hdu 6205 card card card （尺取法）

card card card

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1750    Accepted Submission(s): 780

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

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 41 5 7 9 2WYJ 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 82 1 5 7 9WYJ 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]

 

题意:

按顺序取牌，每次取一堆牌a[i]，再把b[i]张牌扔掉，当手上的牌小于要扔的牌数b[i]时，你可以得到所有的之前取过的所有牌

让你将开头的几堆排放在结尾，使得你可以得到最多的牌，输出要放到结尾的牌的堆数

解析:

一开始就只要把预处理c[i]=a[i]-b[i]，用尺取法，

#include<cstdio>
#include<cstring>
#define INF 1999999999
using namespace std;

const int MAXN = 1e6+10;

int cc[MAXN*2],a[MAXN*2];



int main()
{
	int t,n,s;
	//while(n=read()!=-1)
	while(scanf("%d",&n)!=EOF)
	{
		for(int i=0;i<n;i++)
		{
			scanf("%d",&a[i]);
			//a[i]=read();
			a[i+n]=a[i];     //扩展它，因为最多只能放n-1堆到结尾
		}
		int tmp;
		for(int i=0;i<n;i++)
		{
			scanf("%d",&cc[i]);
			//cc[i]=read();
			cc[i]=a[i]-cc[i];
			cc[i+n]=cc[i];
		}
		long long sum=0;
		int i,j;
		j=0;
		for(i=0;i<n;i++)
		{
			for(j=i;j<i+n;j++)
			{
             	sum+=cc[j];
				if(sum<0)
				{
					sum=0;
					i=j;
					break;
				}
				
			}
			if(j-i==n) break;
		}
		printf("%d\n",i);
		/*for(i=0;i<2*n;i++)
		{
			sum+=cc[i];
			if(i-j+1==n)break;
			if(sum<0)
			{
				sum=0;
				j=i+1;
			}
		}
		printf("%d\n",j);*/
	}
	return 0;
}

另外一种大神的方法是

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

http://blog.youkuaiyun.com/fire_to_cheat_/article/details/77927787

#include<cstdio>
#include<cstring>
#define INF 1999999999
using namespace std;

const int MAXN = 1e6+10;

int cc[MAXN],a[MAXN];

int main()
{
	int t,n,s;
	while(scanf("%d",&n)!=EOF)
	{
		for(int i=0;i<n;i++)
			scanf("%d",&a[i]);
		int tmp;
		for(int i=0;i<n;i++)
		{
			scanf("%d",&cc[i]);
			cc[i]=a[i]-cc[i];
		}
		long long sum=0;
		long long end=-1;
		for(int i=0;i<n;i++)
		{
			sum+=cc[i];
			if(sum<0)
			{
				end=i;
				sum=0;
			}

		}
		printf("%lld\n",end+1);
	}
	return 0;
}

最后再加一个加输入挂的代码

#include<cstdio>
#include<cstring>

#define INF 1999999999
using namespace std;

const int MAXN = 1e6+10;

int cc[MAXN*2],a[MAXN*2];

inline 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()
{
    int t,n,s;
    //while(n=read()!=-1)
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=0;i<n;i++)
        {
		scan_d(a[i]);
            //scanf("%d",&a[i]);
            //a[i]=read();
            a[i+n]=a[i];
        }
        int tmp;
        for(int i=0;i<n;i++)
        {
			scan_d(cc[i]);
            //scanf("%d",&cc[i]);
            //cc[i]=read();
            cc[i]=a[i]-cc[i];
            cc[i+n]=cc[i];
        }
        long long sum=0;
        int i,j;
        j=0;
        for(i=0;i<n;i++)
        {
            for(j=i;j<i+n;j++)
            {
                 sum+=cc[j];
                if(sum<0)
                {
                    sum=0;
                    i=j;
                    break;
                }
                
            }
            if(j-i==n) break;
        }
        printf("%d\n",i);
        /*for(i=0;i<2*n;i++)
        {
            sum+=cc[i];
            if(i-j+1==n)break;
            if(sum<0)
            {
                sum=0;
                j=i+1;
            }
        }
        printf("%d\n",j);*/
    }
    return 0;
}