2018年ACM-ICPC亚洲青岛区域竞赛 - J:Books

http://acm.zju.edu.cn/onlinejudge/showContestProblem.do?problemId=5837

Time Limit: 1 Second      Memory Limit: 65536 KB

Problem solving report:

Description: 在n个书买m本书，买的规则如下:从左到右一次扫n本书,如果当前钱包里面的钱大于书的价钱,就买下来(必须买),问最多能带多少钱,钱数必须是非负整数,如果没有方案,就输出Impossible,如果可以带无限的钱,就输出Richman.

Problem solving: 贪心

1. 先数数共有多少个0(0是必须要买的),假设ans个,如果ans>m,则没有方案,输出impossible;
2. 然后再在剩下n-ans本书中买前m-ans本,得钱数sum;
3. 对于剩下的求最小值,则最多可以带的钱为sum+min-1.
4. 还有一种情况就是n与m相等的时候,则可以带无限的钱,输出Richman.

#include 
int a[100010];
int main()
{
	long long sum;
	int t, x, k, n, m, ans, minn;
	scanf("%d", &t);
	while (t--)
	{
		minn = 1e9;
		sum = ans = k = 0;
		scanf("%d%d", &n, &m);
		for (int i = 0; i < n; i++)
		{
			scanf("%d", &a[k]);
			if (!a[k])
				ans++;
			else k++;
		}
		if (ans > m)
			printf("Impossible\n");
		else if (n == m)
			printf("Richman\n");
		else
		{
			m -= ans;
			for (int i = 0; i < m; i++)
				sum += a[i];
			for (int i = m; i < k; i++)
				if (a[i] < minn)
					minn = a[i];
			printf("%lld\n", sum + minn - 1);
		}
	}
	return 0;
}