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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#include <iostream> #include <algorithm> using namespace std; int a[1000000]; int main() { int n, e = 0; scanf("%d", &n); for(int i = 0; i < n; i++) scanf("%d", &a[i]); sort(a, a+n, greater<int>()); while(e < n && a[e] > e+1) e++; printf("%d", e); return 0; } |
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