WOJ1037-Circular Road

最新推荐文章于 2019-07-15 16:00:10 发布

暴走777

最新推荐文章于 2019-07-15 16:00:10 发布

阅读量754

点赞数

CC 4.0 BY-SA版权

分类专栏： WOJ

本文链接：https://blog.youkuaiyun.com/LXQ1071717521/article/details/77744032

WOJ 专栏收录该内容

144 篇文章

订阅专栏

本文介绍了一个基于汽车环路挑战的问题，需要确定是否可以从特定加油站出发完成一圈旅行。问题包括多个测试案例，每个案例中包含加油站数量及各站的汽油供应量和到下一站的消耗量。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Jacob bought a new car last week, and he is going to test the capability of the car on a circular road this weekend. Now it's time to make a plan.

There are n gas stations lying on the circular road, labeled by 1, 2, 3, ... , n clockwise (as Fig.1 shows where n = 4). The ith station can supply c[i]
liter of gas to the car. Meanwhile, it costs d[i] liter of gas to drive from the ith station to its next station on the clockwise direction.

Jacob can choose any gas station to start and he is planning to travel one lap along the road clockwise and back to the starting station.
Obviously, choosing the proper starting station is critical. Otherwise the car may have to come to a halt in the midway because of the lack of
gas. But in some cases, no matter which gas station Jacob chooses to start, the trip can not be completed. Now here comes the problem: Given
the descriptions of the gas stations(c[i] and d[i]), you, his best friend, tell him whether he can complete the trip by choosing a proper starting
gas station. You may assume that the gasoline tank in the car is empty at the beginning and its capability is unlimited.

输入格式

The input contains several test cases. Each case begins with an integer n(1<=n<=50000), indicating the number of gas stations. The following

n integers are c[i] and the next n integers are d[i]. 1<=c[i],d[i]<=1000. All integers are separated by a single space or a newline character. The
input is terminated by n=0.

输出格式

For each test case, output "YES" if Jack can complete the trip, otherwise "NO" in a single line

样例输入

样例输出

YES
NO

#include<stdio.h>
int c[50001],d[50001];
int main(){
	int n,i,sumc,sumd;
	while(scanf("%d",&n)&&n!=0){
		sumc=0;sumd=0;
		for(i=1;i<=n;i++){
			scanf("%d",&c[i]);
			sumc=sumc+c[i];
		}
		for(i=1;i<=n;i++){
			scanf("%d",&d[i]);
			sumd=sumd+d[i];
		}
		if(sumc>=sumd)
			printf("YES\n");
		else
			printf("NO\n");
	}
	return 0;
}