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最新推荐文章于 2024-11-21 15:51:53 发布

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#include <bits/stdc++.h>
using namespace std;

const int N = 1e5 + 10;
int a[N];
int f[N][16];
int n;

void rmq_init() //建立:dp(i,j) = min{dp(i,j-1),dp(i+2^(j-1),j-1) O(nlogn)
{
    for (int i = 1; i <= n; i++)
        f[i][0] = a[i];
    int k = floor(log((double)n) / log(2.0)); //C/C++取整函数ceil()大,floor()小
    for (int j = 1; j <= k; j++)
        for (int i = 1; i <= n; i++)
        {
            if (i + (1 << (j - 1)) <= n) //f(i,j) = min{f(i,j-1),f(i+2^(j-1),j-1)
                f[i][j] = max(f[i][j - 1], f[i + (1 << (j - 1))][j - 1]);  ///改为 min 可求最小值
        }
}

int rmq(int i, int j) //查询：返回区间[i,j]的最小值
{
    int k = floor(log((double)(j - i + 1)) / log(2.0));
    return max(f[i][k], f[j - (1 << k) + 1][k]);  ///改为 min 可求最小值
}

int main()
{
    int icase = 0;
    while(scanf("%d", &n)!=EOF)
    {
        memset(a,0,sizeof(a));
        memset(f,0,sizeof(f));
        for (int i = 1; i <= n; i++)
            scanf("%d", &a[i]);
        rmq_init();
        int m;
        printf("Case %d:\n",++icase);
        scanf("%d",&m);
        for(int i = 0; i < m; i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            printf("%d\n",rmq(x,y));
        }
    }
}