card card card
Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1230 Accepted Submission(s): 549
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 题意：每一次按顺序从第一排拿一个数字，当得到的数字和小于第二排的“惩罚数”之和时，停止取数。在进行操作之前能够把一、二排对应数字放到数列最后。问最少几次把数字放到最后，最终取得的数字和最大。 思路：按照最大字段和的思路，先把数列扩展两倍，在扫的时候记录最区间左端点，每次更新a数列区间和时记录最优解的左端点，输出即可。 AC代码：
1 #include<iostream>
2 #include<cstring>
3 #include<cstdio>
4 using namespace std;
5 const int MAXN=2e6+10;
6 const int INF=1e9+7;
7 int a[MAXN],b[MAXN],c[MAXN];
8 int min(int a, int b)
9 {
10 return (a>b)?b:a;
11 }
12 int main()
13 {
14 int n;
15 while(~scanf("%d", &n))
16 {
17 for(int i=0;i<n;i++){
18 scanf("%d", &a[i]);
19 a[i+n]=a[i];
20 }
21 for(int i=0;i<n;i++){
22 scanf("%d", &b[i]);
23 c[i]=a[i]-b[i];
24 c[i+n]=c[i];
25 }
26 int p=0,res=-INF;
27 int sum=0,suma=0;
28 int l=0;
29 for(int i=0;i<2*n;i++){
30 sum+=c[i];
31 suma+=a[i];
32 if(suma>res)
33 {
34 res=suma;
35 p=l;
36 }
37 if(sum<0)
38 {
39 sum=0;
40 suma=0;
41 l=i+1;
42 if(l>=n) break;
43 }
44 }
45 printf("%d\n", p);
46 }
47
48 return 0;
49 }