2017 Multi-University Training Contest - Team 2&&hdu 6047 Maximum Sequence

Maximum Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1450    Accepted Submission(s): 673

Sample Input

4

8 11 8 5

3 1 4 2

Sample Output

27

Hint

For the first sample:
1. Choose 2 from {bi}, then a_2…a_4 are available for a_5, and you can let a_5=a_2-2=9;
2. Choose 1 from {bi}, then a_1…a_5 are available for a_6, and you can let a_6=a_2-2=9;

Source

2017 Multi-University Training Contest - Team 2

题意：给你一个a数组  代表这个数组里的数值  给你一个b数组  代表a数组的位序  在b数组中选一个值  进行奇葩（解释不清楚）操作

8 11 8 5

3 1 4 2  选2   a变成-》8 11 8 5 9  在选1-》 8 11 8 5 9 9  再选3变成8 11 8 5 9 9 5  在选4-》8 11 8 5 9 9 5 4

求增加的数字和最大为多少

题解：其实将b数组排个序 从小到大选取 用一个优先队列维护一下是不是可以取到最大值  因为b数组的值不确定

 #include<iostream>
 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #include<cstdlib>
 #include<string.h>
 #include<set>
 #include<vector>
 #include<queue>
 #include<stack>
 #include<map>
 #include<cmath>
 typedef long long ll;
 typedef unsigned long long LL;
 using namespace std;
 const double PI=acos(-1.0);
 const double eps=0.0000000001;
 const int N=+;
 const ll mod=1e9+;
 int a[N],b[N];
 struct node{
     int num;
     int pos;
     friend bool operator<(node aa,node bb){
         return aa.num<bb.num;
     }
 };
 priority_queue<node>q;
 int main(){
     int n;
     while(scanf("%d",&n)!=EOF){
        // memset(c,0,sizeof(c));
         while(!q.empty())q.pop();
         node c;
         for(int i=;i<=n;i++){
             scanf("%d",&a[i]);
             c.num=a[i]-i;
             c.pos=i;
             q.push(c);
         }
         //cout<<q.top().num<<endl;
         for(int i=;i<=n;i++)scanf("%d",&b[i]);
         sort(b+,b++n);
         ll ans=;
         node t;
         int tt=n+;
         for(int i=;i<=n;i++){
             while(b[i]>q.top().pos)q.pop();
             if(q.empty()==)break;
             c=q.top();
             ans=(ans+c.num)%mod;
             t.num=c.num-tt;
             t.pos=tt;
             q.push(t);
             tt++;
             //cout<<c.num<<endl;
         }
         ans%=mod;
         printf("%I64d\n",ans);
     }
 }