一道RMQ板子题，分别维护最大值和最小值，不解释。
1 #include<cstdio> 2 #include<iostream> 3 #include<algorithm> 4 #include<cmath> 5 #include<cstring> 6 #include<cstdlib> 7 #include<cctype> 8 #include<stack> 9 #include<queue> 10 #include<vector> 11 using namespace std; 12 #define enter puts("") 13 #define space putchar(' ') 14 #define Mem(a) memset(a, 0, sizeof(a)) 15 typedef long long ll; 16 typedef double db; 17 const int INF = 0x3f3f3f3f; 18 const db eps =1e-8; 19 const int maxn = 5e4 + 5; 20 inline ll read() 21 { 22 ll ans = 0; 23 char ch = getchar(), last = ' '; 24 while(!isdigit(ch)) {last = ch; ch = getchar();} 25 while(isdigit(ch)) {ans = ans * 10 + ch - '0'; ch = getchar();} 26 if(last == '-') ans = -ans; 27 return ans; 28 } 29 inline void write(ll x) 30 { 31 if(x < 0) putchar('-'), x = -x; 32 if(x >= 10) write(x / 10); 33 putchar(x % 10 + '0'); 34 } 35 36 int n, m, a[maxn]; 37 38 int dp[maxn][20][2], b[maxn]; 39 void RMQ() 40 { 41 for(int i = 1; i <= n; ++i) dp[i][0][0] = dp[i][0][1] = a[i]; 42 for(int j = 1; (1 << j) <= n; ++j) 43 for(int i = 1; i + (1 << j) - 1 <= n; ++i) 44 { 45 dp[i][j][0] = max(dp[i][j - 1][0], dp[i + (1 << (j - 1))][j - 1][0]); 46 dp[i][j][1] = min(dp[i][j - 1][1], dp[i + (1 << (j - 1))][j - 1][1]); 47 } 48 int x = 0; 49 for(int i = 1; i <= n; ++i) 50 { 51 b[i] = x; 52 if((1 << (x + 1)) <= (i + 1)) x++; 53 } 54 } 55 int query(int L, int R, bool flag) 56 { 57 int k = b[R - L + 1]; 58 if(flag) return min(dp[L][k][flag], dp[R - (1 << k) + 1][k][flag]); //应该是R - (1 << k) + 1,不是R - k + 1…… 59 else return max(dp[L][k][flag], dp[R - (1 << k) + 1][k][flag]); 60 } 61 62 63 int main() 64 { 65 n = read(); m = read(); 66 for(int i = 1; i <= n; ++i) a[i] = read(); 67 RMQ(); 68 for(int i = 1; i <= m; ++i) 69 { 70 int L = read(), R = read(); 71 write(query(L, R, 0) - query(L, R, 1)); enter; 72 } 73 return 0; 74 }