hdu 5349 MZL's simple problem
2015-08-07 14:06
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hdu 5349 的传送门
Problem Description
A simple problem
Problem Description
You have a multiple set,and now there are three kinds of operations:
1 x : add number x to set
2 : delete the minimum number (if the set is empty now,then ignore it)
3 : query the maximum number (if the set is empty now,the answer is 0)
Input
The first line contains a number N (N≤106),representing the number of operations.
Next N line ,each line contains one or two numbers,describe one operation.
The number in this set is not greater than 109.
Output
For each operation 3,output a line representing the answer.
Sample Input
6
1 2
1 3
3
1 3
1 4
3
Sample Output
3
4
题目大意:给你一个数,表示有几行,然后给你一个a和x,如果a等于1的话,就向里面增加一个数,如果a等于2的话 ,就删除集合中最小的一个数,如果等于a==3 的话,就输出最大的数;
解题思路:STL,集合;
具体见代码:
Problem Description
A simple problem
Problem Description
You have a multiple set,and now there are three kinds of operations:
1 x : add number x to set
2 : delete the minimum number (if the set is empty now,then ignore it)
3 : query the maximum number (if the set is empty now,the answer is 0)
Input
The first line contains a number N (N≤106),representing the number of operations.
Next N line ,each line contains one or two numbers,describe one operation.
The number in this set is not greater than 109.
Output
For each operation 3,output a line representing the answer.
Sample Input
6
1 2
1 3
3
1 3
1 4
3
Sample Output
3
4
题目大意:给你一个数,表示有几行,然后给你一个a和x,如果a等于1的话,就向里面增加一个数,如果a等于2的话 ,就删除集合中最小的一个数,如果等于a==3 的话,就输出最大的数;
解题思路:STL,集合;
具体见代码:
[code]#include <iostream> #include <cstdio> #include <set> using namespace std; set<long long >::iterator it; set<long long > s; int main() { long long n, a, x; scanf("%lld",&n); s.clear(); while(n--) { scanf("%lld",&a); if(a == 1) { scanf("%lld",&x); s.insert(x); } else if(a == 2) { if(!s.empty()) s.erase(s.begin()); } else if(a == 3) { if(s.empty()) puts("0"); else { it=s.end(); it--; printf("%lld\n",*it); } } } return 0; }
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