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// P3264-Balanced Lineup.cpp : 定义控制台应用程序的入口点。 | ||
//This problem is TLE if using cin and cout, but ac while changing to scanf and printf. | ||
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#include "stdafx.h" | ||
#include <stdio.h> | ||
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#define N 50005 | ||
#define Q 200005 | ||
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int n, q; | ||
int a[N], mini[4 * N], maxi[4 * N]; | ||
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void init(int node, int l, int r) { | ||
if(l == r) { | ||
mini[node] = l; | ||
maxi[node] = l; | ||
} else { | ||
init(2 * node, l, (l + r)/2); | ||
init(2 * node + 1, (l + r)/2 + 1, r); | ||
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if(a[mini[2 * node]] <= a[mini[2 * node + 1]]) | ||
mini[node] = mini[2 * node]; | ||
else | ||
mini[node] = mini[2 * node + 1]; | ||
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if(a[maxi[2 * node]] >= a[maxi[2 * node + 1]]) | ||
maxi[node] = maxi[2 * node]; | ||
else | ||
maxi[node] = maxi[2 * node + 1]; | ||
} | ||
} | ||
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int minquery(int node, int l, int r, int i, int j){ | ||
int p1, p2; | ||
if(i > r || j < l) | ||
return -1; | ||
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if(l >= i && r <= j) | ||
return mini[node]; | ||
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p1 = minquery(2 * node, l, (l + r)/2, i, j); | ||
p2 = minquery(2 * node + 1, (l + r)/2 + 1, r, i, j); | ||
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if(p1 == -1) | ||
return p2; | ||
if(p2 == -1) | ||
return p1; | ||
if(a[p1] <= a[p2]) | ||
return p1; | ||
return p2; | ||
} | ||
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int maxquery(int node, int l, int r, int i, int j){ | ||
int p1, p2; | ||
if(i > r || j < l) | ||
return -1; | ||
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if(l >= i && r <= j) | ||
return maxi[node]; | ||
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p1 = maxquery(2 * node, l, (l + r)/2, i, j); | ||
p2 = maxquery(2 * node + 1, (l + r)/2 + 1, r, i, j); | ||
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if(p1 == -1) | ||
return p2; | ||
if(p2 == -1) | ||
return p1; | ||
if(a[p1] >= a[p2]) | ||
return p1; | ||
return p2; | ||
} | ||
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int _tmain(int argc, _TCHAR* argv[]) | ||
{ | ||
while(scanf("%d%d", &n, &q) != EOF) { | ||
for(int i = 1; i <= n; ++i) | ||
scanf("%d", &a[i]); | ||
init(1, 1, n); | ||
for(int i = 0; i < q; ++i) { | ||
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int j, k; | ||
scanf("%d%d", &j, &k); | ||
printf("%d\n", a[maxquery(1, 1, n, j, k)] - a[minquery(1, 1, n, j, k)]); | ||
} | ||
} | ||
return 0; | ||
} | ||
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