RMQ问题3.2.1

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#define MAXN 50010
#define Max(x,y) (x>y?x:y)
#define Min(x,y) (x>y?y:x)
using namespace std;  ///8540K	3422MS	G++

int maxsum[MAXN][20],minsum[MAXN][20];///表示从第i个数起连续2^j个数中的最大值/最小值

 void RMQ(int num) {
     for(int j=1; j<20; j++) {
         for(int i=1; i<=num; i++) {
             if(i+(1<<j)-1 <= num) {
                 maxsum[i][j]=Max(maxsum[i][j-1], maxsum[i+(1<<(j-1))][j-1]);
               /// printf("maxsum[%d][%d]=Max(maxsum[%d][%d], maxsum[%d][%d]),%d, %d\n", i, j, i, j-1, i+(1<<(j-1)), j-1, maxsum[i][j-1], maxsum[i+(1<<(j-1))][j-1]);
                 minsum[i][j]=Min(minsum[i][j-1], minsum[i+(1<<(j-1))][j-1]);
             }
         }
     }
 }

 int main()
 {
     int i, num, query;
     while(scanf("%d%d",&num, &query) != EOF) {
         for(i=1; i<=num; i++) {
             scanf("%d", &maxsum[i][0]);
             minsum[i][0] = maxsum[i][0];
         }
         RMQ(num);
         int st, en, maxl, minl;
         while(query--) {
             scanf("%d%d", &st, &en);
             int k=(int)((log(en-st+1))/log(2.0));
             maxl=Max(maxsum[st][k], maxsum[en-(1<<k)+1][k]);
             minl=Min(minsum[st][k], minsum[en-(1<<k)+1][k]);
             printf("%d\n",maxl-minl);
         }
     }
     return 0;
 }
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参考资料：
http://www.cnblogs.com/Missa/archive/2012/10/01/2709686.html
http://blog.youkuaiyun.com/niushuai666/article/details/7400587
http://blog.youkuaiyun.com/niushuai666/article/details/6624672
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