PAT1117. Eddington Number
思路:搞懂题意是关键–E满足有共有E天骑车的距离超过E米,求最大的E!
将数组排序,我们假设最大的E是e,e满足条件有e天骑车超过e米,并且e+1不满足有e+1天骑车超过e+1米。那么我们可以逆序统计所有满足的就是答案。
AC代码
#include <stdio.h>
#include <algorithm>
using namespace std;
const int maxn = 1e5 + 5;
int a[maxn];
int main() {
int n;
scanf("%d", &n);
for(int i = 0; i < n; i++) {
scanf("%d", &a[i]);
}
sort(a, a+n);
int ans = 1;
for(int i = n-1; i >= 0 && ans < a[i]; i--) {
ans++;
}
printf("%d\n", ans-1);
return 0;
}
如有不当之处欢迎指出!
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