1056 Mice and Rice

题目描述
思路：本题为纯模拟题，按题目意思就好了。主要题目看的比较难受。然后如何用代码简洁实现比较关键。

#include
#include
#include
using namespace std;
const int N = 1100;
int weight[N],r[N];
vector<int> cur;
int main(){
    int n,m;
    cin>>n>>m;
    for(int i = 0 ;i < n;i++) cin>>weight[i];
    for(int i = 0;i < n;i++){
        int a;
        cin>>a;
        cur.push_back(a);
    }
    while(cur.size() != 1){
        int remain = (cur.size() - 1) / m + 1;
        vector<int> next;
        for(int k = 0;k < cur.size();){//k代表新一组开始的下标
            int j = min(m+k,(int)cur.size());//新一组结束的下标
            int t = k;
            for(;k < j;k++){
                if(weight[cur[k]] > weight[cur[t]])
                    t = k;
            }
            next.push_back(cur[t]);
        }     
        for(auto i : cur)
            r[i] = remain+1;
        cur = next;
    }    
    r[cur[0]] = 1;
    cout<<r[0];
    for(int i = 1;i < n;i++) cout<<" "<<r[i];
    return 0;
}
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