您的位置：首页 > 其它

1117. Eddington Number(25)-PAT甲级真题

2016-09-11 17:52 357 查看

1117. Eddington Number(25)
British astronomer Eddington liked to ride a bike. It is said that in order to show off his skill, he has even defined an "Eddington number", E -- that is, the maximum integer E such that it is for E days that one
rides more than E miles. Eddington's own E was 87.

Now given everyday's distances that one rides for N days, you are supposed to find the corresponding E (<=N).

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N(<=105), the days of continuous riding. Then N non-negative integers are given in the next line, being the riding distances
of everyday.

Output Specification:

For each case, print in a line the Eddington number for these N days.

Sample Input:
10
6 7 6 9 3 10 8 2 7 8
Sample Output:
6

题目大意：英国天文学家爱丁顿很喜欢骑车。据说他为了炫耀自己的骑车功力，还定义了一个“爱丁顿数”E，即满足有E天骑车超过E英里的最大整数E。据说爱丁顿自己的E等于87。

分析：从下标1开始存储n天的公里数在数组a中，对n个数据从大到小排序，i表示了骑车的天数，那么满足a[i] > i的最大值即为所求

#include <cstdio>
#include <algorithm>
using namespace std;
int a[1000000];
bool cmp1(int a, int b) {
return a > b;
}
int main() {
int n;
scanf("%d", &n);
for(int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
}
sort(a+1, a+n+1, cmp1);
int ans = 0;
int p = 1;
while(ans <= n && a[p] > p) {
ans++;
p++;
}
printf("%d", ans);
return 0;
}

内容来自用户分享和网络整理，不保证内容的准确性，如有侵权内容，可联系管理员处理

标签：

相关文章推荐

新的分享

章节导航