主席树
用主席数来维护区间第k大，然后不能构成三角形的最小条件就是斐波那契数列。
题目给的范围只有1e9，最坏情况下也只有44项斐波那契数列，也就是说我们从第1大，第2大，第3大依次往下判断，也最多遍历44次，肯定可以在规定时间内求解。
#include#define INF 0x3f3f3f3f#define full(a, b) memset(a, b, sizeof a)#define FAST_IO ios::sync_with_stdio(false)using namespace std;typedef long long LL;inline int lowbit(int x){ return x & (-x); }inline int read(){ int ret = 0, w = 0; char ch = 0; while(!isdigit(ch)){ w |= ch == '-', ch = getchar(); } while(isdigit(ch)){ ret = (ret << 3) + (ret << 1) + (ch ^ 48); ch = getchar(); } return w ? -ret : ret;}inline int lcm(int a, int b){ return a / __gcd(a, b) * b; }template inline A fpow(A x, B p, C lyd){ A ans = 1; for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd; return ans;}const int N = 200005;int n, m, tot, k, a[N], b[N], tree[N*20], lc[N*20], rc[N*20], root[N];void init(){ full(a, 0), full(b, 0); full(tree, 0), full(root, 0); full(lc, 0), full(rc, 0); k = tot = 0;}int buildTree(int l, int r){ int cur = ++tot; if(l == r) return cur; int mid = (l + r) >> 1; lc[cur] = buildTree(l, mid); rc[cur] = buildTree(mid + 1, r); return cur;}int insert(int rt, int l, int r, int p){ int cur = ++tot; tree[cur] = tree[rt] + 1, lc[cur] = lc[rt], rc[cur] = rc[rt]; if(l == r) return cur; int mid = (l + r) >> 1; if(p <= mid) lc[cur] = insert(lc[rt], l, mid, p); else rc[cur] = insert(rc[rt], mid + 1, r, p); return cur;}int query(int a, int b, int l, int r, int k){ int p = tree[lc[b]] - tree[lc[a]]; if(l == r) return l; int mid = (l + r) >> 1; if(k <= p) return query(lc[a], lc[b], l, mid, k); else return query(rc[a], rc[b], mid + 1, r, k - p);}int main(){ while(~scanf("%d%d", &n, &m)){ init(); for(int i = 1; i <= n; i ++){ scanf("%d", &a[i]); b[i] = a[i]; } sort(b + 1, b + n + 1); k = unique(b + 1, b + n + 1) - b - 1; root[0] = buildTree(1, k); for(int i = 1; i <= n; i ++){ int p = lower_bound(b + 1, b + k + 1, a[i]) - b; root[i] = insert(root[i - 1], 1, k, p); } int l = 0, r = 0; while(m --){ scanf("%d%d", &l, &r); int cnt = r - l + 1; if(cnt < 3){ printf("-1\n"); continue; } bool good = false; int fst = b[query(root[l - 1], root[r], 1, k, cnt)]; int sed = b[query(root[l - 1], root[r], 1, k, cnt - 1)]; for(int i = cnt - 2; i >= 1; i --){ int val = b[query(root[l - 1], root[r], 1, k, i)]; if(1LL * val + sed > fst){ printf("%lld\n", 1LL * val + sed + fst); good = true; break; } fst = sed, sed = val; } if(!good) printf("-1\n"); } } return 0;}