[ABC446C] Omelette Restaurant 题解

AT_abc446_c [ABC446C] Omelette Restaurant

题目描述

AtCoder Restaurant was open for $ N $ days after opening.

On day $ i $ ( $ 1\leq i\leq N $ ) after opening, the following actions were performed.

• In the morning of day $ i $ , $ A_i $ eggs are purchased.
• At noon on day $ i $ , $ B_i $ eggs are used. Here, eggs in stock are used in the order they were purchased.
• In the evening of day $ i $ , all eggs that have been stocked for $ D $ or more days are discarded.

There were no eggs in stock before the morning of day $ 1 $ , and eggs never ran out at noon on any day.

Find how many eggs remain in the restaurant after the evening action on day $ N $ .

$ T $ test cases are given; solve each.

输入格式

The input is given from Standard Input in the following format:

$ T $ $ \mathrm{case}_1 $ $ \mathrm{case}_2 $ $ \vdots $ $ \mathrm{case}_T $

$ \mathrm{case}_i $ represents the $ i $ -th test case.
Each test case is given in the following format:

$ N $ $ D $ $ A_1 $ $ A_2 $ $ \ldots $ $ A_N $ $ B_1 $ $ B_2 $ $ \ldots $ $ B_N $

输出格式

Output $ T $ lines.
The $ i $ -th line ( $ 1\leq i\leq T $ ) should contain the answer for the $ i $ -th test case.

输入输出样例 #1

输入 #1

3
3 1
7 2 3
1 3 2
3 2
7 2 3
1 3 2
2 1
2 1
1 2

输出 #1

3
5
0

说明/提示

Sample Explanation 1

In the first test case, the following actions are performed.

• Initially, AtCoder Restaurant has no eggs.
• In the morning of day $ 1 $ , $ 7 $ eggs are purchased. The restaurant has $ 7 $ eggs stocked on day $ 1 $ .
• At noon on day $ 1 $ , $ 1 $ egg is used. The restaurant has $ 6 $ eggs stocked on day $ 1 $ remaining.
• In the evening of day $ 1 $ , no eggs are discarded. The restaurant has $ 6 $ eggs stocked on day $ 1 $ remaining.
• In the morning of day $ 2 $ , $ 2 $ eggs are purchased. The restaurant has $ 6 $ eggs stocked on day $ 1 $ and $ 2 $ eggs stocked on day $ 2 $ .
• At noon on day $ 2 $ , $ 3 $ eggs are used. The restaurant has $ 3 $ eggs stocked on day $ 1 $ and $ 2 $ eggs stocked on day $ 2 $ remaining.
• In the evening of day $ 2 $ , the eggs stocked on day $ 1 $ are discarded. The restaurant has $ 2 $ eggs stocked on day $ 2 $ remaining.
• In the morning of day $ 3 $ , $ 3 $ eggs are purchased. The restaurant has $ 2 $ eggs stocked on day $ 2 $ and $ 3 $ eggs stocked on day $ 3 $ .
• At noon on day $ 3 $ , $ 2 $ eggs are used. The restaurant has $ 3 $ eggs stocked on day $ 3 $ remaining.
• In the evening of day $ 3 $ , no eggs are discarded. (This is because all eggs stocked on day $ 2 $ have already been used.) The restaurant has $ 3 $ eggs stocked on day $ 3 $ remaining.

Thus, $ 3 $ eggs remain after the evening action on day $ 3 $ , so output $ 3 $ on the first line.

For the second test case, remember to output the number of eggs after discarding the eggs stocked on day $ 1 $ in the evening of day $ 3 $ .

Constraints

• $ 1 \leq T\leq 2\times 10^5 $
• $ 1 \leq D \leq N \leq 2\times 10^5 $
• $ 1 \leq A_i,B_i \leq 10 $
• Eggs never run out at noon on any of the $ N $ days.
• For each input, the sum of $ N $ over all test cases is at most $ 2\times 10^5 $ .
• All input values are integers.

思路

水题，直接AC。

代码见下

#include<bits/stdc++.h> 
using namespace std;
long long t,n,d,a[200005],b[200005],c[2000006],l=1,r=0;
int main(){
	cin>>t;
	while(t--){
		cin>>n>>d;
		for(int i=1;i<=n;i++){
			cin>>a[i];
		}
		for(int i=1;i<=n;i++){
			cin>>b[i];
		}
		l=1;
		r=0;
		for(int i=1;i<=n;i++){
			for(int j=1;j<=a[i];j++){
				c[++r]=i;
			}
			l+=b[i];
			while(l<=r&&c[l]+d<=i){
				l++;
			}
		}
		cout<<r-l+1<<endl;
	}
	return 0; 	
}