HDU 6205 card card card（展开字符串思想+思维）

card card card

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14    Accepted Submission(s): 5

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

54 6 2 8 41 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]

题解：
本次比赛a的第二题。。一开始还以为做不出来，后来尝试了一下居然ac了
思路：
由于是可以挪动的，所以我们先扩展一下字符串，也就是将数组从0到n-1复制一遍到n到2*n-1，然后我们扫一遍这个长度为2*n-1的数组，ans初始为0，每次加上上面的数组和下面的数组的差，如果ans是一个负数，那么就开始计算当前的最长长度和最长的值，否则继续。。然后扫到最后都还符合条件还要特判一下，还有就是如果当前的i大于了n，如果ans小于了0可以直接退出，或者是当前选择长度已经达到了n也可以退出
代码：

#include<iostream>#include<cstring>#include<stdio.h>#include<math.h>#include<string>#include<stdio.h>#include<queue>#include<stack>#include<map>#include<vector>#include<deque>#include<algorithm>using namespace std;int a[2000005];int b[2000005];int main(){    int i,j,n,ans,d,tag,maxx,len,num;    while(scanf("%d",&n)!=EOF)    {        for(i=0;i<n;i++)        {            scanf("%d",&a[i]);            a[n+i]=a[i];        }        for(i=0;i<n;i++)        {            scanf("%d",&b[i]);            b[n+i]=b[i];        }        ans=0;        maxx=0;        num=0;        len=0;        d=-1;        for(i=0;i<2*n;i++)        {            num+=a[i];            ans+=(a[i]-b[i]);            len++;            if(len>=n)            {                maxx=num;                d=i-len+1;                break;            }            if(ans<0)            {                if(num>maxx)                {                    maxx=num;                    d=i;                }                if(i>n)                    break;                ans=0;                num=0;                len=0;            }        }        printf("%d\n",d);    }    return 0;}