HDU

Steph is extremely obsessed with “sequence problems” that are usually seen on magazines: Given the sequence 11, 23, 30, 35, what is the next number? Steph always finds them too easy for such a genius like himself until one day Klay comes up with a problem and ask him about it.

Given two integer sequences {ai} and {bi} with the same length n, you are to find the next n numbers of {ai}: an+1…a2nan+1…a2n. Just like always, there are some restrictions on an+1…a2nan+1…a2n: for each number aiai, you must choose a number bkbk from {bi}, and it must satisfy aiai≤max{ajaj-j│bkbk≤j

题意

给出长度为n的两个数组a，b
其中1<=bi<=n
每次从b中取一个数（不重复取同一个数）
然后在 a数组中 将第j个数 设置为 a[bi]-bi~a[n+j-1]-(n+j-1) 中的最大值。
求，最大的sigm(a[n+1]~a[n+n]).

思路

应该有更简单的思路
我当时想的是
1.贪心：先把将b排序，然后从小到大取b中的数
因为肯定要越大的数尽量放在前面才可以保证 再后面安排的数会可能取到更大
2.然后就是一个优先队列，将(a[i]-i，i) push进去，大的a[i]-i 在前面。每次取b[i]的时候，判断一下下标，如果b[i]小于等于队首值的下标那么说明可以取到队首这个最大值；否则说明当前这个队首值取不到了，就需要将队首值弹出去，直到找到一个队首值的下标大于b[i] 说明可以取。每次取完需要把新的(a[i+n]-(i+n),i+n)push进去 ，因为可能后加进去的值会成为最大值。

注意

ans要开longlong 这个是肯定的
bug：最开始push(Pair(a[i]-i,i))push错了
写成了(a[i]-i,b[i])…….手残
嗯。。答案还要取余 不要忘了

#include <iostream>#include <cstdio>#include <cmath>#include <algorithm>#include <cstring>#include <vector>#include <queue>using namespace std;const long long MOD=1e9+7;int a[250005],b[250005];struct Pair{    int id;    long long v;    Pair(long long V,int ID){        v=V;        id=ID;    }    friend bool operator <(const Pair a,const Pair b){        if(a.v!=b.v) return a.v<b.v;        else return a.id>b.id;    }};priority_queue<Pair> que;int main() {    freopen("1003.in","r",stdin);//  freopen("1003.txt","w",stdout);    int n;    while(~scanf("%d",&n)){        for(int i=1;i<=n;i++) scanf("%d",&a[i]);        for(int i=1;i<=n;i++) scanf("%d",&b[i]);        sort(b+1,b+1+n);        while(!que.empty()) que.pop();        long long ans=0;        for(int i=1;i<=n;i++){            que.push(Pair((long long )a[i]-i,i));        }        for(int i=1;i<=n;i++){            struct Pair f=que.top();            if(b[i]>f.id){                while(1){                    que.pop();                    f= que.top();                    if(f.id>=b[i])break;                }            }            ans= (ans+f.v) % MOD;//          cout<<f.v<<"**"<<endl;            que.push(Pair(f.v-(n+i),n+i));        }        printf("%lld\n",ans);    }}/*90 11 01 22 -12 12 33 24 13 0*/