给出以下定义：

1.若子序列[L,R]的极差（最大值-最小值）<=M，则子序列[L,R]为一个均匀序列。

2.均匀序列[L,R]的权值为Sum(L,R)即序列的元素和。

现在给你一个长度为N的整数序列A，请你求出权值前K大的均匀序列，输出K行为它们的权值。

第一行为两个整数N,M,K。
第二行为N个整数Ai。

输出K行，第i行为第i大的均匀序列的权值。

9 3 105 1 3 5 8 6 6 9 10

29252120191915141312

#include <iostream>#include <cstdio>#include <algorithm>#include <cmath>#include <stack>#include <vector>#include <queue>#include <cstring>#include <string>#include <map>#include <set>using namespace std;const int BufferSize = 1 << 16;char buffer[BufferSize], *Head, *Tail;inline char Getchar() {    if(Head == Tail) {        int l = fread(buffer, 1, BufferSize, stdin);        Tail = (Head = buffer) + l;    }    return *Head++;}int read() {    int x = 0, f = 1; char c = getchar();    while(!isdigit(c)){ if(c == ‘-‘) f = -1; c = getchar(); }    while(isdigit(c)){ x = x * 10 + c - ‘0‘; c = getchar(); }    return x * f;}#define maxn 100010#define maxlog 21#define LL long longint n, m, k;LL S[maxn], A[maxn];LL mx2[maxlog][maxn];int mx[maxlog][maxn], mn[maxlog][maxn], Log[maxn], mxp[maxlog][maxn];void rmq_init() {	Log[1] = 0;	for(int i = 2; i <= n; i++) Log[i] = Log[i>>1] + 1;	for(int i = 1; i <= n; i++) mx[0][i] = mn[0][i] = A[i];	for(int j = 1; j < maxlog; j++)		for(int i = 1; i + (1 << j) - 1 <= n; i++)			mx[j][i] = max(mx[j-1][i], mx[j-1][i+(1<<j-1)]),			mn[j][i] = min(mn[j-1][i], mn[j-1][i+(1<<j-1)]);	return ;}void rmq_init2() {	for(int i = 1; i <= n; i++) mx2[0][i] = S[i], mxp[0][i] = i;	for(int j = 1; j < maxlog; j++)		for(int i = 1; i + (1 << j) - 1 <= n; i++)			if(mx2[j-1][i] > mx2[j-1][i+(1<<j-1)]) mx2[j][i] = mx2[j-1][i], mxp[j][i] = mxp[j-1][i];			else mx2[j][i] = mx2[j-1][i+(1<<j-1)], mxp[j][i] = mxp[j][i] = mxp[j-1][i+(1<<j-1)];	return ;}int qmx(int l, int r, int tp) {	int t = Log[r-l+1], len = (1 << t);	if(tp == 1)		return max(mx[t][l], mx[t][r-len+1]);	return mx2[t][l] > mx2[t][r-len+1] ? mxp[t][l] : mxp[t][r-len+1];}int qmn(int l, int r) {	int t = Log[r-l+1], len = (1 << t);	return min(mn[t][l], mn[t][r-len+1]);}struct Node {	int x, l, r; LL v;	bool operator < (const Node& t) const { return v < t.v; }} ;priority_queue <Node> Q;int R[maxn];int main() {	n = read(); m = read(); k = read();	for(int i = 1; i <= n; i++) A[i] = read(), S[i] = S[i-1] + A[i];		rmq_init(); rmq_init2();	int nl = 1, nr = 0;	for(int i = 1; i <= n; i++) {		nr++;		while(qmx(nl, nr, 1) - qmn(nl, nr) > m) nl++;		R[nl] = nr;//		printf("%d %d %d %lld\n", nl, nr, qmx(nl, nr, 2), S[qmx(nl, nr, 2)] - S[nl-1]);//		Q.push((Node){ nl, nl, nr, S[qmx(nl, nr, 2)] - S[nl-1] });	}	for(int i = 1; i <= n; i++) if(!R[i]) R[i] = R[i-1];	for(nl = 1; nl <= n; nl++) {		LL tmp = S[qmx(nl, R[nl], 2)] - S[nl-1];		Q.push((Node){ nl, nl, R[nl], tmp });//		printf("%d %d %lld\n", nl, R[nl], tmp);	}	while(k--) {		Node u = Q.top(); Q.pop();		printf("%lld\n", u.v);		int p = qmx(u.l, u.r, 2);		if(u.l < p) Q.push((Node){ u.x, u.l, p - 1, S[qmx(u.l, p - 1, 2)] - S[u.x-1] });		if(p < u.r) Q.push((Node){ u.x, p + 1, u.r, S[qmx(p + 1, u.r, 2)] - S[u.x-1] });	}		return 0;}