HDU2275 Kiki & Little Kiki 1

题解:

multiset的应用

代码:

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<map>
#include<set>
using namespace std;
using namespace std;
#define pb push_back
#define mp make_pair
#define se second
#define fs first
#define ll long long
#define CLR(x) memset(x,0,sizeof x)
#define SZ(x) ((int)(x).size())
#define FOR(it,c) for(__typeof((c).begin()) it=(c).begin();it!=(c).end();it++)
typedef pair<int,int> P;
const double eps=1e-9;
const int maxn=100010;
const int N=1000+10;
const int mod=9901;

ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    return x*f;
}
//-----------------------------------------------------------------------------

int main(){
    char op[6];
    int n,x;
    multiset<int> s;
    multiset<int>::iterator it;
    while(~scanf("%d",&n)){
        s.clear();
        for(int i=1;i<=n;i++){
            scanf("%s%d",op,&x);
            if(strcmp(op,"Push")==0) s.insert(x);
            else{
                if(*s.begin()>x||s.empty()) printf("No Element!\n");
                else{
                    it=s.lower_bound(x);
                    if(*it==x) printf("%d\n",x);
                    else printf("%d\n",*(--it));
                    s.erase(it);
                }
            }
        }
        printf("\n");
    }
    return 0;
}