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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#685372#3565. Beer Pressurefatchuan0 0ms0kbC++202.1kb2024-10-28 19:05:272024-10-28 19:05:27

Judging History

This is the latest submission verdict.

  • [2024-10-28 19:05:27]
  • Judged
  • Verdict: 0
  • Time: 0ms
  • Memory: 0kb
  • [2024-10-28 19:05:27]
  • Submitted

answer

#include<bits/stdc++.h>
  #define LL long long
  using namespace std;
  const int inf = 1e9 + 7;
  const int maxn = 8;
  double possible[maxn];
  int temp[maxn];
  const LL P[] = {1LL,100LL,10000LL,1000000LL,100000000LL};
  int n, k;
 int bfs(LL start){
     map<LL, double> m;//记录状态
     m[start] = 100.0;//开始状态
     queue<LL> q;
     q.push(start);
     while(!q.empty()){
         LL u = q.front(); q.pop();
         double pp = m[u];
         int tot = 0;
         LL tmp = u;
         for(int i = n; i > 0; i--){//将long long 还原成数组
             temp[i] = int(tmp%100LL);
             tmp /= 100LL;
             tot += temp[i];
         }
        if(tot == k){//已经投完票, 选出票数最多的将概率分享。
            int winner[maxn] = {0};
             int Max = temp[1];
             int chosen = 0;
             winner[chosen++] = 1;
             for(int i = 2; i <= n; i++){
                 if(Max == temp[i])winner[chosen++] = i;
                 else if(temp[i] > Max){
                     Max = temp[i];
                     chosen = 0;
                     winner[chosen++] = i;
                 }
             }
             for(int i = 0; i < chosen; i++)
                 possible[winner[i]] += pp/(double)chosen;
         }
         else{
             for(int i = n; i >= 1; i--){
                 LL v = u + P[n - i];
                 double c = (double)temp[i] / (double)tot;
 
                 map<LL, double> :: iterator it = m.find(v);
                 if(it == m.end())
                     m[v] = pp * c, q.push(v);//如果没出现这个状态就入队
                 else
                     it->second += pp * c;//出现过直接在这个状态上面加
             }
         }
     }
 }
 int main(){
     while(~scanf("%d %d", &n, &k)){
         memset(possible,0,sizeof(possible));
         LL u = 0;
         for(int i = 0; i < n; i++){int t;scanf("%d", &t); u *= 100LL, u += t;}//映射成一个10位long long
         bfs(u);
         for(int i = 1;i <= n; i++) printf("pub %d: %.2f %%\n",i, possible[i]);
     }
 }

详细


Pretests


Final Tests

Test #1:

score: 0
Runtime Error

input:

3 7
3 1 1
1 1
1
1 50
1
2 2
1 1
2 50
1 1
2 50
1 2
3 3
1 1 1
3 50
1 1 1
3 50
1 1 2
4 4
1 1 1 1
4 50
1 1 1 1
4 50
1 1 1 2
5 5
1 1 1 1 1
5 50
1 1 1 1 1
5 50
1 1 1 1 2
4 49
2 10 4 2
4 36
4 3 2 4
3 27
2 6 10
1 16
9
3 43
3 6 3
2 28
6 5
3 37
5 7 9
1 44
1
5 44
8 1 9 8 7
3 46
8 2 6
4 28
4 10 4 9
4 50
1 8 6 4
...

output:


result: