ZOJ3202-Second-price Auction

Second-price Auction

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Time Limit: 1 Second      Memory Limit: 32768 KB

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Do you know second-price auction? It's very simple but famous. In a second-price auction, each potential buyer privately submits, perhaps in a sealed envelope or over a secure connection, his (or her) bid for the object to the auctioneer. After receiving all the bids, the auctioneer then awards the object to the bidder with the highest bid, and charges him (or her) the amount of the second-highest bid.

Suppose you're the auctioneer and you have received all the bids, you should decide the winner and the amount of money he (or she) should pay.

Input

There are multiple test cases. The first line of input contains an integer T(T <= 100), indicating the number of test cases. Then T test cases follow.

Each test case contains two lines: The first line of each test case contains only one integer N, indicating the number of bidders. (2 <= N <= 100) The second line of each test case contains N integers separated by a space. The i-th integer P_i indicates the i-th bidder's bid. (0 < P_i <= 60000) You may assume that the highest bid is unique.

Output

For each test case, output a line containing two integers x and y separated by a space. It indicates that the x-th bidder is the winner and the amount of money he (or she) should pay is y.

Sample Input

233 2 124 9

Sample Output

1 22 4

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Author: CAO, Peng
Source: The 6th Zhejiang Provincial Collegiate Programming Contest

题意：有n个投标人，最高者获得物品，并且支付第二高投标价，问中标者和他需要支付的钱

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <algorithm>#include <queue>#include <vector>#include <set>#include <stack>#include <map>#include <climits>using namespace std;#define LL long longstruct node{    int id;    int sum;}x[105];bool cmp(node a,node b){    return a.sum>b.sum;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        int n;        scanf("%d",&n);        for(int i=1;i<=n;i++)        {            scanf("%d",&x[i].sum);            x[i].id=i;        }        sort(x+1,x+1+n,cmp);        printf("%d %d\n",x[1].id,x[2].sum);    }    return 0;}