HDU 6205 card【单调队列思想+暴力】

HDU 6205

card card card

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

Problem Description

As a fan of Doudizhu, WYJ likes collecting playing cards very much.
One day, MJF takes a stack of cards and talks to him: let's play a game and if you win, you can get all these cards. MJF randomly assigns these cards inton heaps, arranges in a row, and sets a value on each heap, which is called "penalty value".
Before the game starts, WYJ can move the foremost heap to the end any times.
After that, WYJ takes the heap of cards one by one, each time he needs to move all cards of the current heap to his hands and face them up, then he turns over some cards and the number of cards he turned is equal to thepenaltyvalue.
If at one moment, the number of cards he holds which are face-up is less than thepenaltyvalue, then the game ends. And WYJ can get all the cards in his hands (both face-up and face-down).
Your task is to help WYJ maximize the number of cards he can get in the end.So he needs to decide how many heaps that he should move to the end before the game starts. Can you help him find the answer?
MJF also guarantees that the sum of all "penalty value" is exactly equal to the number of all cards.

 

Input

There are about 10 test cases ending up with EOF.
For each test case:
the first line is an integer n (1≤n≤106), denoting n heaps of cards;
next line contains n integers, the ith integer ai (0≤ai≤1000) denoting there are ai cards in ith heap;
then the third line also contains n integers, the ith integer bi (1≤bi≤1000) denoting the "penalty value" of ith heap is bi.

 

Output

For each test case, print only an integer, denoting the number of piles WYJ needs to move before the game starts. If there are multiple solutions, print the smallest one.

 

Sample Input

54 6 2 8 41 5 7 9 2

 

Sample Output

4

//菜鸡的个人笔记，没有题解

思路：将权值数组扩充两倍，记录前缀和，然后找到一个利用单调队列思想，维持一个单调递减队列，由于这里只用求出拿到最多牌的移动次数，只用维持长度为二的单调递减数列，所以直接暴力将前缀数组扫一遍，就OK啦。

ACcode

#include<iostream>#include<cstdio>#include<cstring>#include<vector>#include<algorithm>using namespace std;typedef long long  LL;const int INF=0x3f3f3f3f;const int N=1000005;int a[N],s[N*2],ans,mixx,n,x; void scan_d(int &ret){    char c;    ret = 0;    while ((c = getchar()) < '0' || c > '9');    while (c >= '0' && c <= '9')    {        ret = ret * 10 + (c - '0'), c = getchar();    }}int main(){    while(scanf("%d",&n)!=EOF)    {        for(int i=1;i<=n;i++)scan_d(a[i]);        for(int i=1;i<=n;i++)        {            scan_d(x);            a[i]-=x;            s[i]=s[i-1]+a[i];        }        for(int i=n+1;i<=n+n;i++)            s[i]=s[i-1]+a[i-n];            s[n+n+1]=-INF;         int l=0;         mixx=-1;         for(int i=1;i<=n*2+1;i++)         {             if(s[i]<s[l])             {                 int v=i-l-1;                 if(v>n)v=n;                 if(v>mixx)                 {                     mixx=v;                     ans=l;                 }                 l=i;             }         }         printf("%d\n",ans);    }    return 0;}