STL--G - For Fans of Statistics（兩個判斷條件-二分）

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G - For Fans of StatisticsTime Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64uSubmit 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

input	output
51234567 666666 3141593 666666 434343451 5 31415931 5 5782022 4 6666664 4 71356101 1 1234567	10101

沒發現和stl有什麼關係。。。。。用二分過的，先以值得大小二分，如果值相同的話，加一組判斷，以當前的id與給出的l和r進行二分，看最後能不能得出合格值

#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");    }}