分塊查找

/*
分塊查找
*/
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;

int a[] = {5,6,7,62,4,2,53,4,6,2,5};
int n = 11;
const int upNum = 3;        //3個分成一塊
int cnt = 0;                //分塊的數量
int index[11];
struct indx{                //塊
    int a;                  //塊頭
    vector<int>msg;         //塊內其他內容
    operator < (const indx other) const{        //使用靜態的防治修改
        return a < other.a;
    }
}t[101];

//返回大於等於x的第一個
int tupper_bound(int x){
    int left = 0, right = cnt;
    int mid;
    while(left <= right){
        mid = (left + right) >> 1;
        if(x <= t[mid].a)right = mid - 1;
        else left = mid + 1;
    }
    return mid;
}
//建立塊表
void build(){
    int ma;                                 //塊內最大值
    for(int i = 0; i < n; i++){
        ma = max(a[i], ma);
        if(!(i % upNum)){                   //建立表頭，並初始化表內的數組下標,以及塊的表頭值
            cnt++;
            ma = a[i];
        }
        t[cnt].msg.push_back(a[i]);
        t[cnt].a= ma;
    }
    sort(t, t + 1);

}


//查詢
int *get(int x){                //返回在第幾塊的第幾個
    int ans[2];
    int ans1 = tupper_bound(x); //塊間使用二分
    int ans2 = -1, len = t[ans1].msg.size();
    for(int i = 0; i < len; i++){
        if(t[ans1].msg[i] == x)ans2 = i;
    }
    if(ans2 == -1)ans1 = -1;
    ans[0] = ans1;
    ans[1] = ans2;
    return ans;
}



int main(){
    build();
    get(53);
    return 0;
}