For Fans of Statistics——vector

For Fans of Statistics

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status

Description

Have you ever thought about how many people are transported by trams every year in a city with a ten-million population where one in three citizens uses tram twice a day?

Assume that there are n cities with trams on the planet Earth. Statisticians counted for each of them the number of people transported by trams during last year. They compiled a table, in which cities were sorted alphabetically. Since city names were inessential for statistics, they were later replaced by numbers from 1 to n. A search engine that works with these data must be able to answer quickly a query of the following type: is there among the cities with numbers from l to r such that the trams of this city transported exactly x people during last year. You must implement this module of the system.

Input

The first line contains the integer n, 0 < n < 70000. The second line contains statistic data in the form of a list of integers separated with a space. In this list, the ith number is the number of people transported by trams of the ith city during last year. All numbers in the list are positive and do not exceed 10 ⁹ − 1. In the third line, the number of queries q is given, 0 < q < 70000. The next q lines contain the queries. Each of them is a triple of integers l, r, and x separated with a space; 1 ≤ l ≤ r ≤ n; 0 < x < 10 ⁹.

Output

Output a string of length q in which the ith symbol is “1” if the answer to the ith query is affirmative, and “0” otherwise.

Sample Input

inputoutput

51234567 666666 3141593 666666 434343451 5 31415931 5 5782022 4 6666664 4 71356101 1 1234567

10101

题意：有5各城市，每个城市的人口用表格表现出来，接下来有5组测试数据，1，5表示用1城市到5城市的人数有没有后面出现的。有可能两个不同的城市人数是相同的。

解析：利用哈希的思想，对每个城市的人口哈希，有相同的借助链表往后接。具体解析看代码：

#include <iostream>#include <algorithm>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#define MAX 100007using namespace std;int a[700100];struct node{    int id;//表示城市的代号    int zhi;//表示城市的人口数};vector<node> v[100007];//声明一个node型的容器int main(){    int n,m,i,j,k,x,y,z,mid;    struct node c;    scanf("%d",&n);    for(i = 1;i <= n;i++)    {        scanf("%d",&a[i]);        c.id = i;        c.zhi = a[i];        v[a[i]%MAX].push_back(c);//将人口哈希，借助v.push_back在最后加入数据    }    scanf("%d",&m);    for(i = 1;i <= m;i++)    {        scanf("%d%d%d",&x,&y,&z);        if(a[x] == z || a[y] == z)        {            printf("1");            continue;        }        int l = v[z%MAX].size();//判断相同人口的城市的数目，并一一遍历，查找符合条件的城市        int bj = 0;        for(j = 0;j < l;j++)        {            if(v[z%MAX][j].zhi == z && v[z%MAX][j].id > x && v[z%MAX][j].id < y)            {                bj = 1;                break;            }        }        if(bj == 1)            printf("1");        else            printf("0");    }    printf("\n");    return 0;}