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): 535 Accepted Submission(s): 229

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

题意：一个人的收益为，当Σ（a[i]-b[i]）<0时，Σ（a[i])，但是在游戏开始前，他可以将队头的a1和b1放到队尾且次数不限。问最少多少次操作能使他有最大收益。
题解：将ai，bi复制一遍粘到序列后面，用一个长度为n“尺”求最大收入下的最小移动距离。
代码：

#include<iostream>#include<stdio.h>#include<stdlib.h>#include<string.h>#include<vector>#include<queue>#include<set>#include<algorithm>#include<map>#include<math.h>using namespace std;typedef long long int ll;typedef pair<int,int>pa;const int N=2e6+100;const int mod=1e9+7;const ll INF=1e18;int read(){    int x=0;    char ch = getchar();    while('0'>ch||ch>'9')ch=getchar();    while('0'<=ch&&ch<='9')    {        x=(x<<3)+(x<<1)+ch-'0';        ch=getchar();    }    return x;}/***********************************************************/int n;int a[N],b[N];int main(){    while(~scanf("%d",&n))    {        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=a[1]-b[1];        int l=1;        for(int i=2;i<=2*n;i++)        {            ans+=a[i]-b[i];            if(i-l+1==n) break;           if(ans<0)            {                l=i+1;                ans=0;            }        }        printf("%d\n",l-1);    }}