UVA 10940 - Throwing cards away II（规律）

最新推荐文章于 2020-01-30 13:34:27 发布

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Problem G: Throwing cards away II

Given is an ordered deck of n cards numbered 1 to n with card 1 at the top and card n at the bottom. The following operation is performed as long as there are at least two cards in the deck:

Throw away the top card and move the card that is now on the top of the deck to the bottom of the deck.

Your task is to find the last, remaining card.

Each line of input (except the last) contains a positive number n ≤ 500000. The last line contains 0 and this line should not be processed. For each number from input produce one line of output giving the last remaining card. Input will not contain more than 500000 lines.

Sample input

Output for sample input

题意：扑克，先是1-n排序，1在顶，n在底，然后每次抽掉一张，第二张放到底，求最后剩下那张是多少。

思路：先写了个模拟的超时了，然后打出来看规律，出了n=1，ans=1，其他都是按 2,(2,4),(2,4,6,8),(2..2^N)这样排的

代码：

#include <stdio.h>
const int MAXN = 500005;
int ans[MAXN];

void init() {
	ans[1] = 1;
	int n = 1;
	for (int i = 2; i <= 500000;) {
		int save = (1<<n);
		for (int j = i; j < i + save / 2 && j <= 500000; j ++) {
			ans[j] = (j - i + 1) * 2;
		}
		i += save / 2;
		n ++;
	}
}

int main() {
	int n; init();
	while (~scanf("%d", &n) && n) {
		printf("%d\n", ans[n]);
	}
	return 0;
}