PAT (Advanced Level) Practice 1144 The Missing Number (20分)

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发布于 2020-09-28 17:42:29

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发布于 2020-09-28 17:42:29

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1144 The Missing Number (20分)

Given N integers, you are supposed to find the smallest positive integer that is NOT in the given list.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤105). Then N integers are given in the next line, separated by spaces. All the numbers are in the range of int.

Output Specification:

Print in a line the smallest positive integer that is missing from the input list.

Sample Input:

10
5 -25 9 6 1 3 4 2 5 17

Sample Output:

法一：排序枚举法，先排序，然后答案从1开始枚举，模拟

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define rg register ll
#define inf 2147483647
#define lb(x) (x&(-x))
ll sz[200005],n;
template <typename T> inline void read(T& x)
{
    x=0;char ch=getchar();ll f=1;
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch)){x=(x<<3)+(x<<1)+(ch^48);ch=getchar();}x*=f;
}
inline ll query(ll x){ll res=0;while(x){res+=sz[x];x-=lb(x);}return res;}
inline void add(ll x,ll val){while(x<=n){sz[x]+=val;x+=lb(x);}}//第x个加上val
ll a[100005];

int main()
{
	cin>>n;
	for(rg i=1;i<=n;i++)read(a[i]);
	sort(a+1,a+1+n);
	ll ans=1,flag=0;
	for(rg j=1;j<=n;j++)
	{
		if(a[j]>0)
		{
			if(a[j]>ans)
			{
				cout<<ans<<endl;
				flag=1;
				break;
			}
			else if(a[j]==ans)ans++;
		}
	}
	if(!flag)cout<<ans<<endl;
    while(1)getchar();
    return 0;
    
}

法二：数学推导，给定数据范围是n个数，那么如果有大于n的数输入的话是无效的，因为最坏情况也只是1~n的一个随机排列，答案是n+1,所以可设初始答案上界为n+1,当输入的数小于0或者大于等于答案上界时，答案上界-1，因为可用的数变少了一个，很好理解，最后再遍历一遍即可

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define rg register ll
#define inf 2147483647
#define lb(x) (x&(-x))
ll sz[200005],n;
template <typename T> inline void read(T& x)
{
    x=0;char ch=getchar();ll f=1;
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch)){x=(x<<3)+(x<<1)+(ch^48);ch=getchar();}x*=f;
}
inline ll query(ll x){ll res=0;while(x){res+=sz[x];x-=lb(x);}return res;}
inline void add(ll x,ll val){while(x<=n){sz[x]+=val;x+=lb(x);}}//第x个加上val
ll a[100005];
/*
int main()
{
	cin>>n;
	for(rg i=1;i<=n;i++)read(a[i]);
	sort(a+1,a+1+n);
	ll ans=1,flag=0;
	for(rg j=1;j<=n;j++)
	{
		if(a[j]>0)
		{
			if(a[j]>ans)
			{
				cout<<ans<<endl;
				flag=1;
				break;
			}
			else if(a[j]==ans)ans++;
		}
	}
	if(!flag)cout<<ans<<endl;
    while(1)getchar();
    return 0;
    
}
*/

int main()
{
	
	cin>>n;
	ll r=n+1;
	for(rg j=1;j<=n;j++)
	{
		ll x;
		read(x);
		if(x>=1&&x<r)a[x]=1;
		else r--;
	}
	for(rg i=1;i<=r;i++)if(!a[i])
	{
		cout<<i<<endl;
		break;
	}
    while(1)getchar();
    return 0;	
	
}

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