Alice, Bob and Chocolate

最新推荐文章于 2024-06-12 13:14:48 发布

原创最新推荐文章于 2024-06-12 13:14:48 发布 · 1.8k 阅读

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本文介绍了一个有趣的编程挑战，即“巧克力游戏”。在这个游戏中，两名玩家分别从两端开始依次吃掉排成一行的巧克力棒，每根巧克力棒所需消耗的时间已知。文章提供了完整的代码实现，并考虑了特殊边界条件。

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1.题目：

A
 
B
 
C
 
D
 
E
C - Alice, Bob and Chocolate
Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u
Submit
 
Status
 
Practice
 
CodeForces 6C
Description
Alice and Bob like games. And now they are ready to start a new game. They have placed n chocolate bars in a line. Alice starts to eat chocolate bars one by one from left to right, and Bob — from right to left. For each chocololate bar the time, needed for the player to consume it, is known (Alice and Bob eat them with equal speed). When the player consumes a chocolate bar, he immediately starts with another. It is not allowed to eat two chocolate bars at the same time, to leave the bar unfinished and to make pauses. If both players start to eat the same bar simultaneously, Bob leaves it to Alice as a true gentleman.

How many bars each of the players will consume?

Input
The first line contains one integer n (1 ≤ n ≤ 105) — the amount of bars on the table. The second line contains a sequence t1, t2, ..., tn (1 ≤ ti ≤ 1000), where ti is the time (in seconds) needed to consume the i-th bar (in the order from left to right).

Output
Print two numbers a and b, where a is the amount of bars consumed by Alice, and b is the amount of bars consumed by Bob.

Sample Input
Input
5
2 9 8 2 7
Output
2 3
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2.题意：

有两个人分别从左右两个方向吃巧克力，给定各个店铺的时间，若2人时间相同，让给左边的人

3.代码：

#include<stdio.h>
struct node
{
    int visit;
    int v;
}a[100005];
int main()
{
    int n;
    int left=0,right=0;
    int a1=1,a2=1;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        scanf("%d",&a[i].v);
        a[i].visit=0;
    }
    int m=n-2;//注意n-2
    int k=1;
    left=a[0].v;
    right=a[n-1].v;
    for(int i=0;i<n;i++)
    {
        if(left<=right&&a[k].visit==0)
        {
            left+=a[k].v;
            a[k].visit=1;
            k++;
            a1++;
        }
       else if(left>right&&a[m].visit==0)
        {
            right+=a[m].v;
            a[m].visit=1;
            m--;
            a2++;
        }
    }
    if(n==1)
    printf("1 0\n");
    else if(n==2)
    printf("1 1\n");//分别处理
    else
    printf("%d %d\n",a1,a2);
    return 0;
}