HDU 4908 BestCoder Sequence（組合數學)，hdu4908

HDU 4908 BestCoder Sequence（組合數學)，hdu4908
HDU 4908 BestCoder Sequence

題目連結

題意：給定一個序列，1-n的數字，選定一個作為中位元m，要求有多少連續子序列滿足中位元是m

思路：組合數學，記錄下m左邊和右邊一共有多少種情況大於m的數字和小於n數組的差，然後等於左邊乘右邊所有的和，然後最後記得加上左右兩邊差為0的情況。

當時也是比較逗，還用樹狀數組去搞了，其實完全沒必要

代碼：

#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]++;    elserb[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;}