hdu 4455 Substrings(计数)

题目链接:hdu 4455 Substrings


题目大意:给出n,然后是n个数a[1] ~ a[n], 然后是q次询问,每次询问给出w, 将数列a[i]分成若干个连续且元素数量为w的集合,计算每个集合中出现的数字种类,输出总和。


解题思路:一开始想到遍历的算法,保持集合元素为w,每次剔除最前一个,加入一个,移动集合,维护数字种类,这种算法的复杂度为o(n^2), 但是超时了,后来看了下题解,dp[i] = dp[i - 1] + sum[i] - cnt;


http://blog.csdn.net/gotoac/article/details/8188437


 

#include <stdio.h>
#include <string.h>
#include <iostream>

using namespace std;

const int N = 1000005;

int n, vis[N], a[N];
__int64 sum[N], cnt,  dp[N];

void init() {
	memset(vis, -1, sizeof(vis));
	memset(sum, 0, sizeof(sum));

	for (int i = 0; i < n; i++) {
		scanf("%d", &a[i]);
		sum[i - vis[a[i]]]++;
		vis[a[i]] = i;
	}

	for (int i = n - 1; i >= 0; i--)
		sum[i] += sum[i + 1];
}

void solve() {
	memset(dp, 0, sizeof(dp));
	dp[1] = n;

	memset(vis, 0, sizeof(vis));
	vis[a[n - 1]] = cnt = 1;

	for (int i = 2; i <= n; i++) {
		dp[i] = dp[i - 1] - cnt + sum[i];

		vis[a[n - i]]++;
		if (vis[a[n - i]] == 1) cnt++;
	}
}

int main () {
	while (scanf("%d", &n), n) {
		init();
		solve();
		int q, w;
		scanf("%d", &q);
		for (int i = 0; i < q; i++) {
			scanf("%d", &w);
			printf("%I64d\n", dp[w]);
		}
	}
	return 0;
}


 

 

posted on 2013-11-01 19:13  you Richer  阅读(216)  评论(0编辑  收藏  举报