【HDU 6205】 card card card 【尺取】

card card card

Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1280 Accepted Submission(s): 574

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 into n 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 the penaltyvalue.
If at one moment, the number of cards he holds which are face-up is less than the penaltyvalue, 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
5
4 6 2 8 4
1 5 7 9 2

Sample Output
4
Hint

[pre]
For the sample input:

• If WYJ doesn’t move the cards pile, when the game starts the state of cards is:
  4 6 2 8 4
  1 5 7 9 2
  WYJ can take the first three piles of cards, and during the process, the number of face-up cards is 4-1+6-5+2-7. Then he can’t pay the the “penalty value” of the third pile, the game ends. WYJ will get 12 cards.
• If WYJ move the first four piles of cards to the end, when the game starts the state of cards is:
  4 4 6 2 8
  2 1 5 7 9
  WYJ can take all the five piles of cards, and during the process, the number of face-up cards is 4-2+4-1+6-5+2-7+8-9. Then he takes all cards, the game ends. WYJ will get 24 cards.

It can be improved that the answer is 4.

huge input, please use fastIO.
[/pre]

Source
2017 ACM/ICPC Asia Regional Shenyang Online

#include<bits/stdc++.h>  using namespace std;const int MAXN = 1e6+10;const int  MAXM = 1e6;const int inf  = 0x3f3f3f3f;int a[MAXN<<1],b[MAXN<<1];int chiqu(int n){    int ans=0; //记录现在手中的卡片数量    int sum=0;// 记录目前的卡片向上的数量    int mx=0;// 记录过程中手中最大的卡片数量    int cnt=0;// 记录最小的移动个数    for(int l=1,r=1;l<=n&&r<=n;){        if(sum>=0) {            ans+=a[r];            sum+=(a[r]-b[r++]);        }        if(ans>mx) { // 只有严格大于才能够更新cnt，这样才可保持移动数量最小            mx=ans;            cnt=l-1;        }         if(sum<0) {            ans-=a[l];            sum-=(a[l]-b[l++]);        }    }    return cnt;}int main(){    int n;    while(scanf("%d",&n)!=EOF){        for(int i=1;i<=n;i++){            scanf("%d",&a[i]);            a[i+n]=a[i];         }        for(int i=1;i<=n;i++) {            scanf("%d",&b[i]);            b[i+n]=b[i];         }         int ans=chiqu(2*n);         printf("%d\n",ans);    }     return 0;}