1117. Eddington Number(25)-PAT甲级真题

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:在数组a中存储n天的公里数,对n个数据从大到小排序,根据排序后的数组可得知骑车大于等于 a[i] 英里的天数有 i+1 天,因为题目中要求超过E英里且所有的英里数为整数,所以可得知骑车超过 a[i]-1 英里的天数有 i+1 天,如样例,从大到小排序后结果为10、9、8、8、7、7、6、6、3、2,骑车超过9英里的有1天,超过8英里的有2天,超过7英里的有4天/5天,超过6英里的有5天/6天,超过5英里的有7天/8天…英里数不断减少,天数不断增加,既然已知公里数和天数对应的关系,要想满足题意,必须使超过的英里数a[i]-1大于等于天数i+1(比如样例中,超过6英里的有6天,超过5英里的有7天,前者的天数6满足题意这6天都超过了6英里,后者的天数7不满足因为这7天只满足骑车都超过5英里)即a[i]-1 >= i+1,也就是 a[i]  >= i + 2,也就是a[i] > i + 1~从头到尾遍历数组,那么满足a[i] > i+1的最大i+1即为所求

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