Median

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题目挺水的。注意给定的序列是有序的，所以其实只有两种情况，要么给定的两段序列无关，要么出现重叠，这两个情况分开弄一下就好了。
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#include <stdio.h>#include <stdlib.h>#include <string.h>#include <algorithm>using namespace std;const int maxn = 100010;long long a[maxn];int l, r, l1, r1, l2, r2, mergel, merger, sum, n, m;long long find1(int t) {    if (l1+t-1 <= r1) return a[l1+t-1];    t = t-(r1-l1+1);    return a[l2+t-1];}long long find2(int t) {    l = l1;    r = max(r1, r2);    mergel = l2;    merger = min(r1, r2);    if (l+t-1 < mergel) return a[l+t-1];    t = t-(mergel-l);    if (mergel+(t-1)/2 <= merger) return a[mergel+(t-1)/2];    t = t-(merger-mergel+1)*2;    return a[merger+t];}int main() {    int tt;    scanf("%d", &tt);    while (tt--) {        scanf("%d %d", &n, &m);        for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);        while (m--) {            scanf("%d %d", &l1, &r1);            scanf("%d %d", &l2, &r2);            if (l1 > l2) {                swap(l1, l2);                swap(r1, r2);            }            sum = (r1-l1+1)+(r2-l2+1);            if (r1 < l2) {                double ans = find1((sum+1)/2)+find1((sum+2)/2);                printf("%.1lf\n", ans/2.0);            } else {                double ans = find2((sum+1)/2)+find2((sum+2)/2);                printf("%.1lf\n", ans/2.0);            }        }    }}