Black Jack

NC237151. Black Jack

描述

PaiGuDragon participated in the game of fake-blackjack.

Each card is written with a number from $1$ to $10$ . In the game, the PaiGuDragon can choose "shoot" or "stay".

If selecting "shoot", PaiGuDragon will draw the card at the top of the stack.

If selecting "stay", PaiGuDragon will end draw and count the sum of points.

If the total number of cards drawn exceeds $21$ , then PaiGuDragon lose everything!

On the premise that the sum of points is less than or equal to $21$ , the greater the sum of points, the stronger the card power.

As a clairvoyant, PaiGuDragon can see every card in the stack from top to bottom. So how many cards will the smart PaiGuDragon choose to draw to get the best result?

输入描述


The first line contains an interger  - the number of cards.
The second line contains  intergers - card points from top to bottom, each number is an interger between  and .

输出描述

Print a single integer, the number of cards PaiGuDragon will draw.

示例1

输入:

3
10 9 3

输出:

说明:

In the first test case, PaiGuDragon will draw $2$ cards and get $10 + 9 = 19$ points. If it draws $3$ cards, the sum will exceed $21$ .

示例2

输入:

6
1 2 3 4 5 6

输出:

说明:

In the second test case, PaiGuDragon will draw all cards, and get $1 + 2 + 3 + 4 +5 + 6 = 21$ points!

原站题解

上次编辑到这里，代码来自缓存点击恢复默认模板

C 解法, 执行用时: 2ms, 内存消耗: 352K, 提交时间: 2022-06-04 12:33:47

#include <stdio.h>

int main(){
	int n,i,m,t,max;
	t=max=0;
	scanf("%d",&n);
	for(i=0;i<n;i++){
		scanf("%d",&m);
		max+=m;
		if(max<=21)
			t++;
	}
	printf("%d",t);
	return 0;
}

C++ 解法, 执行用时: 4ms, 内存消耗: 460K, 提交时间: 2022-06-04 13:30:26

#include<iostream>
using namespace std;
int main(){int n,ans;cin>>n;for(int i=1,s=0,x;i<=n;i++){cin >> x; s+=x;if(s<=21)ans=i;}cout<<ans;return 0;}

Python3 解法, 执行用时: 45ms, 内存消耗: 4728K, 提交时间: 2022-06-06 09:59:51

i=h=0
input()
for x in map(int,input().split()):
    if h+x<22:
        h+=x
        i+=1
    else:
        break
print(i)

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