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;}