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  • ural1613 For Fans of Statistics

    For Fans of Statistics

    Time limit: 1.0 second
    Memory limit: 64 MB
    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 109 − 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 lr, and x separated with a space; 1 ≤ l ≤ r ≤ n; 0 < x < 109.

    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

    inputoutput
    5
    1234567 666666 3141593 666666 4343434
    5
    1 5 3141593
    1 5 578202
    2 4 666666
    4 4 7135610
    1 1 1234567
    
    10101
    

    分析:map+二分;

    代码:

    #include <iostream>
    #include <cstdio>
    #include <cstdlib>
    #include <cmath>
    #include <algorithm>
    #include <climits>
    #include <cstring>
    #include <string>
    #include <set>
    #include <map>
    #include <queue>
    #include <stack>
    #include <vector>
    #include <list>
    #define rep(i,m,n) for(i=m;i<=n;i++)
    #define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
    #define mod 1000000007
    #define inf 0x3f3f3f3f
    #define vi vector<int>
    #define pb push_back
    #define mp make_pair
    #define fi first
    #define se second
    #define ll long long
    #define pi acos(-1.0)
    const int maxn=1e5+10;
    const int dis[4][2]={{0,1},{-1,0},{0,-1},{1,0}};
    using namespace std;
    ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
    ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
    int n,m,k,t;
    string ans;
    map<int,set<int> >a;
    set<int>::iterator it;
    int main()
    {
        int i,j;
        scanf("%d",&n);
        rep(i,1,n)scanf("%d",&j),a[j].insert(i);
        int q;
        scanf("%d",&q);
        while(q--)
        {
            int l,r,p;
            scanf("%d%d%d",&l,&r,&p);
            if((it=a[p].lower_bound(l))!=a[p].end()&&*it<=r)
                ans+='1';
            else ans+='0';
        }
        cout<<ans<<endl;
        //system("pause");
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/dyzll/p/5785154.html
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