Ted is a employee of Always Cook Mushroom (ACM). His boss Matt gives him a pack of mushrooms and ask him to grade each mushroom according to its weight. Suppose the weight of a mushroom is w, then it’s grade s is 

s = 10000 - (100 - w)^2
What’s more, Ted also has to report the mode of the grade of these mushrooms. The mode is the value that appears most often. Mode may not be unique. If not all the value are the same but the frequencies of them are the same, there is no mode.

3
6
100 100 100 99 98 101
6
100 100 100 99 99 101
6
100 100 98 99 99 97

Case #1:
10000
Case #2:
Bad Mushroom
Case #3:
9999 10000

这个题实际就是hash查找，单他竟然卡了cin，应该是数据加强了，或者我的算法不是很好

#include <bits/stdc++.h>
using namespace std;
int cnt[10005];
int main() {
    int T,k=1;
    scanf("%d",&T);
    while(T--){
        memset(cnt,0,sizeof(cnt));
        int n;scanf("%d",&n);
        for(int i=0;i<n;i++){
            int x;scanf("%d",&x);
            cnt[10000-(100-x)*(100-x)]++;
        }
        int ma=INT_MIN;
        for(int i=0;i<=10000;i++)
            ma=max(cnt[i],ma);
        int f=0;
        for(int i=0;i<=10000;i++)
        if(cnt[i]&&cnt[i]<ma){
            f=1;break;
        }
         printf("Case #%d:\n",k++);
        if(ma<n&&!f)printf("Bad Mushroom\n");
        else{
            int f1=0;
            for(int i=0;i<=10000;i++){
                if(cnt[i]==ma){
                    if(f1)printf(" ");
                    printf("%d",i);
                    f1=1;
                }
            }
            printf("\n");
        }
    }
    return 0;
}

Once upon a time, there is a little frog called Matt. One day, he came to a river. 

The river could be considered as an axis.Matt is standing on the left bank now (at position 0). He wants to cross the river, reach the right bank (at position M). But Matt could only jump for at most L units, for example from 0 to L. 

As the God of Nature, you must save this poor frog.There are N rocks lying in the river initially. The size of the rock is negligible. So it can be indicated by a point in the axis. Matt can jump to or from a rock as well as the bank. 

You don‘t want to make the things that easy. So you will put some new rocks into the river such that Matt could jump over the river in maximal steps.And you don‘t care the number of rocks you add since you are the God. 

Note that Matt is so clever that he always choose the optimal way after you put down all the rocks.

2
1 10 5
5
2 10 3
3
6

Case #1: 2
Case #2: 4

贪心下就可以了

#include <bits/stdc++.h>
using namespace std;
const int N=1000010;
int a[N];
int main() {
    int t,n,m,l,k=1;
    scanf("%d",&t);
    while(t--) {
        scanf("%d%d%d",&n,&m,&l);
        for(int i=0; i<n; i++)
            scanf("%d",&a[i]);
        sort(a,a+n);
        int fr=0,ans=0,pre=-l,now;
        a[n]=m;
        for (int i=0; i <=n; i++) {
            now=a[i];
            int t2=(now-fr)/(l + 1);
            pre+=t2*(l + 1);
            ans+=t2*2;
            if (now-pre>l) {
                pre=fr+t2*(l+1);
                fr=now;
                ans++;
            } else fr=now;
        }
        printf("Case #%d: %d\n",k++,ans);
    }
    return 0;
}