if the sum of its digits is exactly 10. Given a positive integer k, your task is to find the k-th smallest perfect positive integer.

输入

A single line with a positive integer k (1≤k≤10000).

输出

A single number, denoting the k-th smallest perfect integer.

Input

1

Output

19

Input

2

Output

28

提示

The first perfect integer is 19 and the second one is 28.

代码

#include<iostream>
#include<string.h>
#include<algorithm>
#include<stdio.h>
#include<queue>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long long ll;
typedef pair<int, int> PII;
#define mod 1000000007
#define INF 0x3f3f3f3f
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
//head
int k;
int bit[11];//存放mid的各个元素
int dp[20][11];//表示到第i位，数位和为j的个数
int dfs(int pos, int sum, bool limit) {
	if (pos == -1) return sum == 10;
	if (sum > 10) return 0;
	if (!limit&&dp[pos][sum] != -1) return dp[pos][sum];
	int ans = 0;
	int up = limit ? bit[pos] : 9;
	for (int i = 0; i <= up; i++) {
		ans += dfs(pos - 1, sum + i, limit && (i == up));
	}
	if (!limit) dp[pos][sum] = ans;
	return ans;
}

int calc(int x) {
	int len = 0;
	while (x) {
		bit[len++] = x % 10;
		x /= 10;
	}
	return dfs(len - 1, 0, true);
}

int main() {
	memset(dp, -1, sizeof(dp));
	int lb = 0, ub = INF;
	int ans = 0;
	cin >> k;
	while (ub - lb > 1) {
		int mid = (lb + ub) / 2;
		if (calc(mid) < k) lb = mid;//找到mid是第几个
		else ans = mid, ub = mid;
	}
	printf("%d\n", ans);
	return 0;
}

本人也是新手，也是在学习中，勿喷

转载请注明出处

欢迎有问题的小伙伴一起交流哦~