STL--G - For Fans of Statistics（两个推断条件-二分）

G - For Fans of Statistics

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

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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

input	output
5 1234567 666666 3141593 666666 4343434 5 1 5 3141593 1 5 578202 2 4 666666 4 4 7135610 1 1 1234567	10101

没发现和stl有什么关系。。。。。用二分过的，先以值得大小二分，假设值同样的话，加一组推断，以当前的id与给出的l和r进行二分，看最后能不能得出符合条件的值

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

struct node{

    int id , k ;

} p[80000];

bool cmp(node a,node b)

{

    return a.k<b.k || ( a.k==b.k && a.id < b.id );

}

void f(int l,int r,int x,int n)

{

    int low = 0 , top = n-1 ;

    while(low <= top)

    {

        int mid = (low+top)/2 ;

        if( p[mid].k == x && p[mid].id >= l && p[mid].id <= r )

        {

            printf("1");

            return ;

        }

        else if( p[mid].k < x || ( p[mid].k == x && p[mid].id < l ) )

            low = mid+1 ;

        else

            top = mid - 1 ;

    }

    printf("0");

    return ;

}

int main()

{

    int i , n , m , l , r , x ;

    while(scanf("%d", &n)!=EOF)

    {

        for(i = 0 ; i < n ; i++)

        {

            scanf("%d", &p[i].k);

            p[i].id = i+1 ;

        }

        sort(p,p+n,cmp);

        scanf("%d", &m);

        while(m--)

        {

            scanf("%d %d %d", &l, &r, &x);

            f(l,r,x,n);

        }

        printf("\n");

    }

}

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STL--G - For Fans of Statistics（两个推断条件-二分）

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