• UVA 11991 Easy Problem from Rujia Liu(map,vector的使用)


    Input
    There are several test cases. The rst line of each test case contains two integers n, m (1  n;m 
    100; 000), the number of elements in the array, and the number of queries. The next line contains n
    positive integers not larger than 1,000,000. Each of the following m lines contains two integer k and v
    (1  k  n, 1  v  1; 000; 000). The input is terminated by end-of- le (EOF).
    Output
    For each query, print the 1-based location of the occurrence. If there is no such element, output `0'
    instead.
    Sample Input
    8 4
    1 3 2 2 4 3 2 1
    1 3
    2 4
    3 2
    4 2
    Sample Output
    2
    0
    7
    0

    题目大意:

      给你一个长度为n的数组,然后询问你第k个val的下标是多少,n<=1e6, m<=1e6。

    解题思路:

      一看到这个题就想到了mlogn的方法,但是由于STL的水平有限,并没有独立AC,还是看了书上的代码才弄懂。。。

    总结下,map<int,vector<int> >mp;  表示的是一个map类型的变量mp,mp[ ]就相当于一个容器,mp[a][b]表示的是mp[a]这个容器的第b+1个元素是多少。

    有了这个概念,我们就知道了,如果要求n个数字中,第k个val的下标是多少,那么我们直接调用mp[val][k-1]就行了。

    代码:

    # include<cstdio>
    # include<iostream>
    # include<vector>
    # include<map>
    
    using namespace std;
    
    map<int,vector<int> >a;
    
    int main(void)
    {
        int n,m;
        while ( scanf("%d%d",&n,&m)!=EOF )
        {
            a.clear();
            for ( int i = 0;i < n;i++ )
            {
                int x; scanf("%d",&x);
                if ( a.count(x)==0 )
                    a[x] = vector<int>();
                a[x].push_back(i+1);
            }
            while (m--)
            {
                int x, y; scanf("%d%d",&x,&y);
                if ( a.count(y)==0||a[y].size() < x )
                    puts("0");
                else
                    printf("%d
    ",a[y][x-1]);
            }
        }
    
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/wikioibai/p/4742327.html
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