2017 ACM-ICPC 亚洲区-Coconut

最新推荐文章于 2020-12-13 22:07:55 发布

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最新推荐文章于 2020-12-13 22:07:55 发布

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分类专栏： Online Judge 文章标签： acm-icpc 模拟

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船长计划穿越N个城市，每到达一个城市可以获得一定数量的椰子补给，并且在前往下一个城市的途中会消耗椰子。问题是：在整段旅程中，船长能否确保每天都能够喝到椰子汁？

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Discussion

Coconut is Captain Gangplank’s favourite fruit. That is why he needs to drink coconut
juice from b coconuts each day.
On his next trip, he would pass through N citis.
His trip would begin in the 1-st city and end in the N-th city.
The journey from the i-th city to the (i + 1)-th city costs D days.
Initially, there is no coconut on his ship. Fortunately, he could get supply of C coconuts
from the i-th city.
Could you tell him, whether he could drink coconut juice every day during the trip no
not?

Input Format

The first line contains an integer T, indicating that there are T test cases.
For each test case the first line contains two integers N and b as described above.
The second line contains N integers C , C ,⋯, C .
The third line contains N − 1 integers D , D ,⋯, D .
All integers in the input are less than 1000.

Output Format

For each case, output Yes if Captain Gangplank could drink coconut juice every day,
and otherwise output No.

Sample Input

2
4 1
3 2 1 4
1 2 3
4 2
2 4 6 8
3 2 1

Sample Output

Yes
No

题意：

船长将要旅行N个城市，从i到第i+1个城市需要Di天，路过的第i个城市受供Ci个椰子，每天需要吃b个椰子，问船长是否每天都能吃到椰子。

思路：

根据题意，简单模拟。

code：

#include <iostream>
#include <algorithm>
#include <cstdio>

int cd[2][1005];

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,b,sum=0;
        scanf("%d%d",&n,&b);
        for(int i=0;i<n;i++)
        {
            scanf("%d",&cd[0][i]);
        }
        for(int i=0;i<n-1;i++)
        {
            scanf("%d",&cd[1][i]);
        }
        cd[1][n-1]=0;
        int i;
        for(i=0;i<n;i++)//cities
        {
            sum+=cd[0][i];
            sum-=b*cd[1][i];
            if(sum<0)
            {
                break;
            }
        }
        if(i==n)
            printf("Yes\n");
        else
            printf("No\n");
    }
}