codeforces D. Lemonade Line(水)

D. Lemonade Line

time limit per test:1 second
memory limit per test:256 megabytes
input:standard input
output:standard output

It’s a hot summer day out on the farm, and Farmer John is serving lemonade to his NN cows! All NN cows (conveniently numbered 1…N1…N) like lemonade, but some of them like it more than others. In particular, cow ii is willing to wait in a line behind at most wiwi cows to get her lemonade. Right now all NN cows are in the fields, but as soon as Farmer John rings his cowbell, the cows will immediately descend upon FJ’s lemonade stand. They will all arrive before he starts serving lemonade, but no two cows will arrive at the same time. Furthermore, when cow ii arrives, she will join the line if and only if there are at most wiwi cows already in line.

Farmer John wants to prepare some amount of lemonade in advance, but he does not want to be wasteful. The number of cows who join the line might depend on the order in which they arrive. Help him find the minimum possible number of cows who join the line.

Input

The first line contains NN, and the second line contains the NN space-separated integers w1,w2,…,wNw1,w2,…,wN. It is guaranteed that 1≤N≤1051≤N≤105, and that 0≤wi≤1090≤wi≤109 for each cow ii.

Output

Print the minimum possible number of cows who might join the line, among all possible orders in which the cows might arrive.

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水题，意思是有n头牛，每头牛都有忍受的极限——前面多于w只牛它就会溜了；问最短队伍长度；

那直接一个降序排序贪心一下就行了；

代码

#include
#include
#include
#include
#include
#define MS(X) memset(X,0,sizeof(X))
#define MSC(X) memset(X,-1,sizeof(X))
typedef long long LL;
using namespace std;
const int maxn=100005;
int w[maxn],n,i;
bool cmp(int a,int b){
    return a>b;
}
int main(){
    scanf("%d",&n);
    for(i=0;i<n;i++){
        scanf("%d",&w[i]);
    }
    sort(w,w+n,cmp);
    for(i=0;i<n;i++){
        if(i>w[i]) break;
    }
    printf("%d\n",i);
    return 0;
}