2017 ACM-ICPC 亚洲区-Coconut

àì夳堔傛蜴生んèń 2022-06-08 02:07 164阅读 0赞

##  ##

### 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");
        }
    }