NOJ [1358] An Easy Problem

• 问题描述
• Here a simple problem.Give you n numbers.a1,a2...an.Then ask you m times.each time you should answer the the minimum number between ax to ay.
• 输入
• There are several test
  First line input two number n,m(1<=n<=100000,1<=m<=100000)
  then next line follow n integer.
  next m line with two numbers x,y;(1<=x<=y<=n)
• 输出
• for each x,y you should output the minimum number between ax to ay.
  
  
  rmq的裸题啊，不会rmq的话，我简单介绍下，我们用一个数组先把值存好，再开一个DP[i][j]来表示区间[i,i+2^j-1]的最值，
  把区间对分为2块，最值就是左右两块的最值的最值
  
  

  #include<stdio.h>
  #include<math.h>
  
  int min (int a,int b)
  {
    return a<b?a:b;
  }
  
  int num[100010];
  int dp[100010][18];
  int temp;
  int n;
  void st()
  {
    int i,j;
    int k= log((double)(n+1))/log(2.0);
    for(i=1;i<=n;i++)
      dp[i][0]=num[i];
    for(j=1;j<=k;j++)
        for(i=1;i+(1 << j)-1 <= n;i++)
           dp[i][j]=min(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
  }
  
  int rmq(int a,int b)
  {
    int k = (int)(log((double)(b-a+1))/log(2.0));
    return min(dp[a][k],dp[b-(1<<k)+1][k]);
  }
  
  int main()
  {
    int m;
    while(~scanf("%d%d",&n,&m))
    {
      int i;
      for(i=1;i<=n;i++)
        scanf("%d",&num[i]);
      st();
      int start,end;
      for(i=1;i<=m;i++)
      {
        scanf("%d%d",&start,&end);
        printf("%d\n", rmq(start,end));
      }
    }
    return 0;
  }