题目描述
Given is a permutation P1,…,PN of 1,…,N. Find the number of integers i (1≤i≤N) that satisfy the following condition:
·For any integer j (1≤j≤i), Pi≤Pj.
Constraints
·1≤N≤2×105
·P1,…,PN is a permutation of 1,…,N.
·All values in input are integers.
输入
Input is given from Standard Input in the following format:
N
P1 … PN
输出
Print the number of integers i that satisfy the condition.
样例输入 Copy
【样例1】
5
4 2 5 1 3
【样例2】
4
4 3 2 1
【样例3】
6
1 2 3 4 5 6
【样例4】
8
5 7 4 2 6 8 1 3
【样例5】
1
1
样例输出 Copy
【样例1】
3
【样例2】
4
【样例3】
1
【样例4】
4
【样例5】
1
提示
样例1解释
i=1, 2, and 4 satisfy the condition, but i=3 does not - for example, Pi>Pj holds for j=1.
Similarly, i=5 does not satisfy the condition, either. Thus, there are three integers that satisfy the condition.
样例2解释
All integers i (1≤i≤N) satisfy the condition.
样例3解释
Only i=1 satisfies the condition.
#pragma
#pragma
#pragma
#pragma
#pragma
#pragma
#include <bits/stdc++.h>
#include <algorithm>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
using namespace std;
#define
typedef long long ll;
#define
#define
template<class T> inline T min(T &x,const T &y){return x>y?y:x;}
template<class T> inline T max(T &x,const T &y){return x<y?y:x;}
//#define getchar()(p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 21) + 1], *p1 = buf, *p2 = buf;
ll read(){ll c = getchar(),Nig = 1,x = 0;while(!isdigit(c) && c!='-')c = getchar();
if(c == '-')Nig = -1,c = getchar();
while(isdigit(c))x = ((x<<1) + (x<<3)) + (c^'0'),c = getchar();
return Nig*x;}
#define
const ll inf = 1e15;
const int maxn = 2e5 + 7;
const int mod = 1e9 + 7;
#define
#define
int cnt;
start{
int n=read;
int minn=mod;
for(int i=1;i<=n;i++){
int num=read;
minn=min(num,minn);
if(minn==num)
cnt++;
}
cout<<cnt<<endl;
end;
}
/**************************************************************
Language: C++
Result: 正确
Time:24 ms
Memory:2024 kb
****************************************************************/
