hdu 6205 card card card

card card card

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 547    Accepted Submission(s): 235

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  i th  integer  ai  ( 0≤ai≤1000 ) denoting there are  ai  cards in  i th  heap;
then the third line also contains  n  integers, the  i th  integer  bi  ( 1≤bi≤1000 ) denoting the "penalty value" of  i th  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

 

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Statistic |  Submit |  Discuss |  Note 题意：求通过把最前面的一堆牌翻转到最后满足能把所有的牌的数目都取到的最少反转次数 。

思路：根据题意，牌的数量和消耗的费用是相同的，所以总有一种顺序能够取到所有的牌：

我们首先从顺序头开始扫描求和，出现和为负的情况时做一个标记，然后和重置为0，继续向后扫描，最后把所有和为负的长度相加就是需要反转的次数

（来自队友的思路，听完以后豁然开朗，代码不到20行！）

#include<stdio.h>
int a[1000005],b[1000005],c[1000005];
int main(void){
    int test;
    while(scanf("%d",&test)!=EOF){
        int i;
        for(i=1;i<=test;i++)
            scanf("%d",&a[i]);
        for(i=1;i<=test;i++){
            scanf("%d",&b[i]);
            c[i]=a[i]-b[i];
        }
        long long sum=0,head=1;
        for(i=1;i<=test;i++){
            if(sum+c[i]<0){
                head=i+1;
                sum=0;
            }
            else sum+=c[i];
        }
        printf("%lld\n",head-1);
    }
    return 0;
}