poj 3264 Balanced Lineup RMQ
emmm，第一遍的时候用的是C++的cin ，cout 果不其然，超时。
然后全部改为scanf和printf，卡过
1 #include <iostream> 2 #include <algorithm> 3 #include <cstdio> 4 #include <cmath> 5 using namespace std; 6 7 const int maxn = 50005; 8 int a[maxn]; 9 int n, m; 10 int minsum[maxn][50]; 11 int maxsum[maxn][50]; 12 13 void RMQ(int num) //预处理->O(nlogn) 14 { 15 //初始化 16 for (int i = 1; i <= num; i++){ 17 minsum[i][0] = a[i]; 18 maxsum[i][0] = a[i]; 19 } 20 21 for (int j = 1; j < 25; j++) 22 for (int i = 1; i <= num; i++) 23 if (i + (1 << j) - 1 <= num) 24 { 25 maxsum[i][j] = max(maxsum[i][j - 1], maxsum[i + (1 << (j - 1))][j - 1]); 26 minsum[i][j] = min(minsum[i][j - 1], minsum[i + (1 << (j - 1))][j - 1]); 27 } 28 } 29 30 //查询 31 int getmin(int x, int y){ 32 int k = (int)(log((double)(y - x + 1)) / log(2.0)); 33 return min(minsum[x][k], minsum[y - (1 << k) + 1][k]); 34 } 35 36 int getmax(int x, int y){ 37 int k = (int)(log((double)(y - x + 1)) / log(2.0)); 38 return max(maxsum[x][k], maxsum[y - (1 << k) + 1][k]); 39 } 40 41 int main(){ 42 scanf("%d%d", &n, &m); 43 for (int i = 1; i <= n; i++){ 44 scanf("%d", &a[i]); 45 } 46 //预处理O(nlogn) 47 RMQ(n); 48 49 while (m--){ 50 int x, y; 51 scanf("%d%d", &x, &y); 52 int Min = getmin(x, y); 53 int Max = getmax(x, y); 54 int ans = Max - Min; 55 // cout << "*" << Min << " " << Max << endl; 56 printf("%d\n", ans); 57 } 58 //system("pause"); 59 return 0; 60 }
