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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
- 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;
- 然后再在剩下n-ans本书中买前m-ans本,得钱数sum;
- 对于剩下的求最小值,则最多可以带的钱为sum+min-1.
- 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 <stdio.h>
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;
}