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HDU 4908 BestCoder Sequence

题目链接

题意：给定一个序列，1-n的数字，选定一个作为中位数m，要求有多少连续子序列满足中位数是m

思路：组合数学，记录下m左边和右边一共有多少种情况大于m的数字和小于n数组的差，然后等于左边乘右边所有的和，然后最后记得加上左右两边差为0的情况。

当时也是比较逗，还用树状数组去搞了，其实完全没必要

代码：

#include <cstdio>
#include <cstring>

#define lowbit(x) (x&(-x))

const int N = 40005;

int n, m, num[N], bit[N];

void add(int x, int v) {
    while (x < N) {
	bit[x] += v;
	x += lowbit(x);
    }
}

int query(int x) {
    int ans = 0;
    while (x) {
	ans += bit[x];
	x -= lowbit(x);
    }
    return ans;
}

int lb[N], ls[N], rb[N], rs[N];

int main() {
    while (~scanf("%d%d", &n, &m)) {
	memset(bit, 0, sizeof(bit));
	int v;
	for (int i = 1; i <= n; i++) {
	    scanf("%d", &num[i]);
	    if (num[i] == m)
		v = i;
	}
	memset(lb, 0, sizeof(lb));
	memset(ls, 0, sizeof(ls));
	memset(rb, 0, sizeof(rb));
	memset(rs, 0, sizeof(rs));
	for (int i = v - 1; i >= 1; i--) {
	    add(num[i], 1);
	    int small = query(m);
	    int big = v - i - small;
	    if (big >= small)
		lb[big - small]++;
	    else ls[small - big]++;
	}
	memset(bit, 0, sizeof(bit));
	for (int i = v + 1; i <= n; i++) {
	    add(num[i], 1);
	    int small = query(m);
	    int big = i - v - small;
	    if (small >= big)
		rs[small - big]++;
	    else
		rb[big - small]++;
	}
	long long ans = 1;
	ans += lb[0] + rs[0];
	for (int i = 0; i <= 40000; i++) {
	    ans += (long long)lb[i] * rs[i];
	    ans += (long long)ls[i] * rb[i];
	}
	printf("%lld\n", ans);
    }
    return 0;
}

HDU 4908 BestCoder Sequence（组合数学),布布扣,bubuko.com

HDU 4908 BestCoder Sequence（组合数学)

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原文地址：http://blog.csdn.net/accelerator_/article/details/38363197