这是我做过的第十三(不怎么吉利)道 FARMER JOHN 的奶牛题(doge),题目是这样的:
# P2947 [USACO09MAR] Look Up S
## 题目描述
Farmer John's N (1 <= N <= 100,000) cows, conveniently numbered 1..N, are once again standing in a row. Cow i has height H\_i (1 <= H\_i <= 1,000,000).
Each cow is looking to her left toward those with higher index numbers. We say that cow i 'looks up' to cow j if i < j and H\_i < H\_j. For each cow i, FJ would like to know the index of the first cow in line looked up to by cow i.
Note: about 50% of the test data will have N <= 1,000.
约翰的 $N(1\le N\le10^5)$ 头奶牛站成一排,奶牛 $i$ 的身高是 $H_i(1\le H_i\le10^6)$。现在,每只奶牛都在向右看齐。对于奶牛 $i$,如果奶牛 $j$ 满足 $i<j$ 且 $H_i<H_j$,我们可以说奶牛 $i$ 可以仰望奶牛 $j$。 求出每只奶牛离她最近的仰望对象。
Input
## 输入格式
1. \* Line 1: A single integer: N
\* Lines 2..N+1: Line i+1 contains the single integer: H\_i
第 $1$ 行输入 $N$,之后每行输入一个身高 $H_i$。
## 输出格式
\* Lines 1..N: Line i contains a single integer representing the smallest index of a cow up to which cow i looks. If no such cow exists, print 0.
共 $N$ 行,按顺序每行输出一只奶牛的最近仰望对象,如果没有仰望对象,输出 $0$。
## 输入输出样例 #1
### 输入 #1
```
6
3
2
6
1
1
2
```
### 输出 #1
```
3
3
0
6
6
0
```
## 说明/提示
FJ has six cows of heights 3, 2, 6, 1, 1, and 2.
Cows 1 and 2 both look up to cow 3; cows 4 and 5 both look up to cow 6; and cows 3 and 6 do not look up to any cow.
【输入说明】$6$ 头奶牛的身高分别为 3,2,6,1,1,2。
【输出说明】奶牛 #1,#2 仰望奶牛 #3,奶牛 #4,#5 仰望奶牛 #6,奶牛 #3 和 #6 没有仰望对象。
【数据规模】
对于 $20\%$ 的数据:$1\le N\le10$;
对于 $50\%$ 的数据:$1\le N\le10^3$;
对于 $100\%$ 的数据:$1\le N\le10^5,1\le H_i\le10^6$。
由于之前做的Bad Hair Day S,所以我长脑子了(doge)
不废话了,代码奉上:
#include <iostream>
#include <stack>
using namespace std;
const int N=1e5+10;
int a[N],n;
long long c;
int main() {
stack<int> st;
cin>>n;
for(int i=1;i<=n;i++) {
scanf("%d",&a[i]);
}
for(int i=1;i<=n;i++) {
while (!st.empty()&&st.top()<=a[i]) {
st.pop();
}
c+=st.size();
st.push(a[i]);
}
cout<<c;
return 0;
}
不知道为什么突然写了个scanf()()()
好了,本文章到此结束
哦对了,今天是2025高考第二天,祝大家考的都会,蒙的都对!
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